The equation of continuity for probabilities
Now I turn to a second point. An important part of the Schrödinger
equation for a single particle is the idea that the probability to
find the particle at a position is given by the absolute square of the
wave function. It is also characteristic of the quantum mechanics that
probability is conserved in a local sense. When the probability of
finding the electron somewhere decreases, while the probability of the
electron being elsewhere increases (keeping the total probability
unchanged), something must be going on in between. In other words, the
electron has a continuity in the sense that if the probability
decreases at one place and builds up at another place, there must be
some kind of flow between. If you put a wall, for example, in the way,
it will have an influence and the probabilities will not be the
same. So the conservation of probability alone is not the complete
statement of the conservation law, just as the conservation of energy
alone is not as deep and important as the local conservation of
energy.6 If energy
is disappearing, there must be a flow of energy to correspond. In the
same way, we would like to find a “current” of probability such that
if there is any change in the probability density (the probability of
being found in a unit volume), it can be considered as coming from an
inflow or an outflow due to some current. This current would be a vector
which could be interpreted this way—the $x$-component would be the net
probability per second and per unit area that a particle passes in the
$x$-direction across a plane parallel to the $yz$-plane. Passage
toward $+x$ is considered a positive flow, and passage in the opposite
direction, a negative flow. Is there such a current? Well, you know that the probability
density $P(\FLPr,t)$ is given in terms of the wave function by
\begin{equation}
\label{Eq:III:21:7}
P(\FLPr,t)=\psi\cconj(\FLPr,t)\psi(\FLPr,t).
\end{equation}

I am asking: Is there a current $\FLPJ$ such that
\begin{equation}
\label{Eq:III:21:8}
\ddp{P}{t}=-\FLPdiv{\FLPJ}?
\end{equation}

If I take the time derivative of Eq. (21.7), I get two
terms:
\begin{equation}
\label{Eq:III:21:9}
\ddp{P}{t}=\psi\cconj\,\ddp{\psi}{t}+\psi\,\ddp{\psi\cconj}{t}.
\end{equation}

Now use the Schrödinger equation—Eq. (21.3)—for $\ddpl{\psi}{t}$;
and take the complex conjugate of it to get $\ddpl{\psi\cconj}{t}$—each $i$
gets its sign reversed. You get

\begin{equation}
\begin{aligned}
\ddp{P}{t}&=-\frac{i}{\hbar}\biggl[\psi\cconj\,\frac{1}{2m}
\biggl(\frac{\hbar}{i}\,\FLPnabla-q\FLPA\biggr)\cdot
\biggl(\frac{\hbar}{i}\,\FLPnabla-q\FLPA\biggr)\psi+
q\phi\psi\cconj\psi\\[.5ex]
&\hphantom{{}={}}-\psi\,\frac{1}{2m}
\biggl(-\frac{\hbar}{i}\,\FLPnabla-q\FLPA\biggr)\cdot
\biggl(-\frac{\hbar}{i}\,\FLPnabla-q\FLPA\biggr)\psi\cconj
-q\phi\psi\psi\cconj\biggr].
\end{aligned}
\label{Eq:III:21:10}
\end{equation}


\begin{align}
\label{Eq:III:21:10}
&\ddp{P}{t}=\\
&-\frac{i}{\hbar}\biggl[\psi\cconj\!\frac{1}{2m}
\biggl(\!\frac{\hbar}{i}\FLPnabla\!-q\FLPA\!\biggr)\!\cdot\!
\biggl(\!\frac{\hbar}{i}\FLPnabla\!-q\FLPA\!\biggr)\psi\!+\!
q\phi\psi\cconj\psi\notag\\[1ex]
&\kern{2em}-\psi\frac{1}{2m}
\biggl(\!-\frac{\hbar}{i}\FLPnabla\!-q\FLPA\!\biggr)\!\cdot\!
\biggl(\!-\frac{\hbar}{i}\FLPnabla\!-q\FLPA\!\biggr)\psi\cconj\!
\!-\!q\phi\psi\psi\cconj\biggr].\notag
\end{align}

The potential terms and a lot of other stuff cancel out. And it turns
out that what is left can indeed be written as a perfect
divergence. The whole equation is equivalent to

\begin{equation}
\label{Eq:III:21:11}
\ddp{P}{t}=-\FLPdiv{\biggl\{
\frac{1}{2m}\,\psi\cconj
\biggl(\frac{\hbar}{i}\,\FLPnabla-q\FLPA\biggr)\psi+
\frac{1}{2m}\,\psi
\biggl(-\frac{\hbar}{i}\,\FLPnabla-q\FLPA\biggr)\psi\cconj
\biggr\}}.
\end{equation}

\begin{align}
\label{Eq:III:21:11}
\ddp{P}{t}\!=-\FLPdiv{\biggl\{\!
&\frac{1}{2m}\psi\cconj
\biggl(\!\frac{\hbar}{i}\FLPnabla\!-q\FLPA\!\biggr)\psi\,+\\[1ex]
&\frac{1}{2m}\psi
\biggl(\!-\frac{\hbar}{i}\FLPnabla\!-q\FLPA\!\biggr)\psi\cconj
\!\biggr\}}.\notag
\end{align}

It is really not as complicated as it seems. It is a symmetrical
combination of $\psi\cconj$ times a certain operation on $\psi$, plus
$\psi$ times the complex conjugate operation on $\psi\cconj$. It is
some quantity plus its own complex conjugate, so the whole thing is
real—as it ought to be. The operation can be remembered this way: it
is just the momentum operator $\Pcalvecop$ minus $q\FLPA$. I could
write the current in Eq. (21.8) as
\begin{equation}
\label{Eq:III:21:12}
\FLPJ\!=\!\frac{1}{2}\biggl\{\!
\psi\cconj\biggl[\!\frac{\Pcalvecop\!-\!q\FLPA}{m}\!\biggr]\psi\!+\!
\psi\biggl[\!\frac{\Pcalvecop\!-\!q\FLPA}{m}\!\biggr]\cconj\!\!\!\psi\cconj
\!\biggr\}.
\end{equation}

There is then a current $\FLPJ$ which completes Eq. (21.8). Equation (21.11) shows that the probability is conserved
locally. If a particle disappears from one region it cannot appear in
another without something going on in between. Imagine that the first
region is surrounded by a closed surface far enough out that there is
zero probability to find the electron at the surface. The total
probability to find the electron somewhere inside the surface is the
volume integral of $P$. But according to Gauss’s theorem the volume integral of the
divergence $\FLPJ$ is equal to the surface integral of its normal component. If
$\psi$ is zero at the surface, Eq. (21.12) says that
$\FLPJ$ is zero, so the total probability to find the particle inside
can’t change. Only if some of the probability approaches the boundary
can some of it leak out. We can say that it only gets out by moving
through the surface—and that is local conservation.