## Question: Gauge transformed probability current ([1] pr. 2.37 (b))

For the gauge transformed Schrodinger equation

\label{eqn:gaugeTxCurrent:20}
\inv{2m} \BPi(\Bx) \cdot \BPi(\Bx) \psi(\Bx, t) + e \phi(\Bx) \psi(\Bx, t) = i \Hbar \PD{t}{}\psi(\Bx, t),

where

\label{eqn:gaugeTxCurrent:40}
\BPi(\Bx) = -i \Hbar \spacegrad – \frac{e}{c} \BA(\Bx),

find the probability current defined by

\label{eqn:gaugeTxCurrent:60}

Equation \ref{eqn:gaugeTxCurrent:20} and its conjugate are

\label{eqn:gaugeTxCurrent:22}
\begin{aligned}
\inv{2m} \BPi \cdot \BPi \psi + e \phi \psi &= i \Hbar \PD{t}{\psi} \\
\inv{2m} \BPi^\conj \cdot \BPi^\conj \psi^\conj + e \phi \psi^\conj &= -i \Hbar \PD{t}{\psi^\conj}
\end{aligned}

which can be used immediately in a chain rule expansion of the probability time derivative

\label{eqn:gaugeTxCurrent:80}
\begin{aligned}
i \Hbar \PD{t}{\rho}
&=
i \Hbar \psi^\conj \PD{t}{\psi} +
i \Hbar \psi \PD{t}{\psi^\conj} \\
&=
\psi^\conj \lr{ \inv{2m} \BPi \cdot \BPi \psi + e \phi \psi } –
\psi \lr{ \inv{2m} \BPi^\conj \cdot \BPi^\conj \psi^\conj + e \phi \psi^\conj
} \\
&=
\inv{2m} \lr{
\psi^\conj \BPi \cdot \BPi \psi
-\psi \BPi^\conj \cdot \BPi^\conj \psi^\conj
}.
\end{aligned}

We have a difference of conjugates, so can get away with expanding just the first term

\label{eqn:gaugeTxCurrent:100}
\begin{aligned}
\psi^\conj \BPi \cdot \BPi \psi
&=
\psi^\conj
\psi \\
&=
\psi^\conj
\lr{ -i \Hbar \spacegrad – \frac{e}{c} \BA } \cdot \lr{ -i \Hbar \spacegrad – \frac{e}{c} \BA }
\psi \\
&=
\psi^\conj
\lr{
+ \frac{e^2}{c^2} \BA^2
}
\psi.
\end{aligned}

Note that in the directional derivative terms, the gradient operates on everything to its right, including $$\BA$$. Also note that the last term has no imaginary component, so it will not contribute to the difference of conjugates.

This gives

\label{eqn:gaugeTxCurrent:120}
\begin{aligned}
\psi^\conj \BPi \cdot \BPi \psi – \psi \BPi^\conj \cdot \BPi^\conj \psi^\conj
&=
\psi^\conj
\lr{
-\Hbar^2 \spacegrad^2 \psi + \frac{i \Hbar e}{c} \lr{ \BA \cdot \spacegrad \psi + \spacegrad \cdot (\BA \psi) }
} \\
\psi
\lr{
-\Hbar^2 \spacegrad^2 \psi^\conj – \frac{i \Hbar e}{c} \lr{ \BA \cdot \spacegrad \psi^\conj + \spacegrad \cdot (\BA \psi^\conj) }
} \\
&=
\frac{i \Hbar e}{c}
\lr{
\psi^\conj
+
\psi
}
\end{aligned}

The first term is recognized as a divergence

\label{eqn:gaugeTxCurrent:140}
\begin{aligned}
&=
+

\end{aligned}

The second term can also be factored into a divergence operation

\label{eqn:gaugeTxCurrent:160}
\begin{aligned}
\psi^\conj
+
\psi
&=
}
+\lr{
} \\
&= 2 \spacegrad \cdot \lr{ \BA \psi \psi^\conj } \\
\end{aligned}

Putting all the pieces back together we have

\label{eqn:gaugeTxCurrent:180}
\begin{aligned}
\PD{t}{\rho}
&=
\inv{2m i \Hbar} \lr{
\psi^\conj \BPi \cdot \BPi \psi
-\psi \BPi^\conj \cdot \BPi^\conj \psi^\conj
} \\
&=
\inv{2m i \Hbar} \lr{
-\Hbar^2
+ \frac{ i \Hbar e}{c} 2 \BA \psi \psi^\conj
} \\
&=
\lr{
\frac{i \Hbar}{2 m} \lr{ \psi^\conj \spacegrad \psi – \psi \spacegrad \psi^\conj }
+ \frac{e}{m c} \BA \psi \psi^\conj
}.
\end{aligned}

From \ref{eqn:gaugeTxCurrent:60}, the probability current must be

\label{eqn:gaugeTxCurrent:200}
\Bj
=
\frac{\Hbar}{2 i m} \lr{ \psi^\conj \spacegrad \psi – \psi \spacegrad \psi^\conj }
– \frac{e}{m c} \BA \psi \psi^\conj,

or
\label{eqn:gaugeTxCurrent:220}
\boxed{
\Bj
=
\frac{\Hbar}{m} \textrm{Im} \lr{ \psi^\conj \spacegrad \psi }
– \frac{e}{m c} \BA \psi \psi^\conj.
}

# References

[1] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.