In the previous tutorial we learned all about the Schrodinger equation. We derived different
forms of this equation, and talked about what all of them represent. Now it’s time to
apply this knowledge in order to solve our first quantum problem. This will be to find
the eigenenergies and eigenfunctions of a particle in an infinite square well. This
is also famously known as the problem of the particle in a box. We assume a potential which
is as follows. V of x equals infinity everywhere except between x = 0 and x = a, within which
it equals zero. Let’s unpack what that means a bit. We are
probably familiar with the concept of a potential, like an electrical potential, or gravitational
potential, and we know that an object will have a tendency to go towards a lower potential.
A ball will fall from some height, where it has greater gravitational potential energy,
down towards the ground, its potential decreasing as it goes. A compressed spring will relax
from a state of higher elastic potential to attain its equilibrium position. So here,
when we say that V of x equals infinity for these intervals, this is the mathematical
way of saying that the particle is definitely not there. That may seem counterintuitive,
as it sounds like infinite potential should mean that it certainly is there, but remember
that objects proceed towards a lower potential, and infinity is the opposite of that. So we
know the particle is sitting in this square well. The only thing is that since we are
looking specifically at a quantum particle, we have to use the Schrodinger equation to
figure out “where”, or more accurately “how” the particle lives within these walls. So now what we want to do is to find the wavefunctions
psi of x representing this particle which satisfy the Schrodinger equation. This is
the same as asking: What are the allowed states for a particle that lives between these walls?
What does its probability density function look like under these conditions? And also,
what are its allowed energies? All of these answers will arrive from determining which
wavefunctions satisfy the boundary conditions imposed by the potential, and the parameters
imposed by the Schrodinger equation. The boundary conditions tell us something
right away. As we said, the particle will surely not be located wherever the potential
is infinity. That means that for those regions, the probability density function must equal
zero. If the particle is not there, the probability of finding it there is zero, and the probability
density function is equal to zero. The probability density function is psi star times psi, so
the only way for this to equal zero is for psi of x to equal zero. Therefore, we are
looking for a wavefunction that will satisfy the following requirements, where psi of x
equals zero when x is less than or equal to zero, and also when x is greater than or equal
to a. But between x = 0 and x = a the wavefunction does not equal zero, and its value can be
determined using the Schrodinger equation. Let’s use our new time-independent form
of the equation, which as we know, looks like this. We will find that isolating the partial
derivative will make things easier, so let’s move all of these constants to the other side,
which will entail multiplying both sides by negative 2m over h bar squared. So now, the
second derivative of psi with respect to x is equal to negative 2m E over h bar squared
times psi. Let’s do another trick to simplify further. Let’s define k as the square root
of 2m E over h bar, so that k squared will be equal to 2m E over h bar squared. We can
plug that into our equation to simplify the right side, which will now be negative k squared
psi. And finally let’s bring this term to the left side by adding it to both sides.
So we have the second derivative of psi with respect to x, plus k squared psi, equals zero.
This is a second order differential equation, which we need to solve in order to find psi,
and as we know, psi depends on x. Solving this differential equation entails finding
all the forms that psi can take to satisfy this equation, just like solving a quadratic
algebraic equation entails finding the values that x can take to satisfy the equation, there
are simply different rules that must be used to find the solution.
For example, if psi of x is equal to x, then when we take the second derivative of x with
respect to x, the first derivative is one, and the second derivative is zero. That leaves
us with k squared x equals zero, which can only be true if either k or x equals zero.
K can’t be zero, because k depends on energy, and if energy is zero we don’t even have
a particle to begin with. And in this case x can’t be zero, because psi equals x, therefore
if x is zero, psi is zero, and the probability density function equals zero, which once again
means that the particle does not exist. So this is not a useful solution, because we
know that there is a particle. This is a condition for this problem. So we would need to try
something else. Now for brevity, rather than continuing to
go through many possibilities, let’s just state outright that the solution to this differential
equation is of the form psi of x equals A sine (kx) plus B cosine (kx), where A and
B are constants that we don’t know yet, but will determine by using the boundary conditions,
and k is as we defined earlier. This may seem out of left field, but this general solution
is arrived at through some clear-cut rules that are best omitted here, to be covered
over in the mathematics series for those who are interested. But this expression represents
the solution of the time-independent Schrodinger equation between the walls of the well.
Just to quickly prove that this is the solution, we can plug this expression into our rearranged
version of the Schrodinger equation. This first term involves taking the second derivative.
So let’s take the first derivative first. Remembering that the derivative of sine is
cosine, and the derivative of cosine is negative sine, and also that the chain rule requires
that we bring k out front for each term, that will give us kA cosine kx minus kB sine kx.
Then we take the derivative again, and we get negative k squared A sine kx minus k squared
B cosine kx. Then we add to that k squared times the original expression, and distributing
k squared, we get k squared A sine kx plus k squared B cosine kx. So as we can see, everything
cancels out to give us zero, as should be the case for any solution according to this
version of the equation, which means this does indeed satisfy the Schrodinger equation.
So combining this with what we said before, psi equals zero when x is less than or equal
to zero, psi equals this expression when x is between zero and a, and psi again equals
zero when x is greater than or equal to a. Let’s now try to understand this graphically,
as it will provide us with some understanding of the solution. As we said, the wavefunction
must equal zero at the boundaries. This means that psi of zero equals zero, and psi of a
equals zero. Now since we have two different trigonometric functions to consider, it will
be easier to examine them separately to get a sense of how they combine to give us the
value for the wavefunction. Let’s first remind ourselves of what sine of x and cosine
of x look like. One important thing to notice is that sine of zero equals zero, and cosine
of zero equals one. This means that no matter what value for k we have in our expression,
sine of k times zero is still sine of zero, which will always be zero, and cosine of k
times zero is still cosine of zero, which will always be one. This means that in order
for their sum in our expression to equal zero when x equals zero, the constant B must equal
zero. That means we can simplify our expression and get rid of the cosine term entirely. That
leaves us with psi of x equals A sine kx. Now let’s examine the other boundary. Psi
of a must also equal zero. This could be achieved by setting the constant A equal to zero, but
that would mean that psi of x equals zero for all values of x, which again means that
the probability density function would equal zero for all values of x, and there is no
particle, so that is not a useful solution. The other possibility is that sine of kx equals
zero, and we are specifically examining the boundary where x equals a, so sin of ka must
equal zero. Looking back at our graph of sin of x, we know that it is equal to zero at
x equals zero, and again at pi, and two pi, and every successive multiple of pi. Therefore,
sine of ka equals zero when ka equals n pi, where n is any integer. This also means that
k equals n pi over a. Let’s again examine this graphically. To make things simple, let’s
set a equal to one, and then examine the graphs of sin of kx, which will therefore be sin
of n pi x, and see what happens with different integer values for n. Each of these n values
gives us a solution to the Schrodinger equation. We can see that in each case, the wavefunction
has a value of zero at both boundaries. Recalling that k depends on energy, if k also depends
on n, then energy depends on n, and each value of n is giving us a different wavefunction,
or eigenfunction, and a different eigenenergy for the particle. These are the allowed states
for the particle. We can see how quantization comes into play here, as n must be an integer
in order for the sine function, and therefore the wavefunction, to satisfy the boundary
conditions. The fact that n must be an integer is what imposes quantization on this system.
We have just one more thing to do. We have not solved for the constant A, so we still
don’t know precisely what the wavefunction looks like. In order to do this, the wavefunction
must be normalized from 0 to a. As we recall from a previous tutorial, we will use this
bra ket notation to represent the normalization of psi, which will equal the integral from
0 to a of psi star times psi dx, and because we are normalizing, this must equal one. Now
in this case, psi is a real function, there are no imaginary terms, so psi star times
psi is the same thing as psi squared, so let’s just square our expression for psi and put
that in the integral. That gives us A squared, which is a constant that we can pull out of
the integral, and inside the integral we have sine squared kx dx, which again, must equal
one. Now we have to solve this integral, and if
we are a little rusty with our integration we will have to remember that we can use a
trigonometric identity here. Sine squared theta can be expressed as one minus cosine
2 theta, over two. So let’s apply that identity here, and change our expression into one minus
cosine 2kx over two. Let’s make this simpler by splitting it into two terms, and giving
each its own integral, so the integral of one half dx minus the integral of one half
cosine 2kx dx, and we can evaluate these separately. The integral of one half is simply x over
two, and plugging in our values we get a over two minus zero, or a over two. And the other
integral isn’t much harder, since we know that the antiderivative of cosine is sine,
so we get sine 2kx over two, but then we have to put an additional 2k in the denominator,
because we are basically doing the chain rule in reverse, and that will cancel out the additional
2k in the numerator that would occur if this were being differentiated. So now we plug
in our values. First we plug in a for x, and then subtract from that the same expression
but with zero for x. Anything times zero is zero, and the sine of zero is zero, so this
whole second term is simply zero. Now let’s put the results of these two integrals in
parentheses so that we can bring back A squared, and then we can make another substitution.
We know that ka equals n pi, so let’s make that change. We know that n must be an integer,
and we know that the sine of any multiple of pi must be zero. So this entire term will
actually be equal to zero. That leaves us with simply upper case A squared times lower
case a over two equals one. Just a bit more algebra and we find that upper case A must
equal the square root of two over lower case a. And now that we have solved for A, we can express our solution as a set of wavefunctions,
which are the eigenfunctions that satisfy the Schrodinger equation for the boundary
conditions given by the square well. So psi sub n equals root two over a sine k sub n
times x, where k sub n equals n pi over a, where each integer value of n gives us a different
wavefunction. So when n is 1, we get psi 1, when n is 2 we get psi 2, and so forth. Each
of these wavefunctions has an associated eigenenergy. Using our definition of k, we can now express
E sub n as being equal to h bar squared k sub n squared over 2m, and given that k sub
n equals n pi over a, this will equal h bar squared n squared pi squared over 2m a squared.
Again, each integer value of n will give us an eigenenergy which describes the corresponding
wavefunction. And there we have it. So that was definitely a bit of work, but
we have successfully solved our first quantum problem. We determined the allowed energies
of a particle in an infinite square well given certain boundary conditions. Even though the
problem has been solved there is still a lot to discuss in terms of the implications of
the solution, which will require an entire tutorial unto itself, so let’s move forward
and get some more perspective on the particle in a box.