In the two previous tutorials we learned about a number of quantum operators as well as the
wavefunction. We also learned how to calculate expectation values, which represent the way
to determine the results we can expect when we measure specific variables for a quantum
system. Now it’s time to put it all together, so that we can revisit the Schrodinger equation
in more depth. In one of the first tutorials in this modern
physics series, we displayed the Schrodinger equation in roughly the following form, although
we barely discussed it at all, since we were not equipped to do so. Now we can read this
properly as i times h bar times the partial derivative of psi of x and t with respect
to time, being equal to the Hamiltonian operator acting on psi of x and t. So we will now be
more precise in having psi depend on both position and time. The only thing in this
equation that remains unfamiliar at this point is the Hamiltonian operator. We know what
operators are in general, so let’s define this one. The Hamiltonian operator has two
components. There is a kinetic energy operator, T, and a potential energy operator, V, and
the Hamiltonian operator is the sum of these two operators. In the previous tutorial, we
defined the kinetic energy operator, and found that it was equivalent to this expression,
negative h bar squared over 2M times a second partial derivative with respect to x, so let’s
go ahead and make that substitution now. The potential energy operator is a little trickier,
as it will depend on what type of potential we have, so let’s just leave that as it
is for the time being. With the Hamiltonian operator defined, let’s now plug this into
the Schrodinger equation in place of H. And that leaves us with the explicit expression
of the Schrodinger equation. So, we finally have this important equation
well-defined with terms that we understand, but what does the equation mean? Well, one
straightforward way to find meaning in the Schrodinger equation is to think of it kind
of like Newton’s second law for quantum mechanics. For a given set of initial conditions,
which can be written down in the initial wavefunction psi of x and t at t equals zero, the Schrodinger
equation can tell us how the system will evolve with time. That is, what future iterations
of psi of x and t will look like. This is precisely the way that knowing the initial
values for position, velocity, and acceleration with regards to a classical particle, and
being aware of any forces acting upon the particle, we can determine what any of these
parameters will be at some future time. But in the tutorial following this one, we
will use the Schrodinger equation in a way that does not really have a counterpart in
classical physics. We will use it to find what the wavefunction looks like for a static
condition. This would be like saying that we will use Newton’s equation to find out
what a tennis ball looks like, which sounds absurd. But in quantum mechanics, objects
are described by wavefunctions, which, as we saw, are not point-like entities. The spatial
distribution of quantum objects is determined by the boundary conditions of the system at
hand, and it is those conditions that constrain the actual physical shape of the probability
density function that governs any quantum object, as we have discussed previously.
This is a really abstract concept, but grasping this idea will allow us to really understand
things like atomic orbitals with exceptionally greater depth than before. We have to internalize
the notion that a quantum particle is delocalized, and the probability density function is a
measure of the range of positions that the particle occupies, which is determined by
the boundary conditions of the system. This is where the wave-like nature of quantum particles
becomes indispensable. Imagine a flat, calm lake. Then imagine that some object has perturbed
the water, generating waves. These will propagate freely in the lake. We can think of this as
open boundary conditions, which makes these waves free-traveling. Now instead of a lake,
think of a dam with some walls that cause waves to reflect, thereby acting as boundaries.
The wave is trapped by these boundary conditions. In the next tutorial we will examine a specific
example of such boundary conditions on a quantum system.
Now, this approach of examining the wavefunction under static conditions seems to make things
a little easier, as it involves ignoring the time-evolution of the wavefunction. But we
just saw that the Schrodinger equation does indeed depend on time, so we actually have
to deal with this. To do so, we need to find an expression of the Schrodinger equation
that does not depend on time. This is called the time-independent Schrodinger equation.
Let’s go ahead and derive this now. Imagine that you have psi of x and t. Let’s
instead write this function as being equal to the product of phi of x and csi of t. So
the way we are breaking down psi is that phi of x contains the spatial dependencies of
psi of x and t, and csi of t contains the temporal dependencies of psi of x and t. We
call this separation of variables, and the idea is to simplify calculations by separating
a function into partial functions that depend, independently, on different variables. So
again, in this case, phi only depends on x, and csi only depends on t. Now we can plug
this into the Schrodinger equation in place of psi, for both places where psi shows up.
That leaves us with this, and from here, we can now derive the time-independent Schrodinger
equation. First, let’s expand the portion with the
Hamiltonian, so that the kinetic energy operator acts on phi times csi, and we add to that
the potential energy operator acting on phi times csi. Next, looking at this portion containing
the partial derivative with respect to time, phi does not depend on t, but only on x, which
means that when differentiating with respect to t, phi is effectively a constant. So we
can pull phi out front like this. We can actually do the same thing on the other side where
we have the second derivative with respect to x. Csi is in terms of t, so here it acts
as a constant, and we can pull it out front. The next thing we are going to do is to divide
both sides by phi of x times csi of t. On this side, we lose csi and get a phi in the
denominator. Note that the other phi does not cancel because it is being operated upon
by this second partial derivative. And on the other side, we lose phi and get a csi
in the denominator. Again, these don’t cancel because of the partial derivative.
Now here’s the interesting part. At this point, we will notice that the left side of
the equation depends only on t, and the right side of the equation depends only on x. What
this means is that if one side depends only on x and the other on t, changes in only one
variable will affect only one side and not the other. So let’s take the side that depends
on t and label it a constant, E. Since the two sides of any equation must be equal, this
means that the other side is also equal to the constant, E. So let’s set each side
equal to E instead, to get two separate equations. We are going to do something important with
each of these, but let’s work with the one containing phi first. For this equation, let’s
multiply both sides by phi of x. That gets rid of this term, and puts phi of x by both
V and E. This is the time-independent Schrodinger equation. As you can see, nothing in here
depends on time, and this gives the equation some important applications, which we will
be investigating as we move forward, though since we are used to representing wavefunctions
with psi, we will be seeing this with psi instead of phi.
Now let’s return to the time-dependent equation involving csi. Let’s see if we can solve
this equation for csi. First let’s do something that may seem strange, let’s multiply both
sides by dt. It may seem counterintuitive to break up this notation signifying a partial
derivative, but we will see in a moment precisely what this allows us to do mathematically.
To simplify a bit further let’s divide both sides by i h bar. That gets rid of them on
the left, and on the right we get E over i times h bar, dt. Preferring to have imaginary
terms in the numerator, let’s multiply by i over i. That puts i on top and i squared
on the bottom, and i squared equals negative one, so let’s just make the whole term negative.
Now let’s integrate both sides. Remember that we split up the partial derivative notation,
which means we will be integrating with respect to different things for each expression. Doing
the left side first, we can think of d csi of t as being just like dx, which means we
have an expression that resembles one over x, dx. We know that the integral of one over
x is the natural log of x, so that’s essentially what we get here. Integrating the inverse
of csi of t, d csi of t, gives us the natural log of csi of t. On the other side we have
the integral of negative i E over h bar, dt. These are all treated as constants and can
be pulled out, leaving us with the integral of dt. This will simply be t, leaving us with
negative i E over h bar times t. We can now return these expressions to being equal to
each other, and solve for csi. Recalling the definition of a natural log, we know it has
base e, so csi of t will be equal to e raised to this exponent.
This is a very important result. We just found that the temporal part of the Schrodinger
equation will always look like this, with csi of t being equal to this expression, where
E is the energy of the system. In fact, E is what we call an eigenenergy of the particle.
Let’s write this out explicitly, bringing back the equation psi of x and t equals phi
of x times csi of t, but now plug this new result in for csi, which means that psi of
x and t equals phi of x times e to the negative i E t over h bar.
Let’s try to get a little perspective on this equation. Clearly we have a product of
two terms, one being a spatial part which depends on x, and the other being a temporal
part which depends on t. This shows us that solving the time-independent Schrodinger equation
is enough to know about the time-evolution of a particle. When we solve the Schrodinger
equation, we find all allowed wavefunctions, psi of x, with their associated eigenenergies,
E. This means that in order to find the time-evolution of psi of x and t, we just need to plug those
energies into the exponential part of this equation.
Let’s now return to the time-independent Schrodinger equation and see what else we
can take from it. Recall that we have E times psi equals the kinetic energy operator acting
on psi plus the potential energy operator acting on psi. But this sum of kinetic and
potential energy operators is the Hamiltonian operator. This leads us to an interesting
conclusion. The time-independent Schrodinger equation can also be written very simply like
this, where the Hamiltonian operator acting on the wavefunction equals the eigenenergy
of the particle times the wavefunction. What this means is that energy, E, is the eigenvalue
of psi when the Hamiltonian operator acts on psi.
In this way, the Schrodinger equation is similar to some equations we have already learned.
We saw that when the position operator acts on psi, we get an eigenvalue of x. When the
momentum operator acts on psi, we get an eigenvalue of p, which for a plane wave, will be equal
to h bar k. And now, when the Hamiltonian operator acts on psi, the eigenvalue is E,
which is the total energy associated with that wavefunction.
So again, it is this energy that, according to this equation, will determine the time-evolution
of the system. That is, if you know the eigenenergy of a quantum state, and you know its wavefunction,
you also know how it will evolve over time, and you can make actual predictions about
it, just like the way you can predict the time-evolution of the classical motion of
a particle. The only difference is that now you have to deal with uncertainty relations
which are not a factor for macroscopic objects. And with that, we should have a better understanding
of the Schrodinger equation in all of its forms, as well as precisely what these forms
represent, and what we can do with them.