## 21The Schrödinger Equation in a Classical Context: A Seminar on Superconductivity

(There was no summary for this lecture.)

(There was no summary for this lecture.)

### 21–1Schrödinger’s equation in a magnetic field

This lecture is only for entertainment. I would like to give the
lecture in a somewhat different style—just to see how it works
out. It’s not a part of the course—in the sense that it is not
supposed to be a last minute effort to teach you something new. But,
rather, I imagine that I’m giving a seminar or research report on the
subject to a more advanced audience, to people who have already been
educated in quantum mechanics. The main difference between a seminar
and a regular lecture is that the seminar speaker does not carry out
all the steps, or all the algebra. He says: “If you do such and such,
this is what comes out,” instead of showing all of the details. So in
this lecture I’ll describe the ideas all the way along but just give
you the *results* of the computations. You should realize that
you’re not supposed to understand everything immediately, but believe
(more or less) that things would come out if you went through the
steps.

All that aside, this is a subject I *want* to talk about. It is
recent and modern and would be a perfectly legitimate talk to give at
a research seminar. My subject is the Schrödinger equation in a
classical setting—the case of superconductivity.

Ordinarily, the wave function which appears in the Schrödinger
equation applies to only one or two particles. And the wave function
itself is not something that has a classical meaning—unlike the
electric field, or the vector potential, or things of that kind. The
wave function for a single particle *is* a “field”—in the
sense that it is a function of position—but it does not generally
have a classical significance. Nevertheless, there are some situations
in which a quantum mechanical wave function *does* have classical
significance, and they are the ones I would like to take up. The
peculiar quantum mechanical behavior of matter on a small scale
doesn’t usually make itself felt on a large scale except in the
standard way that it produces Newton’s laws—the laws of the
so-called classical mechanics. But there are certain situations in
which the peculiarities of quantum mechanics can come out in a special
way on a large scale.

At low temperatures, when the energy of a system has been reduced very, very low, instead of a large number of states being involved, only a very, very small number of states near the ground state are involved. Under those circumstances the quantum mechanical character of that ground state can appear on a macroscopic scale. It is the purpose of this lecture to show a connection between quantum mechanics and large-scale effects—not the usual discussion of the way that quantum mechanics reproduces Newtonian mechanics on the average, but a special situation in which quantum mechanics will produce its own characteristic effects on a large or “macroscopic” scale.

I will begin by reminding you of some of the properties of the
Schrödinger equation.^{1} I want to describe the behavior of a particle
in a magnetic field using the Schrödinger equation, because the
superconductive phenomena are involved with magnetic fields. An
external magnetic field is described by a vector potential, and the
problem is: what are the laws of quantum mechanics in a vector
potential? The principle that describes the behavior of quantum
mechanics in a vector potential is very simple. The amplitude that a
particle goes from one place to another along a certain route when
there’s a field present is the same as the amplitude that it would go
along the same route when there’s no field, multiplied by the
exponential of the line integral of the vector potential, times the
electric charge divided by
Planck’s constant^{2} (see Fig. 21–1):
\begin{equation}
\label{Eq:III:21:1}
\braket{b}{a}_{\text{in $\FLPA$}}=\braket{b}{a}_{A=0}\cdot
\exp\biggl[\frac{iq}{\hbar}\int_a^b\FLPA\cdot d\FLPs\biggr].
\end{equation}
It is a basic statement of quantum mechanics.

Now without the vector potential the Schrödinger equation of a
charged particle (nonrelativistic, no spin) is
\begin{equation}
\label{Eq:III:21:2}
-\frac{\hbar}{i}\,\ddp{\psi}{t}=\Hcalop\psi=
\frac{1}{2m}\biggl(\frac{\hbar}{i}\,\FLPnabla\biggr)\cdot
\biggl(\frac{\hbar}{i}\,\FLPnabla\biggr)\psi+q\phi\psi,
\end{equation}
\begin{gather}
\label{Eq:III:21:2}
-\frac{\hbar}{i}\,\ddp{\psi}{t}=\Hcalop\psi=\\[1ex]
\frac{1}{2m}\biggl(\frac{\hbar}{i}\,\FLPnabla\biggr)\cdot
\biggl(\frac{\hbar}{i}\,\FLPnabla\biggr)\psi+q\phi\psi,\notag
\end{gather}
where $\phi$ is the electric potential so that $q\phi$ is the
potential energy.^{3} Equation (21.1) is equivalent
to the statement that in a magnetic field the gradients in the
Hamiltonian are replaced in each case by the gradient minus $q\FLPA$,
so that Eq. (21.2) becomes
\begin{equation}
\label{Eq:III:21:3}
-\frac{\hbar}{i}\,\ddp{\psi}{t}=\Hcalop\psi=
\frac{1}{2m}\biggl(\frac{\hbar}{i}\,\FLPnabla-q\FLPA\biggr)\cdot
\biggl(\frac{\hbar}{i}\,\FLPnabla-q\FLPA\biggr)\psi+q\phi\psi.
\end{equation}
\begin{gather}
\label{Eq:III:21:3}
-\frac{\hbar}{i}\,\ddp{\psi}{t}=\Hcalop\psi=\\[1ex]
\frac{1}{2m}\biggl(\frac{\hbar}{i}\,\FLPnabla-q\FLPA\biggr)\cdot
\biggl(\frac{\hbar}{i}\,\FLPnabla-q\FLPA\biggr)\psi+q\phi\psi.\notag
\end{gather}
This is the Schrödinger equation for a particle with charge $q$
moving in an electromagnetic field $\FLPA,\phi$ (nonrelativistic, no
spin).

To show that this is true I’d like to illustrate by a simple example
in which instead of having a continuous situation we have a line of
atoms along the $x$-axis with the spacing $b$ and we have an
amplitude $iK\!/\hbar$ per unit time for an electron to jump from one atom to another when there is no
field.^{4} Now according to Eq. (21.1) if
there’s a vector potential in the $x$-direction $A_x(x,t)$, the
amplitude to jump will be altered from what it was before by a
factor $\exp[(iq/\hbar)\,A_xb]$, the exponent being $iq/\hbar$ times the vector
potential integrated from one atom to the next. For simplicity we will
write $(q/\hbar)A_x\equiv f(x)$, since $A_x$, will, in general, depend
on $x$. If the amplitude to find the electron at the atom “$n$”
located at $x$ is called $C(x)\equiv C_n$, then the rate of change of
that amplitude is given by the following equation:
\begin{align}
-\frac{\hbar}{i}\ddp{}{t}C(x)=E_0C(x)&\!-\!Ke^{-ibf(x+b/2)}C(x\!+\!b)\notag\\
\label{Eq:III:21:4}
&-\!Ke^{+ibf(x-b/2)}C(x\!-\!b).
\end{align}

There are three pieces. First, there’s some energy $E_0$ if the electron is located at $x$. As usual, that gives the term $E_0C(x)$. Next, there is the term $-KC(x+b)$, which is the amplitude for the electron to have jumped backwards one step from atom “$n+1$,” located at $x+b$. However, in doing so in a vector potential, the phase of the amplitude must be shifted according to the rule in Eq. (21.1). If $A_x$ is not changing appreciably in one atomic spacing, the integral can be written as just the value of $A_x$ at the midpoint, times the spacing $b$. So $(iq/\hbar)$ times the integral is just $ibf(x+b/2)$. Since the electron is jumping backwards, I showed this phase shift with a minus sign. That gives the second piece. In the same manner there’s a certain amplitude to have jumped from the other side, but this time we need the vector potential at a distance $(b/2)$ on the other side of $x$, times the distance $b$. That gives the third piece. The sum gives the equation for the amplitude to be at $x$ in a vector potential.

Now we know that if the function $C(x)$ is smooth enough (we take the long wavelength limit), and if we let the atoms get closer together, Eq. (21.4) will approach the behavior of an electron in free space. So the next step is to expand the right-hand side of (21.4) in powers of $b$, assuming $b$ is very small. For example, if $b$ is zero the right-hand side is just $(E_0-2K)C(x)$, so in the zeroth approximation the energy is $E_0-2K$. Next comes the terms in $b$. But because the two exponentials have opposite signs, only even powers of $b$ remain. So if you make a Taylor expansion of $C(x)$, of $f(x)$, and of the exponentials, and then collect the terms in $b^2$, you get \begin{align} -\frac{\hbar}{i}\,\ddp{C(x)}{t}&=E_0C(x)-2KC(x)\notag\\ \label{Eq:III:21:5} &\quad-Kb^2\{C''(x)-2if(x)C'(x)-if'(x)C(x)-f^2(x)C(x)\}. \end{align} \begin{align} \label{Eq:III:21:5} -\frac{\hbar}{i}\,\ddp{C(x)}{t}&=E_0C(x)-2KC(x)\\[1ex] &-Kb^2\bigl\{C''(x)-2if(x)C'(x)\,-\notag\\[1ex] &\phantom{-Kb^2\{}\;if'(x)C(x)-f^2(x)C(x)\bigr\}.\notag \end{align} (The “primes” mean differentiation with respect to $x$.)

Now this horrible combination of things looks quite complicated. But
mathematically it’s exactly the same as
\begin{equation}
\label{Eq:III:21:6}
-\frac{\hbar}{i}\,\ddp{C(x)}{t}=(E_0-2K)C(x)-Kb^2
\biggl[\ddp{}{x}-if(x)\biggr]
\biggl[\ddp{}{x}-if(x)\biggr]C(x).
\end{equation}
\begin{align}
\label{Eq:III:21:6}
-\frac{\hbar}{i}\ddp{C(x)}{t}&=(E_0-2K)C(x)\\
&-Kb^2 \biggl[\ddp{}{x}-if(x)\biggr]
\biggl[\ddp{}{x}-if(x)\biggr]C(x).\notag
\end{align}
The second bracket operating on $C(x)$ gives $C'(x)$
minus $if(x)C(x)$. The first bracket operating on these two terms gives the
$C''$ term and terms in the first derivative of $f(x)$ and the first
derivative of $C(x)$. Now remember that the solutions for zero
magnetic field^{5} represent a
particle with an effective mass $m_{\text{eff}}$ given by
\begin{equation*}
Kb^2=\frac{\hbar^2}{2m_{\text{eff}}}.
\end{equation*}
If you then set $E_0=2K$, and put back $f(x)=(q/\hbar)A_x$, you can
easily check that Eq. (21.6) is the same as the first
part of Eq. (21.3). (The origin of the potential energy
term is well known, so I haven’t bothered to include it in this
discussion.) The proposition of Eq. (21.1) that the
vector potential changes all the amplitudes by the exponential factor
is the same as the rule that the momentum operator,
$(\hbar/i)\FLPnabla$ gets replaced by
\begin{equation*}
\frac{\hbar}{i}\,\FLPnabla-q\FLPA,
\end{equation*}
as you see in the Schrödinger equation of (21.3).

### 21–2The 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.

### 21–3Two kinds of momentum

The equation for the current is rather interesting, and sometimes causes a certain amount of worry. You would think the current would be something like the density of particles times the velocity. The density should be something like $\psi\psi\cconj$, which is o.k. And each term in Eq. (21.12) looks like the typical form for the average-value of the operator \begin{equation} \label{Eq:III:21:13} \frac{\Pcalvecop-q\FLPA}{m}, \end{equation} so maybe we should think of it as the velocity of flow. It looks as though we have two suggestions for relations of velocity to momentum, because we would also think that momentum divided by mass, $\Pcalvecop/m$, should be a velocity. The two possibilities differ by the vector potential.

It happens that these two possibilities were also discovered in
classical physics, when it was found that momentum could be defined in
two ways.^{7} One of them is called “kinematic
momentum,” but for absolute clarity I will in this lecture call
it the “$mv$-momentum.” This is the
momentum obtained by multiplying mass by velocity. The other is a more
mathematical, more abstract momentum, sometimes called the “dynamical
momentum,” which I’ll call “$p$-momentum.” The two possibilities are
\begin{equation}
\label{Eq:III:21:14}
\text{$mv$-momentum}=m\FLPv,
\end{equation}
\begin{equation}
\label{Eq:III:21:15}
\text{$p$-momentum}=m\FLPv + q\FLPA.
\end{equation}
It turns out that in quantum mechanics with magnetic fields it is the
$p$-momentum which is connected to the gradient operator $\Pcalvecop$,
so it follows that (21.13) is the operator of a velocity.

I’d like to make a brief digression to show you what this is all
about—why there must be something like Eq. (21.15) in
the quantum mechanics. The wave function changes with time according
to the Schrödinger equation in Eq. (21.3). If I would
suddenly change the vector potential, the wave function wouldn’t
change at the first instant; only its rate of change changes. Now
think of what would happen in the following circumstance. Suppose I
have a long solenoid, in which I can produce a flux of magnetic field
($\FLPB$-field), as shown in Fig. 21–2. And there is a charged
particle sitting nearby. Suppose this flux nearly instantaneously
builds up from zero to something. I start with zero vector potential
and then I turn on a vector potential. That means that I produce
suddenly a circumferential vector potential $\FLPA$. You’ll remember
that the line integral of $\FLPA$ around a loop is the same as the flux
of $\FLPB$ through the loop.^{8} Now what
happens if I suddenly turn on a vector potential? According to the
quantum mechanical equation the sudden change of $\FLPA$ does not make a
sudden change of $\psi$; the wave function is still the same. So the
gradient is also unchanged.

But remember what happens electrically when I suddenly turn on a flux. During the short time that the flux is rising, there’s an electric field generated whose line integral is the rate of change of the flux with time: \begin{equation} \FLPE=-\ddp{\FLPA}{t}. \end{equation} That electric field is enormous if the flux is changing rapidly, and it gives a force on the particle. The force is the charge times the electric field, and so during the build up of the flux the particle obtains a total impulse (that is, a change in $m\FLPv$) equal to $-q\FLPA$. In other words, if you suddenly turn on a vector potential at a charge, this charge immediately picks up an $mv$-momentum equal to $-q\FLPA$. But there is something that isn’t changed immediately and that’s the difference between $m\FLPv$ and $-q\FLPA$. And so the sum $\FLPp=m\FLPv+q\FLPA$ is something which is not changed when you make a sudden change in the vector potential. This quantity $\FLPp$ is what we have called the $p$-momentum and is of importance in classical mechanics in the theory of dynamics, but it also has a direct significance in quantum mechanics. It depends on the character of the wave function, and it is the one to be identified with the operator \begin{equation*} \Pcalvecop=\frac{\hbar}{i}\,\FLPnabla. \end{equation*}

### 21–4The meaning of the wave function

When Schrödinger first discovered his equation he discovered the conservation law of Eq. (21.8) as a consequence of his equation. But he imagined incorrectly that $P$ was the electric charge density of the electron and that $\FLPJ$ was the electric current density, so he thought that the electrons interacted with the electromagnetic field through these charges and currents. When he solved his equations for the hydrogen atom and calculated $\psi$, he wasn’t calculating the probability of anything—there were no amplitudes at that time—the interpretation was completely different. The atomic nucleus was stationary but there were currents moving around; the charges $P$ and currents $\FLPJ$ would generate electromagnetic fields and the thing would radiate light. He soon found on doing a number of problems that it didn’t work out quite right. It was at this point that Born made an essential contribution to our ideas regarding quantum mechanics. It was Born who correctly (as far as we know) interpreted the $\psi$ of the Schrödinger equation in terms of a probability amplitude—that very difficult idea that the square of the amplitude is not the charge density but is only the probability per unit volume of finding an electron there, and that when you do find the electron some place the entire charge is there. That whole idea is due to Born.

The wave function $\psi(\FLPr)$ for an electron in an atom does not,
then, describe a smeared-out electron with a smooth charge
density. The electron is either here, or there, or somewhere else, but
wherever it is, it is a point charge. On the other hand, think of a
situation in which there are an enormous number of particles in
exactly the same state, a very large number of them with exactly the
same wave function. Then what? One of them is here and one of them is
there, and the probability of finding any one of them at a given place
is proportional to $\psi\psi\cconj$. But since there are so many
particles, if I look in any volume $dx\,dy\,dz$ I will generally find
a number close to $\psi\psi\cconj\,dx\,dy\,dz$. So in a situation in
which $\psi$ is the wave function for each of an enormous number of
particles which are all in the same state, $\psi\psi\cconj$ *can*
be interpreted as the density of particles. If, under these
circumstances, each particle carries the same charge $q$, we can, in
fact, go further and interpret $\psi\cconj\psi$ as the density of
*electricity*. Normally, $\psi\psi\cconj$ is given the dimensions
of a probability density, then $\psi$ should be multiplied by $q$ to
give the dimensions of a charge density. For our present purposes we
can put this constant factor into $\psi$, and take $\psi\psi\cconj$
itself as the electric charge density. With this understanding,
$\FLPJ$ (the current of probability I have calculated) becomes
directly the electric current density.

So in the situation in which we can have very many particles in exactly the same state, there is possible a new physical interpretation of the wave functions. The charge density and the electric current can be calculated directly from the wave functions and the wave functions take on a physical meaning which extends into classical, macroscopic situations.

Something similar can happen with neutral particles. When we have the
wave function of a single photon, it is the amplitude to find a photon
somewhere. Although we haven’t ever written it down there is an
equation for the photon wave function analogous to the Schrödinger
equation for the electron. The photon equation is just the same as
Maxwell’s equations for the electromagnetic field, and the wave
function is the same as the vector potential $\FLPA$. The wave
function turns out to be just the vector potential. The quantum
physics is the same thing as the classical physics because photons are
noninteracting Bose particles and many of them can be in the same
state—as you know, they *like* to be in the same state. The
moment that you have billions in the same state (that is, in the same
electromagnetic wave), you can measure the wave function, which is the
vector potential, directly. Of course, it worked historically the
other way. The first observations were on situations with many photons
in the same state, and so we were able to discover the correct
equation for a single photon by observing directly with our hands on a
macroscopic level the nature of the wave function.

Now the trouble with the electron is that you cannot put more than one in the same state. Therefore, it was long believed that the wave function of the Schrödinger equation would never have a macroscopic representation analogous to the macroscopic representation of the amplitude for photons. On the other hand, it is now realized that the phenomenon of superconductivity presents us with just this situation.

### 21–5Superconductivity

As you know, very many metals become superconducting below a certain
temperature^{9}—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.

### 21–6The Meissner effect

Now we can describe some of the phenomena of superconductivity. First, there is no electrical resistance. There’s no resistance because all the electrons are collectively in the same state. In the ordinary flow of current you knock one electron or the other out of the regular flow, gradually deteriorating the general momentum. But here to get one electron away from what all the others are doing is very hard because of the tendency of all Bose particles to go in the same state. A current once started, just keeps on going forever.

It’s also easy to understand that if you have a piece of metal in the superconducting state and turn on a magnetic field which isn’t too strong (we won’t go into the details of how strong), the magnetic field can’t penetrate the metal. If, as you build up the magnetic field, any of it were to build up inside the metal, there would be a rate of change of flux which would produce an electric field, and an electric field would immediately generate a current which, by Lenz’s law, would oppose the flux. Since all the electrons will move together, an infinitesimal electric field will generate enough current to oppose completely any applied magnetic field. So if you turn the field on after you’ve cooled a metal to the superconducting state, it will be excluded.

Even more interesting is a related phenomenon discovered experimentally
by Meissner.^{11} If you have a piece
of the metal at a high temperature (so that it is a normal conductor)
and establish a magnetic field through it, and then you lower the
temperature below the critical temperature (where the metal becomes a
superconductor), *the field is expelled*. In other words, it starts
up its own current—and in just the right amount to push the field out.

We can see the reason for that in the equations, and I’d like to
explain how. Suppose that we take a piece of superconducting material
which is in one lump. Then in a steady situation of any kind the
divergence of the current must be zero because there’s no place for it
to go. It is convenient to choose to make the divergence of $\FLPA$
equal to zero. (I should explain why choosing this convention doesn’t
mean any loss of generality, but I don’t want to take the time.)
Taking the divergence of Eq. (21.18), then gives that
the Laplacian of $\theta$ is equal to zero. One moment. What about the
variation of $\rho$? I forgot to mention an important point. There is
a background of positive charge in this metal due to the atomic ions
of the lattice. If the charge density $\rho$ is uniform there is no
net charge and no electric field. If there would be any accumulation
of electrons in one region the charge wouldn’t be neutralized and
there would be a terrific repulsion pushing the electrons
apart.^{12} So in
ordinary circumstances the charge density of the electrons in the
superconductor is almost perfectly uniform—I can take $\rho$ as a
constant. Now the only way that $\nabla^2\theta$ can be zero
everywhere inside the lump of metal is for $\theta$ to be a
constant. And that means that there is no contribution to $\FLPJ$ from
$p$-momentum. Equation (21.18) then says that the current
is proportional to $\rho$ times $\FLPA$. So everywhere in a lump of
superconducting material the current is necessarily proportional to
the vector potential:
\begin{equation}
\label{Eq:III:21:20}
\FLPJ=-\rho\,\frac{q}{m}\,\FLPA.
\end{equation}
Since $\rho$ and $q$ have the same (negative) sign, and since $\rho$
is a constant, I can set $-\rho q/m=-(\text{some positive constant})$;
then
\begin{equation}
\label{Eq:III:21:21}
\FLPJ=-(\text{some positive constant})\FLPA.
\end{equation}
This equation was originally proposed by London and
London^{13} to explain the
experimental observations of superconductivity—long before the quantum
mechanical origin of the effect was understood.

Now we can use Eq. (21.20) in the equations of electromagnetism to solve for the fields. The vector potential is related to the current density by \begin{equation} \label{Eq:III:21:22} \nabla^2\FLPA=-\frac{1}{\epsO c^2}\,\FLPJ. \end{equation} If I use Eq. (21.21) for $\FLPJ$, I have \begin{equation} \label{Eq:III:21:23} \nabla^2\FLPA=\lambda^2\FLPA, \end{equation} where $\lambda^2$ is just a new constant; \begin{equation} \label{Eq:III:21:24} \lambda^2=\rho\,\frac{q}{\epsO mc^2}. \end{equation}

We can now try to solve this equation for $\FLPA$ and see what happens
in detail. For example, in one dimension Eq. (21.23)
has exponential solutions of the form $e^{-\lambda x}$
and $e^{+\lambda x}$. These solutions mean that the vector potential must
*decrease* exponentially as you go from the surface into the
material. (It can’t increase because there would be a blow up.) If the
piece of metal is very large compared to $1/\lambda$, the field only
penetrates to a thin layer at the surface—a layer about $1/\lambda$
in thickness. The entire remainder of the interior is free of field,
as sketched in Fig. 21–3. This is the explanation of the Meissner
effect.

How big is the distance $1/\lambda$? Well, remember that $r_0$, the “electromagnetic radius” of the electron ($2.8\times10^{-13}$ cm), is given by \begin{equation*} mc^2=\frac{q_e^2}{4\pi\epsO r_0}. \end{equation*} Also, remember that $q$ in Eq. (21.24) is twice the charge of an electron, so \begin{equation*} \frac{q}{\epsO mc^2}=\frac{8\pi r_0}{q_e}. \end{equation*} Writing $\rho$ as $q_eN$, where $N$ is the number of electrons per cubic centimeter, we have \begin{equation} \label{Eq:III:21:25} \lambda^2=8\pi Nr_0. \end{equation} For a metal such as lead there are about $3\times10^{22}$ atoms per cm$^3$, so if each one contributed only one conduction electron, $1/\lambda$ would be about $2\times10^{-6}$ cm. That gives you the order of magnitude.

### 21–7Flux quantization

The London equation (21.21) was proposed to account for
the observed facts of superconductivity including the
Meissner effect.
In recent times, however, there have been some even more
dramatic predictions. One prediction made by London was so peculiar
that nobody paid much attention to it until recently. I will now
discuss it. This time instead of taking a single lump, suppose we take
a *ring* whose thickness is large compared to $1/\lambda$, and
try to see what would happen if we started with a magnetic field
through the ring, then cooled it to the superconducting state, and
afterward removed the original source of $\FLPB$. The sequence of
events is sketched in Fig. 21–4. In the normal state there will be a
field in the body of the ring as sketched in part (a) of the
figure. When the ring is made superconducting, the field is forced
outside of the *material* (as we have just seen). There will then
be some flux through the hole of the ring as sketched in part (b). If
the external field is now removed, the lines of field going through
the hole are “trapped” as shown in part (c). The flux $\Phi$ through
the center can’t decrease because $\ddpl{\Phi}{t}$ must be equal to
the line integral of $\FLPE$ around the ring, which is zero in a
superconductor. As the external field is removed a super current
starts flowing around the ring to keep the flux through the ring a
constant. (It’s the old eddy-current idea, only with zero resistance.)
These currents will, however, all flow near the surface (down to a
depth $1/\lambda$), as can be shown by the same kind of analysis that
I made for the solid block. These currents can keep the magnetic field
out of the body of the ring, and produce the permanently trapped
magnetic field as well.

Now, however, there is an essential difference, and our equations
predict a surprising effect. The argument I made above that $\theta$
must be a constant in a solid block *does not apply for a ring*,
as you can see from the following arguments.

Well inside the body of the ring the current density $\FLPJ$ is zero;
so Eq. (21.18) gives
\begin{equation}
\label{Eq:III:21:26}
\hbar\,\FLPgrad{\theta}=q\FLPA.
\end{equation}
Now consider what we get if we take the line integral of $\FLPA$
around a curve $\Gamma$, which goes around the ring near the center of
its cross-section so that it never gets near the surface, as drawn in
Fig. 21–5. From Eq. (21.26),
\begin{equation}
\label{Eq:III:21:27}
\hbar\oint\FLPgrad{\theta}\cdot d\FLPs=q\oint\FLPA\cdot d\FLPs.
\end{equation}
Now you know that the line integral of $\FLPA$ around any loop is
equal to the flux of $\FLPB$ through the loop
\begin{equation*}
\oint\FLPA\cdot d\FLPs=\Phi.
\end{equation*}
Equation (21.27) then becomes
\begin{equation}
\label{Eq:III:21:28}
\oint\FLPgrad{\theta}\cdot d\FLPs=\frac{q}{\hbar}\,\Phi.
\end{equation}
The line integral of a gradient from one point to another (say from
point $1$ to point $2$) is the difference of the values of the
function at the two points. Namely,
\begin{equation*}
\int_1^2\FLPgrad{\theta}\cdot d\FLPs=\theta_2-\theta_1.
\end{equation*}
If we let the two end points $1$ and $2$ come together to make a
closed loop you might at first think that $\theta_2$ would
equal $\theta_1$, so that the integral in Eq. (21.28) would
be zero. That would be true for a closed loop in a simply-connected
piece of superconductor, but it is not necessarily true for a
ring-shaped piece. The only physical requirement we can make is that
*there can be only one value of the wave function for each
point*. Whatever $\theta$ does as you go around the ring, when you get
back to the starting point the $\theta$ you get must give the same
value for the wave function
\begin{equation*}
\psi=\sqrt{\rho}e^{i\theta}.
\end{equation*}
This will happen if $\theta$ changes by $2\pi n$, where $n$ is any
integer. So if we make one complete turn around the ring the left-hand
side of Eq. (21.27) must be $\hbar\cdot2\pi n$. Using
Eq. (21.28), I get that
\begin{equation}
\label{Eq:III:21:29}
2\pi n\hbar=q\Phi.
\end{equation}
*The trapped flux must always be an integer times $2\pi\hbar/q$!*
If you would think of the ring as a classical object with an ideally
perfect (that is, infinite) conductivity, you would think that
whatever flux was initially found through it would just stay
there—any amount of flux at all could be trapped. But the
quantum-mechanical theory of superconductivity says that the flux can
be zero, or $2\pi\hbar/q$, or $4\pi\hbar/q$, or $6\pi\hbar/q$, and so
on, but no value in between. It must be a multiple of a basic quantum
mechanical unit.

London^{14} predicted that the
flux trapped by a superconducting ring would be quantized and said
that the possible values of the flux would be given by
Eq. (21.29) with $q$ equal to the electronic charge.
According to London the basic unit of flux should be $2\pi\hbar/q_e$,
which is about $4\times10^{-7}$ $\text{gauss}\cdot\text{cm}^2$. To
visualize such a flux, think of a tiny cylinder a tenth of a millimeter
in diameter; the magnetic field inside it when it contains this amount
of flux is about one percent of the earth’s magnetic field. It should be
possible to observe such a flux by a sensitive magnetic measurement.

In 1961 such a quantized flux was looked for and found by Deaver and
Fairbank^{15} at
Stanford University and at about the same time by Doll and
Näbauer^{16} in
Germany.

In the experiment of Deaver and Fairbank, a tiny cylinder of superconductor was made by electroplating a thin layer of tin on a one-centimeter length of No. 56 ($1.3\times10^{-3}$ cm diameter) copper wire. The tin becomes superconducting below $3.8^\circ$K while the copper remains a normal metal. The wire was put in a small controlled magnetic field, and the temperature reduced until the tin became superconducting. Then the external source of field was removed. You would expect this to generate a current by Lenz’s law so that the flux inside would not change. The little cylinder should now have magnetic moment proportional to the flux inside. The magnetic moment was measured by jiggling the wire up and down (like the needle on a sewing machine, but at the rate of $100$ cycles per second) inside a pair of little coils at the ends of the tin cylinder. The induced voltage in the coils was then a measure of the magnetic moment.

When the experiment was done by Deaver and Fairbank, they found that
the flux was quantized, *but that the basic unit was only
one-half as large as London had predicted*. Doll and Näbauer got the
same result. At first this was quite mysterious,^{17} but
we now understand why it should be so. According to the Bardeen,
Cooper, and Schrieffer theory of superconductivity, the $q$ which
appears in Eq. (21.29) is the charge of a *pair*
of electrons and so is equal to $2q_e$. The basic flux unit is
\begin{equation}
\label{Eq:III:21:30}
\Phi_0=\frac{\pi\hbar}{q_e}\approx2\times10^{-7}\text{ gauss}\cdot\text{cm}^2
\end{equation}
or one-half the amount predicted by London. Everything now fits
together, and the measurements show the existence of the predicted
purely quantum-mechanical effect on a large scale.

### 21–8The dynamics of superconductivity

The Meissner effect and the flux quantization are two confirmations of our general ideas. Just for the sake of completeness I would like to show you what the complete equations of a superconducting fluid would be from this point of view—it is rather interesting. Up to this point I have only put the expression for $\psi$ into equations for charge density and current. If I put it into the complete Schrödinger equation I get equations for $\rho$ and $\theta$. It should be interesting to see what develops, because here we have a “fluid” of electron pairs with a charge density $\rho$ and a mysterious $\theta$—we can try to see what kind of equations we get for such a “fluid”! So we substitute the wave function of Eq. (21.17) into the Schrödinger equation (21.3) and remember that $\rho$ and $\theta$ are real functions of $x$, $y$, $z$, and $t$. If we separate real and imaginary parts we obtain then two equations. To write them in a shorter form I will—following Eq. (21.19)—write \begin{equation} \label{Eq:III:21:31} \frac{\hbar}{m}\,\FLPgrad{\theta}-\frac{q}{m}\,\FLPA=\FLPv. \end{equation} One of the equations I get is then \begin{equation} \label{Eq:III:21:32} \ddp{\rho}{t}=-\FLPdiv{\rho\FLPv}. \end{equation} Since $\rho\FLPv$ is first $\FLPJ$, this is just the continuity equation once more. The other equation I obtain tells how $\theta$ varies; it is \begin{equation} \label{Eq:III:21:33} \hbar\,\ddp{\theta}{t}=-\frac{m}{2}\,v^2-q\phi+ \frac{\hbar^2}{2m}\biggl\{ \frac{1}{\sqrt{\rho}}\,\nabla^2(\sqrt{\rho})\biggr\}. \end{equation} Those who are thoroughly familiar with hydrodynamics (of which I’m sure few of you are) will recognize this as the equation of motion for an electrically charged fluid if we identify $\hbar\theta$ as the “velocity potential”—except that the last term, which should be the energy of compression of the fluid, has a rather strange dependence on the density $\rho$. In any case, the equation says that the rate of change of the quantity $\hbar\theta$ is given by a kinetic energy term, $-\tfrac{1}{2}mv^2$, plus a potential energy term, $-q\phi$, with an additional term, containing the factor $\hbar^2$, which we could call a “quantum mechanical energy.” We have seen that inside a superconductor $\rho$ is kept very uniform by the electrostatic forces, so this term can almost certainly be neglected in every practical application provided we have only one superconducting region. If we have a boundary between two superconductors (or other circumstances in which the value of $\rho$ may change rapidly) this term can become important.

For those who are not so familiar with the equations of hydrodynamics,
I can rewrite Eq. (21.33) in a form that makes the
physics more apparent by using Eq. (21.31) to express
$\theta$ in terms of $\FLPv$. Taking the gradient of the whole of
Eq. (21.33) and expressing $\FLPgrad{\theta}$ in terms
of $\FLPA$ and $\FLPv$ by using (21.31), I get
\begin{equation}
\label{Eq:III:21:34}
\ddp{\FLPv}{t}=\frac{q}{m}\biggl(-\FLPgrad{\phi}-\ddp{\FLPA}{t}\biggr)-
\FLPv\times(\FLPcurl{\FLPv})-(\FLPv\cdot\FLPnabla)\FLPv+
\FLPgrad{\frac{\hbar^2}{2m^2}
\biggl(\frac{1}{\sqrt{\rho}}\,\nabla^2\sqrt{\rho}\biggr)}.
\end{equation}
\begin{align}
\ddp{\FLPv}{t}&=\frac{q}{m}\biggl(-\FLPgrad{\phi}-\ddp{\FLPA}{t}\biggr)
-\FLPv\times(\FLPcurl{\FLPv})\notag\\[1ex]
\label{Eq:III:21:34}
&-(\FLPv\cdot\FLPnabla)\FLPv +
\FLPgrad{\frac{\hbar^2}{2m^2}
\biggl(\frac{1}{\sqrt{\rho}}\,\nabla^2\sqrt{\rho}\biggr)}.
\end{align}
What does this equation mean? First, remember that
\begin{equation}
\label{Eq:III:21:35}
-\FLPgrad{\phi}-\ddp{\FLPA}{t}=\FLPE.
\end{equation}
Next, notice that if I take the curl of Eq. (21.31), I
get
\begin{equation}
\label{Eq:III:21:36}
\FLPcurl{\FLPv}=-\frac{q}{m}\,\FLPcurl{\FLPA},
\end{equation}
since the curl of a gradient is always zero. But $\FLPcurl{\FLPA}$ is
the magnetic field $\FLPB$, so the first two terms can be written as
\begin{equation*}
\frac{q}{m}(\FLPE+\FLPv\times\FLPB).
\end{equation*}
Finally, you should understand that $\ddpl{\FLPv}{t}$ stands for the
rate of change of the velocity of the fluid at a point. If you
concentrate on a particular particle, its acceleration is the
*total* derivative of $\FLPv$ (or, as it is sometimes called in
fluid dynamics, the “comoving acceleration”), which is related
to $\ddpl{\FLPv}{t}$ by^{18}
\begin{equation}
\label{Eq:III:21:37}
\left.\ddt{\FLPv}{t}\right|_{\text{comoving}}\kern{-2ex}=
\ddp{\FLPv}{t}+(\FLPv\cdot\FLPnabla)\FLPv.
\end{equation}
This extra term also appears as the third term on the right side of
Eq. (21.34). Taking it to the left side, I can write
Eq. (21.34) in the following way:
\begin{equation}
\label{Eq:III:21:38}
\left.m\ddt{\FLPv}{t}\right|_{\text{comoving}}\kern{-3.5ex}=
q(\FLPE\!+\!\FLPv\!\times\!\FLPB)\!+\!\FLPgrad{\frac{\hbar^2}{2m}
\!\biggl(\!\frac{1}{\sqrt{\rho}}\nabla^2\!\!\sqrt{\rho}\!\biggr)}.
\end{equation}
We also have from Eq. (21.36) that
\begin{equation}
\label{Eq:III:21:39}
\FLPcurl{\FLPv}=-\frac{q}{m}\,\FLPB.
\end{equation}

These two equations are the equations of motion of the superconducting
electron fluid. The first equation is just Newton’s law for a charged
fluid in an electromagnetic field. It says that the acceleration of
each particle of the fluid whose charge is $q$ comes from the ordinary
Lorentz force $q(\FLPE+\FLPv\times\FLPB)$ plus an additional force,
which is the gradient of some mystical quantum mechanical
potential—a force which is not very big except at the junction
between two superconductors. The second equation says that the fluid
is “ideal”—the curl of $\FLPv$ has zero divergence (the divergence
of $\FLPB$ is always zero). That means that the velocity can be
expressed in terms of velocity potential. Ordinarily one writes that
$\FLPcurl{\FLPv}=\FLPzero$ for an ideal fluid, but for an *ideal
charged fluid in a magnetic field*, this gets modified to
Eq. (21.39).

So, Schrödinger’s equation for the electron pairs in a superconductor gives us the equations of motion of an electrically charged ideal fluid. Superconductivity is the same as the problem of the hydrodynamics of a charged liquid. If you want to solve any problem about superconductors you take these equations for the fluid [or the equivalent pair, Eqs. (21.32) and (21.33)], and combine them with Maxwell’s equations to get the fields. (The charges and currents you use to get the fields must, of course, include the ones from the superconductor as well as from the external sources.)

Incidentally, I believe that Eq. (21.38) is not quite correct, but ought to have an additional term involving the density. This new term does not depend on quantum mechanics, but comes from the ordinary energy associated with variations of density. Just as in an ordinary fluid there should be a potential energy density proportional to the square of the deviation of $\rho$ from $\rho_0$, the undisturbed density (which is, here, also equal to the charge density of the crystal lattice). Since there will be forces proportional to the gradient of this energy, there should be another term in Eq. (21.38) of the form: $(\text{const})\,\FLPgrad{(\rho-\rho_0)^2}$. This term did not appear from the analysis because it comes from the interactions between particles, which I neglected in using an independent-particle approximation. It is, however, just the force I referred to when I made the qualitative statement that electrostatic forces would tend to keep $\rho$ nearly constant inside a superconductor.

### 21–9The Josephson junction

I would like to discuss next a very interesting situation that was
noticed by
Josephson^{19} while analyzing what might happen at
a junction between two superconductors. Suppose we have two
superconductors which are connected by a thin layer of insulating
material as in Fig. 21–6. Such an arrangement is now called a
“Josephson junction.” If the insulating layer is thick, the
electrons can’t get through; but if the layer is thin enough, there
can be an appreciable quantum mechanical amplitude for electrons to
jump across. This is just another example of the quantum-mechanical
penetration of a barrier.
Josephson analyzed this
situation and discovered that a number of strange phenomena should
occur.

In order to analyze such a junction I’ll call the amplitude to find an electron on one side, $\psi_1$, and the amplitude to find it on the other, $\psi_2$. In the superconducting state the wave function $\psi_1$ is the common wave function of all the electrons on one side, and $\psi_2$ is the corresponding function on the other side. I could do this problem for different kinds of superconductors, but let us take a very simple situation in which the material is the same on both sides so that the junction is symmetrical and simple. Also, for a moment let there be no magnetic field. Then the two amplitudes should be related in the following way: \begin{align*} i\hbar\,\ddp{\psi_1}{t}&=U_1\psi_1+K\psi_2,\\[1ex] i\hbar\,\ddp{\psi_2}{t}&=U_2\psi_2+K\psi_1. \end{align*}

The constant $K$ is a characteristic of the junction. If $K$ were zero, these two equations would just describe the lowest energy state—with energy $U$—of each superconductor. But there is coupling between the two sides by the amplitude $K$ that there may be leakage from one side to the other. (It is just the “flip-flop” amplitude of a two-state system.) If the two sides are identical, $U_1$ would equal $U_2$ and I could just subtract them off. But now suppose that we connect the two superconducting regions to the two terminals of a battery so that there is a potential difference $V$ across the junction. Then $U_1-U_2=qV$. I can, for convenience, define the zero of energy to be halfway between, then the two equations are \begin{equation} \begin{aligned} i\hbar\,\ddp{\psi_1}{t}&=+\frac{qV}{2}\,\psi_1+K\psi_2,\\[1ex] i\hbar\,\ddp{\psi_2}{t}&=-\frac{qV}{2}\,\psi_2+K\psi_1. \end{aligned} \label{Eq:III:21:40} \end{equation}

These are the standard equations for two quantum mechanical states coupled together. This time, let’s analyze these equations in another way. Let’s make the substitutions \begin{equation} \begin{aligned} \psi_1&=\sqrt{\rho_1}e^{i\theta_1},\\[1ex] \psi_2&=\sqrt{\rho_2}e^{i\theta_2}, \end{aligned} \label{Eq:III:21:41} \end{equation} where $\theta_1$ and $\theta_2$ are the phases on the two sides of the junction and $\rho_1$ and $\rho_2$ are the density of electrons at those two points. Remember that in actual practice $\rho_1$ and $\rho_2$ are almost exactly the same and are equal to $\rho_0$, the normal density of electrons in the superconducting material. Now if you substitute these equations for $\psi_1$ and $\psi_2$ into (21.40), you get four equations by equating the real and imaginary parts in each case. Letting $(\theta_2-\theta_1)=\delta$, for short, the result is \begin{align} &\begin{aligned} \dot{\rho}_1&=+\frac{2}{\hbar}\,K\sqrt{\rho_2\rho_1}\sin\delta,\\[1.5ex] \dot{\rho}_2&=-\frac{2}{\hbar}\,K\sqrt{\rho_2\rho_1}\sin\delta, \end{aligned}\\[3ex] \label{Eq:III:21:42} &\begin{aligned} \dot{\theta}_1&=-\frac{K}{\hbar}\sqrt{\frac{\rho_2}{\rho_1}}\cos\delta- \frac{qV}{2\hbar},\\[1.5ex] \dot{\theta}_2&=-\frac{K}{\hbar}\sqrt{\frac{\rho_1}{\rho_2}}\cos\delta+ \frac{qV}{2\hbar}. \end{aligned} \label{Eq:III:21:43} \end{align}

The first two equations say that
$\dot{\rho}_1=-\dot{\rho}_2$. “But,” you say, “they must both be
zero if $\rho_1$ and $\rho_2$ are both constant and equal
to $\rho_0$.” Not quite. These equations are not the whole story. They
say what $\dot{\rho}_1$ and $\dot{\rho}_2$ would be *if there
were no extra electric forces* due to an unbalance between the
electron fluid and the background of positive ions. They tell how the
densities would *start* to change, and therefore describe the
kind of current that would begin to flow. This current from side $1$
to side $2$ would be just $\dot{\rho}_1$ (or $-\dot{\rho}_2$), or
\begin{equation}
\label{Eq:III:21:44}
J=\frac{2K}{\hbar}\sqrt{\rho_1\rho_2}\sin\delta.
\end{equation}
Such a current would soon charge up side $2$, *except* that we
have forgotten that the two sides are connected by wires to the
battery. The current that flows will not charge up region $2$ (or
discharge region $1$) because currents will flow to keep the potential
constant. These currents from the battery have not been included in
our equations. When they are included, $\rho_1$ and $\rho_2$ do not in
fact change, but the current across the junction is still given by
Eq. (21.44).

Since $\rho_1$ and $\rho_2$ do remain constant and equal to $\rho_0$, let’s set $2K\rho_0/\hbar=J_0$, and write \begin{equation} \label{Eq:III:21:45} J=J_0\sin\delta. \end{equation} $J_0$, like $K$, is then a number which is a characteristic of the particular junction.

The other pair of equations (21.43) tells us about $\theta_1$ and $\theta_2$. We are interested in the difference $\delta=\theta_2-\theta_1$ to use Eq. (21.45); what we get is \begin{equation} \label{Eq:III:21:46} \dot{\delta}=\dot{\theta}_2-\dot{\theta}_1=\frac{qV}{\hbar}. \end{equation} That means that we can write \begin{equation} \label{Eq:III:21:47} \delta(t)=\delta_0+\frac{q}{\hbar}\int V(t)\,dt, \end{equation} where $\delta_0$ is the value of $\delta$ at $t=0$. Remember also that $q$ is the charge of a pair, namely, $q=2q_e$. In Eqs. (21.45) and (21.47) we have an important result, the general theory of the Josephson junction.

Now what are the consequences? First, put on a dc voltage. If you put on
a dc voltage, $V_0$, the argument of the sine
becomes $(\delta_0+(q/\hbar)V_0t)$. Since $\hbar$ is a small number
(compared to ordinary voltage and times), the sine oscillates rather
rapidly and the net current is nothing. (In practice, since the
temperature is not zero, you would get a small current due to the
conduction by “normal” electrons.) On the other hand if you have
*zero* voltage across the junction, you can get a current! With no
voltage the current can be any amount between $+J_0$ and $-J_0$
(depending on the value of $\delta_0$). But try to put a voltage across
it and the current goes to zero. This strange behavior has recently been
observed experimentally.^{20}

There is another way of getting a current—by applying a voltage at a
very high frequency in addition to a dc voltage. Let
\begin{equation*}
V=V_0+v\cos\omega t,
\end{equation*}
where $v\ll V_0$. Then $\delta(t)$ is
\begin{equation*}
\delta_0+\frac{q}{\hbar}\,V_0t+\frac{q}{\hbar}\,\frac{v}{\omega}\sin\omega t.
\end{equation*}
Now for $\Delta x$ small,
\begin{equation*}
\sin\,(x+\Delta x)\approx\sin x+\Delta x\cos x.
\end{equation*}
Using this approximation for $\sin\delta$, I get
\begin{equation*}
J=\!J_0\Bigl[\sin\Bigl(\!\delta_0\!+\!\frac{q}{\hbar}V_0t\!\Bigr)\!+\!
\frac{q}{\hbar}\frac{v}{\omega}\sin\omega t
\cos\Bigl(\!\delta_0\!+\!\frac{q}{\hbar}V_0t\!\Bigr)\Bigr].
\end{equation*}
The first term is zero on the average, but the second term is not if
\begin{equation*}
\omega=\frac{q}{\hbar}\,V_0.
\end{equation*}
There should be a current if the ac voltage has just this frequency.
Shapiro^{21} claims to have observed such a
resonance effect.

If you look up papers on the subject you will find that they often write the formula for the current as \begin{equation} \label{Eq:III:21:48} J=J_0\sin\biggl(\delta_0+\frac{2q_e}{\hbar}\int\FLPA\cdot d\FLPs\biggr), \end{equation} where the integral is to be taken across the junction. The reason for this is that when there’s a vector potential across the junction the flip-flop amplitude is modified in phase in the way that we explained earlier. If you chase that extra phase through, it comes out as given above.

Finally, I would like to describe a very dramatic and interesting experiment which has recently been made on the interference of the currents from each of two junctions. In quantum mechanics we’re used to the interference between amplitudes from two different slits. Now we’re going to do the interference between two junctions caused by the difference in the phase of the arrival of the currents through two different paths. In Fig. 21–7, I show two different junctions, “a” and “b”, connected in parallel. The ends, $P$ and $Q$, are connected to our electrical instruments which measure any current flow. The external current, $J_{\text{total}}$, will be the sum of the currents through the two junctions. Let $J_{\text{a}}$ and $J_{\text{b}}$ be the currents through the two junctions, and let their phases be $\delta_{\text{a}}$ and $\delta_{\text{b}}$. Now the phase difference of the wave functions between $P$ and $Q$ must be the same whether you go on one route or the other. Along the route through junction “a”, the phase difference between $P$ and $Q$ is $\delta_{\text{a}}$ plus the line integral of the vector potential along the upper route: \begin{equation} \label{Eq:III:21:49} \Delta\text{Phase}_{P\to Q}=\delta_{\text{a}}+ \frac{2q_e}{\hbar}\int_{\text{upper}}\kern{-3ex}\FLPA\cdot d\FLPs. \end{equation} Why? Because the phase $\theta$ is related to $\FLPA$ by Eq. (21.26). If you integrate that equation along some path, the left-hand side gives the phase change, which is then just proportional to the line integral of $\FLPA$, as we have written here. The phase change along the lower route can be written similarly \begin{equation} \label{Eq:III:21:50} \Delta\text{Phase}_{P\to Q}=\delta_{\text{b}}+ \frac{2q_e}{\hbar}\int_{\text{lower}}\kern{-3ex}\FLPA\cdot d\FLPs. \end{equation} These two must be equal; and if I subtract them I get that the difference of the deltas must be the line integral of $\FLPA$ around the circuit: \begin{equation*} \delta_{\text{b}}-\delta_{\text{a}}= \frac{2q_e}{\hbar}\oint_\Gamma\FLPA\cdot d\FLPs. \end{equation*} Here the integral is around the closed loop $\Gamma$ of Fig. 21–7 which circles through both junctions. The integral over $\FLPA$ is the magnetic flux $\Phi$ through the loop. So the two $\delta$’s are going to differ by $2q_e/\hbar$ times the magnetic flux $\Phi$ which passes between the two branches of the circuit: \begin{equation} \label{Eq:III:21:51} \delta_{\text{b}}-\delta_{\text{a}}=\frac{2q_e}{\hbar}\,\Phi. \end{equation} I can control this phase difference by changing the magnetic field on the circuit, so I can adjust the differences in phases and see whether or not the total current that flows through the two junctions shows any interference of the two parts. The total current will be the sum of $J_{\text{a}}$ and $J_{\text{b}}$. For convenience, I will write \begin{equation*} \delta_{\text{a}}=\delta_0-\frac{q_e}{\hbar}\,\Phi,\quad \delta_{\text{b}}=\delta_0+\frac{q_e}{\hbar}\,\Phi. \end{equation*} Then, \begin{align} J_{\text{total}} &=J_0\biggl\{\!\sin\biggl(\! \delta_0\!-\!\frac{q_e}{\hbar}\Phi\!\biggr)\!+\sin\biggl(\! \delta_0\!+\!\frac{q_e}{\hbar}\,\Phi\!\biggr)\!\biggr\}\notag\\[1.5ex] \label{Eq:III:21:52} &=2J_0\sin\delta_0\cos\frac{q_e\Phi}{\hbar}. \end{align}

Now we don’t know anything about $\delta_0$, and nature can adjust
that anyway she wants depending on the circumstances. In particular,
it will depend on the external voltage we apply to the junction. No
matter what we do, however, $\sin\delta_0$ can never get bigger
than $1$. So the *maximum* current for any given $\Phi$ is given by
\begin{equation*}
J_{\text{max}}=2J_0\left\lvert
\cos\frac{q_e}{\hbar}\,\Phi\right\rvert.
\end{equation*}
This maximum current will vary with $\Phi$ and will itself have maxima
whenever
\begin{equation*}
\Phi=n\,\frac{\pi\hbar}{q_e},
\end{equation*}
with $n$ some integer. That is to say that the current takes on its
maximum values where the flux linkage has just those quantized values
we found in Eq. (21.30)!

The Josephson current through a double junction was recently
measured^{22} as a
function of the magnetic field in the area between the junctions. The
results are shown in Fig. 21–8. There is a general
background of current from various effects we have neglected, but the
rapid oscillations of the current with changes in the magnetic field are
due to the interference term $\cos q_e\Phi/\hbar$ of
Eq. (21.52).

One of the intriguing questions about quantum mechanics is the question
of whether the vector potential exists in a place where there’s no
field.^{23} This
experiment I have just described has also been done with a tiny solenoid
between the two junctions so that the only significant magnetic
$\FLPB$ field is inside the solenoid and a negligible amount is on the
superconducting wires themselves. Yet it is reported that the amount of
current depends oscillatorily on the flux of magnetic field inside that
solenoid even though that field never touches the wires—another
demonstration of the “physical reality” of the vector
potential.^{24}

I don’t know what will come next. But look what can be done. First, notice that the interference between two junctions can be used to make a sensitive magnetometer. If a pair of junctions is made with an enclosed area of, say, $1$ mm$^2$, the maxima in the curve of Fig. 21–8 would be separated by $2\times10^{-6}$ gauss. It is certainly possible to tell when you are $1/10$ of the way between two peaks; so it should be possible to use such a junction to measure magnetic fields as small as $2\times10^{-7}$ gauss—or to measure larger fields to such a precision. One should be able to go even further. Suppose for example we put a set of $10$ or $20$ junctions close together and equally spaced. Then we can have the interference between $10$ or $20$ slits and as we change the magnetic field we will get very sharp maxima and minima. Instead of a $2$-slit interference we can have a $20$- or perhaps even a $100$-slit interferometer for measuring the magnetic field. Perhaps we can predict that the measurement of magnetic fields will—by using the effects of quantum-mechanical interference—eventually become almost as precise as the measurement of wavelength of light.

These then are some illustrations of things that are happening in modern times—the transistor, the laser, and now these junctions, whose ultimate practical applications are still not known. The quantum mechanics which was discovered in 1926 has had nearly 40 years of development, and rather suddenly it has begun to be exploited in many practical and real ways. We are really getting control of nature on a very delicate and beautiful level.

I am sorry to say, gentlemen, that to participate in this adventure it is absolutely imperative that you learn quantum mechanics as soon as possible. It was our hope that in this course we would find a way to make comprehensible to you at the earliest possible moment the mysteries of this part of physics.

- I’m not really reminding you, because I haven’t shown you some of these equations before; but remember the spirit of this seminar. ↩
- Volume II, Section 15–5. ↩
- Not to be confused with our earlier use of $\phi$ for a state label! ↩
- $K$ is the same quantity that was called $A$ in the problem of a linear lattice with no magnetic field. See Chapter 13. ↩
- Section 13–3. ↩
- Volume II, Section 27–1. ↩
- See, for example, J. D. Jackson, Classical Electrodynamics, John Wiley and Sons, Inc., New York(1962), p. 408. ↩
- Volume II, Chapter 14, Section 14–1. ↩
- First discovered by Kamerlingh-Onnes in 1911; H. Kamerlingh-Onnes, Comm. Phys. Lab., Univ. Leyden, Nos. 119, 120, 122 (1911). You will find a nice up-to-date discussion of the subject in E. A. Lynton, Superconductivity, John Wiley and Sons, Inc., New York, 1962. ↩
- J. Bardeen, L. N. Cooper, and J. R. Schrieffer, Phys. Rev. 108, 1175 (1957). ↩
- W. Meissner and R. Ochsenfeld, Naturwiss. 21, 787 (1933). ↩
- Actually if the electric field were too strong, pairs would be broken up and the “normal” electrons created would move in to help neutralize any excess of positive charge. Still, it takes energy to make these normal electrons, so the main point is that a nearly uniform density $\rho$ is highly favored energetically. ↩
- F. London and H. London, Proc. Roy. Soc. (London) A149, 71 (1935); Physica 2, 341 (1935). ↩
- F. London, Superfluids; John Wiley and Sons, Inc., New York, 1950, Vol. I, p. 152. ↩
- B. S. Deaver, Jr., and W. M. Fairbank, Phys. Rev. Letters 7, 43 (1961). ↩
- R. Doll and M. Näbauer, Phys. Rev. Letters 7, 51 (1961). ↩
- It has once been suggested by Onsager that this might happen (see Deaver and Fairbank, Ref. 15), although no one else ever understood why. ↩
- See Volume II, Section 40–2. ↩
- B. D. Josephson, Physics Letters 1, 251 (1962). ↩
- P. W. Anderson and J. M. Rowell, Phys. Rev. Letters 10, 230 (1963). ↩
- S. Shapiro, Phys. Rev. Letters 11, 80 (1963). ↩
- Jaklevic, Lambe, Silver, and Mercereau, Phys. Rev. Letters 12, 159 (1964). ↩
- Jaklevic, Lambe, Silver, and Mercereau, Phys. Rev. Letters 12, 274 (1964). ↩
- See Volume II, Chapter 15, Section 15–5. ↩