Timeline for Why is the $S_{z} =0$ state forbidden for photons?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| S May 5, 2020 at 18:48 | history | suggested | CommunityBot | CC BY-SA 4.0 |
Fixed ordering of left, right to match the ordering of the presented states, as was suggested in a rejected edit by @VivekanandMohaptra
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| May 5, 2020 at 15:31 | review | Suggested edits | |||
| S May 5, 2020 at 18:48 | |||||
| May 4, 2020 at 7:33 | review | Suggested edits | |||
| May 4, 2020 at 9:41 | |||||
| Dec 12, 2012 at 13:54 | comment | added | KDN | @juanrga gives a very complete answer why this polarization state vanishes. Without gauge fixing, more polarization states at first appear possible, but these just represent spurious degrees of freedom. These non-propagating polarization states do exist, in a mathematical sense, but they are not observable (even in the mathematical sense, i.e., an observable quantity cannot be constructed for these states). The Wikipedia article on the Gupta-Bleuler formalism (en.wikipedia.org/wiki/Gupta%E2%80%93Bleuler_formalism) does a good job of addressing this issue in not-too-complex terms. | |
| Dec 12, 2012 at 6:11 | comment | added | Todd R | Thank you for the response @KDN but this is actually the sort of answer that I have been getting frustrated with. What I am interested in is your remark that because there is no rest frame it removes one of the allowed polarizations. Why is this exactly? Also, it's not clear to me why $S_z = 0$ represents and non-propagating field, and not simply a propagating field with no angular momentum. If possible, clarification would be appreciated. Thank you. | |
| Dec 12, 2012 at 5:58 | review | First posts | |||
| Dec 12, 2012 at 7:46 | |||||
| Dec 12, 2012 at 5:42 | history | answered | KDN | CC BY-SA 3.0 |