Gauge Invariance of the Schrodinger Equation

[Diracobama2181]

TL;DR

Find in a quick way to prove gauge invariance without entailing a ton of messy math

Given the Schrödinger equation of the form $$-i\hbar\frac{\partial \psi}{\partial t}=-\frac{1}{2m}(-i\hbar \nabla -\frac{q}{c}A)^2+q\phi$$
I can plug in the transformations $$A'=A-\nabla \lambda$$ , $$\phi'=\phi-\frac{\partial \lambda}{\partial t}$$, $$\psi'=e^{-\frac{iq\lambda}{\hbar c}}\psi$$
$$-i\hbar\frac{\partial \psi'}{\partial t}=(-\frac{1}{2m}(-i\hbar \nabla -\frac{q}{c}A+\frac{q}{c}\nabla \lambda)^2+q\phi-q\frac{\partial \lambda}{\partial t})\psi'$$.

Now when act on $$\psi'$$ on the right hand side, I come across the term $$(-i\hbar \nabla -\frac{q}{c}A+\frac{q}{c}\nabla \lambda)^2\psi'$$
I now that $$(-i\hbar \nabla -\frac{q}{c}A+\frac{q}{c}\nabla \lambda)\psi'=e^{-\frac{iq\lambda}{\hbar c}}(-i\hbar \nabla -\frac{q}{c}A)\psi$$, but can I just say that $$(-i\hbar \nabla -\frac{q}{c}A+\frac{q}{c}\nabla \lambda)^2\psi'=e^{-\frac{iq\lambda}{\hbar c}}(-i\hbar \nabla -\frac{q}{c}A)^2\psi$$, and if so, why? (In other words, is there a reason why I could simply do this twice rather than multiplying everything out, which gets messy). Thanks.

 

[Christopher Grayce]

Why not? Your exponential factor is just a number, and it commutes with every operator inside the parentheses.

 

[vanhees71]

The exponential is not just a number, because $\lambda=\lambda(t,\vec{x})$. The problem with #1 is the gauge transformation of the em. field and several sign mistakes. The SGE reads (with $\hbar=c=1$)
$$\mathrm{i} \partial_t \psi = -\frac{1}{2m} (-\mathrm{i} \vec{\nabla}-q \vec{A})^2 \psi + q \phi \psi.$$
Now you make
$$\psi'=\exp(\mathrm{i} q \lambda), \quad \vec{A}'=\vec{A}+\vec{\nabla} \lambda, \quad \phi=\phi-\partial_t \lambda.$$
Then $\psi'$ with the potentials $\phi'$ and $\vec{A}'$ fullfills the same SGE as $\psi$ with the potentials $\phi$ and $\vec{A}$, i.e., the physics is invariant under gauge transformations, because $\psi'$ differs from $\psi$ only by a phase factor and the potentials only by a gauge transformation, which doesn't change the physical fields $\vec{E}$ and $\vec{B}$.

 

[Diracobama2181]

I know it is gauge invariant. I suppose it is not clear the issue I am having. I know it is the case that
$$(-i\hbar \nabla -\frac{q}{c}A+\frac{q}{c}\nabla \lambda)\psi'=e^{-\frac{iq\lambda}{\hbar c}}(-i\hbar \nabla -\frac{q}{c}A)\psi$$.
I want to know if it is trivially true that $$(-i\hbar \nabla -\frac{q}{c}A+\frac{q}{c}\nabla \lambda)^2\psi'=e^{-\frac{iq\lambda}{\hbar c}}(-i\hbar \nabla -\frac{q}{c}A)^2\psi$$. In other words could I simply just say that the operator $$(-i\hbar \nabla -\frac{q}{c}A+\frac{q}{c}\nabla \lambda)$$ acts on $$\psi'$$ twice., or do I need to show that $$(-i\hbar \nabla -\frac{q}{c}A+\frac{q}{c}\nabla \lambda)e^{-\frac{iq\lambda}{\hbar c}}(-i\hbar \nabla -\frac{q}{c}A)\psi=e^{-\frac{iq\lambda}{\hbar c}}(-i\hbar \nabla -\frac{q}{c}A)^2\psi$$?.

 

[Dr.AbeNikIanEdL]

If you understand why the first equation in your post #4 holds, it should be obvious that also the last equation holds. I am not really sure what you want to shortcut here...

 

[vanhees71]

You don't need to do anything more to "short cut". If the 1st. equation in #4 holds for all (sic) functions $\psi$, then the 2nd equation follows immediately without any further calculation.

 

[Diracobama2181]

Perhaps it is a silly question. I was just wondering why it held. Thank you.

 

[Dr.AbeNikIanEdL]

Well, did you go through the logic that leads to the first equation? Can you apply the same reasoning with $\Phi = (-i\hbar\nabla - q/c A)\Psi$ as a the function?

 

[Diracobama2181]

So $$(-i\hbar \nabla -\frac{q}{c}A+\frac{q}{c}\nabla \lambda)\psi'=(-i\hbar \nabla e^{-\frac{iq \lambda}{\hbar c}}\psi -\frac{q}{c}Ae^{-\frac{iq \lambda}{\hbar c}}\psi+\frac{q}{c}\nabla \lambda e^{-\frac{iq \lambda}{\hbar c}}\psi)=(-\frac{q}{c}\nabla\lambda e^{-\frac{iq \lambda}{\hbar c}} \psi-i\hbar \nabla\psi e^{-\frac{iq \lambda}{\hbar c}}-\frac{q}{c}Ae^{-\frac{iq \lambda}{\hbar c}}\psi+\frac{q}{c}\nabla \lambda e^{-\frac{iq \lambda}{\hbar c}}\psi)=e^{-\frac{iq \lambda}{\hbar c}}(-i\hbar \nabla -\frac{q}{c}A)\psi$$.

The issue I am having trouble understanding is why this would imply $$(-i\hbar \nabla -\frac{q}{c}A+\frac{q}{c}\nabla \lambda)^2\psi'=e^{-\frac{iq \lambda}{\hbar c}}(-i\hbar \nabla -\frac{q}{c}A)^2\psi$$. Wouldn't $$(-i\hbar \nabla -\frac{q}{c}A+\frac{q}{c}\nabla \lambda)$$ also act on $$(-i\hbar \nabla -\frac{q}{c}A)$$?

 

[Diracobama2181]

Wait. Just figured it out. Thank you.