What is the polarization of a virtual photon

In summary, virtual photons in the context of QED are not real particles but mathematical constructs used in perturbative calculations. Depending on the gauge, the photon propagator can have both longitudinal and transverse components. However, in physical calculations, only the two 3-transverse field-degrees of freedom contribute, as the unphysical degrees of freedom cancel out due to gauge invariance. In non-Abelian gauge theories, this cancellation also involves Faddeev-Popov ghosts. Therefore, it is reasonable to consider only one polarization component of the virtual photon in a 1D wave guide or cavity, as the physical degrees of freedom are the only ones that contribute to physical calculations. This information can be found in Richard
  • #1
maxdil
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I read in few places that the virtual photons have longitudinal polarization while in some papers they say virtual photons have both transverse and longitudinal polarizations.

Can anyone please help me to understand this
 
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  • #2
Virtual photons are not real photons. They are just mathematical building blocks in the perturbative evaluation of S-matrix elements. The internal lines of Feynman diagrams do not represent photons or particles but the propagator of the quantum field (in the vacuum case the socalled Feynman propagator of perturbation theory, i.e., the vacuum expectation value of the time-ordered connected two-point field-correlation function).

Depending on the gauge the propgator can have four-longitudinal components. E.g., in Feynman gauge, which is the most convenient one for perturbative calculations, Feynman propagator for photons reads
$$\Delta_{\mu \nu}(p)=-\frac{g^{\mu \nu}}{p^2 + \mathrm{i} 0^+}.$$
There you seemingly have four components.

In Landau gauge your photon propagator is made four-transverse
$$\Delta_{\mu \nu} (p)=-\left ( g^{\mu \nu} - \frac{p^{\mu} p^{\nu}}{p^2 + \mathrm{i} 0^+} \right) \frac{1}{p^2+\mathrm{i} 0^+}.$$
Here you seemingly have 3 components (two 3-transverse and one 3-longitudinal).

These would-be degrees of freedom are not all observable, because you can only make sense of asymptotic free states in terms of observable objects, and here gauge invariance comes to the rescue! For on-shell S-matrix elements finally only the two physical 3-transverse field-degrees of freedom of the quantized electromagnetic field contribute. The unphysical degrees of freedom in the one or the other gauge are cancelling out. In the Abelian case as is QED, current conservation is necessary and sufficient for this cancellation, which is formally encoded in the so-called Ward-Takahashi identities for the proper vertex functions and the connected Green's functions. In non-Abelian gauge theories you need also Faddeev-Popov ghosts (and for Higgsed models also the would-be Goldstone modes in non-unitary gauges) to make this cancellation happen. Here, the Slavnov-Taylor identities substitute the Ward-Takahashi identies of the Abelian case. An exception is the socalled Background-Field Gauge, where the simple Ward-Takahashi identities become available again, and that simplifies the proof of the (perturbative) renormalizability of non-Abelian gauge theories considerably.
 
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  • #3
maxdil said:
I read in few places...

Where exactly? You will get better and more helpful answers if you provide the source of the statements that you need help with (and providing sources is a PhysicsForums rule).
 
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  • #4
Thank for the replies.

I found about photon polarization in Richard Feynman's book "The Theory of Fundamental Processes" chapter 20. It provides the fact that, coulombic interactions are due to the non-transverse polarization components. However I am trying to modify the EM field in a 1D cavity/1D wave guide based on QED and wondering whether it is reasonable to consider only one (longitudinal) polarization of the virtual photon.
 
  • #5
I guess that's in the Coulomb gauge. As I said, at the end you are left with the physical degrees of freedom!
 
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  • #6
vanhees71 said:
I guess that's in the Coulomb gauge. As I said, at the end you are left with the physical degrees of freedom!
@vanhees71: "at the end you are left with the physical degrees of freedom" ,so does that mean we can consider one polarization component of the virtual photon in 1D wave guide/1D cavity? Thanks
 

Related to What is the polarization of a virtual photon

1. What is the definition of polarization in terms of virtual photons?

Polarization refers to the direction of oscillation of a virtual photon's electric field. It can be thought of as the orientation of the photon's spin.

2. How does the polarization of a virtual photon differ from a real photon?

A virtual photon does not actually exist in the physical sense, but rather represents a mathematical calculation in quantum field theory. As such, it does not have a measurable polarization like a real photon does.

3. Can a virtual photon have multiple polarizations at the same time?

Yes, a virtual photon can have multiple polarizations at the same time due to the uncertainty principle in quantum mechanics. This is known as superposition.

4. What is the significance of polarization in the study of quantum mechanics?

Polarization is an important concept in quantum mechanics because it helps us understand the behavior of particles and their interactions with each other. It also plays a crucial role in many practical applications, such as in quantum computing and telecommunications.

5. How is the polarization of a virtual photon determined?

The polarization of a virtual photon is determined by the properties of the particles and fields it interacts with in a given situation. These interactions are described by mathematical equations in quantum field theory, which can be used to calculate the polarization of the virtual photon.

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