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Derivation of the first and second Josephson relation

In the following discussion, we assume that the quantum states are quasi-static. We solve the time-independent Schrödinger equation supposing that external potentials are independent of time v(\mathbf x), \mathbf A(\mathbf x). When we change them slowly enough, the corresponding quantum states (the effective wavefunction) changes so that they keep following the time-independent Schrödinger equation (with the given potentials) at each point of time.

First, we shows that the time-independent Schrödinger equation for the gauge-invariant wavefuntion \psi_{\rm GI}(\mathbf x) has the following form:

　\displaystyle E\psi_{\rm GI}(\mathbf x) = \left[\frac{1}{2m_s}(-i\hbar\nabla)^2 + q_sv(\mathbf x)\right]\psi_{\rm GI}(\mathbf x) --- (1)

[Proof]

We start from the time-dependent Schrödinger equation of the effective wafefunction:

　 \displaystyle i\hbar\partial_t\psi(\mathbf x, t) = \left[\frac{1}{2m_s}\left\{-i\hbar\nabla-q_s\mathbf A(\mathbf x, t)\right\}^2 + q_sv(\mathbf x,t)\right]\psi(\mathbf x, t) --- (3.5)

When the potentials are static \displaystyle v(\mathbf x), \mathbf A(\mathbf x), the energy eigenstate is given by:

　\displaystyle \psi(\mathbf x,t) = e^{-i\hbar Et}\psi(\mathbf x) --- (2)

and (3.5) becomes:

　 \displaystyle E\psi(\mathbf x) = \left[\frac{1}{2m_s}\left\{-i\hbar\nabla-q_s\mathbf A(\mathbf x)\right\}^2 + q_sv(\mathbf x)\right]\psi(\mathbf x) --- (3)

From (2), the gauge-independent wavefunction has the following from:

　\displaystyle \psi_{\rm GI}(\mathbf x,t) = e^{-i\frac{q_s}{\hbar}\int_{\mathbf x_0}^{\mathbf x}\mathbf A(\mathbf r)\cdot d\mathbf l}\psi(\mathbf x, t) = e^{-i\hbar Et} e^{-i\frac{q_s}{\hbar}\int_{\mathbf x_0}^{\mathbf x}\mathbf A(\mathbf r)\cdot d\mathbf l}\psi(\mathbf x)

So we define the time-independent part of \displaystyle \psi_{\rm GI}(\mathbf x,t) as:

　\displaystyle \psi_{\rm GI}(\mathbf x) = e^{-i\frac{q_s}{\hbar}\int_{\mathbf x_0}^{\mathbf x}\mathbf A(\mathbf r)\cdot d\mathbf l}\psi(\mathbf x)

Equivalently, we have:

　\displaystyle \psi(\mathbf x) = e^{i\frac{q_s}{\hbar}\int_{\mathbf x_0}^{\mathbf x}\mathbf A(\mathbf r)\cdot d\mathbf l}\psi_{\rm GI}(\mathbf x) --- (4)

By substituting (4) into (3), we get (1). ■

Now we apply (1) to the one-dimensional potential barrier model of Josephson Junctions:

　 \displaystyle|x| > \frac{d}{2};\ E\psi_{\rm GI}(x) = \frac{1}{2m_s}(-i\hbar\partial_x)^2\psi_{\rm GI}(x)

　 \displaystyle |x| < \frac{d}{2};\ E\psi_{\rm GI}(x) = \left\{\frac{1}{2m_s}(-i\hbar\partial_x)^2 + U_0\right\}\psi_{\rm GI}(x)

where U_0 > E.

For \displaystyle|x| > \frac{d}{2}, the solution can be plane-wave functions:

　 \displaystyle x>\frac{d}{2};\ \psi_{\rm GI}(x) = \sqrt{n_s}\exp\left[i\frac{\sqrt{2m_sE}}{\hbar}\left(x-\frac{d}{2}\right)+i\varphi_R\right]

　 \displaystyle x<-\frac{d}{2};\ \psi_{\rm GI}(x) = \sqrt{n_s}\exp\left[i\frac{\sqrt{2m_sE}}{\hbar}\left(x+\frac{d}{2}\right)+i\varphi_L\right]

The arbitrariness of the global phase is expressed by the constants \varphi_R,\,\varphi_L. At this point, they can be taken independently, but later, we will show that they are related through the boundary condition of currents \mathbf J.

For \displaystyle|x| < \frac{d}{2}, the solution can be a liner combination of \cosh and \sinh.

　\displaystyle \psi_{\rm GI}(x) = \alpha_+\cosh\left(\frac{x}{\xi}\right) + \alpha_-\sinh\left(\frac{x}{\xi}\right) --- (5)

　　\displaystyle=\frac{\alpha_+}{2}\left(e^{\frac{x}{\xi}}+e^{-\frac{x}{\xi}}\right)+\frac{\alpha_-}{2}\left(e^{\frac{x}{\xi}}-e^{-\frac{x}{\xi}}\right)

where \displaystyle \xi = \frac{\hbar}{\sqrt{2m_s(U_0-E)}}.

From the boundary condition (continuity of \psi_{\rm GI}(x)) at \displaystyle x=\pm\frac{d}{2}, we have:

　\displaystyle\frac{\alpha_+}{2}\left(e^{\frac{d}{2\xi}}+e^{-\frac{d}{2\xi}}\right)+\frac{\alpha_-}{2}\left(e^{\frac{d}{2\xi}}-e^{-\frac{d}{2\xi}}\right) = \sqrt{n_s}e^{i\varphi_R}

　\displaystyle\frac{\alpha_+}{2}\left(e^{-\frac{d}{2\xi}}+e^{\frac{d}{2\xi}}\right)+\frac{\alpha_-}{2}\left(e^{-\frac{d}{2\xi}}-e^{\frac{d}{2\xi}}\right) = \sqrt{n_s}e^{i\varphi_L}

By solving these linear equations, we have:

　\displaystyle\alpha_{\pm} = \sqrt{n_s}\frac{e^{i\varphi_R}\pm e^{i\varphi_L}}{e^{\frac{d}{2\xi}}\pm e^{-\frac{d}{2\xi}}} --- (3.37)

In general, the currents:

　\displaystyle \mathbf J(\mathbf x, t) = \frac{\hbar q_sn_s}{m_s}\nabla \varphi(\mathbf x, t) --- (3.23)

can be calculated from the gauge-invariant wavefunction \psi_{\rm GI}(\mathbf x, t) as:

　\displaystyle \mathbf J(\mathbf x, t) = \frac{q_s}{m_s}{\rm Re}\left[-\psi_{\rm GI}^*(\mathbf x, t)i\hbar\nabla \psi_{\rm GI}(\mathbf x, t)\right]

(You can confirm it with the direct calculation using the general form \displaystyle \psi_{\rm GI}(\mathbf x, t) = e^{i\varphi(\mathbf x,t)}\sqrt{n_s(\mathbf x, t)}.)

In our particular case (5), we have:

　\displaystyle J(x, t) =\frac{q_s}{m_s}{\rm Re}\left[-\left\{e^{-i\hbar Et}\psi_{\rm GI}(x)\right\}^*i\hbar\partial_x\left\{e^{-i\hbar Et}\psi_{\rm GI}(x)\right\}\right]

　　\displaystyle = -\frac{q_s\hbar}{m_s}{\rm Im}\left[\psi_{\rm GI}^*(x)\partial_x\psi_{\rm GI}(x)\right]

　　\displaystyle = -\frac{q_s\hbar}{m_s\xi}{\rm Im}\bigg[ \left\{\alpha_+^*\cosh\left(\frac{x}{\xi}\right)+\alpha_-^*\sinh\left(\frac{x}{\xi}\right)\right\}
　　　　　　　　　\displaystyle\left\{\alpha_+\sinh\left(\frac{x}{\xi}\right)+\alpha_-\cosh\left(\frac{x}{\xi}\right)\right\}\bigg]

　　\displaystyle = -\frac{q_s\hbar}{m_s\xi}{\rm Im}\left[ \alpha_+^*\alpha_-\cosh^2\left(\frac{x}{\xi}\right) + \alpha_-^*\alpha_+\sinh^2\left(\frac{x}{\xi}\right) \right]

　　\displaystyle = -\frac{q_s\hbar}{m_s\xi}\frac{n_s}{e^\frac{d}{\xi}-e^{-\frac{d}{\xi}}}{\rm Im} \bigg[ (e^{-i\varphi_R}+e^{-i\varphi_L})(e^{i\varphi_R}-e^{i\varphi_L})\cosh^2\left(\frac{x}{\xi}\right)

　　　　　　　　 \displaystyle +\,(e^{-i\varphi_R}-e^{-i\varphi_L})(e^{i\varphi_R}+e^{i\varphi_L})\sinh^2\left(\frac{x}{\xi}\right) \bigg]

　　\displaystyle = -\frac{q_s\hbar}{m_s\xi}\frac{n_s}{e^\frac{d}{\xi}-e^{-\frac{d}{\xi}}}{\rm Im} \left[ e^{i(\varphi_R-\varphi_L)}-e^{-i(\varphi_R-\varphi_L)}\right]

　　\displaystyle = -\frac{q_s\hbar}{m_s\xi}\frac{2n_s}{e^\frac{d}{\xi}-e^{-\frac{d}{\xi}}}\sin(\varphi_R-\varphi_L)

Hence, we finally get:

　\displaystyle J(x, t) = J_C\sin(\varphi_R-\varphi_L) --- (6)

where:

　\displaystyle J_C = -\frac{q_s\hbar}{m_s\xi}\frac{n_s}{\sinh\left(\frac{d}{\xi}\right)}

(6) suggests that the currents are constant and unique inside the insulator, and we enforce the boundary condition that it's the same as the currents outside the insulator:

　\displaystyle J = \frac{\sqrt{2n_sE}}{\Lambda} --- (3.34)

From this condition, the phase difference \delta\varphi := \varphi_R-\varphi_L can be decided.

In terms of the current intensity:

　 I = J\times A --- (3.32)

we have the similar relationship:

　I=I_C\sin(\delta\varphi)

Now we assume that we slowly apply an external electric field over the circuit (outside the insulator) so that the voltage (the electric potential difference) between the contact points \displaystyle x=\pm\frac{d}{2} to the insulator becomes V. By applying the discussion:

・Derivation of the gauge-independent relation between the phase and the electric field

the phase difference between two points follows the relationship:

　\displaystyle\frac{d}{dt}\delta\varphi(t) = \frac{2\pi}{\Phi_0}V --- (7)

(We're assuming that V is very small and the change \delta\varphi(t) follows the quasi-static assumption described at the beginning of this note.)

So we conclude that we have the oscillating current inside the insulator:

　\displaystyle I(t) = I_C\sin\left(\frac{2\pi}{\Phi_0}Vt + \delta\varphi(0)\right)

Note

As in the discussion in the following link, I need an assumption that n_s is constant and unique to use the relationship (7).

・Derivation of the gauge-independent relation between the phase and the electric field

So I applied the discussion to the wavefunction outside the insulator that (presumably) forms a closed loop. I'm not sure if this is really a common way to justify using (7) in this model.