Is the photon field a vector field and a gauge field?

[syfry]

TL;DR Summary

Need help understanding what something is saying, or how accurate it is. So that what I'm learning is consistent with knowledge in science.

The info at this link says the flowing:

The photon field is a quantum field theory. It is a vector field because it includes spin-1 photons. This field is not identical to the electromagnetic field although the two originate from the same Maxwell’s equations.

There are several levels of details in describing the photon field, each coming with a different level of accuracy of predicting outcomes of experiments. The semi-classical model is quantizing the electromagnetic field. It describes the electromagnetic field as a wave whose energy is incremented or decremented by one photon. This field is sufficiently accurate to analyze interactions between atoms and electromagnetic fields thus it is used for describing light absorption and emission and lasers. This field is not sufficiently accurate for describing high energy interactions.

The photon field of QFT is a gauge field. This is the more likely “photon field” discussed by physicists. It includes virtual photons that are not observable. It is based on the principle that electric charges can’t interact directly. In a gauge theory the force must be mediated by virtual photons or else the theory is not relativistically invariant, a property that is necessary for high energy interactions.

I'll quote and highlight the confusing parts in bold:

"The photon field is a quantum field theory. It is a vector field because it includes spin-1 photons."

"The photon field of QFT is a gauge field. This is the more likely “photon field” discussed by physicists."

So, is the photon field a vector field? Or instead a gauge field? Or, both? (didn't find any answers in searches online)

"There are several levels of details in describing the photon field, each coming with a different level of accuracy of predicting outcomes of experiments. The semi-classical model is quantizing the electromagnetic field"

Is the photon field 'semi classical'? Had thought the photon field is fully quantum.

Not sure how accurate some parts in the rest are accurate but wanna first start with interpreting the bolded parts.

 

[vanhees71]

This is pretty vague and thus it's no surprise that you have difficulties in understanding.

Before starting to consider relativistic quantum field theory you should get a clear understanding of classical relativistic field theory, i.e., here classical electrodynamics.

To understand the concepts best, it's also good the think about classical field theory from the point of view of the symmetries of special-relativistic space time. This helps to understand, why the relativistic field theories look the way they look. The idea is that any relativistic field theory must be compatible with the mathematical structure of Minkowski space and thus it's symmetries, i.e., the physical laws must be formulated such that they are covariant under Poincare transformations, i.e., space-time translations and Lorentz transformations (including rotations and boosts). In addition the field theory should fulfill causality conditions, i.e., given initial values for the fields the field equations should have unique solutions at later times and influences on the field at a later time $t$ must depend only on field values at times $t'<t$. Since this should hold in any inertial frames of reference, this implies that ther must be no faster-than-light signals.

Now the analysis of realizations of the Poincare group in field theories shows that one possible realization of a field theory is with "massless vector fields", and it turns out that such massless vector fields can only be realized as a gauge theory. It also turns out that the most simple realization of such a field theory ends up in electrodynamics with the four-potential field $A^{\mu}$ being the gauge field. The solutions of the equations for the initial-value problem for $A^{\mu}$ are not unique because of gauge invariance, and you can always choose some arbitrary "gauge-fixing constraint". Different choices of this constraints (e.g., Lorenz or Coulomb gauge, to name the most common choices) lead to different solutions for the $A^{\mu}$ all describing the same physics. This implies that the $A^{\mu}$ are not observable fields, but this is the gauge-invariant field-strength tensor $F_{\mu \nu}=\partial_{\mu} A_{\nu} -\partial_{\nu} A_{\mu}$, which in (1+3)-notation contains the electromagnetic field $(\vec{E},\vec{B})$, which is indeed observable (through it's action on charged matter).

For a introductory formulation of classical electrodynamics in covariant form, see

https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf

Since this manuscript is not yet complete, there's however not the detailed derivation from the theory of the Poincare group, but rather it's using the standard (1+3) formulation of electrodynamics to translate it into manifestly covariant form, but I think that's a good starting point without the advanced mathematical tools of Lie-group theory.

Only after you have gained a very clear understanding of classical electrodynamics you can start with QED, because quantizing a gauge theory is among the most subtle issues of relativistic QFT!

 

[PeterDonis]

The info at this link

The link is to a Quora discussion, not a textbook or a peer-reviewed paper or something equivalent. So you need to be careful about using it as a source.

"The photon field is a quantum field theory. It is a vector field because it includes spin-1 photons."

This is just a definition: "vector" in quantum field theory terms means "spin-1".

"The photon field of QFT is a gauge field.

This means that the QFT of the electromagnetic field is a gauge theory.

So, is the photon field a vector field? Or instead a gauge field? Or, both?

It's both. The two terms are just referring to two different properties of the field.

"There are several levels of details in describing the photon field, each coming with a different level of accuracy of predicting outcomes of experiments. The semi-classical model is quantizing the electromagnetic field"

I'm not sure what this part is talking about.

 

[syfry]

It's both. The two terms are just referring to two different properties of the field.

Thanks! My mind has trouble with technical wording so it's really helpful to be on the same page with accuracy. Makes proceeding easier even when I barely understand the material.

Thanks both of you for explaining more as well.

 

[vanhees71]

This is just a definition: "vector" in quantum field theory terms means "spin-1".

One must be careful though for massless representations of the Poincare group.

In short Wigner's idea to find the unitary representations of the Poincare group is to construct special bases of the Hilbert space. One starts with the realization of the space-time translations. The generators define energy and momentum and a Casimir operator of the Poincare group is $p^2=E^2-\vec{p}^2=m^2$.

Then an irreducible representation is characterized by the fact that the proper orthochronous Lorentz group must act transitively on the energy-momentum eigenstates for fixed $m^2$. This means you can reach all energy-momentum eigenvectors by acting with the proper orthochronous Lorentz transformations (generated by rotations and boosts) on a "standard momentum eigenstate".

For $m^2>0$ the most convenient choice of the "standard momentum" is $p=(m,0,0,0)$. The subgroup of the proper orthochronous Lorentz group that keeps this momentum invariant are obviously the rotations, the "little group" of the $m^2>0$-representionas, and thus the further characterization of the irreducible representation is how the little group acts on the momentum eigenstates with $\vec{p}^2=0$. This must be an irreducible representation of the covering group of the rotation group, SU(2), and the Casimir operator of this little group is the spin (i.e., the eigenvalue $s(s+1)$ of $\vec{s}^2$) with $s \in \{0,1/2,1,\ldots \}$. These are the well-known representations of the corresponding angular-momentum operators (though in relativistic physics "spin" is defined in this way only for the particles at rest). Once chosen this unitary representation of the little group you can lift this to the representation of the proper orthochronous Lorentz group by applying boosts which transfer the standard momentum $(m,0,0,0)$ to any other on-shell momentum $(\sqrt{\vec{p}^2+m^2},\vec{p})$ and define the basis of energy-momentum eigenvectors accordingly.

In addition you want to realize these representations in terms of transformations on local field operators, and then in addition you need the corresponding representations with negative frequencies, starting from $(-m,0,0,0)$ as the standard momentum. In order to have only positive energies you have to realize the corresponding decomposition of the local field operator with an annihilation operator in front of the positive-frequency modes and a creation operator in front of the negative-frequency modes. This leads to the necessity of the introduction of antiparticles. Together with the microcausality condition for local observable-operators it also leads to the spin-statistics realation (half-integer-spin representations represent necessarily fermions, integer-spin representations bosons).

For the massive case you have pretty similar properties of the particles as in non-relativstic QM, i.e., concerning the space-time symmetries elementary particles are uniquely defined by their mass (a Casimir operator in the relativistic case, a central charge in the non-relativistic case) and spin $s$ (the Casimir operator of the rotation group resp. its covering group, SU(2)).

This changes for the massless case, i.e., $m^2=0$. Here the standard momentum of course is light-like and thus usually chosen as $(\pm \kappa,0,0,\kappa)$ with an arbitrary $\kappa>0$. Now the little group is again generated by three independent generators. One is the rotation around the 3-axis, building an Abelian (!!!) subgroup of the little group. The other two are generators of null-rotations, i.e., those proper orthochronous Lorentz transformations that keep the standard momentum unchanged but are not rotations around the three-axis. It turns out that the so generated little group is isomorphic to ISO(2), i.e., the symmetry group of the Euclidean plane. The rotations around the 3-axis correspond to the rotations in this plane, and the translations to the null-rotations.

Now there seem to be no particles with a kind of continuous polarization observable, and thus these translations/null rotations should be acting trivially. This means the irreducible representation of the Poincare group is further characterized by the choice of the representations of the rotations around the 3-axis. Since this is an Abelian group, these representations must be one-dimensional, and the representation is further determined by the choice of the corresponding eigenvalue of $\hat{s}_3$, which on the first glance can be choosen as any real number, but now one must not forget that through the representation of the boosts to other arbitrary three-momenta one generates a representation of the full rotation group, and thus the eigenvalues of $\hat{s}_3$ must be $\lambda \in \{0,\pm 1/2,\pm 1,\ldots \}$. Thus the physical states for each possible $s=|\lambda| \geq 0$ are either characterized by $\lambda=+s$ or $\lambda=-s$. Since $\lambda$ is the projection of the spin in the direction of momentum, that's the helicity. For simplicity one calls $s$ "the spin" also in the case of massless representations, one must however keep in mind that for $s \geq 1/2$ there are only 2 "polarization degrees of freedom", which can be characterized by the helicity being $\pm s$.

If you now want to realize such a representation in terms of local field operators it turns out that for $s \geq 1$ you necessarily must formulate the theory as a gauge theory, such that the transformations generated by the null rotations on these field operators lead to a redundant solution of the equations of motion. This implies that the equations of motion do not fully determine the field operators, but only modulo to a gauge transformation. This has all the known implications for the quantization of gauge theory: the gauge fields (e.g., the em. potentials in the case $s=1$) are not representing observables but only such local operators that can be composed of these fields that are gauge invariant do (as in the em. field case the $F_{\mu \nu}$).

Otherwise the results for a local realization are as in the massive case, i.e., half-integer spin must be quantized as fermions, integer-spin as boson, and the CPT theorem holds too.

That the photon has both polarizations is due to the fact that the em. interaction is also invariant under spatial reflections, and to realize the orthochronous Poincare group, $\text{O}(1,3)^{\uparrow}$ you need to have both helicities $\pm 1$ and the Hilbert space is the orthogonal sum of the irreps. of $\text{SO}(1,3)^{\uparrow}$ of helicity +1 and that of helicity -1.

 

[PeterDonis]

Moderator's note: A subthread about information sources and physics teaching has been spun off into a separate thread:

https://www.physicsforums.com/threads/information-sources-and-physics-teaching.1058727/

 

[PeterDonis]

Moderator's note: Thread level changed to "I" since the discussion is beyond "B" level.