# What are the polarization states of the photons in a polarized and unpolarized light?

The photons are completely polarized, i.e their polarization states can be expressed as $a|R\rangle+b|L\rangle$, where $|R\rangle$ and $|L\rangle$ are two helicity eigenstates of the photon. For example, the $|R\rangle$ photon is right circularly polarized and the $|H\rangle=\frac{1}{\sqrt{2}}(|R\rangle+|L\rangle)$ is horizontally linearly polarized photon.

Is the polarized light a pure state and the unpolarized light a statistical mixture of photons with different polarizations?

Yes, a photon in a polarized light is found in a pure state such as $|H\rangle$, $|V\rangle$, $|L\rangle$, $|R\rangle$, or any complex linear combination of them. A photon in (completely) unpolarized light is described by the density matrix $$\rho = \frac{1}{2} \left( |L\rangle \langle L| + |R\rangle \langle R| \right) = \frac{1}{2} \left( |H\rangle \langle H| + |V\rangle \langle V| \right)$$ Note that you omitted the relationship for the vertically polarized state, $|V\rangle = i(|R\rangle - |L\rangle)/\sqrt{2}$, up to an overall sign which is a convention (well, the whole phase including $i$ is physically inconsequential, so it doesn't matter at all but one must be self-consistent with the conventions).
• Your density matrix corresponds to a completely unpolarized light. Partially unpolarized light density matrix is expressed via projection operators with different "weights" $w_n$ (probabilities), not 1/2. – Vladimir Kalitvianski Nov 7 '11 at 10:37
• Dear @ANKU, to discuss frequency (and/or direction), you need to extend the Hilbert space by tensor-multiplying it with the space of different $\vec k$. Again, one may have pure states and mixed states with respect to frequencies, too. The totally non-monochromatic, unpolarized light is given by the density matrix like mine, but it would also have extra labels $\vec k$ in all the bra,ket states and one would integrated over some interval of $\vec k$. However, one may also have pure states relatively to the position/frequencies. There are many states, pure and mixed; what's your exact question? – Luboš Motl Nov 7 '11 at 22:24