DrDu said:
This is about the paper by Greiter:
https://arxiv.org/pdf/cond-mat/0503400.pdf
Greiter argues that local electromagnetic gauge symmetry cannot change the state of a quantum system. On the other hand, in QED charge or particle conservation (if energy is too low to produce particle-antiparticle pairs) is the result of a global U(1) phase symmetry. This symmetry is broken and the states are not invariant under this symmetry. The latter symmetry is sometimes called a global gauge symmetry, which apparently gives rise to my confusion. So we have a "real" global symmetry, which transforms physically different states into each other, and a "local" gauge theory, which does not transform states. The confusion, Greiter is referring to, hence is probably due to Weinberg calling a global phase transformation also a gauge transformation, which is misleading, as the global transformation is a real symmetry operation?
That's perfectly right!
Gauge symmetry is not a symmetry in the sense of Noether's first theorem, which refers to global symmetries only. Gauge symmetry is rather a local symmetry and is treated by Noether's second theorem. Physically a gauge symmetry refers to the case that you describe a physical system (here charged particles and the electromagnetic field) with quantities that are not uniquely determined by the physical equations of motion (here the electromagnetic potentials, ##\Phi## and ##\vec{A}##).
Now we consider semiclassical quantum mechanics, i.e., a quantum theory, which describe non-relativistic charged particles (electrons) and the electromagnetic field as a classical "external" field. One should now note that only gauge-invariant quantities can be physical, i.e., observable quantities.
The most important question now is, what this means mathematically. For simplicity let's discuss a single non-relativistic scalar particle in a classical relativistic field (this is a lot of physics covered btw: You can describe not too heavy atoms, molecules and condensed matter consisting thereof very well with it, i.e., whenever typical energies become not close to the masses of the particles and typiscal velocities not too close to ##c##).
That's described in first quantization in the position representation by the Hamiltonian
$$\hat{H}=\frac{1}{2m} [\hat{p}-q \vec{A}(t,\hat{\vec{x}})]^2 + q \Phi(t,\hat{\vec{x}})$$
with ##\hat{p}=-\mathrm{i} \vec{\nabla}## and ##\hat{\vec{x}}=\vec{x}##, acting on the wave functions ##\psi(t,\vec{x})##.
The physical situation cannot change under gauge transformations of the electromagnetic potentials,
$$\Phi'(t,\vec{x})=\Phi(t,\vec{x})+\partial_t \chi(t,\vec{x}), \quad \vec{A}'(t,\vec{x})=\vec{A}(t,\vec{x})-\vec{\nabla} \chi(t,\vec{x}),$$
for an arbitray scalar field ##\chi(t,\vec{x})##. In the following I skip the arguments ##(t,\vec{x})## in the fields.
Now the first question is, how to transform the wave function under such a gauge transformation. Since the physics must not change, the state described by the wave function must not change, i.e., there must be a unitary transformation applied to the wave function as well as all the operators representing observables. We thus have to find (a) this unitary transformation and (b) an criterion for the operators that describe observables, i.e., the "gauge-invariance condition" on these operators.
For the wave function it's pretty simple to guess that the unitary transformation can be realized by a multiplication with a phase factor, which is a particularly simple case of a unitary transformation. So we guess that
$$\psi'=\exp(\mathrm{i} \alpha \chi) \psi$$
with ##\alpha=\text{const}## may do the job. To figure out, what's ##\alpha## we look at the Schrödinger equation,
$$\mathrm{i} \partial_t \psi=\hat{H}(\Phi,\vec{A}) \psi.$$
In order that this describes the same physics we must have
$$\mathrm{i} \partial_t \psi'=\hat{H}(\Phi',\vec{A}') \psi'.$$
To analyze this, it's convenient to write
$$\hat{H}_0(\Phi,\vec{A})=\frac{1}{2m} (-\mathrm{i} \vec{\nabla}-q \vec{A})^2:=\frac{1}{2m} (-\mathrm{i} \mathrm{D}_{\vec{x}})^2, \quad \mathrm{D}_{\vec{x}}=\vec{\nabla} + \mathrm{i} q \vec{A}$$
and write the Schrödinger equation as
$$(\mathrm{i} \partial_t - q \Phi) \psi=\mathrm{i} \mathrm{D}_t \psi=\hat{H}_0(\Phi,\vec{A}) \psi, \quad \mathrm{D}_t=\partial_t + \mathrm{i} q \Phi.$$
Now we have
$$\mathrm{D}_t'\psi'=(\partial_t + \mathrm{i} q \Phi') \exp(\mathrm{i} \alpha \chi) \psi = (\partial_t + \mathrm{i} q \Phi + \mathrm{i} q \partial_t \chi) \exp(\mathrm{i} \alpha \chi) \psi = \exp(\mathrm{i} \alpha \chi) (\partial_t + \mathrm{i} q \Phi + \mathrm{i} q \partial_t \chi + \mathrm{i} \alpha \partial_t \chi) \psi.$$
Now setting ##\alpha=-q## gives
$$\mathrm{D}_t' \psi'=\exp(-\mathrm{i} q) \mathrm{D}_t \psi.$$
In the same way you find
$$\mathrm{D}_{\vec{x}}' \psi' =(\vec{\nabla} -\mathrm{i} q \vec{A} +\mathrm{i} q \vec{\nabla} \chi) \exp(-\mathrm{i} q \chi) \psi=\exp(-\mathrm{i} q \chi) [\vec{\nabla} - \mathrm{i} q \vec{A} + \mathrm{i} q \vec{\nabla} \chi - \mathrm{i} q \vec{\nabla} \chi) \psi=\exp(-\mathrm{i} q \chi) \mathrm{D}_{\vec{x}} \psi.$$
This implies that
$$\mathrm{i} \mathrm{D}_t' \psi' = \exp(-\mathrm{i} q \chi) \mathrm{D}_t \psi=\hat{H}_0(\vec{A}',\Phi') \psi'=\exp(-\mathrm{i}q \chi) \hat{H}_0(\vec{A},\Phi) \psi.$$
This indeed shows that when setting
$$\psi'=\exp(-\mathrm{i} q \chi) \psi,$$
the Schrödinger equation describes the time evolution of the same state, when using ##\hat{H}(\vec{A},\Phi)## for the time evolution of ##\psi## and ##\hat{H}(\vec{A}',\Phi')## for the time evolution of ##\psi'##. The physics is in the position-probability density ##|\psi|=|\psi'|##, i.e., this is indeed gauge invariant, as it should be.
Now an operator ##\hat{A}=\hat{A}(\hat{\vec{x}},\hat{\vec{p}},\vec{A},\Phi)## describing an observable must obey
$$\hat{A}'=\hat{A}(\hat{\vec{x}},\hat{\vec{p}},\vec{A}',\Phi')=\exp(-\mathrm{i} q \chi) \hat{A} \exp(+\mathrm{i} q \chi). \qquad (*)$$
Not that, e.g., ##\hat{H}_0## is an observable in this sense but not ##\hat{H}##. Indeed
$$\hat{H}'=\hat{H}_0'+q \Phi'=\exp(-\mathrm{i} \chi) \hat{H}_0 \exp(\mathrm{i} \chi) + q \Phi + q \partial_t \chi' = \exp(-\mathrm{i} \chi \hat{H} \exp(\mathrm{i} \chi)+q \partial_t \chi,$$
i.e., ##\hat{H}'## has in addition to the unitary transformation of ##\hat{H}## the extra term ##q \partial_t \chi##.
Also ##\hat{\vec{p}}=-\mathrm{i} \vec{\nabla}## is not an observable. It represents the canonical momentum of the Hamiltonian formalism, and that's not gauge invariant. Indeed the corresponding classical mechanics description of a point particle in an electromagnetic field leads to ##\vec{p}=m \dot{\vec{x}}+q \vec{A}## which is not gauge invariant, however the "kinetic momentum" ##\vec{\pi}=m\dot{\vec{x}}=\vec{p}-q \vec{A}## of course is gauge invariant. It's easy to check that also quantum mechanically
$$\hat{\vec{\pi}}=m \mathring{\hat{\vec{x}}}=m [\hat{x},\hat{H}]=\hat{\vec{p}}-q \hat{A}=-\mathrm{i} D_{\vec{x}} $$
is an observable. That's why the operators ##\mathrm{D}_t## and ##\mathrm{D}_{\vec{x}}##, the "gauge-covariant" space-time derivatives are so useful in treating the quantum mechanical description of a charged particle in a given external em. field, which has to be described in terms of its potentials in QM, rather then the also gauge-covariant electric and magnetic fields ##\vec{E}## and ##\vec{B}## since QM is based on the Hamiltonian formalism.
Now it's clear that any true observable, which must invariant under gauge transformations, must be represented by a gauge-covariant self-adjoint operator, which is defined by the transformation property (*). On the so constraint self-adjoint operators a gauge transformation in QT is a unitary transformation, which keeps the physics invariant under gauge transformations.
