Josephson energy

In superconductivity, the Josephson energy is the potential energy accumulated in the Josephson junction when a supercurrent flows through it. One can think about a Josephson junction as about a non-linear inductance which accumulates (magnetic field) energy when a current passes through it. In contrast to real inductance, no magnetic field is created by a supercurrent in Josephson junction --- the accumulated energy is a Josephson energy.

Derivation

For the simplest case the current-phase relation (CPR) is given by (aka the first Josephson relation):

$I\_s\; =\; I\_c\; sin(phi)$

where $I\_s$ is the supercurrent flowing through the junction, $I\_c$ is the critical current, and $phi$ is the Josephson phase, see Josephson junction for details.

Imagine that initially at time $t=0$ the junction was in the ground state $phi=0$ and finally at time $t$ the junction has the phase $phi$. The work done on the junction (so the junction energy is increased by)

$U\; =\; int\_0^t\; I\_s\; V,dt=\; frac\{Phi\_0\}\{2pi\}\; int\_0^t\; I\_s\; frac\{dphi\}\{dt\},dt=\; frac\{Phi\_0\}\{2pi\}\; int\_0^t\; I\_csin(phi)\; ,dphi=\; frac\{Phi\_0\; I\_c\}\{2pi\}\; (1-cosphi).$

Here $E\_J\; =\; \{Phi\_0\; I\_c\}/\{2pi\}$sets the characteristic scale of the Josephson energy, and $(1-cosphi)$ sets its dependence on the phase $phi$. The energy $U(phi)$ accumulated inside the junction depends only on the current state of the junction, but not on history or velocities, i.e. it is a potential energy. Note, that $U(phi)$ has a minimum equal to zero for the ground state $phi=2pi\; n$, $n$ is any integer.

Josephson inductance

Imagine that the Josephson phase across the junction is $phi\_0$ and the supercurrent flowing trough the junction is

$I\_0\; =\; I\_c\; sinphi\_0.$

Now imagine that we add little extra current (dc or ac) $delta\; I(t)ll\; I\_c$ through JJ, and want to see how the junction reacts. The phase across the junction changes to become $phi=phi\_0+deltaphi$. One can write:

$I\_0+delta\; I\; =\; I\_c\; sin(phi\_0+deltaphi)$Assuming that $deltaphi$ is small, we make a Taylor expansion in the rhs to arrive at$delta\; I\; =\; I\_c\; cos(phi\_0)\; deltaphi$

The voltage across the junction (we use the 2nd Josephson relation) is

$V\; =\; frac\{Phi\_0\}\{2pi\}dot\{phi\}\; =\; frac\{Phi\_0\}\{2pi\}(underbrace\{dot\{phi\_0\_\{=0\}\; +\; dot\{deltaphi\})=\; frac\{Phi\_0\}\{2pi\}\; frac\{dot\{delta\; I\{I\_c\; cos(phi\_0)\}.$

If we compare this expression with the expresson for voltage across the conventional inductance

$V\; =\; L\; dot\{I\},$

we can define the so-called Josephson inductance

$L\_J(phi\_0)\; =\; frac\{Phi\_0\}\{2pi\; I\_c\; cos(phi\_0)\}\; =\; frac\{L\_J(0)\}\{cos(phi\_0)\}.$

One can see that this inductance is not constant, but depends on the state $(phi\_0)$ of the junction. The typical value is given by $L\_J(0)$ and is determined only by the critical current $I\_c$. Note that, according to definition, the Josephson inductance can even become infinite or negaive! This happens when .

One can also calcuate the change in Josephson energy

$delta\; U(phi\_0)\; =\; U(phi)-U(phi\_0)\; =\; E\_J\; (cos(phi\_0)-cos(phi\_0+deltaphi)$Making Taylor expansion for small $deltaphi$, we get$approx\; E\_J\; sin(phi\_0)deltaphi\; =\; frac\{E\_J\; sin(phi\_0)\}\{I\_c\; cosphi\_0\}delta\; I$

If we now compare this with the expression for increase of the inductance energy $dE\_L\; =\; L\; I\; delta\; I$, we again get the same expression for $L$.

Note, that although Josephson junction behaves like an inductance, there is no associated magnetic field. The corresponding energy is hidden inside the junction.