# 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 of a Josephson junction as 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)

$U = \int_0^t I_s V\,dt = \frac{\Phi_0}{2\pi} \int_0^t I_s \frac{d\phi}{dt}\,dt = \frac{\Phi_0}{2\pi} \int_0^\phi I_c\sin(\phi) \,d\phi = \frac{\Phi_0 I_c}{2\pi} (1-\cos\phi).$

Here $E_J = {\Phi_0 I_c}/{2\pi}$ sets the characteristic scale of the Josephson energy, and $(1-\cos\phi)$ 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 $\phi=2\pi n$, $n$ is any integer.

## Josephson inductance

Imagine that the Josephson phase across the junction is $\phi_0\,$ and the supercurrent flowing through the junction is

$I_0 = I_c \sin\phi_0\,$

(This is the same equation as above, except now we will look at small variations in $I_s\,$ and $\phi\,$ around the values $I_0\,$ and $\phi_0\,$.)

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 $\phi=\phi_0+\delta\phi\,$. One can write:

$I_0+\delta I = I_c \sin(\phi_0+\delta\phi)\,$

Assuming that $\delta\phi\,$ is small, we make a Taylor expansion in the right hand side to arrive at

$\delta I = I_c \cos(\phi_0) \delta\phi\,$

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

$V = \frac{\Phi_0}{2\pi}\dot{\phi} = \frac{\Phi_0}{2\pi}(\underbrace{\dot{\phi_0}}_{=0} + \dot{\delta\phi}) = \frac{\Phi_0}{2\pi} \frac{\dot{\delta I}}{I_c \cos(\phi_0)}.$

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

$V = L \frac{\partial I}{\partial t}$,

we can define the so-called Josephson inductance

$L_J(\phi_0) = \frac{\Phi_0}{2\pi I_c \cos(\phi_0)} = \frac{L_J(0)}{\cos(\phi_0)}.$

One can see that this inductance is not constant, but depends on the phase $(\phi_0)\,$ across 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 negative (if $\cos(\phi_0)<=0\,$).

One can also calculate the change in Josephson energy

$\delta U(\phi_0) = U(\phi)-U(\phi_0) = E_J (\cos(\phi_0)-\cos(\phi_0+\delta\phi)\,$

Making Taylor expansion for small $\delta\phi\,$, we get

$\approx E_J \sin(\phi_0)\delta\phi = \frac{E_J \sin(\phi_0)}{I_c \cos\phi_0}\delta I$

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. The Josephson Inductance is also known as a Kinetic Inductance - the behaviour is derived from the kinetic energy of the charge carriers, not energy in a magnetic field.