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):

Is = Icsin(φ)

where Is is the supercurrent flowing through the junction, Ic 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 EJ = Φ0Ic / 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 = 2πn, n is any integer.

[edit] Josephson inductance

Imagine that the Josephson phase across the junction is φ0 and the supercurrent flowing trough the junction is

I0 = Icsinφ0.

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 = φ0 + δφ. One can write:

I0 + δI = Icsin(φ0 + δφ) Assuming that δφ is small, we make a Taylor expansion in the rhs to arrive at δI = Iccos(φ0)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 (φ0) of the junction. The typical value is given by LJ(0) and is determined only by the critical current Ic. Note that, according to definition, the Josephson inductance can even become infinite or negaive! This happens when cos(φ0) < = 0.

One can also calcuate the change in Josephson energy

δU(φ0) = U(φ) − U(φ0) = EJ(cos(φ0) − cos(φ0 + 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 dEL = LIδI, 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.

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