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.