Let us consider the following matrix element: $$\langle n',m',l'|x| n, m, l \rangle$$ For the corresponding radiative transition we have the selection rule that $\Delta m=\pm 1$. But will the photon emitted be circularly polarized i.e.: $$|\psi \rangle=|\pm m=1 \rangle$$ or linearly polarized (i.e. a combination of two circularly polarized waves) i.e.: $$|\psi \rangle=a(| m=1 \rangle+| m=-1 \rangle)$$


2 Answers 2


Given two states $|n,l,m⟩$ and $|n',l',m'⟩$, the following radiative transition matrix elements are nonzero: $$\left\langle n,l\pm1,m+1 \middle| x+iy \middle| n,l,m \right\rangle,$$ $$\left\langle n,l\pm1,m-1 \middle| x-iy \middle| n,l,m \right\rangle,$$ and $$\left\langle n,l\pm1,m \middle| z\middle| n,l,m \right\rangle,$$ and that's it (for transitions between states of definite angular momentum about the $z$ axis; you can also have transitions to e.g. $|n,l+1,m+1⟩+|n,l+1,m-1⟩$, using a linearly polarized photon in the $x$ direction and propagating in the $y,z$ plane, or to similar superposition states).

The first two correspond to absorption or emission of a circularly polarized photon propagating along the $z$ axis, with selection rule $\Delta m=±1$ on the atom, while the third one corresponds to absorption or emission of a linearly polarized photon polarized along the $z$ axis and propagating on any axis in the $x,y$ plane, with selection rule $\Delta m=0$ on the atom.

For further details, consult any advanced undergraduate textbook on quantum mechanics.

  • $\begingroup$ Do you have any specific recommendations for textbooks?? I cannot see why the transition I have given is zero, as it's just a linear combination of two of yours. $\endgroup$ May 31, 2016 at 17:12
  • $\begingroup$ See edit with clarification. On textbooks it's up to taste, really; I normally recommend Cohen-Tannoudji but I don't have it at hand to confirm it treats this well enough to recommend specifically. $\endgroup$ May 31, 2016 at 17:40
  • $\begingroup$ Shankar has a good section on electromagnetic interactions towards the end of the book. $\endgroup$ May 31, 2016 at 18:25
  • $\begingroup$ No worries, I will take a look in Cohen-Tannoudji. Just to clarify, when we do have a transition to the state e.g. $|n,l+1,m+1⟩+|n,l+1,m-1⟩$ the state of the photon would be $a(|m-1\rangle +|m+1\rangle)$, which is a linearly polarized photon?? $\endgroup$ May 31, 2016 at 18:53
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    $\begingroup$ @Quantumspaghettification Yes, so long as the photon is propagating along the $z$ axis (since otherwise the re-use of $m$ for $m_z$ is misleading). $\endgroup$ May 31, 2016 at 18:57

I assume you are talking of the transition dipole element which is typically defined as $$\langle n',m',l'|qx| n, m, l \rangle$$ with q being the charge of the dipole and x its position.

As you already said, the linear polarization is a superposition of the two circular polarization Eigenstates. And in addition, horizontal and vertical polarization states are an independent basis to describe the polarization of the emission. Your measurement will determine the answer to your question. If you check for linear polarization you will project your emitted photon to the linear polarization basis. Same for circular polarization. The basis for your polarization is arbitrary. This may change if you have conditions where the intrinsic symmetries of your Hamiltonian are broken.


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