Timeline for Proof of the Gauge-invariant momentum operator
Current License: CC BY-SA 3.0
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| Nov 27, 2016 at 21:02 | comment | added | rsaavedra | Of course! I don't know how I forgot considering that, the actual definition of an operator transformation is that $<\psi^{\prime}|\tilde{\boldsymbol{P}}|\psi^{\prime}>=<\psi|\boldsymbol{P}|\psi> $. So it is true that the action on a state ket must be considered. Using that fact I obtain a slightly different result from yours, with a minus sign in the transformation, as proposed by the problem. | |
| Nov 27, 2016 at 20:28 | history | answered | Christophe | CC BY-SA 3.0 |