Superconductivity
As you know, very many metals become superconducting below a certain
temperature9—the temperature is different for different metals. When you
reduce the temperature sufficiently the metals conduct electricity
without any resistance. This phenomenon has been observed for a very
large number of metals but not for all, and the theory of this
phenomenon has caused a great deal of difficulty. It took a very long
time to understand what was going on inside of superconductors, and I
will only describe enough of it for our present purposes. It turns out
that due to the interactions of the electrons with the vibrations of the
atoms in the lattice, there is a small net effective attraction
between the electrons. The result is that the electrons form together,
if I may speak very qualitatively and crudely, bound pairs. Now you know that a single electron is a Fermi particle. But a bound
pair would act as a Bose particle, because if I exchange both
electrons in a pair I change the sign of the wave function twice, and
that means that I don’t change anything. A pair is a Bose
particle. The energy of pairing—that is, the net attraction—is very, very
weak. Only a tiny temperature is needed to throw the electrons apart
by thermal agitation, and convert them back to “normal”
electrons. But when you make the temperature sufficiently low that
they have to do their very best to get into the absolutely lowest
state; then they do collect in pairs. I don’t wish you to imagine that the pairs are really held together very
closely like a point particle. As a matter of fact, one of the great
difficulties of understanding this phenomenon originally was that that is
not the way things are. The two electrons which form the pair are really
spread over a considerable distance; and the mean distance between pairs
is relatively smaller than the size of a single pair. Several pairs are
occupying the same space at the same time. Both the reason why electrons
in a metal form pairs and an estimate of the energy given up in forming
a pair have been a triumph of recent times. This fundamental point in
the theory of superconductivity was first explained in the theory of
Bardeen, Cooper, and Schrieffer,10 but that is not the subject of
this seminar. We will accept, however, the idea that the electrons do,
in some manner or other, work in pairs, that we can think of these pairs
as behaving more or less like particles, and that we can therefore talk
about the wave function for a “pair.” Now the Schrödinger equation for the pair will be more or less like
Eq. (21.3). There will be one difference in that the
charge $q$ will be twice the charge of an electron. Also, we don’t
know the inertia—or effective mass—for the pair in the crystal
lattice, so we don’t know what number to put in for $m$. Nor should we
think that if we go to very high frequencies (or short wavelengths),
this is exactly the right form, because the kinetic energy that
corresponds to very rapidly varying wave functions may be so great as
to break up the pairs. At finite temperatures there are always a few
pairs which are broken up according to the usual Boltzmann theory. The
probability that a pair is broken is proportional
to $\exp(-E_{\text{pair}}/\kappa T)$. The electrons that are not bound in pairs
are called “normal” electrons and will move around in the crystal in
the ordinary way. I will, however, consider only the situation at
essentially zero temperature—or, in any case, I will disregard the
complications produced by those electrons which are not in pairs. Since electron pairs are bosons, when there are a lot of them in a
given state there is an especially large amplitude for other pairs to
go to the same state. So nearly all of the pairs will be locked down
at the lowest energy in exactly the same state—it won’t be
easy to get one of them into another state. There’s more amplitude to
go into the same state than into an unoccupied state by the famous
factor $\sqrt{n}$, where $n-1$ is the occupancy of the lowest
state. So we would expect all the pairs to be moving in the same
state. What then will our theory look like? I’ll call $\psi$ the wave
function of a pair in the lowest energy state. However, since
$\psi\psi\cconj$ is going to be proportional to the charge
density $\rho$, I can just as well write $\psi$ as the square root of the
charge density times some phase factor:
\begin{equation}
\label{Eq:III:21:17}
\psi(\FLPr)=\rho^{1/2}(\FLPr)e^{i\theta(\FLPr)},
\end{equation}

where $\rho$ and $\theta$ are real functions of $\FLPr$. (Any complex
function can, of course, be written this way.) It’s clear what we mean
when we talk about the charge density, but what is the physical
meaning of the phase $\theta$ of the wave function? Well, let’s see
what happens if we substitute $\psi(\FLPr)$ into
Eq. (21.12), and express the current density in terms of
these new variables $\rho$ and $\theta$. It’s just a change of variables
and I won’t go through all the algebra, but it comes out
\begin{equation}
\label{Eq:III:21:18}
\FLPJ=\frac{\hbar}{m}\biggl(
\FLPgrad{\theta}-\frac{q}{\hbar}\,\FLPA\biggr)\rho.
\end{equation}

Since both the current density and the charge density have a direct
physical meaning for the superconducting electron gas, both $\rho$
and $\theta$ are real things. The phase is just as observable as $\rho$;
it is a piece of the current density $\FLPJ$. The absolute
phase is not observable, but if the gradient of the phase is known
everywhere, the phase is known except for a constant. You can define
the phase at one point, and then the phase everywhere is determined. Incidentally, the equation for the current can be analyzed a little
nicer, when you think that the current density $\FLPJ$ is in
fact the charge density times the velocity of motion of the fluid of
electrons, or $\rho\FLPv$. Equation (21.18) is then
equivalent to
\begin{equation}
\label{Eq:III:21:19}
m\FLPv=\hbar\,\FLPgrad{\theta}-q\FLPA.
\end{equation}

Notice that there are two pieces in the $mv$-momentum; one is a
contribution from the vector potential, and the other, a contribution
from the behavior of the wave function. In other words, the
quantity $\hbar\,\FLPgrad{\theta}$ is just what we have called the
$p$-momentum.