D.22 Unique ground state wave function

This derivation completes {A.8}. In particular, it proves that ground states are unique, given a real, noninfinite, potential that only depends on position. The derivation also proves that ground states cannot become zero. So they can be taken to be positive.

The basic idea is first to assume tentatively that there would be two independent ground state wave functions. These could then be taken to be orthonormal as usual. That means that the inner product of the two wave functions would be zero. However, it can be shown, see below, that ground state wave functions cannot cross zero. That means that both wave functions can be taken to be everywhere positive. (The one exception is at impenetrable boundaries, where the wave function is forced to be zero, rather than positive. But that exception does not change the argument here.) Now if you check the definition of the inner product, you see that the inner product of two positive wave functions is positive, not zero. But the orthonormality says that it is zero. So there is a contradiction. That means that the made assumption, that there are two independent ground states, must be wrong. So the ground state must be unique.

To finish the proof then, it must still be shown that ground states
that cross zero are not possible. Tentatively suppose that you had a
ground state whose wave function did cross zero. Near a zero
crossing, the wave function

Next recall from {A.8} that

To see why this wave function has less energy, compare what happens to
the kinetic and potential energies. The energy is the sum of a
kinetic energy integral

In the region that is blunted, the integrand of the kinetic energy integral is now zero, instead of whatever positive value it had in the left figure. The constant

You might at first think that the potential energy can compensate by
increasing more than the kinetic energy decreases. But that does not
work, because the integrand of the potential energy integral is
proportional to

That is basically it. Unfortunately, there are a few loose ends in
the reasoning above. That is even if you ignore that
small

is not a valid mathematical concept. A number
is either zero or not zero; that is all in mathematics. The correct
mathematical statement is: “there is a number small enough

does
not imply small. All positive numbers less than 1 000 are small
enough to be less than 1 000.) But that is so picky.

More importantly, you might object that after blunting, the wave
function will no longer be normalized. But you can correct for that
by dividing the given expression of the expectation energy by the
square norm of the wave function. In particular, using a prime to
indicate a quantity after blunting the wave function, the correct
energy is

Now note that the square norm changes negligibly under the smoothing, because its integrand involves

In fact, there is a trick to avoid the normalization problem
completely. Simply redefine the potential energy by a constant to
make the expectation energy zero. You can always do that; changing
the definition of the potential energy by constant does not make a
difference physically. And if the expectation energy

Comparing start and end, you see that the sign of the change in energy is the same as the sign of the change in the kinetic and potential energy integrals. Regardless of whether

You might further object that the given arguments do not account for the possibility that the wave function could cross zero with zero slope. In that case, the integrand of the original kinetic energy would be vanishingly small too. True.

But in one dimension, you can use the Cauchy-Schwartz inequality of
the notations section on

The first inequality above is the Cauchy-Schwartz inequality. The final equality applies because the change in

where

The same arguments normally apply for the blunted region at positive

However, as seen above, the negative blunted part has kinetic energy of at least

Note that in neither approach, the zero crossing point can be at an
impenetrable boundary. Neither blunted wave function is zero at

Also note the need to assume that the potential does not become positive infinity. If the potential is positive infinity in a finite region, then the wave function is in fact zero inside that region. The particle cannot penetrate into such a region. Its surface acts just like an impenetrable boundary.

How about wave functions in more than one dimension? That is easy, if
you will allow a very minor assumption. The minor assumption is that
there is at least a single crossing point where the gradient of

Accepting that assumption, things are straightforward. Take the
blunted wave function essentially like the middle graph in figure
D.1. The

However, all that does require that one minor assumption. You might
wonder about pathological cases. For example, what if one wave
function is only nonzero where the other is zero and vice-versa? With
zero gradient at every single point of the zero crossings to boot? Of
course, you and I know that ground states are not just stupidly going
to be zero in sizable parts of the region. Why would the electron
stay out of some region completely? Would not its uncertainty in
position at least produce a very tiny probability for the elecron to
be inside them? But proving it rigorously is another matter. Then
there are somewhat more reasonable conjectures, like that a wave
function would become zero at a single point not at the boundary.
(That would still give a unique ground state. But would you not want
to know whether it really could happen?) How about fractal wave
functions? Or just discontinuous ones? In one dimension the wave
function must be continuous, period. A discontinuity would produce a
delta function in the derivative, which would produce infinite kinetic
energy. But in multiple dimensions, things become much less obvious.
(Note however that in real life, a noticeably singularity in

You might guess that you could use the approach of the right graph in
figure D.1 in multiple dimensions, taking the

There is however a method somewhat similar to the one of the right
graph that continues to work in more than one dimensions. In
particular, in three dimensions this method uses a small sphere of
radius

The discussion will use three dimensions as an example. That
corresponds, for example, to the electron of the hydrogen molecular
ion. But the same arguments can be made in any number of dimensions.
For example, you might have a particle confined in a two-dimensional
quantum well. In that case, the sphere around the point of zero wave
function becomes a circle. Similarly, in a one-dimensional quantum wire,
the sphere becomes the line segment hypersphere.

A two-dimensional hypersphere is physically a
circle, and a one-dimensional hypersphere is a line segment. As discussed
in the notations section, a general volume

and surface
area

given by:

For example,

volumeis physically the area of the circle, and the

areais its perimeter. The derivations will need that the

Here

To show that points of zero wave function are not possible, once again
first the opposite will be assumed. So it will be assumed that there
is some point where

To find the contradiction, define a radial coordinate

Function

Take now some small sphere, of some small radius

The blunting inside this sphere will now be achieved by replacing the

Consider now first what the corresponding increase in the potential
energy integral inside the sphere is:

Multiplying out the square, that becomes:

Since

Note that the hypersphere formula for the volume of the sphere has been used. The purpose is to make the final result valid in any number

So finally

The next question is what happens to the kinetic energy. In
three-dimensional spherical coordinates, the kinetic energy after blunting
is

The initial kinetic energy is given by a similar expression, with

Since

The expression between square brackets is just the

Note that the kinetic energy does decrease. The right hand side is
positive. And if the maximum potential

But how about positive

The solution is a trick. You might say that only a mathematician
would think up a trick like that. However, the author insists that he
is an aerospace engineer, not a mathematician. The first thing to
note that there is a constraint on how much

Now the remaining integral can be estimated by the Cauchy-Schwartz inequality as before. Comparing this with the maximum possible increase in potential energy will give a limit on the maximum change in

where

If the above inequality is not satisfied, the kinetic energy decrease
would exceed the potential energy increase and it cannot be a ground
state. Note that if the sphere is chosen small, the relative decrease
in

In that case,

That means that

You might say, Why not?

And indeed, there would be
nothing wrong with the idea that almost all the change would occur in
the inner half of the sphere. But the idea is now to drive the
mathematics into a corner from which eventually there is no escape.
Suppose that you now define the inner half of sphere 1 to be a sphere
2. So the radius

(If there are

Now take the midpoint as the radius of a sphere 3. Then

Keep doing this and for sphere number

This must become zero for infinite

If

The sum within square brackets is a geometric series that has a finite limit, not an infinite one.

So there is a contradiction. At some stage the decrease in kinetic energy must exceed the increase in potential energy. At that stage, the energy can be reduced by applying the blunting. So the assumed wave function cannot be a ground state.

You might still object that a Taylor series approximation is not
exact. But in the region of interest

and the additional ratio is just a constant, about 1.15, that does not make a difference.

Woof.