Is it possible a virtual photon is massive particle?

[ndung200790]

Please teach me this problem:
The classical field of real photon obey Maxwell equations,so the speed of light is c=1,therefore real photon is massless particle.But I think virtual photon maybe not obey Maxwell equations because virtual particle is the fluctuation of vacuum.So maybe virtual photon is massive particle,so that the state of virtual photon has complete four polarizations.
Please give me a favour to teach me about that.
Thank you very much for advanced.

 

[ismaili]

Please teach me this problem:
The classical field of real photon obey Maxwell equations,so the speed of light is c=1,therefore real photon is massless particle.But I think virtual photon maybe not obey Maxwell equations because virtual particle is the fluctuation of vacuum.So maybe virtual photon is massive particle,so that the state of virtual photon has complete four polarizations.
Please give me a favour to teach me about that.
Thank you very much for advanced.

You are right, virtual photons can be massive.
Consider the simplest tree diagram for e^+e^- \rightarrow \mu^-\mu^+,
where the positron and electron combine to form a virtual photon and then this virtual photon becomes \mu^-, \mu^+.

Since we have energy-momentum conservation at each vertex, and momentum of external electron and positron are decided, this means the momentum of this virtual photon is fixed. That is, this virtual photon could be "off shell", i.e. it doesn't satisfy the equation of motion.

As for the polarisations, or degrees of freedom for a massive vector field is seemingly four. But if my memory serves, the degrees of freedom should be three. However, I couldn't come up with the reason why the number of DoF is reduced to 3.

Any comments?

 

[ndung200790]

Thank Mr.Ismaili very much,
If virtual photon is 3-polarization,so it there must be virtual photon with spin Sz=0.Therefore
I don't understand why there isn't the process of innihilation of righthanded electron and positron or lefthanded electron and positron(They have oposite spins).
Please explain it for me.I am grateful for your kindness.

 

[haael]

As for the polarisations, or degrees of freedom for a massive vector field is seemingly four.

No, it's just 3. It's always 2j+1.

 

[tom.stoer]

The number of physical polarizations for massless photons is 2, for massive photons it's 3. It may seem strange that even for a virtual photon (for which m²=0 no longer holds) 2 remains valid; the argument goes as follows:

Constructing the Hamiltonian of QED one has two eliminate unphysical degrees of freedom. First one uses the gauge symmetry to set A°=0 which is an unphysical degree of freedom as its canonical conjugate momentum is zero: there is no time derivative of A° in the Lagrangian, therefore it acts as a Lagrange multiplier. The generated constraint in the A°=0 gauge is the Gauss law div E = 0 (in vacuum), which eliminates another polarization.

So in the physical Hilbert space one has two polarizations for the photon.

The elimination of one polarization via the A°=0 gauge does no longer work for massive photons b/c gauge invariance is explicitly broken by the mass term. But A° is still an unphysical Lagrange multiplier which generates a (modified) Gauss law, so one polarization can be eliminated.

 

[tom.stoer]

One must be careful. If the photon mass is contained in the Lagrangian it's 3; if one constructs "virtual" photons for a Lagrangian with massless photons it's 2

 

[ismaili]

One must be careful. If the photon mass is contained in the Lagrangian it's 3; if one constructs "virtual" photons for a Lagrangian with massless photons it's 2

oh, that's the point that I posted a question lately, but I'm still not very sure about it. :shy:

Though we started from a Lagrangian with massless photon,
we could have some virtual photons in a Feynman diagram.
Such a virtual particle is not really a particle state but just some artificial particle,
it may enable the calculation and make the physical interpretation easy.

But, if we calculate the amplitude for a process with virtual photon,
we could see that the A_{0} component actually contributes to the amplitude too,
as well as other components A_{i}.

So this means that not only two polarizations of these artificial virtual particles involved in the physical process?

 

[ndung200790]

I have read Mandl&Shaw and I know that because gauge condition, the formalism traversed polarization and 4-polarisation are equivalent!

 

[tom.stoer]

Usually Feynman rules in QED are formulated in the covariant Lorentz gauge. So whether A° does contribute or not is not a gauge invariant statement.

I explained the reduction of degrees of freedom in the A°=0 gauge as A° is explicitly unphysical and should be removed in order to make things transparent. In other gauges the number of "effective" degrees of freedom is again 2, but this need not be visible directly.

As far as I remember the photons itself may carry more than 2 polarizations, but the propagator turns out to contain a projector which eliminates one degree of freedom.

 

[ndung200790]

The scalar and longitudinal photons contribute to instantanious Coulomb potention.This one could be excluded(A0).

 

[ndung200790]

It seem that there not exist massive photon despite it is virtual photon because in Lagrangian it doesn't distingush real and virtual photon.Seeming that field of virtual photon also obey Maxwell equations,so that they must be massless.

 

[tom.stoer]

It's a bit trickier than that. Quantizing a field and using perturbation theory based on Feynman diagrams (only then it makes sense to talk about virtual particles) one finds expressions containing internal photon lines which we used to call virtual photons

\int d^4k_1\ldots d^4k_i \prod_v\delta^{(4)}\left(\sum_mk^\mu_m\right)_v \ldots \frac{C}{k_i^2+i\epsilon}\ldots

where I omitted all fermionic contributions.

i labels all internal phootn lines (the virtual photons);
v labels all vertices (at each vertec we have 4-momentum conservation which is formally written as a delta function where the sum over m runs over all lines (momenta) at the vertex.

In addition I omitted the regularization of this integral.

Now one sees that the integration variables (momenta) are not restricted to mass shell which would read

\int d^4k\,\delta(k^2)\ldots

The latter expression is used in canonical quantization when expressing field operators in terms of creation and annihilation operators in covariant form via Fourier transformation.

So one sees that we somehow compare apples with oranges:

1) in canonical quantization one can show that the physical Hilbert space carries only two physical polarizations; unphysical polarizations can be elimininated via gauge fixing explicitly (this works even in QCD where Fadeev-Popov ghosts and/or BRST is more familiar but makes conting of degrees of freedom ore complicated); these "photons" are creation and annhililation operators and m²=0 always holds

2) in Feynman diagrams what we call "virtual photons" are internal lines for which the momentum integration is not restricted to m²=0

But the relation between the field operators and the propagators is rather indirect; so our discussion here is - to a large extend - about the confusion regarding different concepts and interpretation what "physical photons" and "virtual photons" "are". The math itself is free of these problems.

 

[ndung200790]

It seem that the off-shell of virtual particles implying the uncertainty of energy-momentum in quantum mechanics.So it also express the failling of relativity theory in quantum fraimwork!
Please give me a favour to teach me again!
Thanks very much.

 

[tom.stoer]

It seem that the off-shell of virtual particles implying the uncertainty of energy-momentum in quantum mechanics.So it also express the failling of relativity theory in quantum fraimwork!

That's misleading.

Quantization of gravity does not fail "obviously", it fails technically when you want to formulate quantum gravity in the perturbatively - but this is far from obvious; it may very well be that there is a non-perturbative definition of quantum gravity which is directly based on general relativity (the approach is called asymptotic safety); in addition it could be that supergravity makes sense even in the perturbative sense.