polarization and frequency of a single photon

I believe that we can take a single photon state as a tensor product of a frequency Hilbert space (infinite dimension) and a polarization Hilbert space (dim 2). Does this mean we can measure the polarization and the frequency at space-like separated regions a the same time?

My second question is about frequency and polarization entanglement. If we have an entangled state between the frequency and the polarization, but we model the system as having a classical distribution over frequencies, what kinds of mistakes will we make in understanding any joint measurements of polarization and frequency? I am thinking about a channel which entangles the polarization and the frequency which we imagine is a classical channel over frequencies.

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sure, frequency is essentially energy and essentially momentum so you can have $|p\rangle |s\rangle =|p,s\rangle$ since they commute. How are you going to measure the energy of a photon, and then at a spacelike separation, it's spin? –  kηives Aug 27 '12 at 15:29
I am not sure how to simultaneously measure these at spacelike separation. Usually, to measure frequency, you split the path according to frequency (prism or grating) and then do a destructive measurement of position at the end of the path which selects the given frequency. Similarly for polarization, you put a polarizing beam splitter and measure at the end of the path that selects a given polarization....seems like a neat problem. –  Ben Sprott Aug 27 '12 at 16:56

The Helicity operator of a photon commutes with all of its momentum components thus with its energy, since from the commutation relations of the Poincare group, we have:

$[P_i, P_j] =0$

$[J_i, P_j] = \epsilon_{ij}^k P_k$

where $P^i$ and $J^i$ are respectively, the momentum and angular momentum generators. Thus the helicity

$h = \frac{\mathbf{J.P}}{E}$

where, $E = \sqrt{\mathbf{P.P}}$ satisfies the commutation relation

$[h, P_j] = \frac{\epsilon_{ij}^k P_i P_k}{E} = 0$

(The result vanishes due to the complete antisymmmetry of $\epsilon$)

Consequently, we have: $[h, E] = 0$

Thus the helicity and energy (which is proportional to the frequency) can be simultaneously measured.

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