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

Given the schrodinger 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.

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.