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.


2 Answers 2


Actually, Schroedinger equation $$ -i\hbar \partial_t \psi+ \left[ -\frac{1}{2m}\left(\frac{\hbar}{i}\nabla-q\vec{A}\right)^2+qV \right]\psi=0\tag{0}$$ under the gauge transformations

$$ \psi \rightarrow \psi'= e^{iq\Lambda/\hbar} \psi$$ $$ \vec{A} \rightarrow \vec{A}' = \vec{A} + \nabla \Lambda$$ $$ V \rightarrow V'= V - \partial_t\Lambda$$

does not remain invariant, but the left-hand side of (0) gives rise to $$ -i\hbar \partial_t \psi'+ \left[ -\frac{1}{2m}\left(\frac{\hbar}{i}\nabla-q\vec{A}'\right)^2+qV' \right]\psi'= e^{iq\Lambda/\hbar}\left\{-i\hbar \partial_t \psi+ \left[ -\frac{1}{2m}\left(\frac{\hbar}{i}\nabla-q\vec{A}\right)^2+qV \right]\psi\right\}\:.$$

In summary, since $e^{iq\Lambda/\hbar}\neq 0$,

$\qquad\quad$ gauge transformed quantities satisfy Schroedinger equation if untranformed quantities do.

To prove it, avoid brute force computations as yours which give rise to unavoidable mistakes almost certainly and go on as follows. First rewrite the initial equation as $$ -\left[i\hbar \partial_t -qV \right]\psi -\frac{1}{2m}\left(\frac{\hbar}{i}\nabla-q\vec{A}\right)^2\psi=0\tag{1}$$ Next notice that, under the transformations, we have

$$\left[i\hbar \partial_t -qV' \right]\psi' = \left[i\hbar \partial_t -q(V - \partial_t\Lambda)\right]e^{iq\Lambda/\hbar}\psi = e^{iq\Lambda/\hbar}\left[i\hbar \partial_t -qV \right]\psi $$ and $$\left(\frac{\hbar}{i}\nabla-q\vec{A}'\right)\psi' = \left(\frac{\hbar}{i}\nabla-q(\vec{A} +\nabla \Lambda)\right)e^{iq\Lambda/\hbar}\psi = e^{iq\Lambda/\hbar}\left(\frac{\hbar}{i}\nabla-q\vec{A}\right)\psi$$ so that, iterating the second result $$\left(\frac{\hbar}{i}\nabla-q\vec{A}'\right)^2\psi' = e^{iq\Lambda/\hbar}\left(\frac{\hbar}{i}\nabla-q\vec{A}\right)^2\psi\:.$$ Putting all together, under the action of gauge transformations, (1) becomes $$ -\left[i\hbar \partial_t -qV' \right]\psi' -\frac{1}{2m}\left(\frac{\hbar}{i}\nabla-q\vec{A}'\right)^2\psi' = e^{iq\Lambda/\hbar}\left\{-\left[i\hbar \partial_t -qV \right]\psi -\frac{1}{2m}\left(\frac{\hbar}{i}\nabla-q\vec{A}\right)^2\psi\right\}=0$$ as wanted.


Check this video by the great Barton Zwiebach from MIT lectures.

He does exactle what you are doing in a 20 min video, if the link stop working, it is Lecture L14.1 from the MIT course 8.06 Quantum Physics III, Spring 2018, which you will always find on Youtube or in the MIT lectures web.

  • 1
    $\begingroup$ While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. - From Review $\endgroup$
    – Miyase
    Jun 7, 2022 at 12:26
  • $\begingroup$ Sorry, I wanted to write it as a comment in the other answers, which is essentially what the video does, but I don't have enough reputation yet :(. I'lll make a superquick summary in the answer, whenever I can. $\endgroup$ Jun 7, 2022 at 13:21

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