Why no position operator for photon?

[Elemental]

Position and momentum are definitely not on an equal footing in relativistic quantum mechanics. Momentum eigenstates are easily represented, but position representations are fraught with problems.

At least for massive particles one has Newton Wigner states which are strictly localized at one point in time in one reference frame. But there are at least two formidable problems within the context of relativistic quantum mechanics: 1) The states spread out faster than the speed of light as mentioned in Marcus Ludy's lecture (referred to as the Hegerfeldt paradox after, e.g., Hegerfeldt and Ruijsenaars, Physical Review D 22, 377 (1980)), and 2) They are localized in one reference frame only, so we ask why should a moving observe see a particle state to be spatially distributed which a stationary observer finds to be strictly localized?

For photons there is an additional obstacle. Any attempt (that I know of) to define a photon wave function ψ would require the transversality condition ψ.k = 0, where k is the wave vector. One can easily construct momentum eigensates ψ(k). But when one performs a Fourier transform of ψ(k), integrating over all k, the transversality condition introduces a dependence on k within the integrand which prevents the integral from forming a delta function.

Yes, the problems are overcome in quantum field theory. But there remains a lot of interest in understanding these issues in the context of quantum mechanics (evidenced by numerous papers over the years). To me it suggests that momentum is a more fundamental property than position, but I'm not sure why that would be the case. Any other insights welcome.

 

[vanhees71]

So, the violations of causality in QFT also apply to massive particles that are moving fast but still less than c?

There are no violations of causality in QFT. QFT is formulated explicitly such that there are none!

 

[HomogenousCow]

This requires choosing a particular coordinate chart, whose coordinate time the time evolution is with respect to, correct?

Choosing a particular frame does not break lorentz invariance, all quantities still have the same transformation properties. We solve Maxwell’s equations in specific frames all the time and There’s no problem.

 

[PeterDonis]

Choosing a particular frame does not break lorentz invariance

I didn't say it did. I was just making the point that the Hamiltonian formalism hides the Lorentz invariance by expressing the dynamics in terms of the time coordinate of a particular frame. The Lagrangian formalism, by contrast, keeps the Lorentz invariance manifest by writing everything in terms of scalars, 4-vectors, and tensors. See the subsequent exchange between me and @vanhees71 .

 

[meopemuk]

Regardless of the causality issue, there seems to be purely mathematical obstacles for building a position operator with 3 commuting components in the photon Hilbert space:

T. F. Jordan, "Simple proof of no position operator for quanta with zero mass and nonzero helicity", J. Math. Phys. 19 (1978), 1382.

Margaret Hawton is working on this problem for many years. You can start, e.g., from this article

M. Hawton, "Photon position operator with commuting components", Phys. Rev. A 59 (1999), 954.

and check its citations at Google Scholar.

Eugene.

 

[Demystifier]

To me it suggests that momentum is a more fundamental property than position, but I'm not sure why that would be the case. Any other insights welcome.

Or, perhaps, non-relativistic QM is more fundamental than relativistic QFT. See the paper linked in my signature below.