Quantum State of Photon Question

I need to learn more about quantum field theory for my PhD research and I wont be able to take a class until after the summer. I am reading the QFT book from Landau and Lifshitz. I have some conceptual questions which I think is basic and was wondering if anyone could help me out. Please excuse my ignorance.

For a photon in mode $\vec{k}$ and spin $\mu$, the photon state is written as

$\hat{a}^{\dagger}(\vec{k},\mu)|0>=|\vec{k},\mu>$

My question deals with a photons "polarization" which I have come to realize has a different interpretation than in classical EM theory. Where does the polarization come into play with this expression? Does the spin factor $\mu$ account for its polarization? Does polarization only come into play from the EM field operator?

$\hat{E}(\vec{r}) \sim \sum_{\vec{k},\mu}\left( \vec{e}^{\mu}\hat{a}(\vec{k},\mu )e^{i \vec{k}\cdot\vec{r}} + \vec{e}^{-\mu}\hat{a}^{\dagger}(\vec{k},\mu )e^{-i \vec{k}\cdot\vec{r}} \right)$

So $\vec{e}^{\pm \mu}$ correspond to the right and left hand circular polarization vectors correct? But this also relates to the photon spin? So if I applied this operator to the vacuum state for one particular mode $\vec{k}$ would I need to count over all values of $\mu$? Would this be the correct answer?

$\hat{E}|0>=\vec{e}^{(1)}e^{-i\vec{k}\cdot\vec{r}}|\vec{k},\mu> + \vec{e}^{(-1)}e^{-i\vec{k}\cdot\vec{r}}|\vec{k},-\mu>$

So is it correct to say that photon polarization is a property of the electric field from the photon and not the photon itself?

Sorry for all the questions, but I want to understand this! I hope this all makes sense.

Answer

I'm afraid I can't give you an answer based on Landau's book, as I haven't read it. If you are referring to Landau's Theoretical Physics, Volume 4, I have it in my library so feel free to ask for more specific information in the comments. I advice you to have a look at other references too, especially at an elementary level.

Anyways, the photon's polarization is indeed encoded in the label of the creation operators: photons are polarized, i.e. they have spin, and this fact does not depend on the electric field operator. It is the photon which is polarized; the field operators are able to measure its spin, but (unless you refer to a technical definition of spin) they don't have spin. I don't get it when you say that the polarization receives a different interpretation than that of classical field theory. As far as I know, the two are completely equivalent (as far as one can say so when comparing classical to quantum mechanics). What receives a different interpretation in QED is the plane wave solution of Maxwell's equation: a solution of the form $e^{i(\vec{k}\cdot\vec{r}-\omega t)}$ (times an appropriate polarization vector) does not correspond to a photon with the same frequency/energy: the photon state is not an eigenstate of the electric (or magnetic) field operator; the quantum-mean field of a photon is zero, so a single photon does not reproduce the plane-wave solution. I'm not sure that you wrote the correct electric field operator, but many conventions are used in QFT, and Landau's ones are not usually the most common. If your electric field operator is correct, then your result is correct as far as you also sum over all possible three-momenta $\vec{k}$.

The $e^{\mu}(\vec{k},s)$'s (I suggest you don't use the label $\mu$ for spin, as it is commonly used for four-vectors' components) correspond to left and right polarization states if you decide to use a basis in which they do so. This is the usual convention.

I feel I have not answered your question as you would have liked to so, again, feel free to ask for more specific information in the comments and I'll try to see what I can do.