D.32 Integral conservation laws

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

The rules of engagement are as follows:

First it will be shown that according to the Schrö­din­ger equation $\int\vert\Psi\vert^2$ is constant. The Schrö­din­ger equation in free space is

\begin{displaymath}
{\rm i}\hbar \frac{\partial\Psi}{\partial t} =
- \frac{\hbar^2}{2m} \nabla^2 \Psi
\end{displaymath}

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

\begin{displaymath}
{\rm i}\hbar \frac{\partial\Psi}{\partial t} +
\sum_i \frac{\hbar^2}{2m} \Psi_{ii}
= 0
\end{displaymath}

Multiply the left hand side by $\Psi^*$$\raisebox{.5pt}{$/$}$${\rm i}\hbar$ and add the complex conjugate of the same equation to get

\begin{displaymath}
\Psi^* \frac{\partial\Psi}{\partial t} +
\Psi \frac{\par...
...\hbar}{2m{\rm i}} (\Psi^* \Psi_{ii} - \Psi \Psi^*_{ii})
= 0
\end{displaymath}

To show that the integral $\int\vert\Psi\vert^2$ is constant, it must be shown that its time derivative is zero. Now the first two terms above are the time derivative of $\vert\Psi\vert^2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\Psi^*\Psi$. 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 $i$

\begin{displaymath}
\int (\Psi^* \Psi_{ii} - \Psi \Psi^*_{ii})
\end{displaymath}

This equals

\begin{displaymath}
\int (\Psi^* \Psi_i - \Psi \Psi^*_i)_i
\end{displaymath}

as can be seen by differentiating out the parenthetical expression with respect to $r_i$. The above integrand can be integrated with respect to $r_i$. 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 $\Psi$ will then be zero on the surface of the container. It will also be zero in an infinite region provided that $\Psi$ and its derivatives vanish at large distances.

There is another way to see that $\int\vert\Psi\vert^2$ is constant. First recall that any solution of the Schrö­din­ger equation takes the form

\begin{displaymath}
\Psi = \sum_n c_n e^{-{\rm i}E_n t} \psi_n({\skew0\vec r})
\end{displaymath}

Here the $\psi_n$ are the energy eigenfunctions. Then

\begin{displaymath}
\int \vert\Psi\vert^2 = \int \Psi^*\Psi =
\int \sum_{\un...
...vec r})
\sum_n c_n e^{-{\rm i}E_n t} \psi_n({\skew0\vec r})
\end{displaymath}

Now because of orthonormality of the eigenfunctions, the integration only produces a nonzero result when ${\underline n}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $n$, and then the product of the eigenfunctions integrates to 1. So

\begin{displaymath}
\int \vert\Psi\vert^2 = \sum_n c_n^* c_n
\end{displaymath}

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

\begin{displaymath}
\Psi = \sum_n c_n e^{-{\rm i}E_n t} \psi_n({\skew0\vec r})
+ \sum_n d_n e^{{\rm i}E_n t} \psi_n({\skew0\vec r})
\end{displaymath}

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

\begin{displaymath}
\int\vert\Psi\vert^2 = \sum_n \left( c_n^* c_n + d_n^* d_n...
... d_n e^{2{\rm i}E_n t} + d_n^* c_n e^{-2{\rm i}E_n t} \right)
\end{displaymath}

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 $d_n$) or only antiparticle states (no $c_n$). In those two cases, the integral is constant. In general

\begin{displaymath}
\Psi = \Psi_1 + \Psi_2
\qquad \Psi_1 = \sum_n c_n e^{-{\...
...d \Psi_2 = \sum_n d_n e^{{\rm i}E_n t} \psi_n({\skew0\vec r})
\end{displaymath}

where the integrated square magnitudes of $\Psi_1$ and $\Psi_2$ are constant.

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

\begin{displaymath}
\frac{1}{c^2} \frac{\partial^2\Psi}{\partial t^2}
- \sum_i \Psi_{ii} + \left(\frac{mc^2}{\hbar c}\right)^2 \Psi = 0
\end{displaymath}

preserves the sum of integrals

\begin{displaymath}
\int \left\vert \frac{1}{c} \frac{\partial\Psi}{\partial t...
...2 +
\int \left\vert \frac{mc^2}{\hbar c} \Psi \right\vert^2
\end{displaymath}

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 $\partial\Psi^*$$\raisebox{.5pt}{$/$}$$\partial{t}$, 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

\begin{displaymath}
\int \frac{1}{c^2}
\left(
\frac{\partial \Psi^*}{\part...
...}{\partial t} \frac{\partial^2\Psi^*}{\partial t^2}
\right)
\end{displaymath}

or taking one time derivative outside the integral, that is

\begin{displaymath}
\frac{1}{c^2} \frac{{\rm d}}{{\rm d}t} \int
\frac{\partial \Psi^*}{\partial t} \frac{\partial\Psi}{\partial t}
\end{displaymath}

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

\begin{displaymath}
- \sum_i \int
\left(
\frac{\partial \Psi^*}{\partial t...
...i} +
\frac{\partial \Psi}{\partial t} \Psi^*_{ii}
\right)
\end{displaymath}

That equals

\begin{displaymath}
- \sum_i \int
\left(
\frac{\partial \Psi^*}{\partial t...
...t)_i
+ \sum_i \frac{{\rm d}}{{\rm d}t} \int \Psi_i^* \Psi_i
\end{displaymath}

as can be seen by differentiating out the parenthetical expression in the first integral with respect to $r_i$ 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

\begin{displaymath}
\left(\frac{mc^2}{\hbar c}\right)^2 \int
\left(
\frac{...
...t} \Psi +
\frac{\partial \Psi}{\partial t} \Psi^*
\right)
\end{displaymath}

That equals

\begin{displaymath}
\left(\frac{mc^2}{\hbar c}\right)^2 \frac{{\rm d}}{{\rm d}t} \int \Psi^* \Psi
\end{displaymath}

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