Consider a particle in two dimensions with an external magnetic field in the $z$-direction. The vector potential can be chosen to be $$\mathbf{A}=-By\ \hat{x}$$ so that the Hamiltonian given by $$H=\frac{(p_x+eBy)^2}{2m}+\frac{p_y^2}{2m}.$$ The system has rotational symmetry. This makes sense since you have a constant field in $z$-direction, and the rotating thing will not change anything. To see prove this, We need to prove that $$[H,L_z]=0.$$

It's easy to show with little algebra that $$[H,L_z]\not = 0.$$ The reason is that our chosen gauge breaks the rotational symmetry. If we instead choose, $$\vec{A}=(-By/2,Bx/2,0),$$ we can restore this symmetry.

Why does the symmetry of the system depend on the gauge we choose? Since, Physically, of course, Any consequence following the symmetry should be followed. Furthermore, What is the gauge in which there is a translation symmetry? Can we find the gauge which have both of these symmetries?

With a couple of hints and references, It looks like $\Pi$ is the right thing to look at since it gauges invariant but still, I'm not finding $[H,\Pi]=0$.

  • $\begingroup$ I don't have time to work this out, but presumably $[H,L_z]$ can be written as a gauge transformation. $\endgroup$
    – Toffomat
    Feb 14 at 10:30
  • $\begingroup$ @Toffomat Can you suggest any reference where I can look? $\endgroup$ Feb 14 at 11:10
  • $\begingroup$ I guess that should be in most QFT books. The key point is that symmetry is alway up to gauge trafos. $\endgroup$
    – Toffomat
    Feb 14 at 11:16
  • $\begingroup$ The short answer to OP's conceptional question is that the notion of energy/Hamiltonian depends on the choice of gauge. $\endgroup$
    – Qmechanic
    Feb 14 at 11:49
  • $\begingroup$ Possibly related $\endgroup$ Feb 14 at 14:23

2 Answers 2


After hints on comments, I came up with this answer: It looks like symmetry depends on the gauge but it's not. The usual translation operator $p$ doesn't do the job because as shown in this answer, it's not invariant under gauge transformation. Yet the operator that does invariant is $$\Pi=p-eA$$ Further note that in the symmetric gauge where $$A=\frac{1}{2}r\times B$$ usual momentum operator $p$ is the generator of the translation operator. Under gauge transformation $$A\rightarrow A+\nabla \Lambda\\ p\rightarrow p+e\nabla \Lambda \\ \Pi\rightarrow \Pi=p-eA$$

It's clear that the operator $$\xi=\Pi+\frac{e}{2}r\times B$$ has a property that in symmetric gauge, it's simply momentum operator and in other gauges, it doesn't change its form. Therefore we expect $\xi$ to generate translation. It's not hard to show that it actually does.

Similarly for generator of rotation $$\zeta = r\times \Pi+\frac{e}{2} r\times (r\times B)$$

It's easy to verify that we did for the generator of translation.


The symmetries don't vanish, they move to the canonical ones

$\mathbf{\hat p^\textrm{can}}= \mathbf {\hat{p}}-e\mathbf A$

$\mathbf{\hat L}^\textrm{can}_z=\mathbf {\hat r} \times \mathbf{\hat p^\textrm{can}} = \mathbf {\hat r}\times (\mathbf {\hat{p}}-e\mathbf A)$

With these new operators $[\hat H, \mathbf{\hat p^\textrm{can}}]=0$ and $[\hat H, \mathbf{\hat L}^\textrm{can}_z]=0$

  • 1
    $\begingroup$ This is incorrect. The canonical momentums do not commute with each other, so they do not commute with $H$. $\endgroup$
    – Meng Cheng
    Feb 14 at 16:39

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