My understanding is that photon can have spin +-1 along propagation direction, corresponding to two circular polarizations. Linear polarization is superposition of two. Since one can measure linear polarization than it is the eigenstate of some observable. Using analogy with electron I would say that measuring linear polarization is measuring what is the spin projection on other axes, but this has to be 0 for all directions except of propagation direction, so one could not distinguish between horizontal and vertical polarization. My references: http://en.wikipedia.org/wiki/Spin_angular_momentum_of_light and light linear polarization and photon spin

I would say I have complete misunderstanding of the topic, so back to the title: Is the linear polarization the eigenstate of some photon observable(edit: operator)? and which one?


1 Answer 1


The polarization states can be effectively defined in terms of the Stokes operators $\hat{S}_i$ for $i=[0,3]$

$\hat{S}_0 = \hat{a}^{\dagger}_{x}\hat{a}_{x}+\hat{a}^{\dagger}_{y}\hat{a}_{y}$

$\hat{S}_1 = \hat{a}^{\dagger}_{x}\hat{a}_{x}-\hat{a}^{\dagger}_{y}\hat{a}_{y}$

$\hat{S}_2 = \hat{a}^{\dagger}_{x}\hat{a}_{y}+\hat{a}^{\dagger}_{y}\hat{a}_{x}$

$\hat{S}_3 = i\left(\hat{a}^{\dagger}_{y}\hat{a}_{x}-\hat{a}^{\dagger}_{x}\hat{a}_{y} \right)$

These quantum-mechanical observables (Hermitian operators) are defined analogously to their classical counterparts. The annihilation and the creation operators $\hat{a}_{x}$ and $\hat{a}^{\dagger}_{x}$ of the mode (given by the subscript $x$) satisfy the usual commutation relations. The last three of these operators can also be viewed equivalently to the Pauli matrices $\sigma_z$, $\sigma_x$, and $\sigma_y$ respectively.

To answer your question directly, it is the operator $\hat{S}_1$ whose eigenstates are the horizontal and vertical polarization states $|H\rangle$ and $|V\rangle$.

To visualize this from an experimental perspective, think of a photon counting setup and the fact that $\hat{S}_1 = \hat n_x - \hat n_y$ essentially. And to make a connection to the comments in the other answer, $\hat{S}_1 \equiv \sigma_z = |H\rangle\langle H| - |V\rangle\langle V| = 2|H\rangle\langle H| - I$, if polarization states were chosen to represent the basis vectors of the Pauli matrices.

Edit on 04/01/2015: I saw a comment asking for some references on the topic. Unfortunately I didn't have time to respond then and now I can't see that comment anymore. Still thought of writing down a couple of works that deal with the topic in a fairly introductory manner.

  • $\begingroup$ This wasn't a question on quantum optics, but a question of simple QM measurement. A Hilbert space and operators like the Pauli matrices were enough. You could follow the comments. $\endgroup$
    – Sofia
    Dec 18, 2014 at 18:54
  • $\begingroup$ @Phonon I had had a quick glance yesterday. Having gained most of my knowledge of quantum optics as an experimentalist, I admit that I tackled spin only when I really had to (e.g., studying for theory exams :-$ The knowledge of Stokes parameter happens to be a lucky aberration because that stuff is bread and butter for a couple of my colleagues). So before I can make any useful contribution to that question, I would need to brush up my concepts. $\endgroup$
    – jayann
    Jan 4, 2015 at 18:01
  • $\begingroup$ @jayann ah I see :), no worries I understand. Although any experimental insight on the matter would be equally useful, undoubtedly! Look forward to that brush up ;) $\endgroup$
    – Ellie
    Jan 4, 2015 at 18:09

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