D.32 Integral conservation laws

This section derives the integral conservation laws given in addendum {A.14}.

The rules of engagement are as follows:

- The Cartesian axes are numbered using an index , with 1, 2, and 3 for , , and respectively.
- Also, indicates the coordinate in the direction, , , or .
- Derivatives with respect to a coordinate are indicated by a simple subscript .
- Time derivatives are indicated by a subscript t.
- A bare integral sign is assumed to be an integration over all space, or over the entire box for particles in a box. The is normally omitted for brevity and to be understood.
- A superscript indicates a complex conjugate.

First it will be shown that according to the Schrödinger equation
is constant. The Schrödinger equation in free space is

Taking the right hand term to the other side and writing it in index notation gives

Multiply the left hand side by and add the complex conjugate of the same equation to get

To show that the integral is constant, it must be shown that its time derivative is zero. Now the first two terms above are the time derivative of . So integrated over all space, they give the time derivative that must be shown to be zero. And the equation above shows that it is indeed zero provided that the remaining sum in it integrates to zero over all space.

The constant is not important in showing that this is true, so just
examine for any

This equals

as can be seen by differentiating out the parenthetical expression with respect to . The above integrand can be integrated with respect to . It will then be zero for a periodic box since the expression in parenthesis is the same at the upper and lower limits of integration. It will also be zero for an impenetrable container, since will then be zero on the surface of the container. It will also be zero in an infinite region provided that and its derivatives vanish at large distances.

There is another way to see that is constant. First recall
that any solution of the Schrödinger equation takes the form

Here the are the energy eigenfunctions. Then

Now because of orthonormality of the eigenfunctions, the integration only produces a nonzero result when , and then the product of the eigenfunctions integrates to 1. So

That does not depend on time, and the normalization requirement makes it 1.

This also clarifies what goes wrong with the Klein-Gordon equation.
For the Klein-Gordon equation

The first sum are the particle states and the second sum the antiparticle states. That gives:

The final two terms in the sum oscillate in time. So the integral is no longer constant.

The exception is if there are only particle states (no ) or only
antiparticle states (no ). In those two cases, the integral is
constant. In general

where the integrated square magnitudes of and are constant.

Next it will be shown that the rearranged Klein-Gordon equation

preserves the sum of integrals

To do so it suffices to show that the sum of the time derivatives of the three integrals is zero. That can be done by multiplying the Klein-Gordon equation by , adding the complex conjugate of the obtained equation, and integrating over all space. Each of the three terms in the Klein-Gordon equation will then give one of the three needed time derivatives. So their sum will indeed be zero.

To check that, look at what each term in the Klein-Gordon equation
produces separately. The first term gives

or taking one time derivative outside the integral, that is

That is the first needed time derivative, since a number times its complex conjugate is the square magnitude of that number.

The second term in the Klein-Gordon equation produces

That equals

as can be seen by differentiating out the parenthetical expression in the first integral with respect to and bringing the time derivative in the second term inside the integral. The first integral above is zero for a periodic box, for an impenetrable container, and for infinite space for the same reasons as given in the derivation for the Schrödinger equation. The second term above is the needed second time derivative.

The final of the three terms in the Klein-Gordon equation produces

That equals

as can be seen by bringing the time derivative inside the integral. This is the last of the three needed time derivatives.