$A^\mu$ can have multipole expansions in classical electrodynamics. This gives rise to dipole photon, quadrupole photon etc. For dipole photon $j=1$ (In electrodynamics books they write it as $l=1$). Since, $\vec J=\vec L+\vec S$ and $j=|l-s|$ to $|l+s|$, in steps of unity.

  1. Can we apply this formula because I think S is not a good quantum number to specify photons but helicity is. If it propagates in z-direction then $S_z$ is good quantum number. Right? Then, can I directly use this formula or should I instead use $m_j=m_l+m_s$? Does it mean whenever I have gamma-transitions between two nuclear levels I should always use $m_j=m_l+m_s$ and not use $\vec J=\vec L+\vec S$ and $j=|l-s|$ to $|l+s|$?

  2. We know that, the projection $S_z=0$ is not allowed for a photon propagating along z-direction. But is it true that $L_z=0$ projection is also not allowed for photons?


1 Answer 1


There is a confusion in this question between classical electrodynamics and quantum electrodynamics. The multipole expansions of the vector potential are classical. A photon is a single elementary particle, and elementary particles are described with quantum mechanics.

Physics is continuous and quantum mechanical ensembles of photons do build up classical electromagnetic radiation which could have multipole properties, but this cannot happen by mixing formalisms. Have a look at this exposition of how classical waves are built up by photons.

The photon has intrinsic spin 1 and is always + or -1, as it is massless. It cannot be in a quantum mechanical orbital characterized by an angular momentum because it cannot be bound in a potential. It has no meaning to speak of its "angular momentum" as an intrinsic property.

Please look at this answer here about what is called photon orbital angular momentum , which is a classical beam vortex about whose axis the photon can have an angular momentum.

  • $\begingroup$ anna v-But nuclear gamma-transitions from a $j=7$ to $j=4$ is possible by the emission of a single photon. This can't be explained if we assume that a single photon carries only one unit of spin angular momentum. It must be qcquiring some orbital angular momentum. Right? Then, how is it getting orbital angular momentum then? Do you mean that such a photon rotates about some axis? $\endgroup$
    – SRS
    Feb 13, 2014 at 13:44
  • $\begingroup$ Angular momentum is given by rXp ( vectors) Any moving particle with respect to an axis through a point of vector distance r will have an angular momentum about that point. That does not characterize the particle in the way an intrinsic spin does. $\endgroup$
    – anna v
    Feb 13, 2014 at 14:17
  • $\begingroup$ if you could get the nuclei spins oriented all in the same direction then you would see the quadrupole or whatever pattern in the classical ensemble of photons coming out $\endgroup$
    – anna v
    Feb 13, 2014 at 14:19
  • $\begingroup$ Actually, I am asking about the OAM of individual photons and not that of a beam or a collection of them as a whole. $\endgroup$
    – SRS
    Feb 14, 2014 at 8:12
  • $\begingroup$ The angular momentum of individual photons has a meaning only with respect to the vortex axis or any other axis as spin axis. It can only be seen within an ensemble of photons because as everything quantum mechanical it is a probability distribution which one gets from many repeats of the same experiment.In the case of photons the ensemble gives the repeats. $\endgroup$
    – anna v
    Feb 14, 2014 at 10:40

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