Photon energy considering angular momentum components Photon energy is generally given as $hf$. But, the photon also has angular momentum of spin 1 [h] (and an orbital momentum component as well in some polarized cases). Does this angular momentum add to the energy hf or is it contained within it? (Of course, I guess we could be "nebulous" and conceive of a momentum without energy - but that would be rather curious.) Please give a careful answer as the De Broglie relation, i.e., in the form $hf=mc^2$ gets "messy" if energy is added by the angular momentum.
Please give an answer where I can cancel units, i.e., please avoid $c=\hbar=G=k$, etc.
 A: It's tempting to think of the spin as a rotation in which case there would be an associated rotational energy:
$$ E = \tfrac{1}{2}I\omega^2 $$
(though what we'd mean by the moment of inertia of a photon would require some head scratching). However the spin, and its associated angular momentum, is a fundamental property of the photon and not like a macroscopic rotation with some associated rotational energy.
The simple way to see this is to take the limit of $\nu \to 0$ in which case the energy goes to zero. However the spin remains $1$, and its angular momentum $\hbar$, even in the limit of zero energy.
A: A single photon with a sharp value $j=1,2,...$ of total angular momentum is the state,
\begin{equation}
|jmk\lambda\rangle=\frac{(2j+1)}{4\pi}\int d\Omega\left\{(U^{j}(\alpha,\beta,0))^{m}_{\ \lambda}\right\}^{*}|\underline{k},\lambda\rangle
\end{equation}
The units are $\hbar=c=1$. The state is a sum of plane wave states $|\underline{k},\lambda\rangle$ where $\underline k$ is the linear momentum vector and $\lambda=\pm 1$ is the helicity (spin along the linear momentum direction). $d\Omega$ is an element of solid angle. $\alpha,\beta$ are Euler angles (rotate by $\alpha$ about the z axis and then rotate by $\beta$ about the resultant y axis). $(U^{j}(\alpha,\beta,0))^{m}_{\ \lambda}$ is Wigner's D-function for spin $j$. This state is made by acting on the plane wave state with the projection operator onto a state of total angular momentum. This state is equation (8.7-2) on page 147 of "Group Theory in Physics" by Wu-Ki Tung or equation (28.35) on page 218 of "Relativistic Theory of Reactions" by J. Werle.  This is a single photon with orbital angular momentum so that the total angular momentum of the photon is $j$ units. 
The question asks for the energy of the photon. We just need to check that the angular-momentum-helicity eigenstate $|jmk\lambda\rangle$ is also an eigenstate of the Hamiltonian and read off the energy eigenvalue.
The normally-ordered Hamiltonian for the free electromagnetic field is,
\begin{equation}
\hat{H}=\frac{1}{2}\sum_{\lambda}\int d^{3}\underline{k}\ \hat{\eta}_{\underline{k},\lambda}\hat{\eta}_{\underline{k},\lambda}^{\dagger}
\end{equation}
Here, $\hat{\eta}_{\underline{k},\lambda}$ is the emission operator and $\hat{\eta}_{\underline{k},\lambda}^{\dagger}$ is the absorption operator for a photon of linear momentum $\underline{k}$ and helicity $\lambda$. The notation is from Dirac's book "Lectures on Quantum Field Theory", published by Belfer Graduate School of Science, 1966. The commutator is,
\begin{equation}
[\hat{\eta}_{\underline{k},\lambda}^{\dagger},\hat{\eta}_{\underline{k}',\lambda'}]_{-}=2\omega\delta^{3}(\underline{k}-\underline{k}')\delta_{\lambda,\lambda'}
\end{equation}
The last two formulae use the Lorentz invariant measure. For reference, the resolution of unity is,
\begin{equation}
\sum_{\lambda}\int \frac{d^{3}\underline{k}}{2\omega}|\underline{k},\lambda\rangle\langle\underline{k},\lambda|=1
\end{equation}
Now check that the angular-momentum-helicity eigenstate is also an energy eigenstate of the Hamiltonian.
\begin{equation}
\hat{H}|jmk\lambda\rangle=\frac{1}{2}\sum_{\lambda'}\int d^{3}\underline{k}'\ \hat{\eta}_{\underline{k}',\lambda'}\hat{\eta}_{\underline{k}',\lambda'}^{\dagger}
\frac{(2j+1)}{4\pi}\int d\Omega\left\{(U^{j}(\alpha,\beta,0))^{m}_{\ \lambda}\right\}^{*}|\underline{k},\lambda\rangle
\end{equation}
The plane wave state is created from the vacuum $|S\rangle$ as $|\underline{k},\lambda\rangle=\hat{\eta}_{\underline{k},\lambda}|S\rangle$. The energy eigenvalue equation is now,
\begin{equation}
\hat{H}|jmk\lambda\rangle=\frac{(2j+1)}{4\pi}\int d\Omega\left\{(U^{j}(\alpha,\beta,0))^{m}_{\ \lambda}\right\}^{*}\frac{1}{2}\sum_{\lambda'}\int d^{3}\underline{k}'\ \hat{\eta}_{\underline{k}',\lambda'}\hat{\eta}_{\underline{k}',\lambda'}^{\dagger}\hat{\eta}_{\underline{k},\lambda}|S\rangle
\end{equation}
Move the absorption operator $\hat{\eta}_{\underline{k}',\lambda'}^{\dagger}$ to the right by using the commutator. One term kills the vacuum and we are left with,
\begin{equation}
\hat{H}|jmk\lambda\rangle=\frac{(2j+1)}{4\pi}\int d\Omega\left\{(U^{j}(\alpha,\beta,0))^{m}_{\ \lambda}\right\}^{*}\frac{1}{2}\sum_{\lambda'}\int d^{3}\underline{k}'\ \hat{\eta}_{\underline{k}',\lambda'}2\omega\delta^{3}(\underline{k}-\underline{k}')\delta_{\lambda,\lambda'}|S\rangle
\end{equation}
Carry out the integral over linear momentum and the sum over helicity.
\begin{equation}
\hat{H}|jmk\lambda\rangle=\frac{(2j+1)}{4\pi}\int d\Omega\left\{(U^{j}(\alpha,\beta,0))^{m}_{\ \lambda}\right\}^{*} \hat{\eta}_{\underline{k},\lambda}\omega|S\rangle
\end{equation}
The integral over solid angle is over all directions of the momentum vector $\underline{k}$ with the magnitude $k$ fixed. The on-shell condition for the photon is $k=\omega$ so that the constant angular frequency of the photon can be taken through the integral over solid angle.
\begin{equation}
\hat{H}|jmk\lambda\rangle=\omega\frac{(2j+1)}{4\pi}\int d\Omega\left\{(U^{j}(\alpha,\beta,0))^{m}_{\ \lambda}\right\}^{*} \hat{\eta}_{\underline{k},\lambda}|S\rangle
\end{equation}
The plane wave state is $|\underline{k},\lambda\rangle=\hat{\eta}_{\underline{k},\lambda}|S\rangle$.
\begin{equation}
\hat{H}|jmk\lambda\rangle=\omega\frac{(2j+1)}{4\pi}\int d\Omega\left\{(U^{j}(\alpha,\beta,0))^{m}_{\ \lambda}\right\}^{*} |\underline{k},\lambda\rangle
\end{equation}
We recognize the original angular-momentum-helicity eigenstate,
\begin{equation}
\hat{H}|jmk\lambda\rangle=\omega|jmk\lambda\rangle
\end{equation}
So, the state $|jmk\lambda\rangle$ of sharp total angular momentum is an eigenstate of the Hamiltonian with energy eigenvalue $\omega$. So, we could have a single photon with various values of total angular momentum $j=1,2,...$ but the photon always has a single unit of energy $\omega$. In contrast, a classical spinning top increases its energy as the angular momentum of the top increases. 
A: A photon has energy hf and angular momentum $\hbar$. To my knowledge there is no model of a photon that involves a division in rotational and kinetic energy.
A: That spin is an intrinsic property of the photon. It don't have anything to do with the notion of angular momentum,Because of the fact that the spin that we assign to photon is not coming from the classical notion of the rotational angular momentum.This spin concept coming from the quantum mechanical background not from the classical. The photon  will start to have this kind of 'spin energy' only when it pass through some magnetic field. When I call it intrinsic property, It mean that it is some thing inbuilt. It is one of its identity. 
