In the previous tutorial, we solved the time-independent Schrodinger equation for a quantum particle
in an infinite square well, also known as the problem of the particle in a box. As we
learned, the solution to this problem takes the form of a set of eigenfunctions, represented
by psi sub n, and their corresponding energies, E sub n, which both depend on the length of
the well, represented by a, and where k sub n is equal to n pi over a, where n is the
set of positive integers. These expressions should look familiar, so if they aren’t,
make sure to watch particle in a box part one first and get yourself up to speed. Now
what we are going to do is try to understand these results on a deeper level. A good way
to start is to analyze the steps we took to solve the equation.
First, recall that we imposed the condition that the wavefunction must be continuous over
its domain of existence. This was because the wavefunction must equal zero where the
potential is infinity, including the boundaries of the well, and the wavefunction must have
some non-zero value where the potential is zero, or within the well. And second, we imposed
the condition that the wavefunction must be normalized. This is because normalization
ensures that the probability of finding a particle within the boundaries is not greater
than one. In this way, a wavefunction that is not normalized can’t represent physical
reality. Now, there is another property of wavefunctions
that we have discussed but that we haven’t mentioned yet in this context. We must recall
that solutions to the Schrodinger equation, which correspond to different eigenvectors,
are orthogonal. So referring back to our solution set, each value of n defines a solution to
the equation given by a different wavefunction, psi, and corresponding with a different eigenenergy.
Each wavefunction in the solution set is linearly independent with respect to the others, and
as we recall this can be expressed mathematically as follows, with the inner product of psi
sub n and psi sub m being equal to the Kronecker delta, which equals one whenever n equals
m, and equals zero whenever n and m are not equal. Here, the subscripts are simply referring
to different wavefunctions. So if n equals one and m equals two, the inner product of
psi one and psi two equals delta one comma two, which equals zero, since these are two
different wavefunctions from the solution set and are thus orthogonal, so their inner
product must be zero. But if n and m both equal two, then the inner product of psi two
with itself will equal delta two comma two, which equals one, since the inner product
of any wavefunction with itself must be one. Let’s go ahead and make sure that the wavefunctions
in our solution set satisfy this condition. Recalling the definition of the inner product
and the boundaries of our system, this must equal the integral from zero to a of psi sub
n star of x times psi sub m of x dx, which again equals the Kronecker delta. As we already
know, this term equals zero when n and m are not equal, and it equals one when n and m
are equal. We solved this integral for the case where n equals m when we normalized the
wavefunction in the previous tutorial, and saw that it did indeed equal one. Now we must
solve for the case where n and m are not equal, meaning that we are looking at two different
wavefunctions from the solution set. So let’s plug in our expression for psi,
so that we can find out if any two arbitrary wavefunctions are indeed orthogonal. Given
that there are no imaginary terms here, psi star is the same as psi, so we can just take
the integral of psi sub n times psi sub m, which will only be different in that one has
n in the parenthetical, and the other has m. Each has a root two over a term, so those
can be pulled out of the integral as constants, where they multiply to give two over a. And
what we have left is the product of two sine terms, which can’t be combined because the
parentheticals are different. Now as with many other instances in which we are integrating
trigonometric functions, there is a particular trigonometric identity that will be useful.
In this case, we need to use the identity which says that two sine theta sine phi equals
cosine theta minus phi minus cosine theta plus phi. This is useful because it splits
up a product of trigonometric functions into a difference, which makes it easier to integrate.
So if we consider theta to be n pi over a times x, and phi to be m pi over a times x,
we can make this substitution. First, since we don’t have a two multiplying the product
of sines, we will actually get one half of this result, so we can compensate by pulling
one half out of the integral, thereby reducing two over a to one over a. Then inside the
integral, we get the cosine of n pi over a times x minus m pi over a times x, minus the
cosine of n pi over a times x plus m pi over a times x. We can get these parentheticals
to each become one term simply by factoring out the n’s and m’s, so for the first
one we can factor out n minus m, and for the second one we can factor out n plus m. And
now, believe it or not, we have something simple that is ready to integrate, because
we just have cosine terms with some constant times x. This means that in finding the antiderivative,
we will get sine terms with the same parenthetical, but each will be divided by its respective
constant, so that if we were to differentiate, the constant coming out front by the chain
rule would get cancelled out. So that leaves us with the constant one over a from earlier,
times this quantity, a over n minus m times pi, sine n minus m times pi over a times x,
minus a over n plus m times pi, sine n plus m times pi over a times x. Now it’s time
to plug in our limits of integration in place of x, and evaluate. If we plug in a, we can
see that it will cancel out with the other a in each term, which leaves us with the sine
of some multiple of pi. The sine of any multiple of pi is zero, so plugging in a gives us zero
for the whole expression. Likewise, plugging in zero means taking the sine of zero, which
is zero, so again the whole expression is zero. So all together, no matter what values
are to be used for n and m, we get zero minus zero, which is zero. In this way, we have
proven that any two wavefunctions from the solution set must be orthogonal.
Now for the second realization we turn from the wavefunctions, or eigenfunctions, over
to the eigenenergies. This expression contains n as well, and the smallest value n can take
is one. This represents the lowest energy, or ground state energy for the particle in
the box. This gives us a markedly quantum result. The quantum particle can not possess
zero energy. With a classical system we could expect a value of zero for kinetic energy.
If a tennis ball was sitting still at the bottom of the well, we would regard this as
having zero kinetic energy. But the quantum particle is not like this. It is delocalized,
sitting everywhere it is allowed to be all at once, and at all times.
There is actually another way we can prove to ourselves that n can’t be zero. Let’s
return to our generalized expression for psi. If n equals zero, then k sub n equals zero,
and we are taking the sine of zero, which is zero. That means psi sub zero equals zero,
and the corresponding probability density function would equal zero everywhere inside
the well, and the particle therefore does not exist. So in order for the particle to
exist at all, it can never possess zero energy. This is a fascinating aspect of quantum mechanics
that is in stark contrast with classical mechanics, and it is important to internalize this strange
fact as thoroughly as possible, even if it seems to defy common sense.
Now let’s tackle some slightly more nuanced points. We are first going to calculate the
expectation value for momentum, and then the expectation value for kinetic energy. So let’s
start out by calculating the expectation value for the momentum operator for the first solution,
psi one. As we recall, this is written by sandwiching the momentum operator in brackets
between the ket of psi and bra of psi, which equals the integral from zero to a of root
two over a sine k one x, times i h bar d dx, root two over a sine k one x dx.
Let’s pull out the constants. Root two over a and root two over a both come out and multiply
to give two over a, and we can pull out i times h bar as well. Left inside is sine k
one x times the partial derivative with respect to x of sine k one x dx. The derivative of
sine k one x with respect to x is k one cosine k one x, so let’s make that change, and
pull k one out to be with the other constants. Now it’s time to use another one of our
fancy trigonometric identities. It is the case that sine two theta equals two sine theta
cosine theta. Therefore, sine theta cosine theta equals sine two theta over two. This
means that we can take sine k one x cosine k one x and express it as sine two k one x
over two. Allowing this two to cancel out the two out here, that leaves us with this,
and this is no problem to integrate. The antiderivative of this sine term will be negative cosine,
and then we must divide by two k one, due to the chain rule that would occur when differentiating.
We can put the negative sign out front, and these k one terms will cancel out, leaving
us with negative one over a times i h bar times cosine two k one x. Now we can evaluate
this first at a and then subtract from that the evaluation at zero. This entails just
replacing x with a and then zero. Now recall that the k term equals n pi over a, and for
k one, n will be one, so k one equals pi over a. For the second term it doesn’t matter
because it’s being multiplied by zero, so that’s just cosine of zero. But for the
first term, the a’s cancel, and we get cosine of two pi. Cosine of two pi is one, and cosine
of zero is also one. One minus one is zero, which means our answer here is zero.
So what we have determined here is that the expectation value of the momentum of the particle
in a box is always zero. This may seem a bit strange, and we will contextualize this in
a moment, but first let’s perform this same operation for kinetic energy.
So we’re going to do what we just did, but with the kinetic energy operator. As we recall,
this is represented by T, and equal to negative h bar squared over two m times the second
partial derivative with respect to x. So to get the expectation value, this time we will
have to evaluate this integral. We can pull the constants out of the integral, so that’s
negative h bar squared over two m, as well as both of the root two over a terms, so the
twos cancel, leaving us with negative h bar squared over a m, and inside the integral
we are left with sine k one x times the second derivative with respect to x of sine k one
x dx. To continue, we need to evaluate this second
derivative, so let’s do this separately. The first derivative is k one cosine k one
x, so the second derivative is negative k one squared sine k one x. Let’s pull negative
k one squared out of the integral, so the negative signs cancel, and we can put h bar
k one quantity squared in the numerator, since both of these terms are being squared. We
have identical sine terms left in the integral, so we need to integrate sine squared k one
x dx. Now we just so happen to have solved this precise integral in the previous tutorial
using yet another trigonometric identity, so refresh your memory there if you need to.
Otherwise we will just skip to the solution here.
We end up with our constants from earlier, times a over two minus sine of two k one a
over four k one. Recalling that k equals n pi over a, we can plug that in everywhere
we see k one. That gives us h bar n pi over a quantity squared, over a m, times a over
two minus sine of two n pi over four n pi over a. Conveniently, the sine of any multiple
of pi is zero, so this whole sine term will always be zero and we can get rid of it. Let’s
simplify by bringing the a squared down to the denominator where it belongs, to combine
with the other a to get a cubed. And then we can multiply this by the a over two term.
This will cancel out one of the a’s in the denominator and put a two down there as well,
so we end up with h bar n pi quantity squared over two m a squared. This expression is the
expectation value for the kinetic energy operator. The most important thing to notice here is
that n is squared, which means that kinetic energy increases quadratically with n.
These results should initially be puzzling. We saw that the expectation value for momentum
is zero, but we also just found that the expectation value for kinetic energy is not zero. How
is this possible? The easiest way to understand this is to do some mathematical trickery.
Let’s recall the general form of the relevant eigenfunctions. Now what we are going to do
is use something called the Euler equation. This tells us that sine theta equals the complex
exponential of i times theta minus its complex conjugate, divided by two i. Let’s use this
in our eigenfunctions, treating k sub n x as theta, and plugging that in where we previously
used theta. Now recall from a previous tutorial about wavefunctions that e to the i k one
x is a rough representation of a wave travelling to the left, and e to the negative i k one
x is a representation of a wave travelling to the right. So let’s just split up this
term into two individual terms, and set up the following definitions for psi left and
psi right. This means that the parenthetical can be written simply as psi left minus psi
right. This formula tells us that the wavefunction
psi sub n is a superposition of a wave travelling to the left and a wave travelling to the right,
with equal probability. This means that this particle is always travelling in both directions
at the same time. So in this way, the particle doesn’t really go anywhere, which is why
its momentum is zero, but the particle is still technically moving, so its kinetic energy
is not zero. And the last realization we will highlight
is that there is a clear pattern emerging between the odd and even solutions, with odd
solutions being those where n equals an odd number, and even solutions being those where
n equals an even number. The odd solutions are symmetric about x equals a over two, meaning
that the left and right halves are mirror images, and the even solutions are anti-symmetric
about this line, meaning that the halves are vertically flipped, such as the case with
n equals two, with one peak going up and the other going down.
And with that we have solved our first quantum problem, the problem of the particle in a
box, and sufficiently analyzed its solution. So at this point we should have a reasonable
grasp of what eigenfunctions and eigenenergies are. They represent the states that a quantum
particle can live in under a set of boundary conditions. Specifically, the eigenfunctions
tell us the shape of the probability density function for each of the eigenenergies that
satisfy the Schrodinger equation and the boundary conditions.
The square well we have been playing in was a great place to get our feet wet, because
it challenges our intuitions for how particles behave in the quantum world. Even though a
particle in an infinite well seems like a hopelessly abstract blackboard example, the
techniques we used to solve the Schrodinger equation will always be the same, and the
mathematical properties of the solutions are universal, regardless of the difficulty of
the problem. As we move forward, we will continue to increase the complexity of the problems
we tackle, thereby eliciting mathematical models that get us ever closer to elucidating
this quantum realm of photons and electrons.