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

The effective wavefunction $\psi(\mathbf x, t)$ and the charge current $\mathbf J$ are given as:

　 $\displaystyle \psi(\mathbf x, t) = e^{i\theta(\mathbf x, t)}\sqrt{n_s(\mathbf x, t)}$ --- (3.4)

　 $\displaystyle \mathbf J = q_sn_s\left(\frac{\hbar}{m_s}\nabla\theta -\frac{q_s}{m_s}\mathbf A\right)$ ---(3.13)

The wavefunction follows the Schrödinger equation:

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

Without losing the generality, we can take the Coulomb gauge:

　 $\nabla\cdot\mathbf A=0$ --- (1)

Now, we assume that the charge density $n_s = |\psi(\mathbf x, t)|^2$ is constant and uniform:

　 $\displaystyle \psi(\mathbf x,t) = e^{i\theta(\mathbf x,t)}\sqrt{n_s}$

In this case, the currents should be divergence-free $\nabla\cdot\mathbf J=0$ . Then, by applying $\nabla$ on (3.13), we have:

　 $\nabla^2\theta=0$ --- (2)

Also, we have:

　 $\displaystyle\left\{-i\hbar\nabla-q_s\mathbf A(\mathbf x,t)\right\}\psi(\mathbf x,t) = \left\{\hbar\nabla\theta(\mathbf x,t)-q_s\mathbf A(\mathbf x,t)\right\}\psi(\mathbf x,t)$

By using (1) and (2), we also have:

　 $\displaystyle\left\{-i\hbar\nabla-q_s\mathbf A(\mathbf x,t)\right\}^2\psi(\mathbf x,t) = \left\{\hbar\nabla\theta(\mathbf x,t)-q_s\mathbf A(\mathbf x,t)\right\}^2\psi(\mathbf x,t)$

　　 $\displaystyle = \frac{m_s^2}{q_s^2n_s^2}\mathbf J(\mathbf x,t)^2\psi(\mathbf x,t)$

(3.5) becomes:

　 $\displaystyle-\hbar\partial_t\theta(\mathbf x,t) = \frac{m_s}{2q_s^2n_s^2}\mathbf J(\mathbf x,t)^2 + q_sv(\mathbf x,t)$ --- (3)

Since this is valid only on the Coulomb gauge, we will rewrite it as a gauge-independent expression.

The gauge-invariant phase $\varphi(\mathbf x,t)$ is defined as:

　 $\displaystyle \varphi(\mathbf x,t)-\varphi(\mathbf x_0,t) = \left\{\theta(\mathbf x,t)-\theta(\mathbf x_0,t)\right\} -\frac{q_s}{\hbar}\int_{\mathbf x_0}^\mathbf x\mathbf A(\mathbf r,t)\cdot d\mathbf l$ --- (3.16)

As the RHS of (3.16) is gauge-invariant, we can take the Coulomb gauge and use (3) without breaking the gauge-invariance of LHS. By applying $\partial_t$ on (3.16), and assuming that the currents are uniform:

　 $\displaystyle \mathbf J(\mathbf x, t) = \mathbf J(\mathbf x_0, t)$

we have:

　 $\displaystyle\partial_t\varphi(\mathbf x,t) - \partial_t\varphi(\mathbf x_0,t) = -\frac{q_s}{\hbar}\left\{v(\mathbf x,t)-v(\mathbf x_0,t)\right\} - \frac{q_s}{\hbar}\int_{\mathbf x_0}^\mathbf x\partial_t\mathbf A(\mathbf r,t)\cdot d\mathbf l$

　　 $\displaystyle = \frac{q_s}{\hbar}\int_{\mathbf x_0}^\mathbf x\left\{-\nabla v(\mathbf r, t)-\partial_t\mathbf A(\mathbf r,t)\right\}\cdot d\mathbf l = \frac{q_s}{\hbar}\int_{\mathbf x_0}^\mathbf x\mathbf E(\mathbf r, t)\cdot d\mathbf l$

where we used the relationship between the electric field $\mathbf E$ and potentials $(\mathbf A,\,v)$ :

　 $\displaystyle \mathbf E(\mathbf x,t) = -\nabla v(\mathbf x,t)-\partial_t\mathbf A(\mathbf x,t)$

Because $\mathbf E(\mathbf x,t)$ is gauge-invariant, the following relationship is gauge-independent:

　 $\displaystyle \partial_t\varphi(\mathbf x,t) - \partial_t\varphi(\mathbf x_0,t) = \frac{q_s}{\hbar}\int_{\mathbf x_0}^\mathbf x\mathbf E(\mathbf r,t)\cdot d\mathbf l = -\frac{q_s}{\hbar}V$

Where $V := V(\mathbf x)-V(\mathbf x_0)$ is a voltage (electric potential difference) between the two points $(\mathbf x,\, \mathbf x_0)$ .

By defining:

　 $\displaystyle\delta\varphi(t) := \varphi(\mathbf x,t)-\varphi(\mathbf x_0,t)$

　 $\displaystyle\Phi_0 := \frac{h}{|q_s|} = -\frac{2\pi\hbar}{q_s}$

We have:

　 $\displaystyle\frac{d}{dt}\delta\varphi(t) = \frac{2\pi}{\Phi_0}V$ --- (4)

Hence, if we apply a constant voltage $V$ between two points, the phase difference grows as:

　 $\displaystyle\delta\varphi(t) = \frac{2\pi}{\Phi_0}Vt + \delta\varphi(0)$

Note:

In the textbook, it says "We have to complete our derivation to regard more general conditions!" on p.27. In the derivation of (4), we used the following two assumptions:

The charge density $n_s = |\psi(\mathbf x,t)|^2$ is constant and uniform (independent of $(\mathbf x,\, t)$ ).

The currents $\mathbf J(\mathbf x,t)$ are uniform (independent of $\mathbf x$ ).

These are the same assumptions used in the next section "3.7 Josephson Junctions". So we can use the relationship (4) in that section.

めもめも

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

Note: