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If we have the polarization of the electric field in the $z$ direction then we have the selection rule that: $$\Delta m=0$$ (source: slide 7 lecture 19 on this page http://butane.chem.illinois.edu/sohirata/)

What is the physical reason behind this rule?

Since if the emitted photon travels along the $z$-axis we must have $\Delta m =\pm 1$ does this mean that we a photon cannot be emitted in the direction of the polarization of the electric field?

Quantum spaghettification
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    Can you give us more detail about what's going on, and tell us what it says in that link? – garyp Mar 23 '16 at 17:13
  • @garyp I am looking at the case of the dipole approximation, and the allowed transitions of a hydrogen atom. The link basically says what I have said above it, which it derives using matrix components. I am basically looking for a physical reason behind it. – Quantum spaghettification Mar 23 '16 at 17:17

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A linearly polarized photon is an eigenstate of angular momentum $m=0$ along the polarization direction. That succinctly explains the selection rule. In your question, you make the error of assuming that the propagation direction and the polarization are both along $z$. In fact, this is impossible, since electromagnetic waves are transverse. If the the electric field direction is along $z$, then the propagation direction lies in the $xy$-plane. For concreteness, let us say the propagation is along $x$. Then the linearly polarized wave is an equal superposition of $m_{x}=1$ and $m_{x}=-1$ circular polarization states; the superposition of these angular momentum states is exactly an $m_{z}=0$ state.

Buzz
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  • But surly the emitted photon (i.e. the one created as a result of the transition and not causing the electric field) can be propagating in any direction? – Quantum spaghettification Mar 23 '16 at 17:23
  • I don't know what you mean by "not causing the electric field." The photon is the electromagnetic field involved here, and the direction of the electric field is the polarization direction of the photon. – Buzz Mar 23 '16 at 17:26
  • Are you sure? I thought the transition was caused by an external (oscillating) electric field, which then caused the release of another photon? Otherwise you have a 'chicken and egg' type problem, which came first the transition or the photon. – Quantum spaghettification Mar 23 '16 at 17:33
  • When you are talking about stimulated emission (as opposed to spontaneous), the two fields are coherent. They point in the same direction and are in phase. (Moreover, because of Bose statistics, the two parts of the field cannot really be disentangled.) This means that the emitted photon has to be propagating in the same direction as the beam that stimulated its emission, so the answer to the question you posed in your first comment is no. – Buzz Mar 23 '16 at 17:37
  • Just to clarify, are you saying that for spontaneous emission, the transition of the electron is caused by the electric field of the emitted photon? and do you possibly have a source? – Quantum spaghettification Mar 23 '16 at 17:39
  • I wouldn't phrase it like that, no. But the point is that there is not a "separate" field produced by the emitted photon. There is one oscillating electric field, whose amplitude changes at the time of emission (or absorption). – Buzz Mar 23 '16 at 17:41
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    @Quantumspaghettification: see this rigorous answer or for a handwaving answer see this answer and this answer. – John Rennie Mar 23 '16 at 18:02