Proof of the Gauge-invariant momentum operator

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:

$$\tilde{\boldsymbol{P}}=U\boldsymbol{P}U^{\dagger}=e^{iq\Lambda(\boldsymbol{R})/\hbar}\boldsymbol{P}e^{-iq\Lambda(\boldsymbol{R})/\hbar}.$$

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?

• @AccidentalFourierTransform It clearly is Einstein's cosmological constant... kidding, it is just some real function. Already edited the question, thank you. – 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.
• 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 $\boldsymbol{P}=-i\hbar\nabla$. – Saavestro Nov 27 '16 at 21:09
• 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 :-) – 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 $$UPU^+\psi=P\psi-{iq\over\hbar}(P\Lambda)\psi =P\psi+q(\nabla\Lambda)\psi$$
• 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. – Saavestro Nov 27 '16 at 21:02