Understanding the Quantum Hall Effect

[Music] so let me connect with the discussion that we were having having um last uh time uh the previous discussion about the quantization of the hall plateaus so this is the uh effect that we have seen uh that this is the the quantum hall effect or the quantise plateaus which it is famous for uh so uh we have to understand these origin of these plateaus and uh uh how they are so robust to uh disorder and impurities as they are uh and also we we have to understand these green plots which represent the row XX or the magnetude resistance or the longitudinal resistance uh so let me just write that row XY would be called as Hall uh resistivity uh and row XX will be called as uh one can call it as a longitudinal resistance or resistivity uh and or it's also called as a Magneto resistivity basically the resistivity that arises because of the magnetic field and um uh so these are uh as you see the green plots actually vanish most of the time excepting when the red curves show a jump from one Plateau to another uh then it sort of shoots up and then again goes to zero uh and it remains zero till the the plateaus exist in the hall conductivity this is uh uh a Quantum phenomena as uh we have said earlier and we'll be talking about this um so the quantum mechanical nature of the electrons in presence of a magnetic field uh will have to be worked out and we have to understand how these plateaus arise and why they are so robust so much so that they actually give the Benchmark of resistance uh the plateus actually h e Square sets the scale of resistance which is 25.81% or how the hall voltage develops uh in the transverse ages of the sample when there's a longitudinal current being sent uh from one end to another and um that was a analysis that was done and at equilibrium when the force vanishes uh but we haven't talked about uh what is a trajectory of the uh particle uh in presence of an electromagnetic field and this is important because the trajectory uh being actually a path that's followed by a particle uh a Charged particle uh should not really depend upon whether we are talking about classical mechanics or quantum mechanics um of course in quantum mechanics we uh rarely discuss or uh worry about uh the trajectory of the particle for the reason that the trajectory is uh uh it's there is an uncertainty principle which is underlying that and which governs the trajectory so let's look at the classical trajectory of a charge particle and this is done with an intention that we'll be uh dealing with a quantum mechanical system uh soon okay so uh let us um talk about two cases uh the case one being uh where there is no electric field uh and there is only a magnetic field which is uh constant which is not equal to zero it's finite and it's uniform okay uh so this is the situation uh one that's case one say and um the energy is constant so uh please do not confuse between the electric field e and the energy energy is a scalar so energy is a constant uh and um energy is constant because of the reason that the power delivered uh is equal to zero which is U uh nothing but the low range force is V cross B and that's why this is equal to zero so we can write this as MV do a V Dot which is equal to DDT of uh half m v² and this is equal to uh Q into V do V CR B which is equal to zero okay so this is your uh F do V uh because your f is the low range Force when there's no electric field it's only the magnetic field will create a lorange force so uh so V do V cross B equal z uh the reason is simple this V cross B is a vector which is perpendicular to both V and B and that's why if you if you take a DOT product with v it will give you zero and that's why DDT of half MV s equal to 0 which means half MV squ equal to constant okay and this is what we mean by uh the energy being a constant that's the only energy that's uh left in the uh problem so this um you know um results in uh separate equations or separate Dynamics uh for the longitudinal and the transverse Direction so if you write your V to be having a parallel component and a perpendicular component and these Dynamics are uncoupled or decoupled and uh the the motion takes place independent of each other in absence of an electric field okay okay so let's talk about uh the perpendicular motion okay and what do we mean by the perpendicular motion so we have uh mvx dot this is equal to uh Q vyb so this is uh say equation number one and the equation number two will be MV y dot = to Min - Q vxb We are continuing with the same situation where e equal to 0 and uh uh B is uniform and not equal to zero and uh we are just um discussing or rather deliberating upon the nature of the motion and this is quite instructive if you take a double derivative of this so uh differentiate with respect to uh with respect to time T and then of course uh use to so then mvx X double dot so this is equal to Q VY dot by B I mean into B and then I can take this m to be um in the denominator uh this is in the denominator so this now I'll use uh VY dot to be here so this is equal to Q into B by m and then this is equal to - Q into VX B uh / M so this becomes equal to uh minus um QB uh by m squ and VX and uh this is equal to uh nothing but uh minus uh which we have written it earlier that Omega b² VX okay or Omega B is nothing but equal to uh QB over M which we have called as the cyclotron frequency even earlier so just a quick recap we are talking about uh no electric field and a Charged particle is moving with a velocity V which is has components VX and VY and uh this is um uh this a b there that's a magnetic field because of magnetic field there's a lowen force and in this particular case we know that the energy is constant because the F do V equal to Z and now we are analyzing the motion by writing down the equations of motion so MV X do equal to QV YB because uh f is equal to uh QV cross B and that's why the X component or the derivative of the X component of the Velocity is related to the Y component of the velocity and similarly for the Y component it's X component but with a negative sign and now I did some uh little uh manipulation here in order to write this uh VX double dot now the end result is that the VX dou dot becomes equal to Min - Omega b² V ofx and all of you know that this corresponds to a harmonic motion okay where uh VX will harmonically depend uh on time and uh so this is what the uh solution is and the solution is that vx isal to V perpendicular uh which is uh the because we talking about the perpendicular motion and it's cine of Omega BT and uh similarly if you do the same calculation for VY then VY turns out to be m / QB into VX dot which is equal to a Min - Q ided by a Q and A V uh and a sin Omega T Omega BT okay so this is the trajectory for the perpendicular motion so there is the trajectory in the X Direction and this is the trajectory in the y direction and why we are talking about the perpendicular because we are considering uh perpendicular to the magnetic field so if you integrate and want to know that how the trajectory looks like then uh you'll get uh so integrating these equations we have already written one and two so this could be three and four so integrating three and four and using you know some initial conditions so X is equal to x0 plus a v perpendicular divided by Omega B uh and sin Omega b t uh because I integrate it with respect to T so this will give me uh the X motion and uh the Y motion is um uh almost similar harmonic uh but uh out of phase uh by uh an amount Pi by 2 this is a cine uh Omega BT so this is the X uh motion and the Y motion and it's clear that if you superpose this motion you a circular motion and uh let me uh show that circular motion let's see how neatly I can draw a circle uh not very good but uh this is that circular motion that we are talking about about this point x0 y0 it may look slightly uh you know shifted towards the positive side but that's not what is intended and um this uh things are so this is the radius of that that so this is where the uh positive Q uh motion of the particle that is clockwise and this is uh the motion when Q is negative and if you want to know the you know the axis because this is very important so this is the X and this is the Y and this is Zed and this is the direction of B so B is in the Zed Direction and and this is the planer motion so this is the orbit of a charge particle uh about a guiding Center x0 y0 okay so uh this is called as a guiding Center this point about which the motion takes place and U this is pretty much uh the motion of a charge particle only in a magnetic field so let's uh look at case two now when uh we'll talk about e not equal to 0 B not equal to Z okay uh but e is perpendicular to B okay so it's basically a generalized condition of the one that we just saw and uh we can write down the equation of motion which is f which is equal to MV dot which is equal to Q into e + uh V cross B okay now the total Force has a component or rather a contribution coming from the electric field and um earlier of course uh for e equal to Z which is the case one which we saw uh your uh V uh parallel was a constant so the parallel component of V but now uh uh because of this the parallel component actually has an acceleration which is given by uh the Q E uh and the parallel component of e divided by m so what happens is that there is no fixed guiding Center that you are seeing here The Guiding Center remains fixed and the charged particle either in the clockwise direction or anticlockwise Direction depending on its sign uh would just move uh around this circle but now this uh guiding Center keeps getting drifted because of the electric field let's see that okay so uh The Guiding Center which was earlier we have denoted it by x0 y0 so this guiding Center x0 y0 uh it moves uh perpendicular to both e and B okay so this guiding Center moves in that direction so the charge particle drifts the charge particle drifts with a velocity VD the reason that I'm writing this is that there is another component of velocity or rather another contribution to the velocity which also we'll have to worry about now this is uh because of the guiding Center that uh you know uh moves in a direction which is perpendicular to both e and B uh the charge particle has to drift and um this uh drifting actually is represented by like this and so on okay so the Drift actually happens in this direction and uh in particular suppose you have uh B to be uh pointing inside the screen uh so that you know this is usually uh represented because you see an arrow uh the back of the arrow if you see uh it looks like a cross uh and that's why when you write it with a cross a direction which means it's going into the plane of the paper in this case it's a plane of the board or or this uh white board that you're seeing and um we have electric field in this direction so one is going piercing inside uh in inside the board and then the other the electric field is in in this direction and then the drift happens in a direction which is perpendicular to both which in this particular case it happens in this direction that is from left to right okay so um how do we explain this drift velocity or what is the corresponding equation that it satisfies the drift velocity satisfies this equation which is v d cross Bal 0 so this is the the net uh force is uh zero and this uh is the equation that it would satisfy and defines this VD is defined by this equation okay and uh so how do we uh get this so we can actually get uh you can take a cross product uh with B uh cross product with B that gives you uh 0 equal to e cross B and uh v d cross B cross B I'm sure you can you have seen these kind of a cross product triple cross product A cross B Cross C and there's a particular formula that one uses uh which you can look up and this is equal to nothing but uh this is like a e cross B uh and and plus VD dob uh and a b here and uh minus a b² and a VD okay so this is the uh equation for this or rather this after you simplify it becomes this part and uh then your VD becomes equal to e cross B divided by by b² okay so b square you can write it with a square and a mod but this is also fine so this uh is the drift velocity of the charge particle in presence of an uh electromagnetic field where the electric and the magnetic fields are crossed with respect to each other that is they're perpendicular with respect to each other all right so uh there is also another contribution to this velocity in fact this veloc is um coming from even if there is no e term that is the electric field term there'll be a velocity due to the Len Force which is what we have seen earlier and uh so this let's call it as v l and this L stands for lorange uh Force so let's call this as a lorange velocity and this lorange velocity can be found out by taking uh DDT of these uh orbit that we have talked about and this is like exponential I Omega B T minus some initial T KN okay so this is purely because of the lowen force which is uh uh nothing but so FL L is nothing but q v cross B okay so this is a lowen uh velocity uh this will get added to the the drift velocity and uh the net velocity will become equal to uh so it's a v l plus a VD okay um and of course uh because of this um this electric field there's also a velocity which will be there which is the parallel component of the velocity and this is the total velocity of the particle and this total velocity will can be integrated in order to find R as a function of T and this is done in all books classical electrodynamic books and uh especially one can look at Griffith's book on electrodynamics and it is nicely worked out there that it is actually a helix a moving Helix which uh you know uh sort of uh moves uh in a direction which is uh perpendicular to both e and B and uh if you have non-uniform B that uh creates another uh complication and in which um maybe you'll have you know the radius of the moving orbit uh Chang and so on the reason that we have discussed is that we needed to understand the trajectory of the charge particle and uh we are uh really dealing with electrons uh in an electromagnetic field or in a magnetic field so uh and these electrons are um governed by laws of quantum mechanics and uh we'll have to do a a quantum mechanical treatment of this problem uh and the finally the problem would lead to uh giving us this uh Quantum hall effect or the integer Quantum hall effect uh which I showed you um even uh at the beginning of the discussion today all right so uh let me um sort of do a quantum mechanics now this is of course a large number of electrons involved but uh they are mostly semiconductors or metals that we are talking about and they have have electrons which the large number of electrons being there but still uh the interparticle interaction can still be ignored so you can talk about them as free electrons uh and with the drude theory would uh uh explain their uh the conductivities and the resistivities as we have talked about let's write down a quantum mechanics of a Charged particle and first let us uh do it only for the magnetic field and we'll see that the physics is not um radically altered when you include the electric field uh there are some um ramifications of that uh which we'll talk about okay and how do we do this problem since we are talking about sing charge particle because the um the electronic interactions can be neglected here they're weakly interacting or they are not interacting at all uh we can solve a shinger equation in presence of a magnetic field so how does one solve uh a short iner equation in presence of a magnetic field so let's write down the short iner equation in its simplest form it is - h² by 2 m uh and d squ and a v B of say R uh and this R is uh well there's a s here uh which is a function of r and this is a function of r and this if you write it the time independent shorer equation then this is a Formula or rather this is a equation that you need to solve and you have solved it in a variety of situations in which uh you had uh say a the particle in a box for example in which VR equal to zero uh inside the box and it's uh equal to Infinity at the walls or rather outside the box it's Infinity which means that the particle cannot Escape uh or you have done a problem in which these are uh finite steps or their finite Wells or their finite barrier or their infinite uh barrier such as a Delta function barrier and so on so forth which are all part of the quantum mechanics course first course on quantum mechanics that you have learned now we are going to solve the same problem and uh we are going to put this equal to zero because we uh claim uh rather assert that there's no interaction that needed to be taken into account uh even if it's there it's very small and can be ignored and now uh you have to somehow include the presence of the magnetic field okay and um if you remember that the magnetic field for a particle charge particle uh that enters through the momentum so the uh the momentum which is written as P which was without uh any magnetic field or any other thing is equal to M into V which is called as a mechanical momentum this thing changes into in presence of a magnetic field this is uh P Min - EA so it it this changes from P to P minus QA where Q is the charge and a is the vector potential and this Vector potential is an important quantity and we know that uh the B actually is uh uh given by curl of a okay so this fixes the gauge now why I say a gauge because you know when B is perpendicular that is in the Zed Direction then a can be either in the X direction or can be in the y direction or can be uh in uh any of X and Y Direction uh I mean it is in a direction which is is perpendicular to The Zed Direction okay so say uh you you have asserted that b is in the Zed Direction okay so the understanding is that we have uh a sort of planer charges uh so there are charges here okay um and which is uh you know a material like this say for example a thin material uh and uh there is a magnetic field which is perpendicular and acting in the Z Direction so this is the Z Direction all right so uh so B is equal to uh b z cap in that case a can be equal to you know uh minus b y that is the X component 0 0 it it can also be 0 BX 0 and uh it can also be you know half equal to half uh r cross B um and uh this is for a constant magnetic field so this is equal to uh like b y by 2 or minus BX by 2 and zero okay so these are B into Y and B into X okay so all three forms of the vector potential would create the same B uh which is in the which is constant and in the Z Direction you just have to convince yourself by taking the curl of this and um uh so these are called gauge Freedom you know I mean in the sense that it the final result would not depend upon which a you use okay and this a is also you can add or subtract say for example you can add a gradient of a scalar quantity and uh see this scalar quantity uh doesn't matter because when you take a curl of this a curl of a gradient would always be zero and you can understand it in this particular fashion that a gradient has a particular direction in space in which you ask this question that if Lambda is a scalar quantity say it's like a uh heat density or something or there's a sound wave that are or sound field that are produced because of some loudspeakers and so on and uh they are placed in four corners of the room and you may want to ask this question that in which direction the sound intensity changes the fastest and in which case you have to find the gradient in which it changes the fastest and that gradient is actually a direction and because it's a direction if you take a curl of a particular direction it has to give you zero that's why uh there are these uh choices of a along with this uh minus gradient of Lambda that these are called the gate choices finally the results that you get are physical results physical observables so they won't depend upon Which choice of a you have taken or whether what choice of Lambda you have taken and it uh the final result won't depend upon that all right so uh let's take this gauge as let's fix the gauge to be equal to axal to that is the X component of a is minus b y so a is equal to so the first choice that we have talked about let's take that but it really doesn't matter okay and this is called as a landow gauge okay so we do this and then put this in the uh short iner equation and now make sure that your Q is nothing but minus E because we are talking about electrons specifically so we'll have this equation which is 1 / 2 m because it's a p² / 2 m so we have a p x minus E uh b y by uh Square uh and uh plus a p y² / 2 M and A Plus p z s/ 2 m you can actually put uh the 2m outside maybe let's do that so that I don't have to write these uh okay and this the only thing and then this s is equal to e is the equation so this is a short iner equation that you have to solve okay now it's very clear that uh the motion in The Zed direction is exactly like a free particle it's just like a a particle in a box and that direction is quantized and in fact it it really doesn't matter if you're talking about perfectly two-dimensional electron the Z component uh cannot be there even if it's there it's quantized so the electron cannot Escape in the Z Direction because we are talking about a two-dimensional uh problem so that problem then uh let's uh eliminate the Zed part of the kinetic energy or this p z Square / 2m and let's simply write it as uh uh like a 1 / 2 m and then the PX minus EB y SAR plus a p y² and so on S is equal to now the S is a function of X and Y only and this is equal to Let's talk about and energy let's let's say that is uh denoted by Epsilon s of XY I hope you understood that why uh PX minus EB Y is written because uh the uh Vector potential is assumed to be only present uh in the X Direction so it's p minus EA or P plus EA rather uh because P minus QA so P plus ea uh that a has only X component so only the PX changes and nothing else The py Remains as it is okay now this is of course a two-dimensional hamiltonian you can call it some H uh XY so if you write it symbolically it's like hxy s XY equal to Epsilon s XY and this is what you have to solve for so you have to solve for the IG Solutions and the igen solutions are nothing but the SX y um and um uh the energies are Epsilon is are the two things that one has to solve for okay there's something interesting that happens here if you want to ask that which component of the momentum is conserved uh then uh it is easy to see that this PX and hxy is conserved and the reason is that uh there's no X variable here had there been an X variable then there would be no commutation between uh PX and H but because we have decided to take the uh Vector potential in the in the X Direction and with a b y That's why it uh this PX is uh commutes with the hamiltonian and uh from elementary quantum mechanics as you know that if an operator commutes with the hamiltonian then the corresponding quantum number remains conserved or there are quantities that remain conserved here of course the PX that's the X component of the momentum remains conserved and the uh the quantum uh number or the sort of quantity that remains conserved is called as a KX okay so KX is H cross KX so PX is equal to constant uh which tells you that H cross KX is equal to uh constant okay so H cross KX is constant and now this KX uh because we are talking about a quantum mechanical problem this KX is quantized as uh 2 pi over um you know uh NX ided by LX okay it's just that uh just the way we have learned that uh particle in a box the KE vectors are quantized the K vectors are Quant ized because of the presence of boundaries there and that's we have seen that k equal to you know n pi/ l here because of the periodic boundary condition you get a factor of two so this KX is equal to 2 pi NX by LX and NX are 0 1 2 3 and so on okay so these are the values that are sorry the zero can be neglected Ed because zero means that the particle is not there so it's it's actually 1 2 3 and so on okay so any integer the rest of the relation can be written as so if PX is a constant I can absorb it as a constant in the hamiltonian and can write down the resultant uh short iner equation as follows okay so this is equal to P y² / 2m py is not conserved for the reason there is a y in the hamiltonian and py and Y do not commute this is the uncertainty principle that you have learned uh and particularly Y and py is equal to I Cross okay this reason that py is not constant and I can write down then this uh 1 by 2 uh that is half M and eb/ m² and Y - y 0 S and now my wave function actually becomes function of only one variable that is y because you see there's no X anywhere in the hamiltonian so I can write this as simply F of Y and this is equal to Epsilon and F of Y okay and what is the y0 y0 is uh the constant thing which is PX over EB and uh this PX is the X component of the momentum which I argued that Y is constant and of course for a given problem e and B are constants B is a uniform field so this y0 is actually PX by EB and this is uh it has a name called as a magnetic length and written as lb because it depends on B okay now this uh tells you that if this is a constant uh there's a p y s and then there's a half something M into some cyclotron frequency so it's a half M Omega squ and then Yus y0 s and then you have this wave function which is only a function of Y and that's why from s XY we have written it f of Y and then energy into F of Y and this clearly represents that the equation for or the shanger equation for a harmonic oscillator oscillating in the y direction about not about zero but about this y0 or the magnetic length which is given by this uh PX over EB okay so uh if it is a uh harmonic oscillator problem then we don't need to solve any further we can get this the solution to be exactly of the form which is Epsilon = n + half H cross Omega B where Omega B is equal to uh of course Omega B is equal to eb/ M which we have said several times that this is equal to the cyclotron frequency okay so a problem which was simple enough to begin with uh we find the solution at least till now we have found the energy so the energy is nothing but it's uh n + half H CR Omega B just following the solution for the energy for a harmonic oscillator n can have 0 1 2 3 Etc all the integers and here it can be equal to zero because uh the harmonic oscillator allows for a solution with a quantum number n equal to0 so uh this is the energy or the igen values of the problem let's see the igen function so the igen function now let's call it U because we we can solve for of course uh F of Y uh that is what we'll do but then we know that the total wave function which is equal to F of or S of XY s of XY is actually a free particle in the X Direction multiplied by this F of Y okay this is the total wave function for the planer motion of the electron that is in the X direction if you take this as a x Direction it propagates like a free particle because PX is constant or KX is constant so it'll propagate like a a free wave there and will have a harmonic motion in the y direction so in the y direction it will have a harmonic motion uh we make a execute a harmonic motion about a point which is given by uh y0 and this y0 inversely depend upon B okay so uh this is nothing but uh I use a normalization uh for the X Direction uh box normalization uh this is equal to exponential I kxx and um now I don't try to normalize it it's not you know required here I use this as a the normalization constant and use the this uh okay so let me write the normalization not with a capital N because you have small n uh this thing there so let's write it with a n so a n is a normalization constant and the wave functions are for a harmonic oscillator are known to be a convolution of a gausian which is exponential minus alpha x square or alpha y Square multiplied by a polinomial function which is called as a hermite polinomial and this polinomial has a property that when n is even the polinomial is even that is it contains terms such as X to the^ 0 x x² x the^ 4 and so on when n is odd it contains terms which are x x Cube x ^ 5 and so on so this is that form that we have to write down so this is equal to uh EB y - y ² / H cross which is the gaussian part just reminding you that uh this is the harmonic oscillator and uh you have a a gausian like this for the for the ground state and and this is for the first excited state it is like this and so on okay so this is an odd function so this is a ground state and this is the first excited state and so on so and then there is a hermite polinomial and this hermite polinomial is written as e y - y0 uh divided by H cross Etc the exact form is not important for us at this moment but we need to know that at least there there's a hermite polinomial and then there is is a gaussian term which comes with a normalization constant and this part is purely because of the X part and this is because of the Y and as I said uh that the Z component even if you consider it's like a free particle in the Zed Direction in particular if you confine electrons in two Dimension then that uh term does not arise now if you go to this uh last last slide you have n + half H cross Omega and this n = to 0 1 2 3 Etc these are called as the landow levels so the energy levels have a name which are called as a landow levels and landow is a Russian physicist and uh the wave function is actually a freely propagating part in The X Direction multiplied by a har harmonic part which is executing a harmonic oscillation which has a gaussian term and a harite polinomial uh n when n is even the polinomial is even when n is odd the polinomial is odd okay these land levels so if if we really plot it how does it look like the landow levels will simply look like this okay they're equidistant okay for a given value of B so this is say n = to 0 n = 1 n equal to 2 and so on uh but if you draw for another magnetic field it may look like this and so on okay so then n = 1 n = 2 N = 3 and so on okay and the reason is that Omega B uh which is the cyclotron frequency it is directly proportional to B so as you increase B let's say this B1 for a magnetic Field B1 and this is for a for B2 and of course B2 is greater than B1 okay and these are called as the landow levels so uh now what are the properties of the landow levels and uh we would eventually get to that that these landow levels are the most important things in understanding the structure of the plateaus okay and these are not really very sharp levels because of disorder this get broadened we'll come to that there's one very important things that these landow levels are infinitely degenerate or their degeneration is very high so degeneracy of the landow levels so there are many many states corresponding to each one of the N values so n equal to0 has very large number of states which means that these States can be occupied by electrons so each land level is occupied by many many electrons so let's understand the how that degeneracy arises and where does it come from it comes from the value that KX is a constant and this result that you have got is independent of uh KX so KX doesn't play a role here which means any value of KX would give rise to this level and this KX has a NX associated with it which denote the quantum number of the states which means that corresponding to any value of n here a given value of n there could be a very large number of values of NX that are possible okay so this in principle this is infinite but the degeneracy is actually limited by some factors and uh let us you know try to understand that what factors do they depend upon but I hope that it is clear that these uh levels are heavily degenerate and the degeneracy is because they are uh coming from the fact that any value of KX would satisfy this equation or any value of NX would satisfy these energy levels so each of the energy levels do not depend upon on um NX and any value would be then acceptable solution okay so uh if we want to know the degeneracy Let's uh look at this okay so uh we have written this down earlier that the y0 is equal to so we have uh y 0 = to PX over EB we have written this earlier uh and which we called as lb which is a magnetic length and this is nothing but H cross KX uh divided by EB okay because PX is H cross KX and this is equal to H cross um uh KX is 2 pi uh uh NX by LX okay so this is your uh i0 and we want to find that what is the maximum degeneracy so we want to find what is NX Max okay and NX Max all others are constant the y0 is the is basically a length scale in the y direction okay along the y direction there some value of y which is determined by of course the magnetic field now this y0 can at the most be ly that's the uh length of the sample in the y direction that y z cannot be outside the sample so the maximum value of y0 will be ly and so maximum value of NX Max can be obtained if you substitute y0 as l y so this is equal to EB uh LX l y uh because I'm putting y0 as ly and this is equal to 2 pi H cross so that's the maximum degeneracy so this is equal to EB into a a is the area of the sample uh do not confuse it with anything else it's the a is the area of the sample and uh divided by this H because 2 pi H cross is nothing but h H I can write this as a / H over uh e and uh so I can I can take that b uh this thing I mean I can leave that b there and put this electronic charge below the uh in the denominator because H over e is a very important quantity which is called as a flux Quantum which is denoted by some 5 0 okay so this is equal to a flux Quantum which has a particular value um in Weber some 10^ -4 10^ -5 value so you see that let's call this as G the degeneracy so G is equal to uh it depends on the the B the magnetic field the area of the sample and a 5 Z and this tells you that this degeneracy can be infinite because a b if you go back to the first slide that we had and you see here the magnetic field goes all the way up to 14 15 Tesla so there's about 15 Tesla so there's a 15 Tesla magnetic field if you want if you have facilities you can increase that magnetic field even more so and the sample size is up to you okay so this is really a very large degeneracy that is coming out of each one of the landow levels are degenerate in fact this has important um repercussions or imp implications on the fractional Quantum hall effect okay so two things that we notice here so you note that uh the degeneracy is of course what I told you that degeneracy is proportional to uh is proportional to b and a uh I told you B is the magnetic field and a is the area and um the second is that you know the trajectory uh can be U the trajectory is centered about some y0 uh is uh shm in y direction and centered about y0 so uh these are two very important you know outcome of this exercise that we have done that is we have solved uh a charge particle a single charge particle uh in presence of a perpendicular or a transverse magnetic field we have taken a particular gauge and suppose we have taken the gauge in which the magnetic Vector potential is in the y direction in that case you'll have just the X and Y being interchanged that the system the the particle will propagate like a free particle in the y direction and will execute a simple harmonic oscillation in the X Direction so X and Y they have no specific significance and the third gauge which we'll talk about later it's called as a spherical gauge or symmetric gauge and in that case you have both X and Y component and these land levels are actually found out to be circular okay so um now there are some incidental similarities which I'll just point out and we'll discuss them uh as we go along and the similarities are that the hall resistivity it shows plateaus in multiples of H over uh these plateaus are H over e s okay so H over say n square or something where n is an integer okay so uh consider to be n0 okay that's the density of carriers so Define a quantity called as new or uh let's let's call this as uh just what we called here as n let me call this as new where new takes values 1 2 3 Etc we just change this because the too many n's n is used for density as well so consider charge density of the of the carriers to be n0 so new is defined as n0 divided by uh G by a now this G by a is the uh the degeneracy per unit area so let's not have the geometry of the sample coming in because you can have any any geometry but if you divide it by a then it becomes a fundamental quantity which only depends on the value of the magnetic field and we are now talking about a given value of the magnetic field and this is equal to nothing but uh n0 H ided EB and uh this is equal to so this is equal to a number of electrons and I can I can write this as n0 ided H over e into B so that's the flux Quantum so this number of electrons and divided by the you know the number of uh flux quanta and so on okay so if new assumes a value integer that is when the number of electrons and the number of flux Quantum they become commen at fractions or even uh fractions in the fractional Quantum hall effect then there are plateus uh that are seen in the resistivity okay so these depending upon now you you tune in the value of the magnetic field go go back to this uh first slide which we had you see the magnetic field is being continuously tuned from a value 0 to 15 Tesla so magnetic field is continuously ramped up and you are doing this experiment and what you see is that when this quantity which we just found out that this quantity becomes an integer then there are these Hall plateus which are seen which we saw last so when this becomes an integer you have a plateau which means that this ratio becomes an integer uh you see plateus then you again tuned B this goes from being an integer to uh a fraction then you no longer see plateau and then this thing uh the resistivity gives shows a jump and then you go to the next integer such that it it shows another plateau and that's how this new actually would uh be uh you know uh sort of it'll go from one uh integer value to another integer value and it will pull along uh the resistivity of the material to be having plate um and and so on and from one Plateau to another and this is what we are going to see more carefully I'll give you a list of references that uh uh you uh should follow uh some of them are Advanced references and um you can follow them along with my lectures here uh some of them are of the same level um often I don't remember them but next uh discussion I'll provide the you with those references [Music]

[Music] so let me connect with the discussion that we were having having um last uh time uh the previous discussion about the quantization of the hall plateaus so this is the uh effect that we have seen uh that this is the the quantum hall effect or the quantise plateaus which it is famous for uh so uh we have to understand these origin of these plateaus and uh uh how they are so robust to uh disorder and impurities as they are uh and also we we have to understand these green plots which represent the row XX or the magnetude resistance or the longitudinal resistance uh so let me just write that row XY would be called as Hall uh resistivity uh and row XX will be called as uh one can call it as a longitudinal resistance or resistivity uh and or it&#39;s also called as a Magneto resistivity basically the resistivity that arises because of the magnetic field and um uh so these are uh as you see the green plots actually vanish most of the time excepting when the red curves show a jump from one Plateau to another uh then it sort of shoots up and then again goes to zero uh and it remains zero till the the plateaus exist in the hall conductivity this is uh uh a Quantum phenomena as uh we have said earlier and we&#39;ll be talking about this um so the quantum mechanical nature of the electrons in presence of a magnetic field uh will have to be worked out and we have to understand how these plateaus arise and why they are so robust so much so that they actually give the Benchmark of resistance uh the plateus actually h e Square sets the scale of resistance which is 25.81% or how the hall voltage develops uh in the transverse ages of the sample when there&#39;s a longitudinal current being sent uh from one end to another and um that was a analysis that was done and at equilibrium when the force vanishes uh but we haven&#39;t talked about uh what is a trajectory of the uh particle uh in presence of an electromagnetic field and this is important because the trajectory uh being actually a path that&#39;s followed by a particle uh a Charged particle uh should not really depend upon whether we are talking about classical mechanics or quantum mechanics um of course in quantum mechanics we uh rarely discuss or uh worry about uh the trajectory of the particle for the reason that the trajectory is uh uh it&#39;s there is an uncertainty principle which is underlying that and which governs the trajectory so let&#39;s look at the classical trajectory of a charge particle and this is done with an intention that we&#39;ll be uh dealing with a quantum mechanical system uh soon okay so uh let us um talk about two cases uh the case one being uh where there is no electric field uh and there is only a magnetic field which is uh constant which is not equal to zero it&#39;s finite and it&#39;s uniform okay uh so this is the situation uh one that&#39;s case one say and um the energy is constant so uh please do not confuse between the electric field e and the energy energy is a scalar so energy is a constant uh and um energy is constant because of the reason that the power delivered uh is equal to zero which is U uh nothing but the low range force is V cross B and that&#39;s why this is equal to zero so we can write this as MV do a V Dot which is equal to DDT of uh half m v² and this is equal to uh Q into V do V CR B which is equal to zero okay so this is your uh F do V uh because your f is the low range Force when there&#39;s no electric field it&#39;s only the magnetic field will create a lorange force so uh so V do V cross B equal z uh the reason is simple this V cross B is a vector which is perpendicular to both V and B and that&#39;s why if you if you take a DOT product with v it will give you zero and that&#39;s why DDT of half MV s equal to 0 which means half MV squ equal to constant okay and this is what we mean by uh the energy being a constant that&#39;s the only energy that&#39;s uh left in the uh problem so this um you know um results in uh separate equations or separate Dynamics uh for the longitudinal and the transverse Direction so if you write your V to be having a parallel component and a perpendicular component and these Dynamics are uncoupled or decoupled and uh the the motion takes place independent of each other in absence of an electric field okay okay so let&#39;s talk about uh the perpendicular motion okay and what do we mean by the perpendicular motion so we have uh mvx dot this is equal to uh Q vyb so this is uh say equation number one and the equation number two will be MV y dot = to Min - Q vxb We are continuing with the same situation where e equal to 0 and uh uh B is uniform and not equal to zero and uh we are just um discussing or rather deliberating upon the nature of the motion and this is quite instructive if you take a double derivative of this so uh differentiate with respect to uh with respect to time T and then of course uh use to so then mvx X double dot so this is equal to Q VY dot by B I mean into B and then I can take this m to be um in the denominator uh this is in the denominator so this now I&#39;ll use uh VY dot to be here so this is equal to Q into B by m and then this is equal to - Q into VX B uh / M so this becomes equal to uh minus um QB uh by m squ and VX and uh this is equal to uh nothing but uh minus uh which we have written it earlier that Omega b² VX okay or Omega B is nothing but equal to uh QB over M which we have called as the cyclotron frequency even earlier so just a quick recap we are talking about uh no electric field and a Charged particle is moving with a velocity V which is has components VX and VY and uh this is um uh this a b there that&#39;s a magnetic field because of magnetic field there&#39;s a lowen force and in this particular case we know that the energy is constant because the F do V equal to Z and now we are analyzing the motion by writing down the equations of motion so MV X do equal to QV YB because uh f is equal to uh QV cross B and that&#39;s why the X component or the derivative of the X component of the Velocity is related to the Y component of the velocity and similarly for the Y component it&#39;s X component but with a negative sign and now I did some uh little uh manipulation here in order to write this uh VX double dot now the end result is that the VX dou dot becomes equal to Min - Omega b² V ofx and all of you know that this corresponds to a harmonic motion okay where uh VX will harmonically depend uh on time and uh so this is what the uh solution is and the solution is that vx isal to V perpendicular uh which is uh the because we talking about the perpendicular motion and it&#39;s cine of Omega BT and uh similarly if you do the same calculation for VY then VY turns out to be m / QB into VX dot which is equal to a Min - Q ided by a Q and A V uh and a sin Omega T Omega BT okay so this is the trajectory for the perpendicular motion so there is the trajectory in the X Direction and this is the trajectory in the y direction and why we are talking about the perpendicular because we are considering uh perpendicular to the magnetic field so if you integrate and want to know that how the trajectory looks like then uh you&#39;ll get uh so integrating these equations we have already written one and two so this could be three and four so integrating three and four and using you know some initial conditions so X is equal to x0 plus a v perpendicular divided by Omega B uh and sin Omega b t uh because I integrate it with respect to T so this will give me uh the X motion and uh the Y motion is um uh almost similar harmonic uh but uh out of phase uh by uh an amount Pi by 2 this is a cine uh Omega BT so this is the X uh motion and the Y motion and it&#39;s clear that if you superpose this motion you a circular motion and uh let me uh show that circular motion let&#39;s see how neatly I can draw a circle uh not very good but uh this is that circular motion that we are talking about about this point x0 y0 it may look slightly uh you know shifted towards the positive side but that&#39;s not what is intended and um this uh things are so this is the radius of that that so this is where the uh positive Q uh motion of the particle that is clockwise and this is uh the motion when Q is negative and if you want to know the you know the axis because this is very important so this is the X and this is the Y and this is Zed and this is the direction of B so B is in the Zed Direction and and this is the planer motion so this is the orbit of a charge particle uh about a guiding Center x0 y0 okay so uh this is called as a guiding Center this point about which the motion takes place and U this is pretty much uh the motion of a charge particle only in a magnetic field so let&#39;s uh look at case two now when uh we&#39;ll talk about e not equal to 0 B not equal to Z okay uh but e is perpendicular to B okay so it&#39;s basically a generalized condition of the one that we just saw and uh we can write down the equation of motion which is f which is equal to MV dot which is equal to Q into e + uh V cross B okay now the total Force has a component or rather a contribution coming from the electric field and um earlier of course uh for e equal to Z which is the case one which we saw uh your uh V uh parallel was a constant so the parallel component of V but now uh uh because of this the parallel component actually has an acceleration which is given by uh the Q E uh and the parallel component of e divided by m so what happens is that there is no fixed guiding Center that you are seeing here The Guiding Center remains fixed and the charged particle either in the clockwise direction or anticlockwise Direction depending on its sign uh would just move uh around this circle but now this uh guiding Center keeps getting drifted because of the electric field let&#39;s see that okay so uh The Guiding Center which was earlier we have denoted it by x0 y0 so this guiding Center x0 y0 uh it moves uh perpendicular to both e and B okay so this guiding Center moves in that direction so the charge particle drifts the charge particle drifts with a velocity VD the reason that I&#39;m writing this is that there is another component of velocity or rather another contribution to the velocity which also we&#39;ll have to worry about now this is uh because of the guiding Center that uh you know uh moves in a direction which is perpendicular to both e and B uh the charge particle has to drift and um this uh drifting actually is represented by like this and so on okay so the Drift actually happens in this direction and uh in particular suppose you have uh B to be uh pointing inside the screen uh so that you know this is usually uh represented because you see an arrow uh the back of the arrow if you see uh it looks like a cross uh and that&#39;s why when you write it with a cross a direction which means it&#39;s going into the plane of the paper in this case it&#39;s a plane of the board or or this uh white board that you&#39;re seeing and um we have electric field in this direction so one is going piercing inside uh in inside the board and then the other the electric field is in in this direction and then the drift happens in a direction which is perpendicular to both which in this particular case it happens in this direction that is from left to right okay so um how do we explain this drift velocity or what is the corresponding equation that it satisfies the drift velocity satisfies this equation which is v d cross Bal 0 so this is the the net uh force is uh zero and this uh is the equation that it would satisfy and defines this VD is defined by this equation okay and uh so how do we uh get this so we can actually get uh you can take a cross product uh with B uh cross product with B that gives you uh 0 equal to e cross B and uh v d cross B cross B I&#39;m sure you can you have seen these kind of a cross product triple cross product A cross B Cross C and there&#39;s a particular formula that one uses uh which you can look up and this is equal to nothing but uh this is like a e cross B uh and and plus VD dob uh and a b here and uh minus a b² and a VD okay so this is the uh equation for this or rather this after you simplify it becomes this part and uh then your VD becomes equal to e cross B divided by by b² okay so b square you can write it with a square and a mod but this is also fine so this uh is the drift velocity of the charge particle in presence of an uh electromagnetic field where the electric and the magnetic fields are crossed with respect to each other that is they&#39;re perpendicular with respect to each other all right so uh there is also another contribution to this velocity in fact this veloc is um coming from even if there is no e term that is the electric field term there&#39;ll be a velocity due to the Len Force which is what we have seen earlier and uh so this let&#39;s call it as v l and this L stands for lorange uh Force so let&#39;s call this as a lorange velocity and this lorange velocity can be found out by taking uh DDT of these uh orbit that we have talked about and this is like exponential I Omega B T minus some initial T KN okay so this is purely because of the lowen force which is uh uh nothing but so FL L is nothing but q v cross B okay so this is a lowen uh velocity uh this will get added to the the drift velocity and uh the net velocity will become equal to uh so it&#39;s a v l plus a VD okay um and of course uh because of this um this electric field there&#39;s also a velocity which will be there which is the parallel component of the velocity and this is the total velocity of the particle and this total velocity will can be integrated in order to find R as a function of T and this is done in all books classical electrodynamic books and uh especially one can look at Griffith&#39;s book on electrodynamics and it is nicely worked out there that it is actually a helix a moving Helix which uh you know uh sort of uh moves uh in a direction which is uh perpendicular to both e and B and uh if you have non-uniform B that uh creates another uh complication and in which um maybe you&#39;ll have you know the radius of the moving orbit uh Chang and so on the reason that we have discussed is that we needed to understand the trajectory of the charge particle and uh we are uh really dealing with electrons uh in an electromagnetic field or in a magnetic field so uh and these electrons are um governed by laws of quantum mechanics and uh we&#39;ll have to do a a quantum mechanical treatment of this problem uh and the finally the problem would lead to uh giving us this uh Quantum hall effect or the integer Quantum hall effect uh which I showed you um even uh at the beginning of the discussion today all right so uh let me um sort of do a quantum mechanics now this is of course a large number of electrons involved but uh they are mostly semiconductors or metals that we are talking about and they have have electrons which the large number of electrons being there but still uh the interparticle interaction can still be ignored so you can talk about them as free electrons uh and with the drude theory would uh uh explain their uh the conductivities and the resistivities as we have talked about let&#39;s write down a quantum mechanics of a Charged particle and first let us uh do it only for the magnetic field and we&#39;ll see that the physics is not um radically altered when you include the electric field uh there are some um ramifications of that uh which we&#39;ll talk about okay and how do we do this problem since we are talking about sing charge particle because the um the electronic interactions can be neglected here they&#39;re weakly interacting or they are not interacting at all uh we can solve a shinger equation in presence of a magnetic field so how does one solve uh a short iner equation in presence of a magnetic field so let&#39;s write down the short iner equation in its simplest form it is - h² by 2 m uh and d squ and a v B of say R uh and this R is uh well there&#39;s a s here uh which is a function of r and this is a function of r and this if you write it the time independent shorer equation then this is a Formula or rather this is a equation that you need to solve and you have solved it in a variety of situations in which uh you had uh say a the particle in a box for example in which VR equal to zero uh inside the box and it&#39;s uh equal to Infinity at the walls or rather outside the box it&#39;s Infinity which means that the particle cannot Escape uh or you have done a problem in which these are uh finite steps or their finite Wells or their finite barrier or their infinite uh barrier such as a Delta function barrier and so on so forth which are all part of the quantum mechanics course first course on quantum mechanics that you have learned now we are going to solve the same problem and uh we are going to put this equal to zero because we uh claim uh rather assert that there&#39;s no interaction that needed to be taken into account uh even if it&#39;s there it&#39;s very small and can be ignored and now uh you have to somehow include the presence of the magnetic field okay and um if you remember that the magnetic field for a particle charge particle uh that enters through the momentum so the uh the momentum which is written as P which was without uh any magnetic field or any other thing is equal to M into V which is called as a mechanical momentum this thing changes into in presence of a magnetic field this is uh P Min - EA so it it this changes from P to P minus QA where Q is the charge and a is the vector potential and this Vector potential is an important quantity and we know that uh the B actually is uh uh given by curl of a okay so this fixes the gauge now why I say a gauge because you know when B is perpendicular that is in the Zed Direction then a can be either in the X direction or can be in the y direction or can be uh in uh any of X and Y Direction uh I mean it is in a direction which is is perpendicular to The Zed Direction okay so say uh you you have asserted that b is in the Zed Direction okay so the understanding is that we have uh a sort of planer charges uh so there are charges here okay um and which is uh you know a material like this say for example a thin material uh and uh there is a magnetic field which is perpendicular and acting in the Z Direction so this is the Z Direction all right so uh so B is equal to uh b z cap in that case a can be equal to you know uh minus b y that is the X component 0 0 it it can also be 0 BX 0 and uh it can also be you know half equal to half uh r cross B um and uh this is for a constant magnetic field so this is equal to uh like b y by 2 or minus BX by 2 and zero okay so these are B into Y and B into X okay so all three forms of the vector potential would create the same B uh which is in the which is constant and in the Z Direction you just have to convince yourself by taking the curl of this and um uh so these are called gauge Freedom you know I mean in the sense that it the final result would not depend upon which a you use okay and this a is also you can add or subtract say for example you can add a gradient of a scalar quantity and uh see this scalar quantity uh doesn&#39;t matter because when you take a curl of this a curl of a gradient would always be zero and you can understand it in this particular fashion that a gradient has a particular direction in space in which you ask this question that if Lambda is a scalar quantity say it&#39;s like a uh heat density or something or there&#39;s a sound wave that are or sound field that are produced because of some loudspeakers and so on and uh they are placed in four corners of the room and you may want to ask this question that in which direction the sound intensity changes the fastest and in which case you have to find the gradient in which it changes the fastest and that gradient is actually a direction and because it&#39;s a direction if you take a curl of a particular direction it has to give you zero that&#39;s why uh there are these uh choices of a along with this uh minus gradient of Lambda that these are called the gate choices finally the results that you get are physical results physical observables so they won&#39;t depend upon Which choice of a you have taken or whether what choice of Lambda you have taken and it uh the final result won&#39;t depend upon that all right so uh let&#39;s take this gauge as let&#39;s fix the gauge to be equal to axal to that is the X component of a is minus b y so a is equal to so the first choice that we have talked about let&#39;s take that but it really doesn&#39;t matter okay and this is called as a landow gauge okay so we do this and then put this in the uh short iner equation and now make sure that your Q is nothing but minus E because we are talking about electrons specifically so we&#39;ll have this equation which is 1 / 2 m because it&#39;s a p² / 2 m so we have a p x minus E uh b y by uh Square uh and uh plus a p y² / 2 M and A Plus p z s/ 2 m you can actually put uh the 2m outside maybe let&#39;s do that so that I don&#39;t have to write these uh okay and this the only thing and then this s is equal to e is the equation so this is a short iner equation that you have to solve okay now it&#39;s very clear that uh the motion in The Zed direction is exactly like a free particle it&#39;s just like a a particle in a box and that direction is quantized and in fact it it really doesn&#39;t matter if you&#39;re talking about perfectly two-dimensional electron the Z component uh cannot be there even if it&#39;s there it&#39;s quantized so the electron cannot Escape in the Z Direction because we are talking about a two-dimensional uh problem so that problem then uh let&#39;s uh eliminate the Zed part of the kinetic energy or this p z Square / 2m and let&#39;s simply write it as uh uh like a 1 / 2 m and then the PX minus EB y SAR plus a p y² and so on S is equal to now the S is a function of X and Y only and this is equal to Let&#39;s talk about and energy let&#39;s let&#39;s say that is uh denoted by Epsilon s of XY I hope you understood that why uh PX minus EB Y is written because uh the uh Vector potential is assumed to be only present uh in the X Direction so it&#39;s p minus EA or P plus EA rather uh because P minus QA so P plus ea uh that a has only X component so only the PX changes and nothing else The py Remains as it is okay now this is of course a two-dimensional hamiltonian you can call it some H uh XY so if you write it symbolically it&#39;s like hxy s XY equal to Epsilon s XY and this is what you have to solve for so you have to solve for the IG Solutions and the igen solutions are nothing but the SX y um and um uh the energies are Epsilon is are the two things that one has to solve for okay there&#39;s something interesting that happens here if you want to ask that which component of the momentum is conserved uh then uh it is easy to see that this PX and hxy is conserved and the reason is that uh there&#39;s no X variable here had there been an X variable then there would be no commutation between uh PX and H but because we have decided to take the uh Vector potential in the in the X Direction and with a b y That&#39;s why it uh this PX is uh commutes with the hamiltonian and uh from elementary quantum mechanics as you know that if an operator commutes with the hamiltonian then the corresponding quantum number remains conserved or there are quantities that remain conserved here of course the PX that&#39;s the X component of the momentum remains conserved and the uh the quantum uh number or the sort of quantity that remains conserved is called as a KX okay so KX is H cross KX so PX is equal to constant uh which tells you that H cross KX is equal to uh constant okay so H cross KX is constant and now this KX uh because we are talking about a quantum mechanical problem this KX is quantized as uh 2 pi over um you know uh NX ided by LX okay it&#39;s just that uh just the way we have learned that uh particle in a box the KE vectors are quantized the K vectors are Quant ized because of the presence of boundaries there and that&#39;s we have seen that k equal to you know n pi/ l here because of the periodic boundary condition you get a factor of two so this KX is equal to 2 pi NX by LX and NX are 0 1 2 3 and so on okay so these are the values that are sorry the zero can be neglected Ed because zero means that the particle is not there so it&#39;s it&#39;s actually 1 2 3 and so on okay so any integer the rest of the relation can be written as so if PX is a constant I can absorb it as a constant in the hamiltonian and can write down the resultant uh short iner equation as follows okay so this is equal to P y² / 2m py is not conserved for the reason there is a y in the hamiltonian and py and Y do not commute this is the uncertainty principle that you have learned uh and particularly Y and py is equal to I Cross okay this reason that py is not constant and I can write down then this uh 1 by 2 uh that is half M and eb/ m² and Y - y 0 S and now my wave function actually becomes function of only one variable that is y because you see there&#39;s no X anywhere in the hamiltonian so I can write this as simply F of Y and this is equal to Epsilon and F of Y okay and what is the y0 y0 is uh the constant thing which is PX over EB and uh this PX is the X component of the momentum which I argued that Y is constant and of course for a given problem e and B are constants B is a uniform field so this y0 is actually PX by EB and this is uh it has a name called as a magnetic length and written as lb because it depends on B okay now this uh tells you that if this is a constant uh there&#39;s a p y s and then there&#39;s a half something M into some cyclotron frequency so it&#39;s a half M Omega squ and then Yus y0 s and then you have this wave function which is only a function of Y and that&#39;s why from s XY we have written it f of Y and then energy into F of Y and this clearly represents that the equation for or the shanger equation for a harmonic oscillator oscillating in the y direction about not about zero but about this y0 or the magnetic length which is given by this uh PX over EB okay so uh if it is a uh harmonic oscillator problem then we don&#39;t need to solve any further we can get this the solution to be exactly of the form which is Epsilon = n + half H cross Omega B where Omega B is equal to uh of course Omega B is equal to eb/ M which we have said several times that this is equal to the cyclotron frequency okay so a problem which was simple enough to begin with uh we find the solution at least till now we have found the energy so the energy is nothing but it&#39;s uh n + half H CR Omega B just following the solution for the energy for a harmonic oscillator n can have 0 1 2 3 Etc all the integers and here it can be equal to zero because uh the harmonic oscillator allows for a solution with a quantum number n equal to0 so uh this is the energy or the igen values of the problem let&#39;s see the igen function so the igen function now let&#39;s call it U because we we can solve for of course uh F of Y uh that is what we&#39;ll do but then we know that the total wave function which is equal to F of or S of XY s of XY is actually a free particle in the X Direction multiplied by this F of Y okay this is the total wave function for the planer motion of the electron that is in the X direction if you take this as a x Direction it propagates like a free particle because PX is constant or KX is constant so it&#39;ll propagate like a a free wave there and will have a harmonic motion in the y direction so in the y direction it will have a harmonic motion uh we make a execute a harmonic motion about a point which is given by uh y0 and this y0 inversely depend upon B okay so uh this is nothing but uh I use a normalization uh for the X Direction uh box normalization uh this is equal to exponential I kxx and um now I don&#39;t try to normalize it it&#39;s not you know required here I use this as a the normalization constant and use the this uh okay so let me write the normalization not with a capital N because you have small n uh this thing there so let&#39;s write it with a n so a n is a normalization constant and the wave functions are for a harmonic oscillator are known to be a convolution of a gausian which is exponential minus alpha x square or alpha y Square multiplied by a polinomial function which is called as a hermite polinomial and this polinomial has a property that when n is even the polinomial is even that is it contains terms such as X to the^ 0 x x² x the^ 4 and so on when n is odd it contains terms which are x x Cube x ^ 5 and so on so this is that form that we have to write down so this is equal to uh EB y - y ² / H cross which is the gaussian part just reminding you that uh this is the harmonic oscillator and uh you have a a gausian like this for the for the ground state and and this is for the first excited state it is like this and so on okay so this is an odd function so this is a ground state and this is the first excited state and so on so and then there is a hermite polinomial and this hermite polinomial is written as e y - y0 uh divided by H cross Etc the exact form is not important for us at this moment but we need to know that at least there there&#39;s a hermite polinomial and then there is is a gaussian term which comes with a normalization constant and this part is purely because of the X part and this is because of the Y and as I said uh that the Z component even if you consider it&#39;s like a free particle in the Zed Direction in particular if you confine electrons in two Dimension then that uh term does not arise now if you go to this uh last last slide you have n + half H cross Omega and this n = to 0 1 2 3 Etc these are called as the landow levels so the energy levels have a name which are called as a landow levels and landow is a Russian physicist and uh the wave function is actually a freely propagating part in The X Direction multiplied by a har harmonic part which is executing a harmonic oscillation which has a gaussian term and a harite polinomial uh n when n is even the polinomial is even when n is odd the polinomial is odd okay these land levels so if if we really plot it how does it look like the landow levels will simply look like this okay they&#39;re equidistant okay for a given value of B so this is say n = to 0 n = 1 n equal to 2 and so on uh but if you draw for another magnetic field it may look like this and so on okay so then n = 1 n = 2 N = 3 and so on okay and the reason is that Omega B uh which is the cyclotron frequency it is directly proportional to B so as you increase B let&#39;s say this B1 for a magnetic Field B1 and this is for a for B2 and of course B2 is greater than B1 okay and these are called as the landow levels so uh now what are the properties of the landow levels and uh we would eventually get to that that these landow levels are the most important things in understanding the structure of the plateaus okay and these are not really very sharp levels because of disorder this get broadened we&#39;ll come to that there&#39;s one very important things that these landow levels are infinitely degenerate or their degeneration is very high so degeneracy of the landow levels so there are many many states corresponding to each one of the N values so n equal to0 has very large number of states which means that these States can be occupied by electrons so each land level is occupied by many many electrons so let&#39;s understand the how that degeneracy arises and where does it come from it comes from the value that KX is a constant and this result that you have got is independent of uh KX so KX doesn&#39;t play a role here which means any value of KX would give rise to this level and this KX has a NX associated with it which denote the quantum number of the states which means that corresponding to any value of n here a given value of n there could be a very large number of values of NX that are possible okay so this in principle this is infinite but the degeneracy is actually limited by some factors and uh let us you know try to understand that what factors do they depend upon but I hope that it is clear that these uh levels are heavily degenerate and the degeneracy is because they are uh coming from the fact that any value of KX would satisfy this equation or any value of NX would satisfy these energy levels so each of the energy levels do not depend upon on um NX and any value would be then acceptable solution okay so uh if we want to know the degeneracy Let&#39;s uh look at this okay so uh we have written this down earlier that the y0 is equal to so we have uh y 0 = to PX over EB we have written this earlier uh and which we called as lb which is a magnetic length and this is nothing but H cross KX uh divided by EB okay because PX is H cross KX and this is equal to H cross um uh KX is 2 pi uh uh NX by LX okay so this is your uh i0 and we want to find that what is the maximum degeneracy so we want to find what is NX Max okay and NX Max all others are constant the y0 is the is basically a length scale in the y direction okay along the y direction there some value of y which is determined by of course the magnetic field now this y0 can at the most be ly that&#39;s the uh length of the sample in the y direction that y z cannot be outside the sample so the maximum value of y0 will be ly and so maximum value of NX Max can be obtained if you substitute y0 as l y so this is equal to EB uh LX l y uh because I&#39;m putting y0 as ly and this is equal to 2 pi H cross so that&#39;s the maximum degeneracy so this is equal to EB into a a is the area of the sample uh do not confuse it with anything else it&#39;s the a is the area of the sample and uh divided by this H because 2 pi H cross is nothing but h H I can write this as a / H over uh e and uh so I can I can take that b uh this thing I mean I can leave that b there and put this electronic charge below the uh in the denominator because H over e is a very important quantity which is called as a flux Quantum which is denoted by some 5 0 okay so this is equal to a flux Quantum which has a particular value um in Weber some 10^ -4 10^ -5 value so you see that let&#39;s call this as G the degeneracy so G is equal to uh it depends on the the B the magnetic field the area of the sample and a 5 Z and this tells you that this degeneracy can be infinite because a b if you go back to the first slide that we had and you see here the magnetic field goes all the way up to 14 15 Tesla so there&#39;s about 15 Tesla so there&#39;s a 15 Tesla magnetic field if you want if you have facilities you can increase that magnetic field even more so and the sample size is up to you okay so this is really a very large degeneracy that is coming out of each one of the landow levels are degenerate in fact this has important um repercussions or imp implications on the fractional Quantum hall effect okay so two things that we notice here so you note that uh the degeneracy is of course what I told you that degeneracy is proportional to uh is proportional to b and a uh I told you B is the magnetic field and a is the area and um the second is that you know the trajectory uh can be U the trajectory is centered about some y0 uh is uh shm in y direction and centered about y0 so uh these are two very important you know outcome of this exercise that we have done that is we have solved uh a charge particle a single charge particle uh in presence of a perpendicular or a transverse magnetic field we have taken a particular gauge and suppose we have taken the gauge in which the magnetic Vector potential is in the y direction in that case you&#39;ll have just the X and Y being interchanged that the system the the particle will propagate like a free particle in the y direction and will execute a simple harmonic oscillation in the X Direction so X and Y they have no specific significance and the third gauge which we&#39;ll talk about later it&#39;s called as a spherical gauge or symmetric gauge and in that case you have both X and Y component and these land levels are actually found out to be circular okay so um now there are some incidental similarities which I&#39;ll just point out and we&#39;ll discuss them uh as we go along and the similarities are that the hall resistivity it shows plateaus in multiples of H over uh these plateaus are H over e s okay so H over say n square or something where n is an integer okay so uh consider to be n0 okay that&#39;s the density of carriers so Define a quantity called as new or uh let&#39;s let&#39;s call this as uh just what we called here as n let me call this as new where new takes values 1 2 3 Etc we just change this because the too many n&#39;s n is used for density as well so consider charge density of the of the carriers to be n0 so new is defined as n0 divided by uh G by a now this G by a is the uh the degeneracy per unit area so let&#39;s not have the geometry of the sample coming in because you can have any any geometry but if you divide it by a then it becomes a fundamental quantity which only depends on the value of the magnetic field and we are now talking about a given value of the magnetic field and this is equal to nothing but uh n0 H ided EB and uh this is equal to so this is equal to a number of electrons and I can I can write this as n0 ided H over e into B so that&#39;s the flux Quantum so this number of electrons and divided by the you know the number of uh flux quanta and so on okay so if new assumes a value integer that is when the number of electrons and the number of flux Quantum they become commen at fractions or even uh fractions in the fractional Quantum hall effect then there are plateus uh that are seen in the resistivity okay so these depending upon now you you tune in the value of the magnetic field go go back to this uh first slide which we had you see the magnetic field is being continuously tuned from a value 0 to 15 Tesla so magnetic field is continuously ramped up and you are doing this experiment and what you see is that when this quantity which we just found out that this quantity becomes an integer then there are these Hall plateus which are seen which we saw last so when this becomes an integer you have a plateau which means that this ratio becomes an integer uh you see plateus then you again tuned B this goes from being an integer to uh a fraction then you no longer see plateau and then this thing uh the resistivity gives shows a jump and then you go to the next integer such that it it shows another plateau and that&#39;s how this new actually would uh be uh you know uh sort of it&#39;ll go from one uh integer value to another integer value and it will pull along uh the resistivity of the material to be having plate um and and so on and from one Plateau to another and this is what we are going to see more carefully I&#39;ll give you a list of references that uh uh you uh should follow uh some of them are Advanced references and um you can follow them along with my lectures here uh some of them are of the same level um often I don&#39;t remember them but next uh discussion I&#39;ll provide the you with those references [Music]

Transcript for:Understanding the Quantum Hall Effect

Transcript for:
Understanding the Quantum Hall Effect