Electron EM Field as Local Gauge Variation of Electron Matter Field

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CSnowden
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Given the bound EM field associated with any higher energy state of the electron matter field (ie. electron), does the local gauge symmetry relationship between the matter and EM fields (local phase variance of matter field requiring EM field for local gauge symmetry) imply there is a corresponding phase change occurring in the matter field.
The EM field seems to be required for for local gauge symmetry of the electron matter field under local phase variation. Following is a description (not my verbiage):

There is a symmetry in physics which we might call the Local Phase Symmetry in quantum mechanics. In this symmetry we change the phase of the (electron) wavefunction by a different amount everywhere in spacetime. To compensate for this change, we need to also make a gauge transformation of the electromagnetic potentials. The local phase symmetry requires that Electromagnetism exist and have a gauge symmetry so that we can keep the Schrödinger Equation invariant under this phase transformation.

Does the fact that there is a local EM field associated with any higher energy state of the electron matter field (ie. electron, QFT fields coupled) imply that there is a corresponding phase shift in the electron matter field?
 

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  • #3
gentzen
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Does the fact that there is a local EM field associated with any higher energy state of the electron matter field (ie. electron, QFT fields coupled) imply that there is a corresponding phase shift in the electron matter field?
It depends on the exact details of what you want to know. Gauge freedom in the equations arises quite naturally in phenomenological descriptions of local near-equilibrium states, see for example page 5 and 7 of Phenomenological thermodynamics in a nutshell (arXiv:1404.5273) for an easy to understand instance of that phenomenon.

Nevertheless, you could try to eliminate complex numbers and their phase from your equations. Will that help? I once wrote that:
you may think that this discussion is all academic, because there is no reasonable natural formulation of QM without complex numbers anyway. But things are not necessarily so clear cut. See my (rather long) comment on the formulation of QM in section “2.1 The Ehrenfest picture of quantum mechanics” via (6), (7), and (8) in A. Neumaier’s Foundations of quantum physics II. The thermal interpretation:
… appreciate the beauty and depth of (6), (7), and (8). That beauty goes deeper than my parenthetical remark. Note for example that those equations don’t contain i. (How could they, given that they unify classical and quantum mechanics?)

After appreciating the beauty, the difficult next step would be to understand why this is still not enough. Even so i no longer occurs explicitly in the equations, and all beables are real valued, complex numbers will continue to play a key role behind the scenes. How physical are those real valued beables? I once wrote: “I admit that it is often easier to compute with the vector potential instead of the actual fields. But the actual fields are measurable, at least in principle, while the vector potential is not.” Are those real valued beables more like the actual fields than like the vector potential? They still share properties with the vector potential, i.e. some gauge freedom is still left. Only the complex phase which explained the gauge freedom is more hidden now.
 
  • #5
CSnowden
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I found a closely related question with extensive discussion titled

How do electrons couple to gauge field?​

, of which I found The Duck's post on that thread very enlightening.
 
  • #6
CSnowden
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Its perhaps best to simplify/restrict my question to QED as the electron is the topic of interest. It seems (from The Duck’s clear explanation in the thread above) that there are terms in the QED Lagrangian which link the electron matter field to the vector potential A (describing the EM field) in ways that vary the phase of the matter field and associated phase of the EM field in response to higher energy states.
 
  • #7
vanhees71
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The idea of a gauge symmetry is that you make a "global symmetry" local. In the case here you take the Dirac equation of a free particle,
$$(\mathrm{i} \gamma^{\mu} \partial_{\mu}-m) \psi=0.$$
I use natural units with ##\hbar=c=1##. Physical quantities are represented as functions/functionals of the sesquilinear forms, ##\bar{\psi} \Gamma \psi##, where ##\Gamma## is some expression formed by the Dirac matrices (like ##\gamma^{\mu}## itself or ##\gamma_5=\mathrm{i} \gamma^0 \gamma^1 \gamma^2 \gamma^3##, etc.); also you have ##\bar{\psi}=\psi^{\dagger} \gamma^0##.

This implies that the physics is invariant under the global transformations,
$$\psi(x) \rightarrow \psi'(x)=\exp(\mathrm{i} \alpha) \psi(x).$$
"Global" means that ##\alpha=\text{const}##, i.e., independent of the spacetime vector ##x##.

You can show, using Noether's theorem, that as a result the "current"
$$j^{\mu} = \bar{\psi} \gamma^{\mu} \psi$$
is conserved, i.e.,
$$\partial_{\mu} j^{\mu}=0$$
for the solutions of the free Dirac equation (it's also easy to confirm this directly by using the equation).

If you now want to make the transformation also a symmetry if ##\alpha \rightarrow \alpha(x)##, i.e., if you make the phase factor a function of ##x## and thus the symmetry a "local symmetry", you see that you have to introduce a "gauge field" ##A^{\mu}## and define a "gauge-covariant derivative",
$$\mathrm{D}_{\mu} = \partial_{\mu} + \mathrm{i} e A_{\mu}.$$
It is constructed such that
$$\mathrm{D}_{\mu}' \psi'(x)=\mathrm{D}_{\mu}' \exp(\mathrm{i} \alpha(x)) \psi(x) = \exp(\mathrm{i} \alpha(x)) [\partial_{\mu} +\mathrm{i} e A_{\mu}' +\mathrm{i} \partial_{\mu} \alpha] \psi(x) = \exp(\mathrm{i} \alpha(x)) \mathrm{D}_{\mu} \psi(x) = \exp(\mathrm{i} \alpha(x)) (\partial_{\mu} + \mathrm{i} e A_{\mu}).$$
This implies that
$$e A_{\mu}' = e A_{\mu} -\mathrm{i} \partial_{\mu} \alpha,$$
i.e., ##A_{\mu}'## is a gauge-transformed four-potential for an electromagnetic field, and the coupling you get from "minimal substitution", i.e., writing ##\mathrm{D}_{\mu}## instead of ##\partial_{\mu}## in the Dirac equation, is the coupling of a charged particle to the electromagnetic field. The electromagnetic current turns out to be
$$j^{\mu} =e \bar{\psi} \gamma^{\mu} \psi.$$
 
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