# Transformation of Wave function under gauge transformation

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?

• 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. Commented Jun 29, 2020 at 16:02
• You shouldn't have to make any assumptions about $f$ Commented Jun 29, 2020 at 16:03

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.