I like to tell people interested in light polarization that the photon is a vector boson for which the third spin axis, the one in the direction of travel, is suppressed due to photons being massless and traveling at c.

It's an argument that kind of makes sense for classical physics.

However, mulling over quantum physics makes me a bit more cautious. In QED for example, a photon that travels over a sufficiently short distance can have non-zero amplitudes for both superluminal and subluminal velocities.

But that leads to an interesting question: Does a photon with a non-zero amplitude for subluminal travel also have a non-zero amplitude for exhibiting a real and measurable $\pm$spin axis pointed along its direction of travel? That is, does such a photon include an amplitude to behave like a true three-axis vector boson? And if so, what kind of experiment might detect that amplitude?

Also, what does all of this imply for a superluminal photon amplitude? Does that amplitude imply the existence of some sort of mirror-image vector boson version of the photon? What would that even mean?

And finally: How does any of this relate to fully classical circularly polarized photons?

While circularly polarized photons have same symmetry as particles with spin along the direction of travel, they behave like fully classical particles that can travel indefinite distances. But by the same arguments I just made, that means they cannot possibly be true vector boson states, which would only exist as small amplitudes over very short distances! Are circularly polarized photons perhaps more accurately understood as some kind of spin-cancelling superposition of the subluminal and superluminal true vector boson states?

  • $\begingroup$ Related: physics.stackexchange.com/q/46643/2451 $\endgroup$ – Qmechanic Oct 30 '13 at 20:29
  • 2
    $\begingroup$ Qmechanic, as always I am highly appreciative and somewhat amazed at times at how good your cross references are. That's a great one. I didn't immediately see within it the answer for my question -- it's an intentionally odd combination of perspectives, I confess -- but I will enjoy reading those answers and notations, and will look for an answer there. $\endgroup$ – Terry Bollinger Oct 31 '13 at 1:13

The photon polarization directions are only transversal when it is free in space . The polarization of an interacting photon or a photon with nonfree boundary conditions is not transversal in general. One example is that electromagnetic waves possess longitudinal polarizations in waveguides.

Another relatively simple example where (a certain combination of) the transversal and the scalar polarizations becomes physical when the photon is a mediator of a Coulomb interaction. This can be seen as follows:

The covariant (off-shell) photon propagator without gauge fixing has the form:

$$D^{\mu\nu}(k) = \frac{1}{k^2}\epsilon^{\mu}_{\lambda}(\mathbf{k}) \epsilon^{\nu\lambda}(\mathbf{k}) = \frac{g^{\mu\nu}}{k^2}$$

Where the Minkowski summation convention is adopted:

$$a.b = a^{\lambda} b_{\lambda} = - a^0b^0+a^1b^1+a^2b^2 + a^3b^3$$

The four polarization vectors can be chosen to be orthonormal:

$$\epsilon_{\mu}^{\lambda}(\mathbf{k}) \epsilon_{\nu\lambda}(\mathbf{k}) = g^{\mu\nu}$$

Therefore three of them should be spacelike and one of them timelike.

Without loss of generality they can be chosen as:

The transversal polarization vectors

$$ \epsilon_r^{\mu}= [0, \mathbf{\epsilon}_r], \ \ r=1,2$$

Which are defined to be orthogonal to the momentum

$$\mathbf{\epsilon}_r . \mathbf{k} = 0$$

The scalar polarization vector

$$\epsilon_0^{\mu} \equiv n^{\mu} = [1, 0, 0, 0]$$

The longitudinal polarization vector:

$$\epsilon_3^{\mu} = [0,\frac{\mathbf{k}}{k}]$$

The longitudinal polarization vector can be written covariantly as:

$$\epsilon_3^{\mu} = \frac{k^{\mu} - (k.n) n^{\mu}}{[ (k.n) ^2- k^2]^{\frac{1}{2}}}$$

Substituting into the propagator we get

$$D^{\mu\nu}(k) = \frac{1}{k^2}\big [\sum_{r=1,2} \epsilon^{\mu}_{r}(\mathbf{k}) \epsilon^{\nu}_{r}(\mathbf{k}) + \frac{[k^{\mu} - (k.n) n^{\mu}][k^{\nu} - (k.n) n^{\nu}]}{[ (k.n) ^2-k^2]} + n^{\mu}n^{\nu} \big ]$$

The propagator can be arranged as:

$$D^{\mu\nu}(k) = D_{\mathrm{trans}}^{\mu\nu} + D_{\mathrm{Coulomb}}^{\mu\nu} + D_{\mathrm{resid}}^{\mu\nu} $$


$$D_{\mathrm{trans}}^{\mu\nu}(k) = \frac{1}{k^2}\sum_{r=1,2} \epsilon^{\mu}_{r}(\mathbf{k}) \epsilon^{\nu}_{r}(\mathbf{k}) $$

$$D_{\mathrm{Coulomb}}^{\mu\nu}(k) = \frac{n^{\mu}n^{\nu}}{[ (k.n)^2-k^2]}$$

$$D_{\mathrm{resid}}^{\mu\nu}(k) = \frac{k^{\mu} k^{\nu} - (k^{\mu} n^{\nu} + k^{\nu} n^{\mu})(k.n)}{k^2[ (k.n) ^2-k^2]}$$

The Coulomb and the residual parts are combinations of the scalar and the longitudinal parts of the propagator.

The residual part always gives a vanishing magnitude when sandwitched between two conserved currents ($k^{\mu}J_{\mu} = 0$):

$$D_{\mathrm{resid}}^{\mu\nu} J^{(1)}_{\mu}J^{(2)}_{\nu} = 0$$

Thus it does not contribute to physical observables. The Coulomb part is just the Fourier transform of the instantaneous Coulomb potential:

$$D_{\mathrm{Coulomb}}^{\mu\nu}(x-y) = \delta^{\mu0} \delta^{\nu0}\frac{\delta(x_0-y_0)}{4 \pi| \mathbf{x}-\mathbf{y}|}$$

Thus free photons can never result a Coulomb interaction.

  • $\begingroup$ "One example is that electromagnetic waves possess longitudinal polarizations in waveguides." - certainly the fields can have components in the axial direction, but the fields are still solenoidal so can we not still represent them as superpositions of (generally skewed) left and right hand transversely circularly polarized waves? $\endgroup$ – Selene Routley Oct 31 '13 at 13:28
  • $\begingroup$ @WetSavannaAnimal aka Rod Vance Electromagnetic waves in waveguides propagate in TE and TM modes where either the magnetic the electric field have components along the direction of propagation. $\endgroup$ – David Bar Moshe Oct 31 '13 at 14:17
  • $\begingroup$ Should be $\epsilon_{\mu}^{\lambda}\epsilon_{\nu\lambda}=g_{\mu\nu}$. I admit it doesn't matter much in this context though. $\endgroup$ – TeeJay Oct 31 '13 at 23:12
  • $\begingroup$ Would the following be a reasonable way of thinking then? Thinking of a simple, hollow waveguide, I agree that the TE/TM modes have components along the axial direction, but the plane wave components making them up in a momentum space decomposition are skewed and still solenoidal. One cannot, however, do a momentum space decomoposition over all of space, only for the restricted volume inside the waveguide. There is singular behaviour at the boundaries owing to charge distributions where the $\vec{E}$ lines end, so there is where the longitudinal behaviour arises: .... $\endgroup$ – Selene Routley Nov 1 '13 at 2:57
  • $\begingroup$ ... from this shuttling free charge on the waveguide's inner surface. $\endgroup$ – Selene Routley Nov 1 '13 at 2:57

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.