Josephson energy

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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.

[edit] Derivation

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

$I s = I c sin(φ)$

where $I s$ is the supercurrent flowing through the junction, $I c$ is the critical current, and $φ$ is the Josephson phase, see Josephson junction for details.

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

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

Here $E J = Φ 0 I c / 2π$ sets the characteristic scale of the Josephson energy, and $(1 - cosφ)$ sets its dependence on the phase $φ$ . The energy $U (φ)$ 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 $U (φ)$ has a minimum equal to zero for the ground state $φ = 2π n$ , $n$ is any integer.

[edit] Josephson inductance

Imagine that the Josephson phase across the junction is $φ 0$ and the supercurrent flowing trough the junction is

$I 0 = I c sinφ 0 .$

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 $φ = φ 0 + δφ$ . One can write:

$I 0 + δ I = I c sin(φ 0 + δφ)$ Assuming that $δφ$ is small, we make a Taylor expansion in the rhs to arrive at $δ I = I c cos(φ 0)δφ$

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 expresson for voltage across the conventional inductance

$V = L \dot{I},$

we can define the so-called Josephson inductance

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

One can see that this inductance is not constant, but depends on the state $(φ 0)$ 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 $cos(φ 0) < = 0$ .

One can also calcuate the change in Josephson energy

$δ U (φ 0) = U (φ) - U (φ 0) = E J (cos(φ 0) - cos(φ 0 + δφ)$ Making Taylor expansion for small $δφ$ , 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 $d E L = L I δ I$ , 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.