The Josephson junction
I would like to discuss next a very interesting situation that was
noticed by
Josephson19 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.
Shapiro21 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
measured22 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.