# Photon Angular Momentum

Essentially I am wanting to evaluate $$\langle j\, m \mid a^\dagger(\mathbf{k}, \lambda) \mid 0 \rangle \,,$$ where $\lambda$ indicates the circular polarization (about $\mathbf{k}$). We have that $\mathbf{J}= \mathbf{L} + \mathbf{S}$. It's straightforward to show that circular polarization corresponds to definite spin projections along the $\hat{\mathbf{k}}$ axis (the direction of propagation). However, I don't yet know how to find the $\mid \ell\, m_\ell \rangle$ projections. I don't know how how to express $\mid \ell\, m_\ell \rangle$ states in terms of Fock states, for example. With $\hbar = c = 1$,

$$\mathbf{L} = \frac{1}{4\pi}\int \mathop{d^3r} \sum_i E_i\left(\mathbf{r} \times \boldsymbol{\nabla} \right)A_i$$ $$\mathbf{S} = \frac{1}{4\pi}\int \mathop{d^3r} \mathbf{E} \times \mathbf{A} = -i \int \mathop{d^3k} \mathbf{a}^\dagger\left(\mathbf{k}\right) \times \mathbf{a}\left(\mathbf{k}\right)\,.$$

The above uses the following definition $$\mathbf{a}\left( \mathbf{k}\right) = \sum_{\lambda=\pm1} \boldsymbol{\epsilon}_\lambda\left( \mathbf{k}\right) a_\lambda\left( \mathbf{k}\right)$$

So, any help determining the angular momentum of a photon would be appreciated (incuding mention of references dealing with this subject). I have been using the Coulomb gauge.

For instance the canonical quantization of the plane-wave expansion which you've referenced starts off by expanding the vector potential $\mathbf A$ (in the Coulomb gauge) as $$\mathbf A(\mathbf r) \propto \sum_{\lambda, \mathbf k} \epsilon_\lambda a_\lambda(\mathbf k) e^{i(\mathbf{k\cdot r}-\omega t)} + c.c. \rightarrow \hat{\mathbf{A}}(\mathbf r) \propto \sum_{\lambda, \mathbf k} \epsilon_\lambda \hat a_\lambda(\mathbf k) e^{i(\mathbf{k\cdot r}-\omega t)} + h.c.$$ So now each mode designated by the polarization $\lambda$ and the wavevector $\mathbf k$ represents an eigenmode of linear momentum i.e. this mode has momentum $\mathbf p = \hbar\mathbf k$ per photon, and the number of excitations are the only quantum parts i.e. the (average) number of photons in this mode is $$N_{\lambda,\mathbf k}=\left<\hat a^\dagger_\lambda(\mathbf k) \hat a_\lambda(\mathbf k)\right>.$$
In order to find projections on orbital angular momentum (OAM) eigenstates, or any other type of eigenstate, you need a new representation which means a new modal expansion of $\hat{\mathbf{A}}$. This also means you'll have different operators $\hat a$ that in general can be written as a linear combination of $\hat a_\lambda(\mathbf k)$. For instance if you wanted to treat polarization in the H/V basis (where $\epsilon_{H/V}\propto\epsilon_{\lambda}\pm i\epsilon_{-\lambda}$) then $$\hat a'_{H/V}\propto\hat a_\lambda\pm i\hat a_{-\lambda}.$$ For a paraxial beam around $\mathbf k = \mathbf k_0$ with transverse wavevector $\mathbf{k_\perp}=(\rho,\phi)$ an OAM mode will have the form $e^{i\ell\phi}$, so $$\hat a(\mathbf{k_\perp,k_0})=\sum_{\ell,p} e^{i\ell\phi}f_{\ell,p}(\rho)\hat a_{\ell,p},$$ and where $f_{\ell,p}(\rho)$ is the radial modes (as an example one could use Laguerre Gaussian modes to have a complete set of radial modes or depending on what you want to compute just remember to sum/average/etc over the $\rho$ component in the end)). Inverting gives $$\hat a_{\ell,p} = \int \text d\mathbf{k_\perp} e^{-i\ell\phi}f^*_{\ell,p}(\rho)\hat a(\mathbf{k_\perp,k_0}).$$
For some of the details on QFT of OAM see . For nonparaxial beams $\mathbf L$ and $\mathbf S$ don't commute i.e. are coupled which is discussed in . A more complete discussion of this problem is given in .