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I am trying to obtain the transformed wavefunction $\psi$ under gauge transformation in the presence of the E-M field. So Schrödinger's equation is (in the units $c=1$ and $\hbar$ = 1)

$$i\frac{d\psi}{dt} = H\psi \text{,} \qquad \text{where } H=\frac{(p -qA)^2}{2m} + q\phi.$$

consider the gauge transformation

$$A\rightarrow A + \nabla f, \qquad \qquad \phi\rightarrow\phi - \frac{df}{dt}.$$

To conserve probability $\psi$ must change only via a phase, I put $\psi\rightarrow\alpha$ $\psi$, where modulus square of $\alpha$ is 1. Inserting all those transformations in the Schrodinger's equation gives me (After some simplifications)

$$i\frac{d\alpha}{dt} = -q\alpha\frac{df}{dt} + \alpha (q\nabla f)^2 - \alpha\frac{(p-qA)(q\nabla f)}{m}.$$

Now, How do I proceed further to find $\alpha$, it's evident that I get the standard phase factor, i.e $\alpha$ = $e^{ifq}$ if only all the space-dependent terms vanish, i.e all $\nabla f$ terms vanish, but that doesn't seem right, also it suggests that the transformed $\psi\rightarrow e^{iqf}\psi$ only works for specific cases of function f.

Any suggestions here?

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    $\begingroup$ You're correct, this gauge transformation will give you the gauge transformed wavefunction $e^{iqf}\psi$ without any assumptions about terms vanishing. Remember how $p$ is written as a derivative in position space to simply further. $\endgroup$
    – 4xion
    Commented Jun 29, 2020 at 16:02
  • $\begingroup$ You shouldn't have to make any assumptions about $f$ $\endgroup$
    – 4xion
    Commented Jun 29, 2020 at 16:03

1 Answer 1

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Under a Gauge transformation you have:

$\phi \rightarrow \phi - \frac{\partial f}{\partial t}$

$\vec A \rightarrow \vec A + \nabla f$

And we want to prove that this implements a phase transformation Schrödinger's Equation.

$\psi \rightarrow e^{iqf}\psi$

The time derivative term transforms as:

$i\frac{\partial \psi}{\partial t} \rightarrow -qe^{iqf}\frac{\partial f}{\partial t}\psi +i e^{iqf}\frac{\partial \psi}{\partial t}$

The first term of this cancels with the change in $\phi$. Now the spatial pieces, for this we need to evaluate:

$(p-qA-q\nabla f)(p-qA -q\nabla f)e^{iqf}\psi$

To evaluate this you need to know that $[G(x),p]=i\nabla G$, to commute the exponential factor with each of the parenthesis (I do assume here that both $A$ and $f$ depend only on coordinates and not on momentum):

$(p-qA -q\nabla f)e^{iqf}=e^{iqf}(p-qA)$

Commuting with the parenthesis one by one you can see that the spatial term transforms as:

$\frac{(p-qA)^2}{2m}\psi\rightarrow e^{iqf}\frac{(p-qA)^2}{2m}\psi$

Which means that Schrödinger's Equations is unchanged.

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