Transcript for:
Introduction to Quantum Mechanics

sleep and study don't forget to subscribe welcome to quantum mechanics my name is brent carlson since this is the first lecture on quantum mechanics um we had to have some sort of an introduction and what i want to do to introduce quantum mechanics is to explain first of all why it's necessary and second of all to put it in historical context to um well i'll show one of the most famous photographs in all of physics that really gives you a feel for the brain power that went into the construction of this theory and hopefully we'll put it in some historical context as well so you can understand where it fits in the broader philosophy of science but the the main goal of this lecture is about the need for quantum mechanics which i really ought to just have called why do we need quantum mechanics uh this subject has a reputation for being a little bit annoying so why do we bother with it well first off for some historical context imagine yourself back in 1900 turn of the century science has really advanced a lot we have electricity we have all this fabulous stuff that electricity can do and even almost 100 years before that physicists thought they had things figured out there's a famous quote from laplace given for one instant an intelligence which could comprehend all the forces by which nature is animated and the respective position of the beings which compose it nothing would be uncertain in the future as the past would be present to its eyes now maybe you think uh intelligence which can comprehend all the forces of nature is a bit of a stretch and maybe such a being which can know all the respective positions of everything in the universe is a bit of a stretch as well but the feeling at the time was that if you could do that you would know everything if you had perfect knowledge of the present you could predict the future and of course you can infer what happened in the past and everything is connected by one unbroken chain of causality now in 1903 albert michelson another famous quote from that time period said the more important fundamental laws and facts of physical science have all been discovered our future discoveries must be looked for in the sixth place of decimals now this sounds rather audacious this is 1903 and he thought that the only thing that we had left to nail down was the part in a million level precision well to be fair to him he wasn't talking about never discovering new fundamental laws of physics he was talking about really astonishing discoveries like the discovery of uranus on the basis of orbital perturbations of neptune never having seen the planet uranus before they figured out that it had to exist just by looking at things that they had seen that's pretty impressive and michelson was really onto something precision measurements are really really useful especially today but back in 1903 it wasn't quite so simple and michelson probably regretted that remark for the rest of his life the attitude that i want you guys to take when you approach quantum mechanics though is not this sort of 1900s notion that everything is predicted it comes from shakespeare horatio says one oh day and night but this is wondrous strange to which hamlet replies one of the most famous lines in all of shakespeare and therefore as a stranger give it welcome there are more things in heaven and earth horratio than are dreamt of in your philosophy so that's the attitude i want you guys to take when you approach quantum mechanics it is wondrous strange and we should give it welcome there are some things in quantum mechanics that are deeply non-intuitive but if you approach them with an open mind quantum mechanics is a fascinating subject there's a lot of really fun stuff that goes on now to move on to the necessity for quantum mechanics there were some dark clouds on the horizon even at the early 20th century michelson wasn't quite having a big enough picture in his mind when he said that everything was down to the sixth place of decimals the dark clouds on the horizon at least according to kelvin here were a couple of unexplainable experiments one the black body spectrum now a black body you can just think of as a hot object and a hot object like for example the coils on an electrical stove when they get hot will glow and the question is what color do they glow do they glow red they go blue what is the distribution of radiation that is emitted by a hot object another difficult to explain experiment is the photoelectric effect if you have some light and it strikes a material electrons will be ejected from the surface and as we'll discuss in a minute the properties of this experiment do not fit what we think we know about or at least what physicists thought they knew about the physics of light in the physics of electrons at the turn of the 20th century the final difficult experiment to explain is bright line spectra for example if i have a flame coming from say bunsen burner and i put a chunk of something perhaps sodium in that flame it will emit a very particular set of frequencies that looks absolutely nothing like a black body we'll talk about all of these experiments in general or in a little bit more detail in a minute or two but just looking at these experiments now these are all experiments that are very difficult to explain knowing what we knew at the turn of the 20th century about classical physics they're also also experiments that involve light and matter so we're really getting down to the details of what stuff is really made of and how it interacts with the things around it so these are some pretty fundamental notions and and that's where quantum mechanics really got its start so let's pick apart these experiments in a little more detail the black body spectrum as i mentioned you can think of as the light that's emitted just by a hot object and while hot objects have some temperature associated with them let's call that t the plot here on the right is showing very qualitatively i'll just call it the intensity of the light emitted as a function of the wavelength of that light so short wavelengths high energy long wavelengths low energy now if you look at t equals 3 500 kelvin curve here it has a long tail to long wavelengths and it cuts off pretty quickly as you go to short wavelength so it doesn't emit very much high energy light whereas if you have a much hotter object 5500 kelvin it emits a lot more high energy light the red curve here is much higher than the black curve now if you try to explain this knowing what early 20th century physicists knew about radiation and about electrons and about atoms and how they could possibly emit light you get a prediction and it works wonderfully well up until about here at which point it blows up to infinity infinities are bad in physics this is the the rayleigh gene's law and it works wonderfully well for long wavelengths but does not work at all for short wavelengths that's called the ultraviolet catastrophe if you've heard that term on the other end of things if you look at what happens down here well it's not so much a prediction but an observation but there's a nice formula that fits here so on one side we have a prediction that works well on one end but doesn't work on the other and on the other hand we have a sort of empirical formula called veen's law that works really well at the short wavelengths but well also blows up to infinity at the long wavelengths both of these blowing up things are a problem the question is how do you get something that explains both of them this is the essence of the the black body spectrum and how it was difficult to interpret in the context of classical physics the next experiment i mentioned is the photoelectric effect this is sort of the opposite problem it's not how a material emits light it's how light interacts with the material so you have light coming in and the experiment is usually done like this you have your chunk of material typically a metal and when light hits it electrons are ejected from the surface hence the electric part of the photoelectric effect and you do all this in a vacuum and the electrons are then allowed to go across a gap to some other material another chunk of metal where they strike this metal and the experiment is usually done like this you connect it up to a battery so you have your material on one side and your material on the other you have light hitting one of these materials and ejecting electrons and you tune the voltage on this battery such that your electrons when they're ejected never quite make it so the electric field produced by this voltage is opposing the motion of the electrons when that voltage is just high enough to stop the motion of the electrons keep them from completely making it all the way across we'll call that the stopping voltage now it turns out that what classical e m predicts as i mentioned doesn't match what actually happens in reality but let's think about what does classical e m predict here well classical electricity and magnetism says that electromagnetic waves here have electric fields and magnetic fields associated with them and these are propagating waves if i increase the intensity of the electromagnetic wave that means the magnitude of the electric field involved in the electromagnetic wave is going to increase and if i'm an electron sitting in that electric field the energy i acquire is going to increase that means the stop is going to increase because i'll have to have more voltage to stop a higher energy electron as would be produced by a higher intensity beam of light the other parameter of this incoming light is its frequency so we can think about varying the frequency if i increase the frequency i have more intense light now that doesn't say anything about the string sorry if i increase the frequency i don't necessarily have more intense light the electric field magnitude is going to be the same which means the energy and the stopping voltage will also be the same now it turns out what actually happens in reality does not match this at all in reality when the intensity increases the energy which i should really write as v stop the stopping voltage necessary doesn't change and when i increase the frequency the voltage necessary to stop those electrons increases so this is sort of exactly the opposite what's going on here that's the puzzle in explaining the photoelectric effect just to briefly check your understanding consider these plots of stopping voltage as a function of the parameters of the incident light and check off which you think shows the classical prediction for the photoelectric effect the third experiment that i mentioned is bright line spectra and as i mentioned this is what happens if you take a flame or some other means of heating a material like the bar of sodium i mentioned earlier this will emit light and uh in this case the spectrum of light from red to blue of sodium looks like this actually i'm sorry that's not sodium that's mercury the these are four different elements hydrogen mercury neon and xenon and instead of getting a broad continuous distribution like you would from a black body under these circumstances where you're talking about gases you get these very bright regions it's the spectrum instead of looking like a smooth curve like this looks like spikes those bright lines are extraordinarily difficult to explain with classical physics and this is really the uh the straw that broke the camel's back broke classical physics is back that really kicked off quantum mechanics how do you explain this this is that famous photograph that i mentioned this is really the group of people who first built quantum mechanics now i mentioned three key experiments the black body spectrum this guy figured that out this is plonk the photoelectric effect this guy who i hope needs no introduction this is einstein figured that out this is the paper that won einstein the nobel prize and as far as the bright line spectra of atoms it took a much longer time to figure out how all of that fit together and it took a much larger group of people but they all happened to be present in this photograph there's this guy and this guy and these two guys and this guy this photograph is famous because these guys worked out quantum mechanics but that's not the only these aren't the only famous people in this photograph you know this lady as well this is marie curie this is lorenz which if you studied special relativity you know einstein used the lorentz transformations pretty much everyone in this photograph is a name that you know i went through and looked up who these people were these were all of the names that i recognized which doesn't mean that the people whose names i didn't recognize weren't also excellent scientists for example ctr wilson here one of my personal favorites inventor of the cloud chamber this is the brain trust that gave birth to quantum mechanics and it was quite a brain trust you had some of the most brilliant minds of the century working on some of the most difficult problems of the century and what's astonishing is they didn't really like what they found they discovered explanations that made astonishingly accurate predictions but throughout the history you keep seeing them disagreeing like no that can't possibly be right not necessarily because the predictions were wrong or they thought there was a mistake somewhere but because they just disliked the nature of what they were doing they were upending their view of reality einstein in particular really disliked quantum mechanics to the day that he died just because it was so counter-intuitive and so with that introduction to a counter-intuitive subject i'd like to remind you again of that shakespeare quote there are more things in heaven and earth horatio than are dreamt of in your philosophy uh try to keep an open mind and hopefully we'll have some fun at this knowing that quantum mechanics has something to do with explaining the interactions of light and matter for instance in the context of the photoelectric effect or blackbody radiation or brightline spectra of atoms and molecules one might be led to the question of when is quantum mechanics actually relevant the domain of quantum mechanics is unfortunately not a particularly simple question when does it apply well on the one hand you have classical physics and on the other hand you have quantum physics and the boundary between them is not really all that clear on the classical side you have things that are certain whereas on the quantum side you have things that are uncertain what that means in the context of physics is that on the classical side things are predictable they may be chaotic and difficult to predict but in principle they can be predicted well on the quantum side things are predictable too but with a caveat in the classical side you determine everything basically every property of the system can be known with perfect precision whereas in quantum mechanics what you predict are probabilities and learning to work with probabilities is going to be the first step to getting comfortable with quantum mechanics the boundary between these two realms when the uncertain and probabilistic effects of quantum mechanics start to become relevant is really a dividing line between things that are large and things that are small and that's not a particularly precise way of stating things doing things more mathematically quantum mechanics applies for instance when angular momentum l is on the scale of planck's constant or the reduced flux constant h bar now h bar is the fundamental scale of quantum mechanics and it appears not only in the context of angular momentum planck's constant has units of angular momentum so if your angular momentum is of order planck's constant or smaller you're in the domain of quantum mechanics we'll learn more about uncertainty principles later as well but uncertainties in this context have to do with products of uncertainties uh for instance the uncertainty and the momentum of a particle times the uncertainty in the position of the particle this if it's comparable to planck's constant is also going to give you the realm of quantum mechanics energy and time also have an uncertainty relation again approximately equal to planck's constant most fundamentally the classical action when you get into more advanced studies of classical mechanics you'll learn about a quantity called the action which has to do with the path the system takes as it evolves in space and time if the action of the system is of order planck's constant then you're in the quantum mechanical domain now planck's constant is a really small number it's 1.05 times 10 to the negative 34 kilogram meters squared per second times 10 to the negative 34 is a small number so if we have really small numbers then we're in the domain of quantum mechanics in practice these guys are the most useful whereas this is the most fundamental but we're more interested in useful things than we are in fundamental things after all um for example the electron in the hydrogen atom now you know from looking at the bright line spectra that this should be in the domain of quantum mechanics but how can we tell well to use one of the uncertainty principles as a calculation consider the energy the energy of an electron in a hydrogen atom is you know let's say about 10 electron volts if we say that's p squared over 2m using the classical kinetic energy relation between momentum and kinetic energy that tells us that the momentum p is going to be about 1.7 times 10 to the minus 24th kilogram meter square sorry kilogram where'd it go where's my eraser kilogram meters per second now this suggests that the momentum of the electron is you know non-zero but if the hydrogen atom itself is not moving we know the average momentum of the electron is zero so if the momentum of the electron is going to be zero with still some momentum being given to the electron this is more the uncertainty in the electron momentum than the electron momentum itself next quantity if we're looking at the uncertainty relation between momentum and position is we need to know the size of or the uncertainty in the position of the electron which has to do with the size of the atom now the size of the atom that's about 0.1 nanometers which if you don't remember the conversion from nanometers is 10 to the minus 10th meters so let's treat this as delta x our uncertainty in position because we don't really know where the electron is within the atom so this is a reasonable guess at the uncertainty now if we calculate these two things together delta p delta x you get something i should say this is approximate because this is very approximate 1.7 times 10 to the negative 34th and if you plug through the units it's kilogram meter squared per second this is about equal to h bar so this tells us that quantum mechanics is definitely important here we have to do some quantum in order to understand this system as an example of another small object that might have quantum mechanics relevant to it this is one that we would actually have to do a calculation i don't know intuitively whether a speck of dust in a light breeze is in the realm of fun mechanics or classical physics now um i went online and looked up some numbers for a speck of dust let's say the mass is about 10 to the minus sixth kilograms a microgram has a velocity in this light breeze of let's say one meter per second and make myself some more space here the size of this speck of dust is going to be about 10 to the minus 5 meters so these are the basic parameters of this speck of dust in a light breeze now we can do some calculations with this for instance momentum well the momentum is just the mass times the velocity so p is going to be about equal to 10 to the minus 9 kilogram meters per second better make that ten to the minus sixth kilogram meters per second my notes are backwards here um the uncertainty in the momentum then is we could say it's ten to the minus six kilogram meters per second but let's say it's a little smaller than ten to the minus 6 kilograms per second let's say 10 to the minus 8 kilogram meters per second now the position uncertainty that's going to be a function of the size of the object if we know the size of the object the position uncertainty is probably not all that much larger than the size of the object in fact it's probably smaller we can measure where a speck of dust is to better than the diameter of the speck of dust just by putting it in a microscope for example so let's say the position uncertainty here is going to be about 10 to the minus sixth meters now if we run that calculation delta p delta x comes out to be ten to the sixth times ten to the minus eighth which is ten to the minus fourteenth kilogram meter squared per second this is a good factor of 10 to the 20th larger than hbar so this is solidly in the realm of classical physics so even something really small like a speck of dust in a light breeze is still going to be classical so the the size of something here the smallness of it is something that you might have to calculate until you start getting a feel for it just some basic examples to put things in context a little further quantum mechanics is most likely going to be important if you're dealing with single particles atoms molecules single electrons single photons small systems of electrons and photons it's also going to be very very relevant if you're talking about semiconductors the quantum mechanical properties of semiconductors are what makes them into such spectacularly useful electronic devices lasers are another situation where quantum mechanics is crucially important without quantum mechanics there would be no lasers and finally if you're talking about very low temperature physics uh temperature is less than about 100 kelvin well those are going to be quantum mechanical as well so single particles weird materials crystals lasers low temperatures they're kind of an exotic set of uh phenomena but we're adding more all the time um quantum mechanics allows us to do things that we wouldn't be able to do in the classical world consequently it's in our best interest to try and push quantum mechanics as far as we can take it so to check your understanding this is a short question about the uncertainty in uh well the relevant parameters of interaction between two helium atoms and what temperature scale these interactions become important or become quantum mechanical at in order to understand quantum mechanics there's some basic vocabulary that needs to that i need to go over so let's talk about the key concepts in quantum mechanics thankfully there are only a few there's really only three and the first is the wave function the wave function is and always has been written as psi the greek letter my handwriting gets a little lazy sometimes and it'll end up just looking like this but technically it's supposed to look something like that details are important provided you recognize the symbol psi is a function of position potentially in three dimensions x y and z and time and the key facts here is that psi is a complex function which means that while x y z and t here are real numbers psi evaluated at a particular point in space will potentially be a complex number with both a real and imaginary part what is subtle about the wave function and we'll talk about this in great detail later is that it while it represents the state of the system it doesn't tell you with any certainty what the observable properties of the system are it really only gives you probabilities so for instance if i have coordinate system something like this where say this is position in the x direction psi with both real and imaginary parts might look something like this this could be the real part of psi and this could be say the complex or the imaginary part of psi what is physically meaningful is the squared magnitude of psi which might look something like this in this particular case and that is related to the probability of finding the particle at a particular point in space as i said we'll talk about this later but the key facts that you need to know about the wave function is that it's complex and it describes the state of the system but not with certainty the next key concept in quantum mechanics is that of an operator now operators are what connect psi to observable quantities that is one thing operators can do just a bit of notation usually we use hats for operators for instance x-hat or p-hat are operators that you'll encounter shortly operators act on psi so if you want to apply for instance the x hat operator to psi you would write x hat psi as if this were something that were as it appears on the left of psi the assumption is that x acts on psi if i write psi x hat doesn't necessarily mean that x hat acts on psi you assume operators act on whatever lies to the right likewise of course p hat psi now we'll talk about this in more detail later but x hat the operator can be thought of as just multiplying by x so if i have psi as a function of x x hat psi is just going to be x times psi of x so if psi was a polynomial you could multiply x by that polynomial the the p operator p hat is another example is a little bit more complicated this is just an example now and technically this is the momentum operator but we'll talk more about that later it's equal to minus h bar times the derivative with respect to x so this is again something that needs a function needs the wave function to actually give you anything meaningful now the important thing to note about the operators is that they don't give you the observable quantities either but in quantum mechanics you can't really say the momentum of the wave function for instance p hat psi is not and i'll put this in quotes because you won't hear this phrase very often the momentum of psi it's the momentum operator acting on psi and that's not the same thing as the momentum of psi the final key concept in quantum mechanics is the schrodinger equation and this is really the big equation so i'll write it big i h bar partial derivative of psi with respect to time is equal to h hat that's an operator acting on psi now h hat here is the hamiltonian which you can think of as the energy operator so the property of the physical system that h is associated with is the energy of the system and the energy of the system can be thought of as a kinetic energy so we can write a kinetic energy operator plus a potential energy operator together acting on psi and it turns out the kinetic energy operator can be written down this is going to end up looking like minus h bar squared over 2m partial derivative of psi with respect to sorry second partial derivative of side with respect to position plus and then the potential energy operator is going to look like the potential energy as a function of position just multiplied by psi so this is the schrodinger equation typically you'll be working with it in this form so i h bar times the partial derivative with respect to time is related to the partial derivative with respect to space and then multiply multiplied by some function the basic quantum mechanics that we're going to learn in this course mostly revolves around solving this function and interpreting the results so to put these in a bit of a roadmap we have operators we have the schrodinger equation and we have the wave function now operators act on the wave function and operators are used in the schrodinger equation now the wave function that actually describes the state of the system is going to be the solution to the schrodinger equation now i mentioned operators acting on the wave function what they give you when they act on the wave function is some property of the system some observable perhaps and the other key fact that i mentioned so far is that the wave function doesn't describe the system perfectly it only gives you probabilities so that's our overall concept map to put this in the context of the course outline the probabilities are really the key feature of quantum mechanics and we're going to start this course with the discussion of probabilities we'll talk about the wave function after that and how the wave function is related to those probabilities and we'll end up talking about operators and how those operators and the wave functions together give you probabilities associated with observable quantities that will lead us into a discussion of the schrodinger equation which will be most of the course really the bulk of the material before the first exam will be considered with very concerned with various examples solution to the schrodinger equation under various circumstances this is really the main meat of quantum mechanics in the beginning after that we'll do some formalism and what that means is we'll learn about some advanced mathematical tools that make keeping track of all the details of how all of this fits together a lot more straightforward and then we'll finish up the course by doing some applications so those are our key concepts and a general road map through the course hopefully now you have the basic vocabulary necessary to understand phrases like the momentum operator acts on the wave function or the solution to the schrodinger equation describes the state of the system and that sort of thing don't worry too much if these concepts haven't quite clicked in order to really understand quantum mechanics you have to get experience with them these are not things that you really have any intuition for based on anything you've seen in physics so far so bear with me and this will all make sense in the end i promise complex numbers or numbers involving conceptually you can think about it as the square root of negative 1 i are essential to understanding quantum mechanics since some of the most fundamental concepts in quantum mechanics for instance the wave function are expressed in terms of complex numbers complex analysis is also one of the most beautiful subjects in all of mathematics but unfortunately in this course i don't have the time to go into the details lucky you perhaps here's what i think you absolutely need to know to understand quantum mechanics from the perspective of complex analysis first of all there's basic definition i squared is equal to negative 1 which you can think of also as i equals the square root of negative 1. a in general a complex number z then can be written as a the sum of a purely real part x and a purely imaginary part i times y note in this expression z is complex x and y are real where i times y is purely imaginary the terms purely real or purely imaginary in the context of this expression like this x plus i y something is purely real if y is zero something is purely imaginary if x is zero as far as some notation for extracting the real and imaginary parts typically mathematicians will use this funny calligraphic font to indicate the real part of x plus i y or the imaginary part of x plus i y and that just pulls out x and y note that both of these are real numbers when you pull out the imaginary part you get x and y you don't get i y for instance another one of the most beautiful results in mathematics is e to the i pi plus one equals zero this formula kind of astonished me when i first encountered it but it is a logical extension of this more general formula that e raised to a purely imaginary power i y is equal to the cosine of y plus i times the sine of y this can be shown in a variety of ways in particular involving the taylor series if you know the taylor series for the exponential the taylor series for cosine of y and the taylor series for sine of y you can show quite readily that the taylor series for complex exponential is the taylor series of cosine plus the taylor series of sine and while that might not necessarily constitute a rigorous proof it's really quite fun if you get the chance to go through it at any rate the trigonometric functions here cosine and sine should should be suggestive and there is a geometric interpretation of complex numbers that we'll come back to in a minute but for now know that while we have rectangular forms like this x plus i y where x and y the nomenclature there is chosen on purpose you can also express this in terms of r e to the i theta where you have now a radius and an angle the angle here by the way is going to be the arc tangent of y over x and we'll see why that is in a moment when we talk about the geometric interpretation but given these rectangular and polar forms of complex numbers what do the basic operations look like how do we manipulate these things well addition and subtraction in rectangular form is straightforward if we have two complex numbers a plus ib plus and we want to add to that the second complex number c plus id we just add the real parts a and c and we add the imaginary parts b and d this is just like adding in any other sort of algebraic expression multiplication is a little bit more complicated you have to distribute and you distribute in the usual sort of draw smiley face kind of way a times c and b times d are going to end up together in the real part the reason for that is well a times c a and c both being real numbers a times c will be real whereas ib times id both being purely complex numbers you'll end up with b times d times i squared and i squared is minus one so you just end up with minus bd which is what we see here otherwise the complex part is perhaps a little more easy to understand you have i times b times c and you have a times i times d both of which end up with plus signs in the complex part division in this case is like rationalizing the denominator except instead of involving radicals you have complex numbers if i have some number a plus ib divided by c plus id i can simplify this by both multiplying and dividing by c minus id note the sign change in the denominator here c plus id is then prompting me to multiply by c minus id over c minus id now when you do the distribution there for instance let's just do it in the denominator c plus id times c minus id my top eyebrows here of the smiley face c squared minus sorry c squared times id or c squared plus now id times minus id which is well i'll just write it out i times minus id which is going to be d squared times i times minus i so i squared times minus one and i squared is minus one so i have minus one times minus one which is just one so i can ignore that i've just got d squared so what i end up with in the denominator is just c squared plus d squared what i end up with in the numerator well that's the same sort of multiplication thing that we just discussed so the simplified form of this has no complex part in the denominator which helps keep things a little simple and a little easier to interpret now in polar form addition and subtraction well they're complicated under most circumstances if you have two complex numbers given in polar form it's easiest just to convert to rectangular form and add them together there multiplication and division though in polar form have very nice expressions q e to the i theta times r e to the i phi well these are just all real numbers multiplying together and then i can use the rules regarding multiplication of exponentials meaning if i have two things like e to the i theta and e to the i phi i can just add the exponents together it's like taking x squared times x to the fourth and getting x to the sixth but q are e to the i theta plus v so that was easy we didn't have to do any distribution at all the key factor is that you add the angles together in the case of division it's also quite easy you simply divide the radii q over r and instead of adding you subtract the angles so polar complex numbers expressed in polar form are much easier to manipulate in multiplication and division while complex numbers represented in rectangular form are much easier to manipulate for addition and subtraction taking the magnitude of complex number usually we'll write that as something like z if z is a complex number just using the same notation for absolute value of a real number usually is expressed in terms of the complex conjugate the complex conjugate notationally speaking is usually written by whatever complex number you have here in this case x plus y with a star after it and what that signifies is you flip the sign on the complex part on the imaginary part x plus i y becomes x minus i y the squared magnitude then which is always going to be a real and positive number this absolute value squared notation is what you get for multiplying a number by its complex conjugate and that's what we saw earlier with c plus id say i take the complex conjugate of c plus id and multiply it by c plus i d well the complex conjugate of c plus id is c minus id times c plus id and doing the distribution like we did when we calculated the denominator when we were simplifying the division of complex numbers in rectangular form just gave us c squared plus d squared this should be suggestive if you have something like x plus i y that's really messy x plus i y and i want to know the squared absolute magnitude thinking about this as a position in cartesian space should make this formula c squared plus d squared in this case just make make a little more sense you can also of course write that in terms of real and imaginary parts but let's do an example if w is three plus four i and z is minus one plus two i first of all let's find w plus z well w plus z is three plus four i plus minus 1 plus 2i that's straightforward if you can keep track of your terms 3 minus 1 is going to be our real part so that's 2 and 4i plus 2i which is plus 6i is going to be our complex part sorry our imaginary part now w times z 3 plus 4 i times minus 1 plus 2i for this we have to distribute like usual so from our top eyebrow terms here we've got three times -1 which is minus 3 and 4i times 2i both positive so i have 4 times 2 which is 8 and i times i which is minus 1 minus 8. then for my imaginary part the i guess the mouth and the chin if you want to think about it that way i have 4i times minus 1 minus 4 with the i out front will just be minus 4 inside the parentheses here and 3 times 2i is going to give me 6i plus 6 inside the end result you get here is 8 or minus 8 minus 3 is minus 11. and minus 4 plus 6 is going to be 2. so i get minus 11 plus 2i for my multiplication here i guess if i'm going to circle that answer i should circle this answer as well now slightly more complicatedly w over z w is three plus four i and z is minus one plus two i and you know when you want to simplify an expression like this you multiply by the complex conjugate of the denominator divided by the complex conjugate of the denominator so minus 1 minus 2i divided by -1 minus 2i and if we continue the same sort of distribution i'll do the numerator first same sort of multiplication we just did here only the signs will be flipped a little bit we'll end up with minus three plus eight instead of minus three minus eight and for the complex sorry for the imaginary part we'll end up with minus four minus six instead of minus four plus six and you can work out the details of that distribution on your own if you want the denominator is not terribly complicated since we know we're taking the absolute magnitude of a complex number by multiplying a complex number by its complex conjugate we can just write this out as the square of the real part 1 plus the square of the imaginary part minus 2 which squared is 4. so if i continue this final step this is going to be 5 this is going to be minus 10 i and our denominator here is just going to be 5. so in the end what i'll end up with is going to be 1 minus 2 i so it actually ended up being pretty simple in this case now for the absolute magnitude of w 3 plus 4 i you can think of this as w times w star square root you can think of this as the square root of the real part of w plus the imaginary part of w sorry square root of the squared of the real real part plus the square of the imaginary part which is perhaps a little easier to work with in this case so you don't have to distribute out complex numbers in that in that way real part is 3 imaginary part is 4. so we end up with the square root of 3 squared plus 4 squared which is 5. now this was all in rectangular form let me move this stuff out of the way a little bit and let's do it again at least a subset of it in polar form in polar form w three plus four we know the magnitude of w that's five so that's going to be our radius five and our e to the i theta where theta is like i said the arctan of in this case not four fourths four thirds so that's the polar form of the complex number w now if you plug this into calculator to figure out what the arctan of four-thirds is you'll get five e to the i 0.927 if you do the same thing for z you'll end up with the square root of 5 times e to the i 2.03 and for instance if i wanted to calculate w over z i would just have well the radius associated with w divided by the radius associated with z times e to the i difference of the angles and if you take 0.927 and subtract 2.03 you end up with minus 1.1 give or take and if you actually go through and check you will find out that these two numbers are equal to each other so that's an example of manipulating complex numbers just in a very simple way in order to check your understanding here's another example for you to work through on your own now the geometric interpretation i've been alluding to is oops on how to do this in black is in two dimensions hopefully not too surprising i kept mentioning the rectangular form and the polar form well instead of the rectangular form being rectangular coordinates x and y we're going to talk about the rectangular form being the real the real and imaginary parts of a complex number so if we have some complex number here let's call it z we can think of that as being composed of some real part of z and some imaginary part of z if that's the rectangular form of complex number z if you want to think about the polar form of z well this would be the radius associated with that and this would be the angle associated with that z so if i want to say z equals r e to the i theta this is the distance i'm talking about for r and this is the angle i'm talking about for theta now in the geometric interpretation addition and multiplication do pretty much what you would expect sorry addition and subtraction do pretty much what you would expect if i have a complex number here say w and another complex number here say z i can treat these as vectors and just add them tip to tail so one vector is going in this direction one vector is going in this direction the vector sum will just put me out here w plus z same thing for subtraction but with the sine flip on whatever is being subtracted this is for addition if you want to multiply the multiplication rule or the geometric interpretation also has a nice way of treating multiplication if i have some complex number again say w some complex number say z the best way of thinking about this is in polar form where i have say a vector with one angle here i'll call that theta let's call it theta one and another vector here with an angle theta two theta two sorry for the small font i hope this is legible when you multiply these two complex numbers together you know you have to multiply the radii maybe that will put you out you know somewhere at large radii and you have to add the angles together in this case theta 1 is going upwards theta 2 is going downwards so if i go up theta 1 and then add going back down theta 2 i'll end up somewhere out here say or the distance i've gone is the magnitude of w times the magnitude of z and the angle i'm at now here is theta one plus theta two where theta two is negative so addition and multiplication have nice geometric interpretations as well especially useful in visualizing multiplication is this notion of rotating your complex vector i have effectively rotated down theta 2 from the vector to w in order to get in the direction from the origin towards the product w times z so hopefully that's reasonably clear now the geometric interpretation can help a lot when it comes to visualizing what actually happens with um well with complex numbers especially in the context of quantum mechanics when you're dealing with complex functions at any rate here's another example where i'm going to draw out what these things actually look like so i have two complex numbers 3 plus 4i and minus 1 plus 2i let me draw myself a nice big coordinate system here and i'll put some tick marks on it one two three four five six seven eight one two three four five six seven eight one two three four five six seven eight and one two three four five six seven eight more or less now first of all draw w z and w plus z so w is three plus four i so my real part is three one two three my imaginary part is four so i go one two three four so i am up here that's where w would be now z is minus one plus two i so minus one plus two i is z so z is going to be there now if i treat these both as vectors i have a vector to w and i have a vector to z and i add the vectors the way i would normally add vectors tip to tail i'll end up here at w plus z now it's easy to see this in cartesian coordinates as well um if i'm adding w and z i'm going to end up with 3 minus 1 for the real part or 2 up here and 4i plus 2i or 6i one two three four five six for the imaginary part now w times z if you actually go through and work out what w times z is you'll find out it has quite a large magnitude it has a magnitude a little bit larger than 11. 11 point something but what i really want to emphasize here is the geometric interpretation and in that case you need to know the angles involved so i have an angle here theta 1 and i have an angle here theta 2. and it turns out if i add these two angles together say going up theta 1 and then going over theta 2 i end up somewhere along this in this direction and you actually have to go out quite a ways you have to go all the way to minus 11 plus what was it plus 2i i think yes minus 11 plus 2i is the answer for w times z and if you just look at the angles here adding theta 1 and theta 2 is going to point you in that direction so you know it's going to have a negative real part and a positive imaginary part but perhaps a small positive imaginary part just by eyeballing the angles which is not really all that bad w divided by z is the same sort of situation except now instead of adding theta two to theta one you're going to subtract theta two from theta one so i'm going to go up theta one and then back down theta two and where you end up is actually here at one minus two i but just by eyeballing the angles you can get a good feel for what part of the complex plane you're in where typically come on this is the real axis of the complex plane and this is the imaginary axis of the complex plane in cartesian coordinates you have two degrees of freedom x and y in the complex plane you have two degrees of freedom the real part and the imaginary part the last thing that i need you to understand about complex numbers from the perspective of quantum mechanics is complex functions now the theory of complex functions is a lot of fun but beyond the scope of a quantum course to give you an idea for how complex these things can be f z is a complex function now z as a complex number sorry f of z if f is a complex function is going to be complex meaning i can think about this as some real part function plus imaginary part function times i so my one function of a complex variable can actually be complex function of a complex variable can actually be thought of as two separate functions separate real functions mind you so this is a slightly simpler way of thinking about things but you have a real part an imaginary part now in the case of true complex functions like f of z z itself has two degrees of freedom so you can think of this as f real of the real part of z and then the imaginary part of z has two separate arguments to that function and then plus i times f imaginary times the same thing the real part of z and the imaginary part of z so we have instead of one function of one variable we have two functions of two variables and that makes things very difficult to visualize thankfully in quantum mechanics what we're typically working with is the wave function psi and psi for a lot of the problems that we're going to be thinking about is only a function of one coordinate so you can think about this as psi of x where x is a coordinate so psi well psi of x is complex x is real so while we have to think about the real and imaginary parts of psi we don't have to worry about the real and imaginary parts of x we can just think about this as x a single argument in terms of how to visualize complex functions like this you can think about plotting both the real and imaginary parts of this function as a function of whatever well whatever you're working with in this case let's say we have to deal with psi of x so this is going to be the x-axis and we'll have something maybe it looks like this for the real part of psi of x and something maybe it looks like this for the imaginary part of psi so think about plotting two separate functions the real and imaginary parts of this effectively single function as a function of in the case of this x it's uh rather simple to uh to visualize now i have some more advanced visualizations that use color to represent the angle in the polar interpretation of psi as a complex function and they get a little bit psychedelic and you start hallucinating after a while if you look at them too much but for now try to keep in mind that complex numbers have real and imaginary parts and can be interpreted both as x and y in the complex plane the real part of the imaginary part in a two-dimensional cartesian plane or a radius and an angle in a polar representation of the same two-dimensional complex plane that's about it for quantum mechanics like i said quantum sorry that's about it for complex analysis as needed for quantum mechanics like i said complex analysis is a very deep very beautiful topic and i encourage you to study it further in the future but for now i think that's all you need to know since complex numbers are so important to quantum mechanics let's do a few more examples in this case i'm going to demonstrate how to manipulate complex numbers in a more general way not so much just doing examples with numbers first example simplify this expression you have two complex numbers multiplied in the numerator and then a division first of all the first thing to simplify is this multiplication you have x plus iy times ic this is pretty easy it's a simple sort of distribution we're going to have x times ic that's going to be a complex part so i'm going to write that down a little bit to the right i x c and then we're going to have i y times i c which is going to be minus y c that's going to be real we also have a real part in the numerator from d here so i'm going to write this as d minus y c plus i c that's the result of multiplying this out that's then going to be divided by f plus i g now in order to simplify this you have a complex number in the denominator you know you need to multiply by the complex conjugate and divide by the complex conjugate so f minus i g divided by f minus ig now expanding this out is a little bit messier but fundamentally you've seen this sort of thing before you have real part real part an imaginary part imaginary part in the numerator and then you're going to have imaginary part real part and real part imaginary part and what you're going to end up with from this first term you get f times d minus yc from the second term you have minus ig times ixc which is going to give you xcg we have a minus i times an i which is going to give us a plus incidentally if you're having trouble figuring out something like minus i times i think about it in the geometric interpretation this is i in the complex plane this is minus i in the complex plane so i have one angle going up one angle going down if i'm multiplying them together i'm adding the angles together so i essentially go up and back down and i just end up with 1 equals i times minus i otherwise you can keep track of i squared equals minus 1s and just count up your minus signs this then is the real part suppose i should write that in green unless my fonts get too confusing excuse me so that's the real part the imaginary part then is what you get from these terms here i'm going to write an i out front and now we have x c times f so x c f with an i from here and then we have d minus y c times i g which i'll just write as g d minus y c in the denominator we're now multiplying a number by its complex conjugate you know what to do there f squared plus g squared this is just the magnitude of this complex number sorry squared magnitude now this doesn't necessarily look more simple than what we started with but this is effectively fully simplified you could further distribute this and distribute this but it's not really going to help you very much the thing to notice about this is that the denominator is purely real we've also separated out the real part of the numerator and the imaginary part of the numerator my handwriting is getting messier as i go imaginary part of the numerator so we can look at this numerator now and say ah this is the complex number real part imaginary part and then it's just divided by this real number which effectively is just a scaling it's it's a relatively simple thing to do to divide by a real number as a second example consider solving this equation for x now this is the same expression that we had in the last problem only now we're solving it for it equal to zero so from the last page i'm going to borrow that first simplification step we did distributing this through we had d minus y c for the real part plus i x c for the imaginary part and that was divided by f plus i g if we're setting this equal to zero the nice part about dealing with complex expressions like this is that zero treated as a complex number is 0 plus 0 i it has a real part and an imaginary part as well it's just kind of trivial and in order for this complex number to be equal to 0 the real part must be 0 and the imaginary part must be zero so we can think of this as d minus y c plus i x c this has to equal zero and this has to equal zero separately so we effectively have two equations here not just one which is nice we have d minus y c equals zero and x c equals zero which unless c equals zero just means x equals zero that's the only way that this equation can hold is if x equals zero the key factor is to keep in mind that the in order for two complex numbers to be equal both the real parts and the imaginary parts have to be equal as a slightly more involved example consider finding this the cubed roots of one now you know one cubed is one that's a good place to start we'll see that fall out of the algebra pretty quickly what we're trying to do is solve the equation z cubed equals one which you can think of as x plus i y where x and y are real numbers cubed equals one now if we expand out this cubic you get x cubed plus three x squared times i y plus three x times i y squared plus i y cubed and this is going to have to equal 1. excuse me equal 1. now looking at these expressions here we have an i y here we have an i y squared this is going to give me an i squared which is going to be a minus sign and here i have an i y cubed this is going to give me an i cubed which is going to be minus i so i have two complex parts and two real parts so i'm going to rewrite that x cubed and then now a minus sign from the i squared 3 x y squared plus pulling an i out front the imaginary part then is going to come from this 3x squared y and this y cubed so i've got a 3 x squared y here and then a minus y cubed minus coming from the i squared and this is also going to have to equal one now in order for this complex number to equal this complex number both the real parts and the imaginary parts have to be equal so let's write those two separate equations x cubed minus 3 x y squared equals the real part of this is the real part of the left-hand side has to equal the real part of the right-hand side one and the imaginary part of the left-hand side three x squared y minus y cubed has to equal the imaginary part of the right hand side 0. so those are our two equations this one in particular is pretty easy to work with we can simplify this this is you know we can factor a y out this is y times three x squared minus y squared equals zero one possible solution then is going to come from this you know you have a product like this equals zero either this is equal to 0 or this is equal to 0 and saying y equals to 0 is rather straightforward so let's say y equals 0 and let's substitute that into this expression that's going to give us x cubed equals 1 which might look a lot like the equation we started with z cubed equals 1 but it's subtly different because z is a general complex number whereas our assumption in starting the problem this way is that x is a purely real number so a purely real number which when cubed gives you one that means x equals one so x equals one y equals zero that's one of our solutions z equals one plus zero i or just zero z equals one now we could have told me that right off the bat z z cubed equals one well z one possible solution is that z equals one since one cubed is one the other thing we can do here is we can say three x squared minus y squared is equal to zero this means that i'll just cheat a little bit and simplify this 3x squared equals y squared now i can substitute this in this y squared into this expression as well and what you end up with is x cubed minus three x and then y squared was equal to three x squared so three x squared is gonna go in there that has to equal one now let's move up here what does that leave us with that says x cubed minus nine x cubed equals one so minus eight x cubed equals one this means x again being a purely real number is equal to minus one half minus one half times minus one half times minus one half times eight times minus one is equal to one you can check that pretty easily now where does that leave us where did i go that leaves us substituting this back in to this expression which tells us that 3 x squared equals y squared x equals minus one half so three minus one half squared equals y squared which tells you that y equals plus or minus the square root of three fourths if you finish your solution so now we have two solutions for y here coming from one value for x and that gives us our other two solutions to this cubic we have a cubic equation we would expect there to be three solutions especially when we're working with complex numbers like this this is our other solution z equals minus one half plus or minus the square root of three fourths i so those are our three solutions now finding the cubed roots of one to be these complex numbers is not necessarily particularly instructive however there's a nice geometric interpretation the cubed roots of unity like this the nth roots of unity doesn't have to be a cubed root all lie on a circle of radius 1 in the complex plane and if you check the complex magnitude of this number the complex magnitude of this number you will find that it is indeed unity to check your understanding of this slightly simpler problem is to find the square roots of i um put another way you've got z some generic complex number here equals to x squared plus x plus i y quantity squared if that's going to equal y we'll expand this out solve for x and y in the two equations that would result from setting real and imaginary parts equal to each other and same as with the cubed roots of one the square roots of i will also fall on a circle of radius 1 in the complex plane so those are a few examples of how complex numbers can actually be manipulated in particular finding the roots of unity there are better formulas for that than the approach that we took here but i feel this was hopefully instructive if probability is at the heart of quantum mechanics what does that actually mean well the fundamental source of probability in quantum mechanics is the wave function psi psi tells you everything that you can in principle know about the state of the system but it doesn't tell you everything with perfect precision how that actually gives rise to probability distributions in observable quantities like position or energy or momentum is something that we'll talk more about later but from the most basic perspective psi can be thought of as related to a probability distribution but let's take a step back and talk about probabilistic measurements in general first if i have some space let's say it's position space say this is the floor of a lab and i have a ball that is somewhere on in the floor somewhere on the floor i can measure the position of that ball maybe i measure the ball to be there on the floor if i prepare the experiment in exactly the same way attempting to put the ball in the same position on the floor and measure the position of the ball again i won't always get the same answer because of perhaps some imprecision in my measurements or some imprecision in how i'm reproducing the system so i might make a second measurement there or a third measurement there if i repeat this experiment many times i'll get a variety of measurements at a variety of locations and maybe they cluster in certain regions or maybe they're very unlikely in other regions but this distribution of measurements we can describe that mathematically with the probability distribution a probability distribution for instance i could plot p of x here and p of x tells you roughly how many or how likely you are to make a measurement so i would expect p of x as a function to be larger here where there's a lot of measurements and zero here where there's no measurements and relatively small here where there's few measurements so p of x might look something like this so the height of p of x here tells us how likely we are to make a measurement in a given location this concept of a probability distribution is intimately related to the wave function so the most simple way that you can think of probability in quantum mechanics is to think of the wave function psi of x now psi of x you know is a complex function and a complex number can never really be observable what would it mean for example to measure a position of say 2 plus three i meters this isn't something that's going to occur in the physical universe but the fundamental interpretation of quantum mechanics that most that your book and this book in particular that most physicists think of is the interpretation that psi in the context of a probability distribution the absolute magnitude of psi squared is related to the probability of finding the particle described by psi so if the squared magnitude of psi is large at a particular location that means it is likely that the particle will be found at that location now the squared magnitude here means that we're not that we have to say well we have to take the squared magnitude of psi we can't just take psi itself so for instance in the context of the plot that i just made on the last page if this is x and our y-axis here is psi psi has real and imaginary parts so the real part of psi might look something like this and the imaginary part might look something like this and the squared magnitude would look something like well what you can imagine the square magnitude of that function looking like you can think of the squared magnitude of size the probability distribution let me move this up a little bit give myself some more space the squared magnitude of psi then can be thought of as a probability distribution in the likelihood of finding the particle at a particular location like i said now what does that mean mathematically mathematically suppose you had two positions a and b and you wanted to know what the probability of finding the particle between a and b was given a probability distribution you can find that by integrating the probability distribution so the probability that the particle is between a and b is given by the integral from a to b of the squared absolute magnitude of psi dx you can think of this as a definition you can think of this as an interpretation but fundamentally this is what the physical meaning of the wave function is it is related to the probability distribution of position associated with this particular state of the system now what does that actually mean and that's a bit of a complicated question it's very difficult to answer suppose i have a wave function which i'm just going to write as the square plot is the square of magnitude of psi now suppose it looks something like this now that means i'm perhaps likely to measure the position of the particle somewhere in the middle here so suppose wrong color so suppose i do that suppose i measure the position of the particle here so i've made a measurement now messy handwriting i've made a measurement and i've observed the particle b here what does that mean in the context of the wave function now everything that i can possibly know about the particle has to be encapsulated in the wave function so after the measurement when i know the particle is here you can think of the wave function is looking something like this it's not going to be infinitely narrow because there might be some uncertainty the width of this is related to the precision of the measurement but the wave function before the measurement was broad like this and the wave function after the measurement is narrow what actually happened here what about the measurement caused this to happen this is one of the deep issues in quantum mechanics that is quite difficult to interpret so what do we make of this well one thing that you could think just intuitively is that while this probability distribution wasn't really all the information that was there really the particle was there let's say this is point c one interpretation is that the particle really was at c all along that means that this distribution reflects ignorance on our part as physicists not fundamental uncertainty in the physical system this turns out to not be true and you can show mathematically and in experiments that this is not the case the main interpretation that physicists use is to say that this wave function psi here also shown here collapses now that's a strange term collapses but it's hard to think of it any other way suppose you were concerned with the wavefunction's value here before the measurement it's non-zero whereas after the measurement it's zero so this decrease in the wavefunction out here is uh well it's reasonable to call that a collapse what that wave function collapse means is subject to some debate and there are other interpretations one interpretation that i'll mention very briefly but we won't really discuss very much is the many worlds interpretation and that's that when you make a measurement like this the universe splits so it's not that the wave function all of a sudden decreases here it's that for us in our tiny little chunk of the universe the wave function is now this and there's another universe somewhere else where the wave function is this because the particle is observed to be here um don't worry too much about that but the interpretation issues in quantum mechanics are really fascinating once you start to get into them you can think about this as the universe splitting into sorry splits the universe you can think about this as the universe splitting into many little subuniverses where the probability of uh observable where the particle is observed at a variety of locations one location per universe really this question of how measurements take place is really fundamental but hopefully this explains a little bit of where probability comes from in quantum mechanics the wave function itself can be thought of as a probability distribution for position measurements and unfortunately the measurement process is not something that's particularly easy to understand but that's the fundamental origin of probability in quantum mechanics to check your understanding here is a simple question about probability distributions and how to interpret them since probability is so essential to quantum mechanics it's good to go back and review to make sure we're all on the same page as far as probability the fundamental concepts of probability aside from probability itself are probability distributions and what their properties are so let's review probability there are two main kinds of probability distribute discrete and continuous discrete probability is what you get if you have only a specific set of outcomes sort of well hence the word discrete have a discrete set of outcomes and these outcomes can be anything they can be the outcomes of a measurement they can be responses in a survey they can be any sort of data set for example if this is your data set 0 zero one one one two two three four five two zeros three ones two twos and one each of three four and five for instance suppose you go out and you ask ten members of the general public how many sexual partners they've had you might get a data set something like this the way to think of this in the context of a probability distribution well i like to plot things so let's make a plot of this probability distribution we have our outcomes zero one two three four and five step my axis a little there we have two people who answered zero so if we make a y axis here 1 2 i can put a bar here at 0 that goes up to 2. there were 3 people that answered 1 so i should make a bar that goes up to three above the one five three above the one likewise for two there were two answers for two and there were one each for three four and five one there one there and one there now our probability distribution looks like this when you plot it what are we actually plotting here we're plotting the number of people who responded for each particular answer here we can make a table then of the probabilities either we have on the left we have the value and on the right we have the probability now what is the probability under these circumstances we have a discrete set of outcomes six different numbers zero through five and the probability of say three is just the number of threes that appear in this data set divided by the total number of entries in the data set so for instance if i want to know the probability of the value one well there were two sorry let's start with zero if i want to know the probability of the value zero there were two respondents so that would be two divided by ten two people out of ten so if i chose one of these at random not looking at the numbers the likelihood that i would choose zero is well two out of ten two times in ten i would choose zero probability on average and you can do the same thing for one three out of ten two two out of ten again three four and five all one out of ten and some things you should notice about this let me move it up slightly to move everything up the numbers here if i add these up 2 plus 3 plus 2 plus 1 plus 1 plus 1 it adds up to 10 out of 10. so the total probability of drawing any number is one and that is sort of a certainty if you draw a number you will draw a number so if you only have a discrete set of outcomes for instance the integers like this you're dealing with the discrete probability distribution and the sum of the probabilities in the discrete probability distribution have to sum to one and it's a straight sum if i add up these numbers the other class of probability distribution is a continuous probability distribution continuous probability distributions are what you work with when you have a range of values a range of values not a list of values so for instance suppose you got this set of numbers uh this might be what you would get if you talked to 10 people who'd just gotten off the phone with comcast tech support and asked them for instance how long did comcast put you on hold say in hours we can look at this probability distribution graphically as well only it's slightly more complicated suppose the x-axis now here is my time in hours and i have one and two to give a scale on my axis now if i plot this data set i've got one at like 0.7 or 0.07 one at 0.11 one at 0.23 0.55 0.79 1.09 1.7 probably about there 1.8 the 1.88 and 2.16 something like that so these discrete answers each answer here is only a specific number on this axis so if i want to make a meaningful plot i can either pretend that my probability distribution is discrete or i can attempt to guess at what the probability distribution looks like since i know what the answer is here let's call this row of x is typically the name given to a continuous probability distribution and it would look something like this in this case so we have what's called a probability density function or just a probability density rho of x now the reason we have a probability density like this rho of x is because x is a continuous variable now it can be anything in this case say from zero to infinity what does that mean if i evaluate rho of x at say x equals one you might think well okay that's the probability of one but well one of the three fundamental facts of continuous probability distributions is that the probability that x is exactly equal to one is going to be zero the example given in the text is suppose you walked up to a random person on the street and asked is your age 18 years 3 months 12 days 18 minutes 16 seconds 52 microseconds etc if you if you precisely specify the exact answer you want the probability that you'll get that exact answer is going to be zero the only reason or the only way that makes sense for a continuous variable or continuous probability is for the probability to be specified between limits suppose i had a and b and i want to know the probability that the number that you're drawing from this distribution or the random probability of you know amount of time that you would like to have to spend on hold with comcast between say 45 minutes and an hour and 45 or hour and a half you could calculate that knowing this function and the way you calculate it is by integrating put mathematically the probability that say x is between a and b is the integral from a to b of row of x dx you can think of this more or less as the definition of the probability density function if you have a way of calculating probabilities for x's and ranges that is the probability density function probability density functions also have to sum to 1 but instead of having a sum in the case of a discrete probability distribution we have an integral so the other property here is that the integral from minus infinity to infinity the entire possible range of values rho of x dx has to equal not zero has to equal one the answer the the question that this equation here answers is what is the probability of drawing a number in between negative infinity and positive infinity and that had better be unity this is the equivalent of the sum of the probabilities in a discrete probability distribution being equal to 1. now in this case the probability distribution to give you a feel for what these things look like rho of x in this case is zero if x is less than zero and is equal to e to the minus x if x is greater than or equal to zero so this is an example of a probability distribution function a probability density that's actually the distribution i used when i drew these numbers so that's continuous probability if you have a continuous range of outcomes you have to deal with probability distributions and your probability density functions and you have to do some integrals if you want to calculate properties now what are those properties well the most basic properties that you can that you can think of is probably the most likely outcome now in the case of a discrete probability distribution this is just the you know the most probable what that means is that you've got the highest probability and in this case that means it's just the outcome that occurs the most times in this case 1. so our most probable outcome here is 1. in the case of a continuous distribution it's slightly more complicated if i plot the distribution function from before the exponential turn off the ruler if i plot the probability distribution from before look something like this the most probable is actually going to be zero now you might think what what are the odds that comcast puts me on hold for zero minutes um well zero that's our rule that the proper the probability that x is exactly equal to zero is zero it only makes sense to make a statement like the probability that say x is greater than zero and x is less than say five minutes and that you can calculate with your integral like on the last page but in the case of the most probable value the most probable value the peak of the distribution is indeed zero here in spite of the fact that zero will effectively never happen any exact value will effectively never happen but 0 is the most likely the next property is the median median is not something that appears very often in physics but it appears a lot in statistics so we need to mention it the median is essentially the middle of the data set so you've got half are smaller and half are larger than the median so here i have ten numbers the median is going to be something that falls in between one and two half the data set 0 0 1 1 1 is smaller and 2 2 3 4 5 are all larger typically what statisticians do when they define the median here when there's no exact number that is the median is just put it halfway between the numbers so in this case the median would be 1.5 in the context of a continuous probability distribution the median is the number such that your probability adds up its probability adds up to a half so if i integrate from minus infinity the lower possible limit that you could possibly have up to the median rho of x dx i would get 0.5 so in the case of this probability distribution function e to the minus x i can write that out this integral from 0 to i'll just call the median m now e to the minus x is rho of x dx now i constructed this integral from zero to m instead of from minus infinity to m because this probability distribution is zero for x less than zero just to make sure that it's clear why i used zero here that's not actually a six that's a zero so if you do this integral you know the integral of e to the minus x is minus e to the minus x hopefully if you don't you need to review some calculus there will be a lot of integrals like this in quantum mechanics evaluated between 0 and m and that has to be equal to 0.5 plugging in the limits here you get well i'll just simplify it and skip a step here you get 1 minus e to the minus m equals 0.5 which if you solve it for m gives you that m equals the log of 2. now if i get sloppy and say log here i really mean natural log log if i ever need a log base 10 i'll write log base 10. most of the time we'll be using natural logs in quantum mechanics so that's the median the next property and this is one that will be very common is the mean the symbol that we use for the mean is mu and in the case of a discrete probability distribution mu is the sum over all of your outcomes of the outcome x sub i times the probability that say x is x sub i sometimes this is just written as the probability of x sub i which makes sense in the context of a discrete probability distribution so what this looks like in the case of the discrete probability that i gave you earlier is well there's 0 that's x of i times the probability that x equals 0 which was 2 10 plus 1 that's x sub i times the probability that x equals 1 which was three tenths plus two x sub i times probability that x is equal to two which is two tenths and you can keep going plus three times one tenth plus four times one tenth plus five times one tenth for three four and five all of which have probability one tenth and if you run through the numbers you'll find this is equal to one point nine if you remember from an earlier class that the mean is the sum of the values divided by the total number of values that you added up that's the same formula as this if you say factor the one-tenth out all of these numbers have a one-tenth and if we factor that out we'll get zero plus zero plus one plus one plus one plus two plus two plus three plus four plus five which is what you would get if you added all this up and you get the same answer so there's no black magic there the reason i'm writing this this way is this format here where you have something times the probability of that something is something that shows up repeatedly in the context of a continuous distribution for example the mean mu now instead of a sum we have to have an integral and instead of a sum over all possible values we'll have an integral over all possible values minus infinity to infinity of x times the probability density rho of x dx now if i plug in this example we'd be integrating again from zero to infinity so that i don't have to worry about if i integrate from minus infinity to zero the probability density is going to be zero for all that for that entire range so i can ignore that and just start my integral at zero and then i would have x e to the minus x dx now whenever you see an integral like this hopefully you think integration by parts the way i like to do integration by parts is to try and figure out probably just by trial and error which pieces of this equation are going to become which of the parts that i'm going to use in this case let's say u equals x and dv equals e to the minus x dx now you differentiate this d u equals dx and you integrate this v equals minus e to the minus x now in integration by parts the forms that don't have any differentials in them that's your leading term so we would have this integral is equal to x e to the minus x with a minus sign evaluated at the limits 0 and infinity then we have a minus sign and the integral part of the integration by parts result which are these bottom pieces so the integral now of e to the minus x with a minus sign so i'm going to change this minus out front into a plus and then this part just dx from 0 to infinity now what do we do with the endpoints on this well at infinity e to the minus infinity that's one over e to the infinity that's zero and that's that's a really emphatic zero if you think about it in the context of a limit the limit as x e to the minus x of the limit of x e to the minus x as x goes to infinity is zero so our upper limit here is zero and then we subtract what we get if we plug in zero for this now e to the minus zero is just one and x well x is zero so we have effectively zero minus a negative zero zero plus zero whatever it's still zero this term vanishes and what we're left with here is the integral of e to the minus x dx from zero to infinity and hopefully you know how to do that that just in the end is going to equal 1. so that's our mean uh mean is also called expectation value especially in quantum mechanics um it's not necessarily the value that you expect to get it is the well it's this mathematical expression that tells you the value that is sort of in the middle of your distribution of possible values the reason expectation value is perhaps a better term than mean especially in quantum mechanics is that we can apply it more easily to functions for instance f x the notation we use for the expectation value or the mean value of some function is to put that in brackets so this is read as the x the expected value or the expectation value of x or perhaps just the expectation of f excuse me f not x and this in the context of a discrete probability distribution looks a lot like the formula for the mean except instead of x sub i in the sum we have f of x sub i times the probability that x is x sub i so suppose the function i wanted to work with was f of x equals x squared what's the probability of x squared well just applying this formula you would have zero squared from f of x with x sub i being zero times the probability of zero which was two tenths plus one squared times three tenths now this is one squared times the probability of one plus two squared times two two-tenths plus three squared times one-tenth plus four squared times one-tenth plus five squared times one-tenth and if you plug all the numbers in here this is a fairly simple thing to do hopefully you can get the right answer on your calculator it comes out to 6.1 to keep with my format from earlier i should do this in color the expected value f expected value of f is 6.1 you might also see this written as the expected value of x squared just saving or compacting the notation a little bit and ignoring the definition of f of f um same thing for the case of a continuous distribution the pattern by now is hopefully reasonably clear the expected value of f in the continuous distribution case is equal to the integral from minus infinity to infinity of f of x times rho of x dx and the integral in this case for the example function here is going to be equal to integral from zero to infinity again because rho of x is zero for x less than zero f of x is x squared row of x e to the minus x dx now again you see an integral like this i want you to think integration by parts alternatively i want you to think oh i should go plug this integral into wool from alpha or i should go look this up in a table of integrals or whatever tool if you have like a ti-89 or a ti-92 they can do integrals like this whatever technique you want to use to solve integrals is okay with me but in this case integration by parts will work and it's not terribly difficult so again u is going to be x squared and dv is here's my eraser dv equal to e to the minus x dx sorry i'm running out of space there let me do this slightly more legibly dv equals e to the minus x dx now differentiate this d u equals two x d x and v equals minus e to the minus x in our integration by parts then the term out front without the integral is going to be x squared times minus e to the minus x evaluated at the limits zero and infinity and for the same reasons this term vanished in the last case it's going to vanish here as well if i evaluate this at infinity this is going to be zero if i evaluate this at zero this is going to be zero so this term will vanish the second part of our integration by parts is minus the integral from zero to infinity of just for the sake of neatness zero to infinity the parts here with with the differentials in them 2 x dx e to the minus x with a minus sign so i'm going to change this minus sign to a plus sign this integral i'm going to pull the 2 out front 2 times the integral from 0 to infinity of x e to the minus x dx and this integral you hopefully remember because we just did it this integral was equal to 1. that was what we did on the last page so this overall is going to equal 2. so the expected value of x squared under this continuous probability distribution is 2. to check your understanding here's a sample problem the probability density function in this case i'll plot it just to make things a little more intuitive if this is 0 and this is 1 1 and 2. the probability density is 0 if x is less than 0 it's also 0 if x is greater than 1 and for x between 0 and 1 it's equal to 2x so our probability density function looks something like this and your task is to find the mean of x under this probability distribution and the expected value of f of x equals the square root of x so two tasks here to accomplish with this sample probability variance and standard deviation are properties of a probability distribution that are related to the uncertainty since uncertainty is such an important concept in quantum mechanics we need to know how to quantify how uncertainty results from probability distributions so let's talk about the variance and the standard deviation these questions are related to the shape of a probability distribution so if i have a set of coordinates let's say this is the x-axis and i'm going to be plotting then the probability density function as a function of x probability distributions come in lots of shapes and sizes you can have probability distributions that look like this probability distributions that look like this you can even have probability distributions that look like this or probability distributions that look like this these are all different the narrow peak here versus the broad distribution here the distribution with multiple peaks or multiple modes in this case it has two modes so we call this distribution bimodal or multimodal and then this distribution which is asymmetric has a long tail in the positive direction and a short tail in the negative direction we would say this distribution is skewed so distributions have lots of different shapes and if what we're interested in is the uncertainty you can think about that roughly as the width of the distribution for instance if i'm drawing random numbers from the orange distribution the narrow one here they'll come over roughly this range whereas if i'm drawing from the blue distribution they'll come over roughly this range so if this were say the probability density for position say this is the squared magnitude of the wave function for a particle i know where the particle represented by the orange distribution is much more accurately than the particle represented by the blue distribution so this concept of width of a distribution and the uncertainty in the position for instance are closely related the broadness is related to the uncertainty uh this is fundamental to quantum mechanics so how do we quantify it in statistics the the broadness of a distribution is uh called the variance variance is a way of measuring the broadness of distribution for example so suppose this is my distribution the mean of my distribution is going to fall roughly in the middle here let's say that's the expected value of x if this is the x-axis now if i draw a random number from this distribution i won't always get the expected value suppose i get a value here if i'm interested in the typical deviation of this value from the mean that will tell me something about how broad this distribution is so let's define this displacement here to be delta x delta x is going to be equal to x minus the expected value of x and first of all you might think well if i'm looking for the typical values of delta x let's just try the expected value of delta x well what is that unfortunately the expected value of x doesn't really work for this purpose because delta x is positive if you're on this side of the mean and negative if you're on this side of the mean so the expected value of delta x is zero sometimes it's positive sometimes it's negative and they end up cancelling out now if you're interested in only positive numbers the next guess you might come up with is let's use not delta x but let's use the absolute value of delta x what is that well absolute values are difficult to work with since you have to keep track of whether a number is positive or negative and keep flipping signs if it's negative so this turns out to just be kind of painful what is this what statisticians and physicists do in the end then is instead of taking the absolute value of a number just to uh make it positive we square it so you calculate the expected value of the squared deviation sort of the mean squared deviation um this has a name in statistics it's written as sigma squared and it's called the variance to do an example let's do a discrete example suppose i have two probability distributions all with equally likely outcomes say the outcomes of one distribution are one two and three while the outcomes for the second distribution are 0 2 and 4. photographically these numbers are more closely spaced than these numbers so i would expect the broadness of this distribution to be larger than the broadness of this distribution you can calculate this out by calculating the mean squared deviation so first of all we need to know the mean the expected value of x is 2 in this case and also in this case knowing the expected value of x you can calculate the deviations so let's say delta x here is going to be -1 0 and 1 are the possible deviations from the mean for this probability distribution whereas in this case it's minus 2 0 and 2. then we can calculate the delta x squareds that are possible and you get 1 0 and 1 for this distribution and 4 0 and 4 for this distribution now when you calculate the mean of these squared deviations in this case the expected value of the squared deviation is two thirds whereas in this case the expected value of the squared deviation is eight thirds so indeed we did get a larger number for the variance in this distribution so you can think of that as the definition this is not the easiest way of calculating the variance though it's actually much easier to calculate the variance as an expected value of a squared quantity and an expected m minus the square of the expected value of the quantity itself so the mean of the square minus the square of the mean if that helps you to remember it you can see how this results fairly easily by plugging through some basic algebras so given our definition the expected value of delta x squared we're calculating an expected value so suppose we have a continuous distribution now the continuous distribution expected value has an integral in it so we're going to have the integral of delta x squared times rho of x dx now delta x squared we can we know what delta x is delta x is x minus the expected value of x so we can plug that in here and we're going to get the integral of x minus expected value of x squared times rho of x dx i can expand this out and i'll get integral of x squared minus 2 x expected value of x plus expected value of x quantity squared rho of x dx and now i'm going to split this integral up into three separate pieces first piece integral of x squared row of x dx second piece integral of 2 x expected value of x rho of x dx and third piece integral of expected value of x squared rho of x dx now this integral you recognize right away this is the expected value of x squared this integral i can pull this out front since this is a constant this is just a number this is the expected value so this integral is going to become 2 i can pull the 2 out of course as well 2 times the expected value of x and then what's left is the integral of x rho of x dx which is just the expected value of x this integral again this is a constant so i can pull it out front and when i do that i end up with just the integral of rho of x dx and we know the integral of rho of x dx over the entire domain i should specify that this is the integral from minus infinity to infinity now all of these are integrals from minus infinity to infinity the integral of minus infinity to infinity of rho of x dx is one so this after i pull the expected the expected value of x quantity squared out is just going to be the expected value of x quantity squared so this is expected value of x squared this is well i can simplify this as well this is the expected value of x quantity squared as well so i'm going to erase that and say squared there so i have this minus twice this plus this and in the end that gives you expected value of x squared minus the expected value of x squared so mean of the square minus the square of the mean to check your understanding of how to use this formula i'd like you to complete the following table now i'll give you a head start on this if your probability distribution is given by 1 2 4 5 and 8 all equally likely you can calculate the mean now once you know the mean you can calculate the deviations x minus the mean which i'd like you to fill in here then square that quantity and fill it in here and take the mean of that squared deviation same as what we did when we talked about the variance as the mean squared deviation then taking the other approach i'd like you to calculate the squares of all the x's and calculate the mean square you know the mean you know the mean square you can calculate this quantity mean of the square minus the square of the mean and you should get something that equals the mean squared deviation that's about it for variants but just to say a little bit more about this variance is not the end of the story it turns out there's well there's more i mentioned the distributions that we were talking about earlier on on the first slide here keep forgetting to turn my ruler off the distributions that look like this versus distributions that look like this this is a question of symmetry and the mathematical name for this is skew or skewness there's also distributions that look like this versus distributions that look like this and this is what mathematically this is called kurtosis which kind of sounds like a disease or perhaps a villain from a comic book kurtosis has to do with the relative weights of things near the peak versus things in the tails now mathematically speaking you know the variance sorry let me go back a little further you know the mean that was related to the integral of x rho of x dx we also just learned about the variance which was related to the integral of x squared rho of x dx it turns out the skewness is related to the integral of x cubed rho of x dx and the kurtosis is related to the integral of x to the fourth rho of x dx at least those are common ways of measuring skewness and kurtosis these are not exact formulas for skewness and kurtosis nor is this an exact formula for the variance of course so i'm taking some liberties with the math but you can imagine well what happens if you take the integral of x to the fifth row of x dx you could keep going and you would keep getting properties of the probability distribution that are relevant to its shape now you won't hear very much about skewness and kurtosis and physics but i thought you should know that this field does sort of continue on for the purposes of quantum mechanics what you need to know is that variance is related to the uncertainty and we will be doing lots of calculations of variance on the basis of probability distributions derived from wave functions in this class we talked a little bit about the probabilistic interpretation of the wave function psi that's one of the really remarkable aspects of quantum mechanics that there are probabilities rolled up in your description of the physical state we also talked a fair amount about probability itself and one of the things we learned was the probabilities had to be normalized meaning the total sum of all of the probable outcomes the probabilities of all of the outcomes in a probability distribution has to equal 1. that has some implications for the wave function especially in the context of the schrodinger equation so let's talk about that in a little more detail normalization in the context of a probability distribution just means that the integral from minus infinity to infinity of rho of x dx is equal to one you can think about that as the sort of extreme case of the probability that say x is between a and b being given by the prob the integral from a to b of rho of x dx in the context of the wave function that that statement becomes the probability that the particle is between a and b is given by the integral from a to b of the squared magnitude of psi of x integrated between a and b so this is the same sort of statement you're integrating from a to b and in the case of the probability density you have just the probability density in the case of the wave function you have the squared absolute magnitude of the wave function this is our probabilistic interpretation we're sort of making an analogy between psi squared magnitude and a probability density this normalization condition then has to also hold for psi if the squared magnitude of psi is going to is going to be treated as a probability density so integral from minus infinity to infinity of squared absolute magnitude of psi dx has to equal 1. this is necessary for our statistical interpretation of the wave function this brings up an interesting question though because not just any function can be a probability distribution therefore this normalization condition treating size or probability density means there are some conditions on what sorts of functions are allowed to be wave functions this is a question of normalizability suppose for instance i had a couple of functions that i was interested in say one of those functions looks sort of like this keeps on rising as it goes to infinity if i wanted to consider the squared magnitude of this function this is our possible psi this is our possible psi squared sorry about the messy there this function since it's going to you know it's it's continuing to increase as x increases both in the negative direction and in the positive direction its squared magnitude is going to look something like this i can do a little better there sorry if i tried to say calculate the integral from minus infinity to infinity of this function i've got a lot of area out here from say 3 to infinity where the wave function is positive this would go to infinity therefore what that means is that this function is not normalizable not all functions can be normalized if i drew a different function for example something that looked maybe something like this its squared magnitude might look something like this there is a finite amount of area here so if i integrated the squared magnitude of the blue curve i would get something finite what that means is that whatever this function is i could to multiply or divide it by a constant such that this area was equal to one i could take this function and convert it into something such that the integral from minus infinity to infinity of the squared magnitude of psi equaled one and it obeyed our sort of statistical constraint on the probability distribution in order for this to be possible psi has to have this property and the mathematical way of stating it is that psi must be square integrable and all this means is that the integral from minus infinity to infinity of the squared magnitude of psi is finite you don't get zero you don't get infinity in order for this square integrability to hold for example though you need a slightly weaker condition that psi goes to 0 as x goes to either plus or minus infinity it's not possible to have a function that stays non-zero or goes to infinity itself as x goes to infinity and still have things be integrable like i said if this holds if this integral here is finite you can convert any function into something that is normalized by just multiplying or dividing by a constant is that possible though in the schrodinger equation does multiplying or dividing by a constant do anything well the schrodinger equation here you can just glance at it and see that multiplying and dividing by a constant doesn't do anything the short injury equation is i h bar partial derivative with respect to time of psi equals minus h bar squared over 2m second derivative of psi with respect to position plus the potential times psi now if i made the substitution psi went to some multiple or some constant a multiplied by psi you can see what would happen here i would have psi times a here i would have psi times a and here i would have psi times a so i would have an a here an a here and an a here so i could divide through this entire equation by a and all of those a's would disappear and i would just get the original schrodinger equation back what that means is that if psi solves the schrodinger equation a psi does 2 i'll just say a psi works now this is only if a is a constant does not depend on time does not depend on space if a depended on time i would not be able to divide it out of this partial derivative because the partial derivative would act on on that a same goes for if a was a function of space if a was a function of space i wouldn't be able to divide it out of this partial derivative with respect to x so this only holds if a is a constant that means that i might run into some problems with time evolution i can choose a constant and i can multiply psi by that constant such that psi is properly normalized at say time t equals zero but will that hold for future times it's a question of normalization and time evolution what we're really interested in here is the integral from minus infinity to infinity of psi of x and time squared dx if this is going to always be equal to 1 supposing it's equal to one at some initial time what we really want to know is what the time derivative of this is if the time derivative of this is equal to zero then we'll know that whatever the normalization of this is it will hold throughout the evolution of the well throughout the evolution of the wave function now i'm going to make a little bit of simplifying notation here and i'm going to drop the integral limits since it takes a while to write and we're going to multi or sorry we're going to manipulate this expression a little bit we're going to use the schrodinger equation we're going to use the rules of complex numbers i'm going to use the rules of differential calculus i'm going to get something that will show that indeed this does hold so let's step through that manipulations of the short injury equation like this are a little tricky to follow so i'm going to go slowly and if it seems like i'm being extra pedantic please bear with me some of the details are important so the first thing that we're going to do pretty much the only thing that we can do with this equation is we're going to exchange the order of integration and differentiation instead of differentiating with respect to time the integral with respect to x we're going to integrate with respect to x of the time derivative of this psi of x and t quantity squared basically i've just pushed the derivative inside the integral notationally speaking i'm going to move some stuff around here give myself a little more room and notationally oops didn't mean to change the colors notationally speaking here the d dt became a partial derivative with respect to time the total derivative d by dt is now a partial what the notation is keeping track of here is just the fact that this is a function only of time since you've integrated over x and you've substituted in limits whereas this is a function of both space and time so whereas this derivative is acting on something that's only a function of time i can write it as a simple d by dt a total derivative in this case since what the derivative is acting on is a function of both position and time i have to treat this as a partial derivative now so the next thing that we're going to do aside from after pushing this derivative inside and converting it to a partial derivative is rewrite this squared absolute magnitude of psi as psi star times psi now the squared absolute magnitude of a complex number is equal to the complex number times its complex conjugate it's just simple complex analysis rules there so what we've got is the integral of the partial derivative with respect to time of psi star times psi integral dx now we have a time derivative applied to a product we can apply the product rule from differential calculus what we end up with is the integral of the partial derivative with respect to time of psi star times psi plus psi star partial derivative of psi with respect to time that's integrated dx now what i'm going to do is i'm going to notice these partial derivatives with respect to time and i'm going to ask you to bear with me for a minute while i make a little more space it's probably a bad sign if i'm running out of space on a computer where i have effectively infinite space but bear with me the partial derivatives with respect to time appear in the schrodinger equation i h bar d by dt of psi equals minus h bar squared over 2m partial derivative second partial derivative of psi with respect to position plus potential times psi these are the time derivatives that i'm interested in i can use the schrodinger equation to substitute in say the right hand side for these time derivatives both for psi star and for psi so first i'm going to manipulate this by dividing through by i h bar which gives me these partial psi partial time equals i h bar over 2m second partial of psi with respect to x minus uh where did it go i v over h bar psi so that can be substituted in here i also need to know something for the complex conjugate of psi so i'm going to take the complex conjugate of this entire equation what that looks like is partial derivative of psi star with respect to time now i'm taking the complex conjugate of this so i have a complex part here the sign of that needs to be flipped and i have a complex number here that needs to be complex conjugated since the complex conjugate of a product is the product of the complex conjugates what that means is this is going to become minus i h bar over 2m d squared psi star dx squared sorry i forgot the squared there my plus i v over h bar sine so i've just gone through and changed the signs on all of the imaginary parts of all these numbers psi became psi star i became minus i minus i became i this can be substituted in for that what you get when you make that substitution this equation isn't really getting simpler is it it's getting longer what you get is the integral of something i'll put an open square brackets at the beginning here i've got this equation minus i h bar over 2m second partial derivative of psi star partial x squared plus i v over h bar psi star that's multiplied by psi from here so i've just substituted in this expression for this now the next part i have plus psi star and whatever i'm going to substitute in from this which is what i get from this version of the schrodinger equation here i h bar over 2m second partial derivative of psi with respect to x minus i v over h bar psi close parentheses close square brackets and i'm integrating dx now this doesn't look particularly simple but if you notice what we've got here this term if i distributed this psi in would have i v over h bar psi star times psi this term if i distributed this psi star in would have an i v over h bar psi star and psi this term has a plus sign this term has a minus sign so these terms actually cancel out what we're left with then to rewrite things both of the terms that remain have this minus i h bar over 2 m out front so we're going to have equals to i h bar over 2 m and here i have a minus second partial derivative of psi star with respect to x times psi and here i have plus psi star times the corresponding second partial of psi with respect to x and this is integrated dx is that all right yes now what i'd like you to notice here is that we've got d by dx and we've got an integral dx we don't have any time anymore so we're making progress and we're actually almost done where where did we get so far we started with the time derivative of this effective total probability which should have been equal to one if the which would be equal to one if this were proper probability distribution but we're just considered with the time evolution since we know that we whatever psi is we can multiply it by some constant to make it properly normalized at a particular time now we're interested in the time evolution we're looking at the time derivative of this and we've gone to this expression which has complex conjugates of psi and second partial derivatives with respect to x now what i'd like you to do and this is a check your understanding question is think about why this statement is true this is the partial derivative with respect to x of psi star d psi d x minus d psi star dx so sorry i'm saying d i should be saying partial these are partial derivatives this is true and it's up to you to figure out why but since this is true what we're left with is we have our i h bar over 2 m an integral over minus infinity to infinity of this expression partial with respect to x of psi star partial psi partial x minus partial psi star partial x psi we're integrating dx now and this is nice because we're integrating dx of a derivative of something with respect to x so that's easy fundamental theorem of calculus we end up with i h bar over 2m psi star partial psi partial x minus partial psi star partial x psi evaluated at the limits of our integral which are minus infinity to infinity now if psi is going to be normalizable we know something about the value of psi at negative and positive infinity if psi is normalizable psi has to go to zero as x goes to negative and positive infinity what that means is that when i plug in the infinity here psi star d psi d x d psi the x and psi they're all everything here is going to be zero so when i enter in my limits i'm just going to get 0 and 0. so the bottom line here after all of this manipulation is that this is equal to 0. what that means is that the integral from negative infinity to infinity of the squared absolute magnitude of psi as a function of both x and time is equal to a constant put another way time evolution does not effect normalization what that means is that i can take my candidate wave function not normalized integrate it find out what i would have to multiply or divided by to make it normalized and if i'm successful i have my normalized wave function i don't need to worry about how the system evolves in time the schrodinger equation does not affect the normalization so this is that check your understanding question i mentioned the following statement was that crucial step in the derivation and i want you to show that this is true explain why in your own words now to do an example here normalize this wave function what that means is that we're going to have to find a constant and i've already put the constant in the wave function a such that the integral from minus infinity to infinity of the squared absolute magnitude of psi of x in this case i've left the time dependence out is equal to 1. and same as in the last problem the first thing we're going to do is substitute the squared absolute magnitude of psi for psi star times psi the other thing i'm going to do before i get started is notice that my wavefunction is 0 if the absolute value of x is greater than one meaning for x above one or below negative one so instead of integrating from minus infinity to infinity here i'm just going to focus on the part where psi is non-zero and integrate from minus one to one integral from -1 to 1 of psi star which is going to be a e to the ix is going to become e to the minus i x and 1 minus x squared is still going to be 1 minus x squared now i have a complex conjugated a because part of the assumption about normalization constants like this is usually that you can choose them to be purely real so i'm not going to worry about taking the complex conjugate of a just to make my life a little easier psi well that's just right here a e to the ix 1 minus x squared i'm integrating dx this is psi star this is psi integral dx from -1 to 1 should be equal to 1. so let's do this we end up with a squared times the integral from -1 to 1 of e to the minus ix and e to the ix what's e to the minus ix times e to the ix well thinking about this in terms of the geometric interpretation we have e to the ix which is cosine theta plus i sine theta you can think about that as being somewhere on the unit circle at an angle theta minus i x or minus i theta would just be in the exact opposite direction so when i multiply them together i'm going to get something that has the product of the magnitudes the magnitudes are both one [Music] and it's purely real you can see that also by looking at just the the rules for multiplying exponentials like this e to the minus ix times e to the plus ix is e to the minus ix plus ix or e to the 0 which is 1. so i can cancel these out and what i'm left with is 1 minus x squared quantity squared dx plugging through the algebra a little further a squared integral minus 1 to 1 of 1 minus 2x squared plus x to the fourth dx you can do this integral equals a squared two sorry x minus two thirds x cubed plus x to the fifth over five between minus one and one which when you substitute in the limits becomes a squared this x part is going to be 1 minus the other limit minus 1 which is 1 minus minus 1 which is 2 it's going to give me a 2. minus 2 3 of whatever i get from the x cubed which is again going to be 1 minus a minus 1 or 2 plus 1 5 of x to the fifth one minus one to the fifth and one to the fifth again this is all you know basic algebra hopefully this is not too confusing by this point this ends up being a squared times 16 over 15. now going up here if a squared 16 over 15 is going to be equal to 1 a has to be equal to the square root of 15 over 16. that's what it means to normalize a wave function you have something that you think might be your wave function but it's not properly normalized yet so you guess that there's going to be some constant multiplied in and you write down the normalization condition integrating your wave function absolute magnitude squared which usually you will write as the wave function times its complex conjugate typically the complex parts of the wave function will drop out and you'll end up with just something that you can integrate proceeding through the integral you'll end up with some expression that tells you what that constant is so that's the wave function and how it's normalized the time evolution of the normalization and the fact that it's not affected by the short injury equation is really a nice feature of quantum mechanics otherwise quantum mechanics would be completely unworkable a lot of what we've talked about in this lecture has just been how to manipulate the schrodinger equation or how to use the schrodinger equation in manipulations of the wave function and this is the sort of math that quantum mechanics is all about there's calculus there's partial diff partial derivatives total derivatives integrals keeping track of everything is tricky well the result that we've got for normalization here helps i'm rambling though at any rate that's about it for wave functions and normalization and this is nice because all of the pieces that we've described today all fit together and support the statistical interpretation of the wave function as being related to a probability distribution we know in quantum mechanics that all of the information about the physical system is encapsulated in the wave function psi psi then ought to be related to physical quantities for like like example for example position velocity and momentum of the particle we know a little bit about the position we know how to calculate things like the expected value of the position and we know how to calculate the probability that the particles within a particular range of positions but what about other dynamical variables like velocity or momentum the connection with velocity and momentum brings us to the point where we really have to talk about operators operators are one of our fundamental concepts in quantum mechanics and they connect the wave function with physical quantities but let's take a step back first and think about what it means for a quantum system to move the position of the particle we know say the integral from a to b of the squared magnitude of the wave function dx gives us the probability that the particle is between a and b and we know that the expected position is given by a similar expression the integral from minus infinity to infinity of psi star of x times x times psi of x dx now these expressions are related you know by the fact that the squared magnitude of psi is the probability density function describing position and this is really just the calculation of the expected value of x given that probability density function now what if i want to know what the motion of the particle is one way to consider this is suppose i have a box and if i know the particle is say here at time t equals zero what can quantum mechanics tell me about where the particle is later physically speaking you could wait until say t equals one second and then measure the position of the particle and maybe it would be here you could then wait a little longer and measure the particle again maybe at that point it would be here that say t equals two seconds or if i wait a little bit longer and measure the particle yet again at say t equals three seconds maybe the particle would be up here now does that mean that the particle followed a path that looked something like this no we know that the position of the particle is not something that we can observe at any given time with impunity because of the way the observation process affects the wave function back when we talked about measurement we talked about having a wave function that looks something like this a probability density that looks something like that and then after we measure the prob measure the position of the particle the probability density has changed if we say measure the particle to be here the new wave function has to accommodate that new probability density function the fact that measurement affects the system like this means that we really can't imagine repeatedly measuring the position of a particle in the same system what we really need is an ensemble that's the technical term for what we need on what what an ensemble means in this context is that you have many identically prepared systems now if i had many identically prepared systems i could measure the position over and over and over and over again once per system if i have you know 100 systems i could measure this measure the position 100 times and that would give me a pretty good feel for what the probability density for position measurements is at the particular time when i'm making those measurements if i wanted to know about the motion of the particle i could do that again except instead of taking my 100 measurements all at the same time i would take them at slightly different times so instead of this being the same system this would be these would all be excuse me these would all be different systems that have been allowed to evolve for different amounts of time and as such the motion of the particle isn't going to end up looking something like that it's going to end up looking like some sort of probabilistic motion of the wave function in space what we're really interested in here sorry i should make a note of that many sorry single measurement per system this notion of averaging over many identically prepared systems is important in quantum mechanics because of this effect that measurement has on the system so what we're interested in now in the context of something like motion is well can we predict can we predict where the particle is likely to be as a function of time and yes we can and what i'd like to do to talk about that is to consider a quantum mechanical calculation that we can actually do the time derivative of the expected value of position this time derivative tells us how the center of the probability distribution if you want to think about it that way how the center of the wave function moves with time so this time derivative d by dt of the expected value of x that's d by dt of let's just write out the expected value of x integral from minus infinity to infinity of x times psi star of x psi of x where this is the probability density function that describe given by the wave function and this is x we're integrating dx now if you remember when we talked about normalization whether the normalization of the wave function changed as the wave function evolved in time we're going to do the same sort of calculation with this we're going to do some calculus with this expression we're going to apply the schrodinger equation but as before the first thing we're going to do is move this derivative inside the equation this is a total time derivative of something that's a function of in principle position and time i should write these as functions of x and t and what you get when you push that in is as before the integral or the total derivative becomes a partial derivative since x is just the coordinate x in this context of functions of both space and time the total time derivative will not affect the coordinate x even when it comes becomes a partial derivative so what we'll end up with is x times the partial time derivative of psi star psi integral dx i'm not going to write the integral from minus infinity to infinity here just to save myself some time now if you remember this expression the integral or sorry not the not the full integral just the partial time derivative of psi star psi that was what we worked with in the lecture on normalization so if we apply the result from the lecture on normalization and it's equation 126 yes in the book if we apply that you can simplify this down a lot right off the bat and what you end up with is i h bar over 2m times this integral x and then what we substitute in the equation 126 is gives an expression for this highlighted part here in orange and what you get is the partial derivative with respect to x of psi star partial of psi with respect to x minus partial of psi star with respect to x times psi integral still with respect to dx of course now if we look at this equation we're making the same sort of progress we made when we did the normalization derivation we had time derivatives here now we have only space derivatives and we have only space derivatives in an integral over space so this is definitely progress now we can start thinking about what we can do with integration by parts the first integration by parts i'm going to do has the non-differential part just being x and the differential part being dv is equal to you know i'm not going to have space to write this here i've got to move stuff around a little bit so the differential part is dv is the partial derivative well what's left of this equation the partial derivative with respect to x of psi star d psi dx minus d psi dx psi oh sorry d psi star dx psi and then there's the dx from the integral sorry i'm running out of space this differential part here is just this part of the equation now i can take this derivative dudx in my integration by parts procedure d u equals dx and dv here is easy to integrate because this is a derivative so when i integrate the derivative there i'll just end up with v equals psi star d psi dx minus d psi star dx psi now when i actually apply those uh that integration by part the boundary term here with the without the integral in it is going to involve these two so i'm going to have x times psi star partial psi partial x minus partial psi star partial x psi and that's going to be evaluated between minus infinity and infinity the limits on my integral the integral part which comes in with the minus sign is going to be composed of these bottom two terms integral of psi star partial psi partial x minus partial psi star partial x psi and it's integral dx from minus infinity to infinity now what's nice oh you know i forgot something here what did i forget my leading constants i still have this i h bar over 2m out there i h bar over 2m is multiplied by this entire expression now the boundary terms here vanish boundary terms in integration by parts and quantum mechanics will often vanish because if you're evaluating something at say infinity psi has to go to zero at infinity so this term is going to vanish psi star has to go to zero at infinity so this is going to vanish so even though x is going to infinity psi is going to zero and if you dig into the mathematics of quantum mechanics you can show convincingly that the limit as x times psi goes to infinity is going to be zero so this boundary term vanishes both at infinity and at minus infinity and all we're left with is this yes all you're left with is that so i'll write that over i h bar over 2m times the integral of psi star partial psi partial x minus partial psi star partial x psi integral dx i'm actually going to split that up into two separate integrals so i'll stick another integral sign in here and i'll put a dx there and i'll put parentheses around everything so my leading constant gets multiplied in properly and now i'm going to apply integration by parts again but this time just to the second integral here so here we're going to say u is equal to psi and dv is equal to again using the fact that when we do this integral if we can integrate a derivative that potentially simplifies things so this is going to be partial psi star partial x dx so when we derivative take the derivative of this we're going to get d u is equal to partial psi partial x and when we integrate this we're going to get v equals psi star now when we do the integration when we write down the answer from this integration by parts the boundary term here psi star times psi is going to vanish again because we're evaluating it at a region where both psi star and psi well vanish so the boundary term vanishes and you notice i have a minus sign here when we do the integration by parts the integral term has a minus sign in it here so we're going to have the partial psi with respect to x and psi star with a minus sign coming from the integration by parts and a minus sign coming from the leading term here so we're going to end up with the plus sign there so we get a minus from the integral part um what that means though is that i have psi star and partial psi partial x in my integration by parts i end up with partial psi partial x and size star it's the same the fact that i had a minus and another minus means i get a plus so i have two identical terms here the result of this then is i h bar over m i'm adding a half and a half and getting one basically times the integral of psi star partial psi partial x dx and this is going to be something that i'm going to call now the expectation of the velocity vector velocity operator [Music] this is the sort of thing that you get out of operators in quantum mechanics you end up with expressions like this and this i'm sort of equating just by analogy with the expectation of a velocity operator this is not really a probability distribution anymore at least not obviously we started with the probability distribution due to psi the absolute magnitude of psi squared and we end up with the partial derivative on one of the size so it's not obvious that this is a probability distribution anymore and well it's the probability distribution in velocity and it's giving you the expected velocity in some sense in a quantum mechanical sense so this is really a more general sort of thing we have the velocity operator the expectation of the velocity operator oh and operator wise i will try to put hats on things i will probably forget i don't have that much attention to detail when i'm making lectures like this the hat notation means operator if you see something that you really serve as an operator but it doesn't have a hat that's probably just because i made a mistake but this expression for the expectation of the velocity operator is the one we just derived minus i h bar over m times the integral of psi star partial derivative of psi with respect to x integral dx now it's customary to talk about momentum instead of velocity momentum has more meaning because it's a conserved quantity and under you know most physics so we can talk about the momentum operator the expectation of the momentum operator and i'm going to write this momentum operator expression in a slightly more suggestive way the integral of psi star times something in parentheses here which is minus i h bar partial derivative with respect to x i'm going to close the parentheses there put a psi after it and a dx for the integral you have the same sort of expression for the position operator we were just writing that as the expected value of position without the hat earlier but that's going to be the integral of psi star what goes in the parentheses now is just x psi dx so this you recognize is the expectation of the variable x uh subject to the probability distribution given by psi star times psi uh this is slightly more subtle you have psi star and psi which looks like a probability distribution but what you have in the parentheses now is very obviously an operator that does something it does more than just multiply by x it multiplies by minus i h bar and takes the derivative of psi operators in general do that we can write them as say x hat equals x times where there's very obviously something that has to go after the x in order for it to be considered an operator or we can say the same for v hat it's minus i h bar over m times the partial derivative with respect to x where there obviously has to be something that goes here likewise for momentum minus i h bar partial derivative with respect to x something has to go there another example of an operator is the kinetic energy operator usually that's written as t and that's minus h bar squared over 2m you can think of it as the momentum operator squared it's got a second derivative with respect to x and again there very obviously has to be something that goes there the operator acts on the wave function that's what i said back when i talked about the fundamental concepts of quantum mechanics and this is what it means for the operator to act on the wave function the operator itself is not meaningful it's only meaningful in the context when it's acting on away function in general rule color in general the expectation value of some operator which i'll just call q is going to be the integral of psi star q psi dx where q acts on the psi and this is what allows us to say predict the expected value of really any physical quantity you need to know how to write that physical quantity in terms of operators and typically uh typically this q hat here will be written as you know something with x hat the position operator and p hat the momentum operator so the operators can in principle be quite complicated but they're generally expressed in terms of simpler operators like position and momentum to give an example of how this is actually used suppose i want to find the expected value of the momentum for the wave function given by psi psi here is that wave function that i talked about back in the normalization example where i found what the value of a was to normalize this wave function properly i'm just going to leave it as a for now and deal with the wave function part it's also important to note that this wave function has no time dependence and it has no complex part so i'm simplifying things a lot you will work with wave functions that look a lot like this under specific conditions but know that this isn't a general wave function this isn't a solution to any particular schrodinger equation this is just something that i'm i'm cooking up as an example it is a valid wavefunction at least if you consider it at say t equals 0. the goal is to find p-hat expected value and the expected value of something you know expected value of p hat is given by the integral of psi star p hat psi dx which is the integral and this integral will be from minus infinity to infinity but since my wavefunction is zero for x greater than for absolute value of x greater than one i'm going to drop the condition here and make my integral only go from minus one to one and for psi star i have a complex conjugate of a assuming it's real 1 minus x squared complex conjugate of 1 minus x squared x is real no complex conjugate or complex conjugate has no effect here times the momentum operator which is going to be i h bar minus i h bar partial derivative with respect to x of psi without the complex conjugate a 1 minus x squared all integrated dx now i can simplify this a little bit you end up with the integral from minus one to one oh sorry let me pull my ih bar out front minus i h bar integral from minus one to one of i can pull out the a squared as well integral from minus 1 to 1 of 1 minus x squared times the second or sorry the derivative now of 1 minus x squared well the derivative of one and this is a sum so i can push the derivative in so the derivative of one is zero and the derivative of minus x squared is minus two x so i'm gonna have a minus two x here and just write two x and i'll put a plus sign out front to compensate for the minus sign and this is integrated dx this then is i h bar let's say 2 i h bar i can pull the 2 out as well a squared integral from minus 1 to 1 of x minus x cubed dx now right away hopefully you can look at this and say x minus x cubed this is an odd function meaning if i actually plotted this it would look something like this this is my coordinate system x minus x cubed it's going to look something like this where um f of x equals minus f of minus x what that means is when i integrate this from minus one to one over an interval it starts equally far into the negative side as it goes into the positive side i'm going to get zero um so you can just write down just by looking at this that the integral is going to be zero um you can also of course plug this in you'll get x squared over two and x to the fourth over four for these integrals and then you'll plug in minus one and one and find out that minus and plus doesn't make any difference for x squared and x to the fourth and you'll end up with zero that way as well but the bottom line here is that the expected value of momentum is equal to zero and this makes sense the reason this makes sense is that my wave function is symmetric if that's a coordinate system we can say this is the x-axis if this is going to be my wave function it could look i don't want the ruler my wave function would look something like this where it goes from -1 to 1 and it's zero outside of that range now if you had an expected value of momentum that was non-zero here that would suggest the wave function was on average moving and if you look at this wave function psi it doesn't really seem to be moving there's no preferred direction to this if you if you showed me this sign asked is it moving to the left or to the right i wouldn't be able to tell you and it makes sense then that the expected value of the momentum is zero the wave function is effectively not moving or the particle represented by this wave function has no momentum on average now this doesn't mean that if you measure the momentum of the particle as described by this wave function you would always get zero it just means you would get zero on average so that's a little bit about operators and we'll be working with operators much more in the future the basic concepts that i want you guys to take away from this lecture is that the expected value of some general operator is what you get if you take psi star and psi and sandwich your operator in between them and integrate and then know that the operator may actually do something to psi it's not just a multiplication so i cannot say that this q hat is equal to the integral of q hat psi star psi dx i can't just move the q around i'm not allowed to do that because q is acting on this psi really this is just reasoning by analogy right now there's no rigorous formal proof we'll come back to discuss the math of quantum mechanics quite a lot later on but that's it for this lecture as a check your understanding i'd like you to find the expected value of the kinetic energy operator for this wave function remember that the kinetic energy operator t hat is equal to minus h bar squared over 2m times the second derivative with respect to x and i almost wrote a psi there but i'm not supposed to write a psi the operator has to act on psi so there should be some space here for psi to go so go back to that same formula for the expectation use the kinetic energy operator instead and calculate the expected value of the kinetic energy for this wave function as an introduction to the uncertainty principle we're going to talk about waves and how waves are related to each other we'll get into a little bit of the context of fourier analysis which is something we'll come back to later but the overall context of this lecture is the uncertainty principle and the uncertainty principle is one of the key results from quantum mechanics and it's related to what we discussed earlier in the context of the boundary between classical physics and quantum physics quantum mechanics has these inherent uncertainties that are built into the equations built into this state built to the nature of reality that we really can't surmount and the uncertainty principle is one way in which those is the mathematical description uh it's those relationships that i gave you earlier delta p delta x is greater than about equal to h bar over two i think i just said greater than about equal to h bar earlier we'll do things a little more mathematically here and it turns out there's a factor of two there to start off though conceptually think about position and wavelength and this really is now in the context of a wave so say i had a coordinate system here something like this and if i had some wave with a very specific wavelength you can just think about it as a sinusoid if i asked you to measure the wavelength of this wave you could take a ruler and you could plop it down there and say okay well how many inches are there from peak to peak or from zero crossing to zero crossing or if you really wanted to you could get a tape measure and measure many wavelengths one two three four wavelengths in this case that would allow you to very accurately determine what the wavelength was if on the other hand the wave looked more like this give you another coordinate system here the wave looked something like this you wouldn't be able to measure the wavelength very accurately you could as usual put your ruler down on top of the wave for instance and count up the number of inches or centimeters from one side to the other but that's just one wavelength it's not nearly as accurate as say measuring four wavelengths or 10 wavelengths or 100 wavelengths you can think of some limiting cases suppose you had a wave with many many many many many oscillations it looks like i'm crossing out the wave underneath there so i'm going to erase this in a moment but if you had a wave with many wavelengths and you could measure the total length of many wavelengths you would have a very precise measurement of the wavelength of the wave the opposite is the case here you only have one wavelength you can't really measure the wavelength very accurately what you can do however is measure the position very accurately here i can say pretty certainly the wave is there you know plus or minus a very short spread in position the other hand here i cannot measure the position of this wave accurately at all you know if this thing continues i can't really say where the wave is it's not really a sensical question to ask where is this wave this wave is everywhere these are the sorts of built-in uncertainties that you get out of quantum mechanics where is the wave the wave is everywhere it's a wave it doesn't have a local position it turns out if you get into the mathematics of fourier analysis that there is a relationship between the spread of wavelengths and the spread of positions if you have a series of waves of all different wavelengths and they're added up the spread in the wavelength will is related to the spread in positions of the sum and we'll talk more about fourier analysis later but for now just realize that this product is always going to be greater than or equal to about one wavelength is something with units of inverse length and length when the position of course is something with units of length so the dimensions of this equation are sort of a guideline wavelength and position have this sort of relationship and this comes from fourier analysis so how do these waves come into quantum mechanics well waves in quantum mechanics really first got their start with louis de bruy i always thought his name was pronounced de broglie but it's well he's french so there's all sorts of weird pronunciations in french is my best guess at how it would probably be pronounced de voy proposed that matter could travel in waves as well and he did this with a interesting argument on the basis of three fundamental equations that had just recently been discovered when he was doing his analysis this was in his phd thesis by the way e equals m c squared you all know that equation you all hopefully also know this equation e equals h f planck's constant times the frequency of a beam of light is the energy associated with a quanta of light this was another one of einstein's contributions and it has to do with his explanation of the photoelectric effect the final equation that de bruy was working with was c c equals f lambda the speed of light is equal to the frequency of the light times the wavelength of the light and this is really not true just for light this is true for any wave phenomenon the speed the frequency and the wavelength are related now if these expressions are both equal to waves or are both equal to energy then i ought to be able to say m c squared equals h f and this expression tells me something about f it tells me that f equals c over lambda so i can substitute this expression in here and get m c squared equals h c over lambda now i can cancel out one of the c's and i'm left with m c equals h over lambda now what du voy said was this this is like momentum so i'm going to write this equation as p equals h over lambda and then i'm going to wave my hands extraordinarily vigorously and say while this equation is only true for light and this equation is only true for waves this is also true for matter how actually this happened in the context of quantum mechanics and the early historical development of quantum mechanics is deployed noticed that the spectrum of the hydrogen atom this bright line spectra that we were talking about where a hydrogen atom emits light of only very specific wavelengths intensity as a function of wavelength looks something like this that that could be explained if he assumed that the electrons were traveling around the nucleus of the hydrogen atom as waves and that only an integer number of waves would fit the one that i just drew here didn't end up back where it started so that wouldn't work if you had a wavelength that looked something like this going around say three full times in a circle that would potentially count for these allowed emission energies that was quite a deep insight and it was one of the things that really kicked off quantum mechanics at the beginning the bottom line here for our purpose is that we're talking about waves and we're talking about matter waves so that uncertainty relation or the relationship between the spreads of wavelengths and the spreads in positions that i mentioned in the context of fourier analysis will also potentially hold for matter and that gets us into the position momentum uncertainty relation the wave momentum relationship we just derived on the last slide was p equals h over lambda this tells you that the momentum and the wavelength are related from two slides ago we were talking about waves and whether or not you could say exactly where a wave was we had a relationship that was something like delta lambda the spread in wavelengths times the spread and positions of the wave is always greater than about equal to one combining these relationships together in quantum mechanics and this is not something that i'm doing rigorously now i'm just waving my hands gives you delta p delta x is always greater than about equal to h bar over two and this is the correct mathematical expression of the heisenberg uncertainty principle that we'll talk more about and derive more formally in chapter three but for now just realize that the position of a wave the position of a particle are on certain quantities and the uncertainties are related by this which in one perspective results from consideration of adding many waves together in the context of fourier analysis which is something we'll talk about later as well extended through the use of or the interpretation of matter as also a wave phenomenon to check your understanding here are four possible wave packets and i would like to rank i would like you to rank them in two different ways one according to the uncertainties in their positions and two according to the uncertainties in their momentum so if you consider say wave b to have a very certain position you would rank that one highest in terms of the certainty of its position perhaps you think wave b has a very low uncertainty in position you would put it on the other end of the scale i'm looking for something like the uncertainty of b is greater than the uncertainty of a is greater than the uncertainty of d is greater than the uncertainty of c for both position and momentum the last comment i want to make in this lecture is on energy time uncertainty this was the other equation i gave you when i was talking about the boundary between classical physics and quantum physics we had delta p delta x is greater than or equal to h bar over two and now we also had uh excuse me for a moment here delta e delta t greater than about equal to h bar over two same sort of uncertainty relation except now we're talking about spreads in energy and spreads in time i'd like to make an analogy between these two equations delta p and delta x delta p according to dubai is related to the wavelength which is sort of a spatial frequency it's uh the frequency of the wave in space delta x of course is just well i'll just say that's a space and these are related according to this equation in the context of energy and time we have the same sort of thing delta t well that's pretty clear that's time and delta e well that then therefore by analogy here has to have something to do with the frequency of the wave now in time that's simple that's just the frequency the fact that these are also related by an uncertainty principle tells you that there's something about energy and frequency and time and this is something that we'll talk about in more detail in the next lecture when we start digging into the schrodinger equation and the time dependent true in your equation and deriving the time independent schrodinger equation which will give us the relationship exactly but for now position and momentum energy and time we're all talking are both talking about sort of wave phenomenon except in the context of position and momentum you're talking about wavelength frequency of the wave in space whereas energy and time you're talking about the frequency of the wave in time how quickly it oscillates that's about all the uncertainty principle as i've said is something that we'll treat in much more detail uh in chapter three but for now the uncertainty principle is important because you have these equations and these are fundamental properties of the universe if you want to think of them that way and there's something that we're going to be working with as a way of checking the validity of quantum mechanics throughout the rest of the next throughout chapter 2. um that's all for now you just need to conceptually understand how these wave lengths and positions or frequencies and times are interrelated the last few lectures have been all about the wave function psi and since psi is such an important concept in quantum mechanics really the first entire chapter of the textbook is devoted to the wave function and all of its various properties since we've reached the end of chapter one now now is a good opportunity to go and review the key concepts of quantum mechanics in particular the wave function and how it is related to the rest of quantum mechanics the key concepts as i stated them earlier were operators the schrodinger equation and the wave function operators are used in the schrodinger equation and act on the wave function your friend and mine psy what we haven't really talked about a lot yet is how to determine the wave function and the wave function is determined as solutions to the schrodinger equation that's what chapter 2 is all about solving the schrodinger equation for various circumstances the key concepts that we've talked about so far operators and the wave function conspire together to give you observable quantities things like position or momentum or say the kinetic energy of a particle but they don't give us these properties with certainty in particular the wave function really only gives us probabilities and these probabilities don't give us really any certainty about what will happen uncertainty is one of the key concepts that we have to work with in quantum mechanics so let's take each of these concepts in turn and talk about them in a little more detail since now we have some actual results that we can use some mathematics we can put more meat on this concept map than just simply the concept map first the wave function the wave function psi does not tell us anything with uh with certainty and it's a good thing too because psi as a function of position and time is complex it's not a real number and it's hard to imagine what it would mean to actually observe a real number so the wave function is already on somewhat suspect ground here but it has a meaningful connection to probability distributions if we more or less define the squared modulus the absolute magnitude of the wave function to be equal to a probability distribution this is the probability distribution for what well it's the probability distribution for outcomes of measurements of position for instance you can think about this as a probability distribution for where you're likely to find the particle should you go looking for it this interpretation as a probability distribution requires the wave function to be normalized namely that if i integrate the squared magnitude of the wave function over the entire space that i'm interested in i have to get one this means that if i look hard enough for the particle everywhere i have to find it somewhere the probability distributions as i mentioned earlier don't tell you anything with certainty in particular there is a good deal of uncertainty which we express as a standard deviation or a variance for instance if i'm interested in the standard deviation of the uncertainty or standard deviation of the position excuse me it's most easy to express as the variance which is the square of the standard deviation and the square of this standard deviation or the variance is equal to the expectation value of the square of the position minus the square of the expectation value of the position and we'll talk about expectation values in a moment expectation values are calculated using expressions with operators that look a lot like these sorts of integrals in fact i can re-express this as the expectation of the square in terms of a probability distribution is just the x squared times multiply multiplied by the probability distribution with respect to x integrated overall space this is the expectation of x squared i can add to that or subtract from that sorry the square of the expectation of x which has a very similar form and that gives us our variance so our wave function which is complex gives us probability distributions which can be used to calculate expectation values and uncertainties this probabilistic interpretation of quantum mechanics gets us into some trouble pretty quickly i'm going to move this up now give myself some more space namely with the concept of wave function collapse now collapse bothers a lot of people and it should this is really a philosophical problem with quantum mechanics that we don't really have a good interpretation of what quantum mechanics really means for the nature of reality but the collapse of the wave function is more or less a necessary consequence of the interpretation of the wave function as a probability distribution if i have some states some space some coordinate system and i plot on this coordinate system the squared magnitude of psi this is related to our probability distribution with respect to position if i then measure the position of the particle what i'm going to get is say i measure the particle to be here now if i measure the position of the particle again immediately i should get a number that's not too different than the number that i just got this is just sort of to make sure that if i repeat a measurement it's consistent with itself that i don't have particles jumping around truly randomly if i know the position i know the position that's a reasonable assumption what that means is that the new probability distribution for the position of the particle after the measurement is very sharply peaked about the position of the measurement if this transition from a wavefunction for instance that has support here to a wavefunction that has no support here did not happen instantaneously it's imaginable that if i tried to measure the particle's position twice in very rapid succession that i would have one particle measured here and another particle measured here does that really mean i have one particle or do i have two particles these particles could be separated by quite a large distance in space and my measurements could be not separated by very much in time so i might be getting into problems with special relativity in the speed of light and these sorts of considerations are what leads to the copenhagen interpretation of quantum mechanics which centers on this idea of wave functions as probability distributions and wave function collapse as part of the measurement process now i mentioned operators in the context of expectation values operators are our second major concept in quantum mechanics what about operators in the wave function well operators let's just write a general operator as q hat hats usually signify operators operators always act on something you can never really have an operator in isolation and what the operators act on is usually the wave function we have a couple of operators that we've encountered namely the position operator x hat which is defined as x times and what's it multiplied by well it's multiplied by the wave function we also have the momentum operator p hat and that's equal to minus i h bar times the partial derivative with respect to x of what well of the wave function we also have the kinetic energy which i'll write as k e hat you could also write it as t hat that operator is equal to minus h bar squared over 2m times the second derivative with respect to position of what well of the wave function and finally we have h hat the hamiltonian which is an expression of the total energy in the wave function it's a combination of the kinetic energy operator here which you can see first of all as p squared we have a second derivative with respect to position and minus h bar squared this is just p squared divided by 2m p squared over 2m is a classical kinetic energy the analogy is reasonably clear there you add a potential energy term in here and you get the hamiltonian now expectation values of operators like this are calculated as integrals the expectation value of q for instance is the integral of psi star times q acting on psi overall space this bears a striking resemblance to our expression for instance for the expectation of the position which was the integral of just x times rho of x where rho of x is now given by the absolute magnitude of psi squared which is given by psi star times psi now basically the pattern here is you take your operator and you sandwich it between psi star and psi and you can think about this position as being sandwiched between psi star and psi as well because we're just multiplying by it doesn't really matter where i put it in the expression the sandwich between psi star and psi of the operator is more significant when you have operators with derivatives in them but i'm getting a little long-winded about this perhaps it's suffice it to say that operators in the wave function allow us to calculate meaningful physical quantities like x the expectation of position this is more or less where we would expect to find the particle or the expectation of p and i should be putting hats on these since technically they're operators the expectation of p is more or less the expected value of the momentum the sort of sorts of momentum momenta that the system can have or the expectation value of h the typical energy the system has and all of these are tied together in the context of uncertainty for instance if i wanted to calculate the uncertainty in the momentum i can do that with the same sort of machinery we used when we were talking about probability that i calculate the expectation of p squared and i subtract the expectation of p squared so the expectation of the square minus the square of the expectations is directly related to the uncertainty so that's a little bit about operators and a little bit about the wave function and a little bit about how they're used operators acting on the wave function calculating expectations in the context of the wave function being treated as a probability distribution now where are we all going with this we're going towards the schrodinger equation the schrodinger equation to write it out is i h bar partial derivative with respect to time of the wave function and that's equal to minus h-bar squared over 2m second partial derivative with respect to position of the wave function plus some potential function function of x times the wave function now the wave function psi here i've left it off as a function of position and time so this is really the granddaddy of them all this is the equation that we will be working with throughout chapter two we will be writing this equation for various scenarios and solving it and describing the properties of the solutions so hopefully now you have a reasonable understanding of the wave function and the short and enough understanding of operators to understand what to do with the wave function the sorts of questions you can ask of the wavefunction are things like what sorts of energy does this system have how big is the spread in momenta where am i likely to find the particle if i went looking for it but all of that relies on having the function and you get the wave function by solving the schrodinger equation so that's where we're going with this and that's all of the material for chapter one and without further ado moving on to the next lecture we'll start solving the schrodinger equation thanks for watching please subscribe and don't miss out on new videos and lectures you