Timeline for Why is the $S_{z} =0$ state forbidden for photons?
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| Aug 13, 2023 at 20:21 | comment | added | Albert | Kvothe, a proper explanation would delve into representarion theory, again. An illustrative but poorly rigurous one follows. The helicity of a particle is defined as the projection of its spin (quantized along its momentum) over its momentum. Parity inverts the momentum but leaves the spin unchanged (handwavingly one could think of spin is an intrisic angular momentum, which is a pseudovector); as a result, under parity the helicity flips its sign. | |
| Apr 17, 2019 at 10:23 | comment | added | Kvothe | "Now, this single number h flips its sign under parity" any chance you could elaborate on this? How do we see that parity acts this way? (Or should i ask this as a separate question? | |
| Apr 13, 2017 at 12:40 | history | edited | CommunityBot |
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| Jul 9, 2016 at 2:47 | history | bounty ended | CommunityBot | ||
| Jul 2, 2016 at 19:06 | comment | added | Rococo | This is very nice, but if the goal of this question was to provide an intuitive understanding for people not familiar with QFT (a tall order, to be sure), I doubt it will be successful. | |
| Jul 1, 2016 at 11:14 | comment | added | ACuriousMind♦ | @EmilioPisanty: With "horribly difficult" I mean that I think that is an unsolved problem. Interacting QFTs have generically unknown spaces of states - no one knows how to write the state of an atom, so we can't really ask how the interaction looks in QFT. I'm not sure what the best way to do this is - that probably makes for an interesting new question in itself. | |
| Jul 1, 2016 at 11:01 | comment | added | Emilio Pisanty | Hard it may be, but interesting it remains. In the usual non-relativistic QM, you think of photon absorption as going from one state of global well-defined $J_z=+1$ to another such state. Here you say that the photon's state isn't in that representation - so what's going on there? | |
| Jul 1, 2016 at 10:57 | comment | added | ACuriousMind♦ | @EmilioPisanty: Bound states like atoms are horribly difficult to deal with in quantum field theory, so I think one usually models that in the non-relativistic QM regime and just imposes that $S_z = 0$ doesn't exist for a photon by hand. You can, however, have an electron in QFT have absorb or emit a photon in the presence of a nucleus - that's (reverse) bremsstrahlung, but there you don't have $S_z$ levels for the electron, the intrinsic helicity of the photon just goes into angular momentum of the electron. | |
| Jul 1, 2016 at 10:43 | comment | added | Emilio Pisanty | Can you comment on how this relates to photon absorption? You can have an atom in an $S_z=0$ state absorb a photon and transition to an $S_z=±1$ state, so the photon 'spin' does couple to the mechanical angular momentum degrees of freedom. How does that process look like in terms of the group representations in your answer? | |
| Jul 1, 2016 at 10:07 | history | answered | ACuriousMind♦ | CC BY-SA 3.0 |