In classical physics, the vector potential $\mathbf A(\mathbf r)$ is defined by its relation to the magnetic field $\mathbf B(\mathbf r) $ via the equation $\mathbf B = \nabla \times \mathbf A$. In describing the quantum mechanics of a charged particle moving in a fixed, classical field, for example in Landau level calculations, $\mathbf A(\mathbf r) \to \mathbf A(\hat {\mathbf{r}})$, and the momentum operator $\hat{\mathbf p}$ looks in position space like a gradient, $\hat{\mathbf p} = -i \hbar \nabla$. One can define an operator $$\hat{\mathbf{B}} = (i \,\hat{\mathbf{p}} \times \mathbf{A} (\hat{\mathbf{r}} ) - i \, \mathbf A (\hat{\mathbf r}) \times \hat{\mathbf p})/2\hbar ,$$ which formally looks like a "magnetic field operator," but is there any sense in doing this? And what would be the interpretation?

  • $\begingroup$ I'm not sure what the question is. That indeed is the magnetic field operator... As an example, the Dirac equation coupled to EM fields depends only on the four-potential, but its nonrelativistic limit is a Pauli Hamiltonian with a $\mu \cdot B$ term, for the magnetic dipole moment. Where do you think the $B$ comes from? It arises from a structure much like what you wrote down. $\endgroup$
    – knzhou
    Aug 10 at 0:33
  • $\begingroup$ I guess I'm curious if there is more intuition for what this means. For instance, naively this says that the magnetic field felt by a particle depends on that particle's momentum. Somehow the dependence on the position operator feels more intuitive to me. $\endgroup$
    – Munthe
    Aug 10 at 0:39
  • $\begingroup$ Another way to say it is, in classical physics we're used to thinking of A and B equally as being fields defined as functions of position. What does it mean for the eigenstates of this quantum analogue of B to not be position eigenstates? $\endgroup$
    – Munthe
    Aug 10 at 0:42

1 Answer 1


Operator $\mathbf B$ expressed in terms of operators $\mathbf A$ and $\mathbf p$ does not mean magnetic field felt by particle depends on the value of its momentum.

Just as in classical EM theory $\mathbf B = \nabla \times \mathbf A$ does not mean magnetic field depends on the value of $\nabla$.

In both cases, $\nabla$ or $\mathbf p$ is an operator, a rule/algorithm, not a number with a value. The expression involving them is another operator acting on $\psi$ functions which we can use to get value of $\mathbf B$ or probability distribution for $\mathbf B$.

$\mathbf p$ is not even the kinetic momentum operator $\boldsymbol \pi$ (the actual physical momentum), it is the so-called canonical momentum, shifted from $\boldsymbol \pi$ by the vector potential:

$$ \mathbf p = -i \hbar \nabla_{\mathbf r} = \boldsymbol \pi+q\mathbf A(\mathbf r). $$

It is normal for operators to be expressible using other operators. For example, kinetic momentum can be in some scenarios expressed as constant times $\mathbf rH_0 - H_0\mathbf r$. This does not mean physical or canonical momentum value depends on the value of Hamiltonian and position. The latter two quantities are not even compatible (do not have simultaneous values in QT), since their operators do not commute.

Now, if we had an operator that was a product of two commuting operators, e.g. for the hydrogen atom, the operator

$$ \hat{G} = \hat{L}_z \hat{H}_0, $$ then it would make sense to say the value of $G$ is product of values of $L_z$ and $H_0$, because the operators all commute and have common eigenfunctions.

  • $\begingroup$ Still thinking this over, especially your point about $[r,H_0]$, but I'll point out right away that you could use the physical momentum instead of the canonical momentum in my formula and nothing would change, since $A \times A = 0$. So that probably isn't part of the resolution. But interesting to think in terms of commutators! $\endgroup$
    – Munthe
    Aug 10 at 16:22
  • $\begingroup$ Right. I've added further comment on commuting variables. $\endgroup$ Aug 10 at 18:05

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