In NRQM why are gauge equivalent electromagnetic potentials physically equivalent?

Question

In the NRQM'al theory of a charged particle, if we gauge transform the potentials using some arbitrary gauge function $\Lambda(\textbf{X},t)$ then Schrodinger's equation implies that the quantum state $|\psi\rangle$ changes to $$e^{ie\Lambda(\textbf{X},t)/\hbar c}|\psi\rangle$$ where $\textbf{X}$ is the position operator (so this is not a trivial numerical phase factor). I think if we want to establish that this is physically equivalent to $|\psi\rangle$ we need to show that the probability density is unchanged as well as expectation values. The former is trivial but not all expectation values are unchanged, specifically $\langle \textbf{P}\rangle$ will be different due to the fact that $\textbf{X}$ and $\textbf{P}$ do not commute. At this point, the usual explanation I've seen is that this expectation value is not observable so it's fine for it to change since it doesn't correspond to anything physical. Since all observable quantities supposedly don't change then the gauge transformed state is physically equivalent to the original one. My question is the following: are we appealing to experiment here (in telling us which quantities are observable and which are not) to get this conclusion? and is there a way to see that the two states are physically equivalent just from the theory?

In classical E&M the logic is more clear to me: the state of the system is described by its trajectory which from the Lorentz force law depends only on the electric and magnetic fields which are invariant under gauge transformations, so it seems that in classical E&M the fact that gauge equivalent potentials are physically equivalent just comes out of the theory.

Answer 1

I think you are strugling with the concept of observable.

are we appealing to experiment here (in telling us which quantities are observable and which are not) to get this conclusion?

Yes and no. A lot we know and use in NRQM comes from what we know about classical systems, through the Principle of Correspondence. For example, we know that gauge equivalent potentials are physically equivalent in classical E&M, so we would expect that it would remain so in NRQM. Considering an expectation value of a dynamical variable like $\langle \mathbf P \rangle$, it is reasonable to expect that it would not be observable, since its classical counterpart $\mathbf p$ is not observable. It is also reasonable to expect that, on average, it would transform like its classical counterpart, under gauge transform.

However, to be rigorous, we know what is observable through experience. The definition of an observable is something that could be measured, i.e., it is connected with our experience in a straightfoward way. If one finds a way to measure $\mathbf p$ or $\mathbf A$ in an experiment, it would become an observable by definition, and it would imply a scientific revolution, since the consensus is that they are not. Such revolution would show that the Newtonian Mechanics is incomplete, since it describes trajectories using fields, not potentials, and now there are observables that could not be described only through it.

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What is confusing about gauge transform in NRQM, is that the Schrodinger equation relies on the Hamiltonian, an object that depends on the potentials, not the fields, and then it changes under gauge transforms. It implies that the quantum state would change, with at first sight seems that implies all the things becoming gauge dependent. But then you note that the expectation values of the operators should transform like its classical counterparts, and expectation values are all we have access through experiment. On average, we observe what is observed classically, so there is no conflict with classical E&M, and we can rest.

Answer 2

1. Let us for simplicity use the Schrödinger position representation, i.e. $$ \hat{p}_{\mu} ~=~\frac{\hbar}{i}\partial_{\mu} , \qquad \hat{x}^j~=~x^j. \tag{1}$$

2. A sandwich observables $\langle \psi | \hat{O}| \psi \rangle$ should be gauge invariant, i.e.
   $$\text{a bra }\langle \psi | , \qquad \text{ an operator }\hat{O} , \qquad \text{a ket }| \psi \rangle, \tag{2}$$ should gauge-transform in the $$\text{antifundamental}, \qquad \text{adjoint}, \qquad \text{fundamental}, \tag{3}$$ representation, i.e. as$^1$ $$ \langle \psi |\to \langle \psi |\hat{U}^{-1} , \qquad \hat{O}\to\hat{U}\hat{O}\hat{U}^{-1}, \qquad | \psi \rangle\to\hat{U}| \psi \rangle, \tag{4}$$ respectively.

3. Gauge-covariant operators should be build using e.g. gauge-covariant derivatives$^2$ $$D_{\mu}=\partial_{\mu}-\frac{i}{\hbar}qA_{\mu}.\tag{5}$$ Let us now address some of OP's questions: The canonical momentum operator $\hat{p}_{\mu}$ corresponds to a partial derivative $\partial_{\mu}$, and is hence not gauge-covariant. Similar with the gauge potential operator $\hat{A}_{\mu}$.

4. The gauge-covariance of the Schrödinger equation is maintained$^1$ since the Schrödinger equation is on a gauge-covariant form: $$\hat{O}| \psi \rangle~=~0,\tag{6}$$ cf. e.g. this Phys.SE post.

   This fact along with Wigner's theorem serve as important motivations for the transformation rules (4), cf. OP's question.

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$^1$ By the way, Galilean covariance is maintained via similar transformation rules (4), cf. e.g. this and links therein.

$^2$ By the way, the gauge-covariant derivative $D_{\mu}$ corresponds to the kinetic momentum operator, cf. e.g. my Phys.SE answer here.

Answer 3

Since all observable quantities supposedly don't change then the gauge transformed state is physically equivalent to the original one.

The usual thinking (or at least, how I think about this) is different: since we're making a gauge transformation, which is just a mathematical change in representation of EM field and of quantum state, then all predictions must remain unchanged. We're assuming the physical situation does not change, and we're just switching to another equivalent mathematical description.

My question is the following: are we appealing to experiment here (in telling us which quantities are observable and which are not) to get this conclusion?

No, it's just how the theory works when changing gauges.

It's like in classical theory with EM potentials, Hamilton's function $H$ and the Hamilton principal function $W$ (in the Hamilton-Jacobi equation): we can change the EM potentials by some gauge transformation, which changes also Hamilton's function and Hamilton's principal function, but this leaves the EM field and also the trajectory the system goes through unchanged (because that's what gauge transformation is).

Similarly in QT, when we do such change of potentials, then like the principal function, also the Hamiltonian $\hat{H}$ and the function $\psi$ changes correspondingly, but the EM field and all predictions of the theory remain the same.

The quantum-theoretical operator $\hat{\mathbf P}$ is usually introduced in the context of canonical quantization, and it is associated with canonical momentum $\mathbf P$ in classical theory. In classical theory, in Hamiltonian description of a system of one or more particles in external EM field, canonical momentum of a particle is a gauge-variant variable - its value depends on the choice of potentials $\varphi, \mathbf A$. The "actual" momentum $m\mathbf V$ (sometimes called kinetic momentum $\pi$) of the particle in classical theory is, in terms of canonical momentum, and vector potential $\mathbf A$,

$$ \pi = \mathbf P - q\mathbf A(\mathbf R, t). $$

In non-relativistic quantum theory, where we describe the external EM field as a classical field, the corresponding operator for kinetic momentum is $$ \hat{\pi} = \hat{\mathbf P} - q\mathbf A(\hat{\mathbf R}, t). $$

This QT relation is not really derived from anything, it is "motivated" by classical theory, and guessed. One could formulate description of momentum and EM field in a different way. But people just tried the simplest replacement of classical variables by operators and it kind of worked (with limitations, of course).