In my quantum-optics lecture, my Professor wanted to derive some quantities regarding the time evolution of an electromagnetic field, coupled to an electron. He started with the usual Hamiltonian in the Schroedinger Picture: $$ (\hat{\vec{p}}-\frac{e}{c} \hat{\vec{A}}(0))^2\frac{1}{2m} + V(\hat{\vec{x}}) + \sum_{\vec{k}, \lambda} \hat{a}_{\vec{k}, \lambda}^\dagger \hat{a}_{\vec{k}, \lambda} $$ All the operators are being time-independent schroedinger-operators there. In order to simplify calculations, he then applied two Unitary transformations, one being $\hat{U} = e^{\frac{e}{\hbar}\hat{i\vec{x}}\hat{\vec{E}}}$, the other one being $\hat{T}(t) = D(-\{\alpha_{\vec{k}}e^{-i \omega_{\vec{k}}t} \})$ with D being the displacement operator for multiple modes. Applying those transformations, the time depence of states changes to: $$ i \hbar \frac{d}{dt} | \Psi(t) \rangle = \frac{1}{2m}\hat{p}^2 + V(\hat{\vec{x}}) - e \hat{\vec{x}}(\hat{\vec{E}}(0) + \vec{E}_\mathrm{classical}(0, t) ) | \Psi(t) \rangle $$

Of course I can solve the time evolution of a given state with that, but what bothers me is (and what my question is):

what does this state mean? Since I did 2 transformations, and I want to know for example the position of the electron, the operator for that is no longer $\hat{\vec{x}}$, but instead $\hat{U}^{-1} \hat{T}^{-1} \hat{\vec{x}} \hat{U}^{-1} \hat{T}^{-1}$ Is it true to say something like that? Furthermore, what meaning to eigenstates of $\hat{\vec{x}}$ have, since it is no longer $\hat{\vec{x}}$, measuring position, but instead $\hat{U}^{-1} \hat{T}^{-1} \hat{\vec{x}} \hat{U}^{-1} \hat{T}^{-1}$?

  • $\begingroup$ You were presumably missing an $i$ in the definition of \hat U$, but please check the sign of the corrected version. $\endgroup$ Oct 4, 2017 at 0:20

1 Answer 1


Yes, the meaning of operators can change when you do a gauge transformation like this one.

On the other hand, you are fortunate in that the specific example that worries you does not actually change: here $\hat x$ commutes with $\hat U = e^{i \hat x \cdot \hat F}$, since it commutes with itself and with the field operator $\hat F$, and it commutes with $\hat T$ since that only depends on the field. That means, therefore, that $$\hat{U}^{-1} \hat{T}^{-1} \hat{{x}} \hat{U}^{-1} \hat{T}^{-1}=\hat x,$$ since the unitaries cancel out, and you don't need to worry about any of that.

That's a very specific example, though, and in general things do change quite dramatically, and there's a nontrivial number of papers out there in the literature that are wrong (and wrong enough to be useless) because the authors didn't pay enough attention to these issues.

The core operators that change with this transformation are the momentum and the velocity, and one needs to be extremely careful with how those are handled, because as you've noticed, the momentum does change in the transformation: $$\hat{U}^{-1} \hat{{p}} \hat{U}^{-1} =\hat p + \frac{e}{c}A.$$ (I'm ignoring the vector nature of these quantities, which is trivial to put back in. I'm also ignoring the quantized nature of the field, but if you do include that then the change in $\hat p$ is no longer a constant, it is an entangling operator with the field sector.)

In short:

  • In the velocity gauge (i.e. your original hamiltonian), $\hat p$ is the canonical momentum, and $\hat p_\mathrm{kin} = \hat p - \frac ec A$ is the kinematic momentum, which itself is related to the velocity through $\hat p_\mathrm{kin} = m\hat v$.
  • On the other hand, in the length gauge (i.e. your transformed hamiltonian), the kinematic momentum is still related to the velocity via $\hat p_\mathrm{kin} = m\hat v$, but the canonical momentum $\hat p$ has changed, and it now coincides with $\hat p_\mathrm{kin}$.
  • In both frames, it is the canonical momentum that acts as the gradient, $\langle x|\hat p = -i\hbar \frac{\partial}{\partial x}\langle x|$, on the position representation.

As you've noticed, the canonical momentum changes with a gauge transformation, i.e. it is not gauge invariant and it is not physically measurable. (If that alarms you, good. Go and sit down to process that until it no longer does.) Only the kinematic momentum is physically measurable.

So... how does one manage to keep one's head in such a maze of changes? Well, for one, it pays to not panic and work out exactly what the operators transform to. In most real-world cases there's plenty you can say about what an operator transformation $\hat U^\dagger \hat O\hat U$ will come out to, in explicit terms, and you just have to sit down and work them until they simplify. And, once you do, it is crucial to keep in mind what physical quantity, exactly, each operator represents in the different gauges. And once you have that, it's actually relatively easy.

Finally, an absolutely crucial reminder: none of this is unique to quantum mechanics, and none of this can credibly be classed as 'mysterious quantum mumbo-jumbo'. All of these issues $-$ gauge transformations, changes in the physical meaning of quantities, the different relationships between kinematic and canonical momentum, and the non-physicality of the latter until you fix a gauge $-$ have identical versions in hamiltonian mechanics. If the quantum version looks confusing, and you don't have a clear picture of what the question maps into in hamiltonian mechanics and how it's resolved there, then that's the first port of call in resolving those confusions.


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