# Does a quadrupole transition mean emission of one photon with spin 2?

If it's true and spin-2 photons do exist, could you please point to some literature that discusses spin-2 photons?

If not, then how exactly does a selection rule for quadrupole transition make sense in terms of interaction with spin-1 photons?

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No, photons are always spin-one particles. The state of a photon may be described as a linear superposition of states with well-defined momenta $p$ and polarization vectors $\vec e$ and each of these basis vectors corresponds to a spin-one particle because the polarization is given by a transverse vector.

The selection rules – see this summary table – primarily boil down to the angular momentum conservation law. The conservation law says that the angular momentum of the atom before the transition is equal to the angular momentum of the products after the transition (the atom in the final state plus the photon). So for quadrupole transitions, as the table recalls, we may have $\Delta J = 0,\pm 1,\pm 2$ – depending on the relative angle between the angular momenta.

There isn't any need for the photon itself to have $J=2$ just because the operator responsible for the transition is a spin-two (quadrupole) operator. Such an agreement between the spins of the operator causing the transition on one side and the photon on the other hand would only have to hold if the initial and final states were the same – or at least had the same spin.

But because in the transition, the initial and final state of the atom are different and have different values of the spin in general, the spin-two character of the transition operator (quadrupole) means that the photon plus the difference between the final and initial spin of the atom corresponds to spin-two: this conservation law doesn't constrain the spin of the photon itself.

To summarize, I believe that in your argument which was "nearly" right, you forgot to add the difference between the final and initial angular momentum of the atom which is nonzero.

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This seems self-contradictory. In the second paragraph you say that momentum of "the atom in the final state plus the photon" conserves. Then you say that "photon plus the difference between the final and initial spin of the atom corresponds to spin-two". These two equations seem to have no commons solutions. Kind of like. 'J1 = J2 + Photon' together 'Photon + (J2 - J1) = 2'. It's equivalent to '0=2'. –  Klayman Apr 8 '13 at 14:35
Also, it's confusing that this article in wiki implies that photons may have angular momentum larger than one: "The emitted particle carries away an angular momentum λ, which for the photon must be at least 1" –  Klayman Apr 8 '13 at 14:44
No, there's no contradiction. If the atom changes its state from $m_J = 0$ to $m_J = -1$, and the photon has $m_s = +1$, the exchange of angular momentum is still 2$\hbar$, which is all quadrupole transitions need with regards to angular momentum conservation. –  Chay Paterson Apr 9 '13 at 1:34
Dear Klayman, the two conditions you mentioned are not only non-contradictory but completely equivalent. $J_{atom,ini}+J_\gamma = J_{atom, final}$ is the same thing as $J_\gamma = J_{atom, final}-J_{atom, initial}$. The spin-two character of the quadrupole operator doesn't really enter to the angular momentum conservation in the general form. Maybe an important thing to realize is that the total angular momentum of a photon in a momentum state isn't really $1$ or $2$ in general? It also has an "orbital part". The spin-two nature of the quadrupole operator imposes a selection on the whole sum. –  Luboš Motl Apr 9 '13 at 5:31