# Orbital Angular Momentum and Photon Energy

1. Jan 2, 2015

### jkg0

Just a quick question on photon orbital angular momentum.
In the equation for photon energy: E2 = p2c2 + m2c4

Is OAM counted in the p2c2 part? Or does the above equation only apply to photons with normal momentum and there is another term for the angular momentum?

The normal relation for p seems to be p = kħ but isn't k just a vector?

2. Jan 2, 2015

### jfizzix

The energy momentum relation for a photon is, $E=pc=\hbar k c$ since $m=0$.

If you really want to talk about OAM for a single photon, you will have to specify it's quantum state. Now, you can represent the state of a photon as an infinite sum over all possible momenta (i.e., plane waves, each with a different momentum $\hbar \vec{k}$), but you can also represent the state of a photon as an infinite sum over all possible values of angular momentum $L=m\hbar$. The best you can do in finding the angular momentum of a photon is knowing the probabilities of measuring any particular value of the angular momentum, or measuring the average value from that distribution.

Wikipedia has a decent coverage of this.

3. Jan 3, 2015

### jkg0

The problem I am having is that the energy for a single photon is defined by $E=pc=\hbar k c$ with the k vector.
If the quantum state of the photon can also be written as the integration of $L=m\hbar$ over varying probabilities is there a way to write the energy momentum relationship with respect to the m values of angular momentum instead of the k vector?

4. Jan 3, 2015

### vanhees71

There's no sensible notion of a photon's orbital and spin angular momentum, because there's no gauge invariant definition of this splitting. Only total angular momentum has a physical meaning in this case.

5. Jan 3, 2015

### jfizzix

I would say there is a meaningful decomposition of orbital and spin angular momentum, though the exact values depend on your reference frame.

The spin-angular momentum is given by a photon's polarization state,
while the orbital angular momentum is obtained from its linear momentum.

As far as expressing the energy-momentum relation in terms of angular momentum, I don't know that the four-vector for angular momentum would be.

The energy-momentum relation comes from the conservation of the "length" of the momentum four-vector.
$p^{\mu}p_{\mu} = \frac{E^{2}}{c^{2}}-p^{2} =\frac{E'^{2}}{c^{2}}=m^{2}c^{2}$
Rearranged, this gives
$E^{2} = p^{2}c^{2}+m^{2}c^{4}$
which looks more familiar (where again for a photon $m=0$.

Unfortunately, I don't know what the angular-momentum four-vector would be, but I expect it must be possible to represent energy in terms of angular momentum just as it is in Newtonian physics

6. Jan 3, 2015

### vanhees71

Exactly! No only you don't know what the angular-momentum four-vector would be but nobody else. Only total angular momentum is a gauge invariant quantity and defined via the corresponding Belinfante energy-momentum tensor components. Written in 3D-form it reads
$$\vec{J} = \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \vec{x} \times (\vec{E} \times \vec{B}).$$
A basis for the polarization states of the photon are the helicity $\eta=\pm 1$ states. These are the dual and self-dual parts of the Faraday (fielt-strength) tensor or in terms of classical optics the right- and left-circular polarized modes of the electromagnetic field.

7. Jan 3, 2015

### DrDu

I think it is important to note that angular momentum eigenstates (vector spherical harmonics) can be constructed from momentum eigenstates with the same absolute value k but different direction. Hence the relation between E and k remains valid.