4
$\begingroup$

In the quantization of electromagnetic field the physical states $|\psi\rangle$ are found to obey the following relation:

$[a^{(0)}(k)-a^{(3)}(k)]|\psi\rangle=0$

It is explained as the physical states are admixtures of longitudinal and timelike photons. What do longitudinal and timelike photons physically mean? Why the polarizations, $\epsilon^{(0)}$ and $\epsilon^{(3)}$, timelike and longitudinal photons, are called unphysical?

$\endgroup$
3
$\begingroup$

When you change the free field $A_\mu$ by means of a gauge transformation, you can easily see that it affects longitudinal and timelike degrees of feedom. Since observables are gauge invariant, those degrees of freedom cannot be physical.

$\endgroup$
  • $\begingroup$ Do we always deal with a free field? How about Coulomb interaction? $\endgroup$ – Vladimir Kalitvianski Nov 17 '14 at 9:39
  • $\begingroup$ Obviously not. I just explained why the name "unphysical states". It refers to the free quantized electromagnetic field. The one used in covariant perturbation theory in particular, which includes interaction but it is asymptotically pictured as a free field in scattering states. Those states are for instance handled by the so called Gupta-Bleuler formalism. $\endgroup$ – Valter Moretti Nov 17 '14 at 10:27
  • $\begingroup$ If it were about free field, then longitudinal and timelike photons would not come in this admixture, but separately. $\endgroup$ – Vladimir Kalitvianski Nov 17 '14 at 11:18
  • $\begingroup$ I think you should have a look to the Gupta-Bleuler formalism where these mixtures appear for the free field and all the problem is how to get rid of them. I do not intend to discuss further this point since it is standard literature. Sec. 3.2.1 Itzykson Zuber. Bye, Valter $\endgroup$ – Valter Moretti Nov 17 '14 at 11:51
  • $\begingroup$ The "Lorenz gauge" is applied not only to free field. $\endgroup$ – Vladimir Kalitvianski Nov 17 '14 at 12:51
1
$\begingroup$

The total field consists of the "near" field like the Coulomb one and more generally (and loosely) a retarded Coulomb field, which are always "attached" to the charge, and the photon (radiated) field with different polarization orientations. The near field is always present, its "photons" are not created and annihilated. The corresponding "photons", when introduced for "symmetry", should not modify the existing near field. That is why the creation/annihilation of them is restricted in some way.

The "near field" exists between two charges. It means their "virtual photons" are absorbed in the Feynman diagrams. The real photons are emitted or scattered, not completely absorbed.

$\endgroup$
0
$\begingroup$

It seems to me that longitudinal photons are not unphysical. They are responsible for the Coulomb interaction between charged particles.

$\endgroup$
  • 1
    $\begingroup$ Yes, but only virtual photons with $E^2 \ne p^2c^2$ can have longitudinal polarization. Real, physical photons always have transverse polarization and follow light-like paths through spacetime. $\endgroup$ – rob Jun 11 '14 at 1:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.