I am trying to proof explicitly that Schrodinger equation: $$ i\hbar \partial_t \psi = \left[ -\frac{1}{2m}\left(\frac{\hbar}{i}\nabla-q\vec{A}\right)^2+qV \right]\psi$$
remains the same under the following gauge transformation:
$$ \psi \rightarrow e^{iq\Lambda/\hbar} \psi$$ $$ \vec{A} \rightarrow \vec{A} + \nabla \Lambda$$ $$ V \rightarrow V - \partial_t\Lambda$$
where $\partial_t$ stands for the time derivative operator.
However, I am having problems with the algebra, so I will show my procedure with the hopes that someone point to an error:
Left side of equation $$ i\hbar \partial_t (e^{iq\Lambda/\hbar}\psi) = ih \left(e^{iq\Lambda/\hbar} \partial_t\psi+\frac{iq}{\hbar}e^{iq\Lambda/\hbar} \psi \partial_t\Lambda \right) = ihe^{iq\Lambda/\hbar} \partial_t\psi-qe^{iq\Lambda/\hbar} \psi \partial_t\Lambda $$
Right side of equation $$ \left[ \frac{1}{2m}\left(\frac{\hbar}{i}\nabla-q(\vec{A} + \nabla \Lambda)\right)^2+q(V - \partial_t\Lambda) \right]= $$ $$ \frac{1}{2m} \left[ -\hbar^2\nabla^2-\frac{q\hbar}{i}( \nabla \cdot\vec{A} + \nabla^2\Lambda+ \vec{A} \cdot \nabla + \nabla \Lambda \cdot \nabla) + q^2[\vec{A}^2+2(\vec{A}\cdot \nabla \Lambda) + (\nabla \Lambda )^2]\right] e^{iq\Lambda/\hbar} \psi +qV e^{iq\Lambda/\hbar} \psi -qe^{iq\Lambda/\hbar} \psi \partial_t\Lambda $$
It is possible to observe that the last term in both (the right and left) sides cancel each other. Then, using:
$\nabla ( e^{iq\Lambda/\hbar} \psi ) = e^{iq\Lambda/\hbar}\nabla\psi + \frac{iq}{h} \psi \nabla \Lambda$
$ \nabla^2 ( e^{iq\Lambda/\hbar} \psi ) =e^{iq\Lambda/\hbar} \nabla^2\psi + \frac{2iq}{\hbar}e^{iq\Lambda/\hbar}(\nabla \Lambda)(\nabla \psi) + \psi \frac{iq}{\hbar} e^{iq\Lambda/\hbar} \nabla^2 \Lambda - \frac{q^2}{\hbar^2}\psi e^{iq\Lambda/\hbar} (\nabla \Lambda)^2 $
we then obtain (by applying operators and canceling all the $e^{iq\Lambda/\hbar}$ ):
$$ i\hbar \partial_t \psi= \frac{1}{2m} \left[ -\hbar^2 \nabla^2 \psi - 2iqh(\nabla \Lambda)(\nabla \psi)-iq\hbar \psi \nabla^2\Lambda + q^2\psi(\nabla \Lambda)^2 + iq\hbar (\nabla \cdot \vec{A})\psi + iq\hbar\nabla^2\Lambda \psi +iq\hbar (\vec{A}\cdot \nabla\psi) - q^2 \psi (\vec{A}\cdot \nabla \Lambda )+iq\hbar (\nabla \Lambda)(\nabla \psi) - q^2\psi (\nabla\Lambda)^2+q^2\vec{A}^2+2q^2(\vec{A}\cdot \nabla \Lambda)\psi +q^2(\nabla \Lambda)^2 \psi \right] + qV\psi$$
cancelling some terms, and rearranging:
$$ i\hbar \partial_t \psi= \frac{1}{2m} \left[ -\hbar^2 \nabla^2 \psi + iq\hbar (\nabla \cdot \vec{A})\psi +iq\hbar (\vec{A}\cdot \nabla\psi)+q^2\vec{A}^2 - 2iqh(\nabla \Lambda)(\nabla \psi) + q^2\psi(\nabla \Lambda)^2- q^2 \psi (\vec{A}\cdot \nabla \Lambda )+iq\hbar (\nabla \Lambda)(\nabla \psi) +2q^2(\vec{A}\cdot \nabla \Lambda)\psi \right] + qV\psi $$
after more reordering:
$$ i\hbar \partial_t \psi= \frac{1}{2m} \left[ \left(\frac{\hbar}{i}\nabla-q\vec{A}\right)^2 \right] +qV\psi + \frac{1}{2m} \left[ - iqh(\nabla \Lambda)(\nabla \psi) + q^2\psi(\nabla \Lambda)^2 + q^2(\vec{A}\cdot \nabla \Lambda)\psi \right]$$
It is possible to observe that the original schrodinger equation is up there, but with an extra part in the right side, this extra part is: $$ \frac{1}{2m} \left[ - iqh(\nabla \Lambda)(\nabla \psi) + q^2\psi(\nabla \Lambda)^2 + q^2(\vec{A}\cdot \nabla \Lambda)\psi \right]$$
So am wondering, is this extra part some how 0, or am I making a mistake. Also I don't know how to make the algebra "nicer" to follow, if there is anything I can do please comment.
