Photon angular momentum, multipole transitions

Relying on many sources, I can summarize that somehow it is possible to show that a photon, in its plane wave expansion, can carry values of total angular momentum more than $\hbar$. Such photons can be generated in higher multipole atomic transitions. I also think that the formalism might be expressed in terms of vector spherical harmonics.

Now, I have never seen a rigorous proof that a photon can carry total AM $J>\hbar$. I would love to get references, or maybe someone knows a short and elegant proof to post here. Thanks=)

The key to understand why the total angular momentum of a photon can be larger than $$\hbar$$, is to realize that total angular momentum can be separated into spin angular momentum and orbital angular momentum (OAM). While spin angular momentum is bounded to be of magnitude $$\hbar$$ for a single photon, its orbital momentum can be much larger. In 1992 is was shown that the Laguerre-Gauss modes represent spatial modes of light in which each photon carries an OAM of $$\ell\hbar$$, where $$\ell$$ is an integer representing the azimuthal index of the mode, and it equals the topological charge of the optical vortex in the center of the mode.

An unpolarized photon can be directed with its electric (magnetic) field component in every 360° direction perpendicular to its direction of propagation. But if you choose one direction for a instantaneous moment, say 0° for the direction of the positive electric field component than there are only two possibilities for the north direction of the magnetic dipole moment. This moment (in vacuum) could be directed only to 90° or to -90°. This is called the intrinsic spin of the photon.

More see here Visualizing A Light Wave, last paragraph.

• Well, wait, photon spin would give me only $J<=\hbar$, it cannot give more that it's norm no matter how you pick the coordinate system. There has to be an orbital angular momentum component to get higher values... I just need an explicit rigorous proof of the fact that for the photon $J$ in principle can be more than $\hbar$ – MsTais Sep 7 '16 at 20:45
• MsTais, for a circular polarised photon the field components can have (synchron) any direction. The question, is this turn quantized or not is really an interesting question, you are asking this? – HolgerFiedler Sep 7 '16 at 21:00

Set up a state $$|k_{0},\lambda\rangle$$ for a photon of fiducial momentum $$k_{0}$$ moving along the z axis with helicity $$\lambda$$.

Now, instead of working with states of sharp momentum and helicity, we can consider states of sharp total angular momentum by using the projection operator onto the states of a group irrep.

Suppose we have a group $$G$$. The group has a unitary representaion $$\hat{U}(g)$$ which acts on the states of a Hilbert space. The Hilbert space can be decomposed into irreps with states $$|a,i,u\rangle$$ where $$a$$ labels the irreps, $$i$$ labels the states in irrep $$a$$ and $$u$$ labels copies of irrep $$a$$. The projection operator onto the states of the irreps is, $$$$\hat{P}_{aij}=\frac{n_{a}}{N}\sum_{g\in G}(U^{(a)}(g))^{j}_{\ \ i}\hat{U}(g^{-1}) =\sum_{u}|a,i,u\rangle\langle a,j,u|$$$$ Here, $$n_{a}$$ is the dimension of irrep $$a$$, $$N$$ is the size of the group $$G$$, $$(U^{(a)}(g))^{j}_{\ \ i}$$ is the unitary matrix rep of the group on the irrep $$a$$. The definition of the projection operator used here is from equation (1) in section 3 of chapter 7 (page 177) of "Theory of Group Representations and Applications", Barut and Raczka, World Scientific, 1986.

In order to get a photon with sharp total angular momentum $$j$$, the group is $$G=SO(3)$$ and the projection operator becomes, $$$$\hat{P}_{jm_{1}m_{2}}=\frac{2j+1}{N}\sum_{g\in SO(3)} (U^{(j)}(g))^{m_{2}}_{\ \ m_{1}}\hat{U}(g^{-1})$$$$ It helps if the operator is $$\hat{U}(g)$$ instead of $$\hat{U}(g^{-1})$$ so change variable $$g\rightarrow g^{-1}$$. $$$$\hat{P}_{jm_{1}m_{2}}=\frac{2j+1}{N}\sum_{g\in SO(3)} ((U^{(j)}(g))^{m_{1}}_{\ \ m_{2}})^{*}\hat{U}(g)$$$$ The projection operator is defined for a finite group, but SO(3) is compact so, $$$$\frac{1}{N}\sum_{g\in SO(3)}\rightarrow \int_{g\in SO(3)}d\mu(g)$$$$ where the Haar measure is, $$$$d\mu(g)=\frac{\sin{\beta}d\alpha d\beta d\gamma}{8\pi^{2}}$$$$ and $$\alpha,\beta,\gamma$$ are Euler angles (rotate by $$\alpha$$ about z, then $$\beta$$ about resultant y, then $$\gamma$$ about resultant z). The ranges are $$\alpha \in [0,2\pi]$$, $$\beta\in [0,\pi]$$, $$\gamma\in [0,2\pi]$$. The projection operator onto a state of sharp total angular momentum $$j$$ is now, $$$$\hat{P}_{jm_{1}m_{2}}=\frac{2j+1}{8\pi^{2}}\int \sin{\beta}\ d\alpha\ d\beta\ d\gamma \{(U^{(j)}(g))^{m_{1}}_{\ \ m_{2}}\}^{*}\hat{U}(g)$$$$ The matrices $$(U^{(j)}(g))^{m_{1}}_{\ \ m_{2}}$$ are Wigner's D-functions.

Now let's apply the projection operator to the photon state $$|k_{0},\lambda\rangle$$. A photon of sharp total angular momentum $$j$$ is, $$$$\hat{P}_{jm_{1}m_{2}}|k_{0},\lambda\rangle=\frac{2j+1}{8\pi^{2}}\int \sin{\beta}\ d\alpha\ d\beta\ d\gamma \{(U^{(j)}(g))^{m_{1}}_{\ \ m_{2}}\}^{*}\hat{U}(g)|k_{0}\lambda\rangle$$$$ Now, a general rotation $$g\in SO(3)$$ is the matrix $$R_{z}(\alpha)R_{y}(\beta)R_{z}(\gamma)$$. Hence, $$$$\hat{U}(g)|k_{0},\lambda\rangle=\hat{U}(R_{z}(\alpha)R_{y}(\beta)R_{z}(\gamma))|k_{0},\lambda\rangle=\hat{U}(\alpha,\beta,0)\hat{U}(0,0,\gamma)|k_{0},\lambda\rangle=\hat{U}(\alpha,\beta,0)e^{-i\gamma\lambda}|k_{0},\lambda\rangle$$$$ Furthermore, Wigner's D-functions are, $$$$(U^{(j)}(g))^{m_{1}}_{\ \ m_{2}}=\langle j m_{1}|\hat{U}(\alpha,\beta,\gamma)|j m_{2}\rangle=e^{-im_{1}\alpha}\langle j m_{1}|\hat{U}(\theta=\beta/2)|jm_{2}\rangle e^{-im_{2}\gamma}$$$$ where $$\langle j m_{1}|\hat{U}(\theta=\beta/2)|jm_{2}\rangle$$ are the reduced d-functions. We can substitute the last two equations in the integral for the state of sharp total angular momentum and carry out the integral over the Euler angle $$\gamma$$. The result is, $$$$\hat{P}_{jm_{1}\lambda}|k_{0},\lambda\rangle=\frac{2j+1}{4\pi}\int \sin{\beta}\ d\alpha\ d\beta\ \{(U^{(j)}(\alpha,\beta,0))^{m_{1}}_{\ \ \lambda}\}^{*}\hat{U}(\alpha,\beta,0)|k_{0},\lambda\rangle$$$$ The state $$\hat{U}(\alpha,\beta,0)|k_{0},\lambda\rangle$$ is just a photon of momentum $$k$$ travelling in the direction given by the polar angles $$\alpha,\beta$$. In other words, a plane wave state $$|k,\lambda\rangle$$. So, a photon of sharp total angular momentum $$j$$ is, $$$$\hat{P}_{jm_{1}\lambda}|k_{0},\lambda\rangle=\frac{2j+1}{4\pi}\int \sin{\beta}\ d\alpha\ d\beta\ \{(U^{(j)}(\alpha,\beta,0))^{m_{1}}_{\ \ \lambda}\}^{*}|k,\lambda\rangle$$$$ The momentum $$k=R(\alpha,\beta,0)k_{0}$$ and the element of solid angle is $$d\Omega=\sin{\beta}\ d\alpha\ d\beta$$ . Finally, a photon of sharp total angular momentum $$j$$ is the state, $$$$|jmk\lambda\rangle=\hat{P}_{jm\lambda}|k_{0},\lambda\rangle= \frac{2j+1}{4\pi}\int d\Omega\ \{(U^{(j)}(\alpha,\beta,0))^{m}_{\ \ \lambda}\}^{*}|k,\lambda\rangle$$$$ This formula agrees with (8.7-2) on page 147 of Wu-Ki Tung "Group Theory in Physics" and (28.35) on page 218 of Werle, "Relativistic Theory of Reactions", North Holland, 1966.