Orbital Angular Momentum and Photon Energy

[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?

 

[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.

 

[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?

 

[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.

 

[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

 

[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.

 

[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.