It is known that canonical commutation relations do not fix the form of momentum operator. That means that if canonical commutation relations (CCR) are given by

$$[\hat{x}^i,\hat{p}_j]~=~i\hbar~\delta^i_j~ {\bf 1}$$

they can be satisfied by the following choice of momentum operators:

$p_x = -ih\frac{∂}{∂x}+\frac{∂f}{∂x}$

$p_y = -ih\frac{∂}{∂y}+\frac{∂f}{∂y}$

$p_z = -ih\frac{∂}{∂z}+\frac{∂f}{∂z}$

where $f(x,y,z)$ - arbitrary function.

On the other hand, for any choice of $f(x,y,z)$ momentum operators can be transformed to their most frequently used form $(-ih\frac{∂}{∂x})$ (etc for $y$ and $z$) by the following transformation of the wave function $\psi$ and operators $p$:


$p^{'}_x =e^{-\frac{i}{h}f(x,y,z)}p_x e^{+\frac{i}{h}f(x,y,z)}=-ih\frac{∂}{∂x}$

Hence, we obtain $U(1)$ gauge transformation using only canonical commutation relations for momentum and position operators.

Does this mean that $U(1)$ gauge invariance corresponds to conservation of momentum rather than to conservation of electric charge?


1 Answer 1


1) If we interpret OP's transformation$^1$ as a passive transformation, i.e. a mere change of coordinates/description that doesn't alter the system, then there is no conservation law.

2) So in the following, let us interpret OP's transformation as an active transformation.

For a system to have a symmetry, its action $S$ (or more precisely, in this quantum mechanical context, its Hamiltonian operator $\hat{H}$) must respect this symmetry. The corresponding conservation law would depend on the specific form of Hamiltonian $\hat{H}$. This is all we have to say about QM.

3) Finally, in a field theoretic context, we can interpret OP's transformation

$$\tag{1} \frac{\hbar}{i} {\bf \nabla} ~\longrightarrow~ e^{-i\Lambda({\rm r})}\frac{\hbar}{i} {\bf\nabla} e^{i\Lambda({\rm r})} $$

as a pure gauge transformation in an electromagnetic theory with zero electromagnetic field strength.

When interpreted as an EM theory, on one hand, global gauge symmetry leads to electric charge conservation, cf. Noether's first Theorem. See also this Phys.SE question.

On the other hand, there is no conserved quantity associated with local gauge symmetry per se, cf. Noether's second Theorem. (Its off-shell Noether identity is a triviality. See also this Phys.SE question.)


$^1$ OP's transformation is related to the choice of phase factors in overlaps between position and momentum eigenstates in QM, cf. this Phys.SE post.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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