The Gauge invariant momentum operator is said to be $\tilde{\boldsymbol{P}}=\boldsymbol{P}-q\nabla\Lambda(\boldsymbol{R})$, where $\boldsymbol{R}$ is the position operator and $\Lambda(\boldsymbol{R})$ is a real function.

The given unitary transformation is $U=e^{iq\Lambda(\boldsymbol{R})/\hbar}$. So, in order to show the form of $\tilde{\boldsymbol{P}}$ I need to calculate:


I think I can proceed using a Taylor expansion of $U$, so:

$$\begin{align}\tilde{\boldsymbol{P}}&=\bigg(1+\frac{i}{\hbar}q\Lambda(\boldsymbol{R})+\ldots\bigg)\boldsymbol{P}\bigg(1-\frac{i}{\hbar}q\Lambda(\boldsymbol{R})+\ldots\bigg)\\ &=\boldsymbol{P}-\frac{i}{\hbar}q\boldsymbol{P}\Lambda(\boldsymbol{R})+\frac{i}{\hbar}q\Lambda(\boldsymbol{R})\boldsymbol{P}+\frac{q^2}{\hbar^2}\Lambda(\boldsymbol{R})\boldsymbol{P}\Lambda(\boldsymbol{R})+\ldots\\ &\approx \boldsymbol{P}-\frac{i}{\hbar}q\boldsymbol{P}\Lambda(\boldsymbol{R})+\frac{i}{\hbar}q\Lambda(\boldsymbol{R})\boldsymbol{P}, \end{align} $$

where, in the last approximation I have used that $q\ll1$. If I instroduce $\boldsymbol{P}=-i\hbar\nabla$ in the last step, I obtain the desired form of $\tilde{\boldsymbol{P}}$ from the first two terms, but the last term is additional and is equivalent to $\Lambda(\boldsymbol{R})\nabla$.

Why am I obtaining this additional term? Maybe I'm just not following the right procedure. Any Suggestions?

  • $\begingroup$ @AccidentalFourierTransform It clearly is Einstein's cosmological constant... kidding, it is just some real function. Already edited the question, thank you. $\endgroup$ – Saavestro Nov 27 '16 at 20:10

If $\Lambda$ is a function of the position operator, you can't just treat $\Lambda$ as a scalar function. In fact, what you are obtaining is $\tilde{\mathbf{P}} = \mathbf{P} - i q [\mathbf{P},\Lambda(\mathbf{R})]$.

For analytic functions you can always expand $\Lambda$ as a power series in $\mathbf{R}$ and compute the commutator with $\mathbf{P}$ for each power of $\mathbf{R}$ using the canonical commutation relations and the product rule for operators. You will obtain $[\mathbf{P},\Lambda(\mathbf{R})] = - i \Lambda'(\mathbf{R})$. Using this you get the result you were looking for.

  • $\begingroup$ For commutators involving functions of operators, see for instance here physics.stackexchange.com/q/98372 $\endgroup$ – Faser Nov 27 '16 at 20:36
  • $\begingroup$ That makes sense to me. But I think that the commutater must be $[\boldsymbol{P},\Lambda(\boldsymbol{R})]=-i\hbar\nabla\Lambda(\boldsymbol{R})$? with the corresponding $\hbar$ from $\boldsymbol{P}=-i\hbar\nabla$. $\endgroup$ – Saavestro Nov 27 '16 at 21:09
  • $\begingroup$ Yes! that is what I wrote. I have the habit to set $\hbar =1$. Writing $\Lambda'$ instead of $\nabla \Lambda$ is just a matter of notation :-) $\endgroup$ – Faser Nov 27 '16 at 21:10

You should remember that these operators act on wavefunctions: $$UPU^+\psi=P\psi-{i\over\hbar}P\big(q\Lambda\psi\big) +{i\over\hbar}q\Lambda P\psi$$ to first order in $q\Lambda$. Note that in the second term in the r.h.s., $P$ acts on $\Lambda\psi$ and not only on $\Lambda$. Using the fact that $$P\big(\Lambda\psi\big)=(P\Lambda)\psi+\Lambda\psi$$ The last term of the first equation is therefore cancelled by the last term of the second relation. It follows that $$UPU^+\psi=P\psi-{iq\over\hbar}(P\Lambda)\psi =P\psi+q(\nabla\Lambda)\psi$$

  • $\begingroup$ 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. $\endgroup$ – Saavestro Nov 27 '16 at 21:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.