Gauge Invariance of superconducting circuits wavefunction model

Effective wavefunction describing the flow of superconducting elections (Cooper pairs):

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

Schrödinger equation for $\psi(\mathbf x, t)$ :

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

I will show that this model is invariant under the gauge transformation:

　 $\displaystyle\psi(\mathbf x,t) \, \rightarrow\, \displaystyle\psi'(\mathbf x,t) = e^{-i\frac{q_s}{\hbar}\chi(\mathbf x,t)}\psi(\mathbf x,t)$

　 $\displaystyle \mathbf A(\mathbf x, t)\,\rightarrow\, \mathbf A'(\mathbf x, t) = \mathbf A(\mathbf x, t)-\nabla\chi(\mathbf x, t)$

　 $\displaystyle v(\mathbf x,t)\, \rightarrow\, v'(\mathbf x, t) = v(\mathbf x,t)+\partial_t\chi(\mathbf x, t)$

[Proof]

(3.5) is equivalent to

　 $\displaystyle \left[i\hbar\partial_t - q_sv(\mathbf x,t) - \frac{1}{2m_s}\left\{-i\hbar\nabla-q_s\mathbf A(\mathbf x,t)\right\}^2\right] \psi(\mathbf x, t) = 0$ --- (1)

A simple calculation shows the following commutation relations.

　 $\displaystyle \left[i\hbar\partial_t,\ e^{i\frac{q_s}{\hbar}\chi(\mathbf x,t)}\right] =-e^{i\frac{q_s}{\hbar}\chi(\mathbf x,t)}q_s\partial_t\chi(\mathbf x,t)$

　 $\displaystyle \left[-i\hbar\nabla,\ e^{i\frac{q_s}{\hbar}\chi(\mathbf x,t)}\right] =e^{i\frac{q_s}{\hbar}\chi(\mathbf x,t)}q_s\nabla\chi(\mathbf x,t)$

Hence we have:

　 $\displaystyle \left[i\hbar\partial_t-q_sv(\mathbf x,t),\ e^{i\frac{q_s}{\hbar}\chi(\mathbf x,t)}\right] =-e^{i\frac{q_s}{\hbar}\chi(\mathbf x,t)}q_s\partial_t\chi(\mathbf x,t)$ --- (2)

　 $\displaystyle \left[-i\hbar\nabla-q_s\mathbf A(\mathbf x, t),\ e^{i\frac{q_s}{\hbar}\chi(\mathbf x,t)}\right] =e^{i\frac{q_s}{\hbar}\chi(\mathbf x,t)}q_s\nabla\chi(\mathbf x,t)$ --- (3)

By using (2), we have:

　 $\displaystyle \left\{i\hbar\partial_t - q_sv(\mathbf x,t)\right\}\psi(\mathbf x,t) = \left\{i\hbar\partial_t - q_sv(\mathbf x,t)\right\} e^{i\frac{q_s}{\hbar}\chi(\mathbf x,t)}\psi'(\mathbf x,t)$

　 $\displaystyle = e^{i\frac{q_s}{\hbar}\chi(\mathbf x,t)} \left\{i\hbar\partial_t - q_sv(\mathbf x,t)\right\} \psi'(\mathbf x,t)-e^{i\frac{q_s}{\hbar}\chi(\mathbf x,t)}q_s(\partial_t\chi(\mathbf x,t)) \psi'(\mathbf x,t)$

　 $\displaystyle = e^{i\frac{q_s}{\hbar}\chi(\mathbf x,t)} \left\{i\hbar\partial_t - q_sv'(\mathbf x,t)\right\} \psi'(\mathbf x,t)$ --- (4)

Similarly, by using (3), we have:

　 $\displaystyle \left\{-i\hbar\nabla-q_s\mathbf A(\mathbf x,t)\right\}\psi(\mathbf x, t) = \left\{-i\hbar\nabla-q_s\mathbf A(\mathbf x,t)\right\}e^{i\frac{q_s}{\hbar}\chi(\mathbf x,t)}\psi'(\mathbf x,t)$

　 $\displaystyle = e^{i\frac{q_s}{\hbar}\chi(\mathbf x,t)} \left\{-i\hbar\nabla-q_s\mathbf A(\mathbf x,t)\right\} \psi'(\mathbf x,t)+e^{i\frac{q_s}{\hbar}\chi(\mathbf x,t)}q_s(\nabla\chi(\mathbf x,t)) \psi'(\mathbf x,t)$

　 $\displaystyle = e^{i\frac{q_s}{\hbar}\chi(\mathbf x,t)} \left\{-i\hbar\nabla-q_s\mathbf A'(\mathbf x,t)\right\} \psi'(\mathbf x,t)$

Using (3) again:

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

　 $\displaystyle=\left\{-i\hbar\nabla-q_s\mathbf A(\mathbf x,t)\right\}e^{i\frac{q_s}{\hbar}\chi(\mathbf x,t)}\left[ \left\{-i\hbar\nabla-q_s\mathbf A'(\mathbf x,t)\right\} \psi'(\mathbf x,t) \right]$

　 $\displaystyle=e^{i\frac{q_s}{\hbar}\chi(\mathbf x,t)}\left\{-i\hbar\nabla-q_s\mathbf A(\mathbf x,t)+q_s\nabla\chi(\mathbf x,t)\right\}\left[ \left\{-i\hbar\nabla-q_s\mathbf A'(\mathbf x,t)\right\} \psi'(\mathbf x,t) \right]$

　 $\displaystyle=e^{i\frac{q_s}{\hbar}\chi(\mathbf x,t)} \left\{-i\hbar\nabla-q_s\mathbf A'(\mathbf x,t)\right\}^2\psi'(\mathbf x, t)$ --- (5)

From (4) and (5), (1) is equivalent to:

　 $\displaystyle \left[i\hbar\partial_t - q_sv'(\mathbf x,t) - \frac{1}{2m_s}\left\{-i\hbar\nabla-q_s\mathbf A'(\mathbf x,t)\right\}^2\right] \psi'(\mathbf x, t) = 0$
■

Now we can show that the phase $\varphi(\mathbf x,t)$ defined by the following equation is gauge-invariant:

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

First, in terms of the phase $\theta(\mathbf x, t)$ , the gauge transformation of $\psi(\mathbf x,t)$ can be written as:

　 $\displaystyle \theta(\mathbf x,t)\,\rightarrow\,\theta'(\mathbf x,t) = \theta(\mathbf x,t) - \frac{q_s}{\hbar}\chi(\mathbf x,t)$

So, by combining with the gauge transformation of $\mathbf A(\mathbf x,t)$ , the combination $\displaystyle \nabla \theta(\mathbf x,t)-\frac{q_s}{\hbar}\mathbf A(\mathbf x, t)$ becomes gauge-invariant. By integrating it through a path from $\mathbf x_0$ to $\mathbf x$ , the following combination gives a gauge-invariant phase $\varphi(\mathbf x,t)$ up to a constant $C$ .

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

By setting $C=\varphi(\mathbf x_0,t)$ , we get the expression (3.16)

めもめも

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Gauge Invariance of superconducting circuits wavefunction model