Timeline for Electron-Positron annihilation: helicity combinations and spin of photon
Current License: CC BY-SA 4.0
14 events
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| Feb 27, 2023 at 18:34 | history | edited | Qmechanic♦ | CC BY-SA 4.0 |
added 1 character in body; edited tags; edited title
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| Feb 27, 2023 at 15:59 | history | edited | David Bailey | CC BY-SA 4.0 |
Fixed misspelling of "annihilation" in title.
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| Nov 19, 2017 at 17:26 | vote | accept | kbg | ||
| Nov 15, 2017 at 18:07 | answer | added | Cosmas Zachos | timeline score: 3 | |
| Nov 15, 2017 at 8:33 | comment | added | kbg | @CosmasZachos Sir, could you explain the reason again. I did not get this point. | |
| Nov 15, 2017 at 1:50 | comment | added | kbg | @CosmasZachos Just a small question: In case I have LH electron and LH positron, am I in a total spin state 1 or 0? (Of course total $S_{z}=0$) | |
| Nov 15, 2017 at 1:47 | history | edited | kbg | CC BY-SA 3.0 |
added 648 characters in body
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| Nov 15, 2017 at 0:04 | history | edited | kbg | CC BY-SA 3.0 |
added 2 characters in body
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| Nov 14, 2017 at 23:48 | comment | added | kbg | @CosmasZachos I edited the question! My question is: yes, the initial spin state is 1. But, my query is- Does the total initial state has spin 1 because photon has spin 1? Or because mathematically, other hellicity combinations cancel out and you are left with total of spin1 state. i.e. Are the other helicity combinations zero because they do not give Spin 1 total state?(Like LH electron and LH positron which have S=1, S_z=0 which goes against what I think that photon being spin 1 dhould allow this hellicity combination) | |
| Nov 14, 2017 at 23:39 | history | edited | kbg | CC BY-SA 3.0 |
edited title
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| S Nov 14, 2017 at 21:07 | history | suggested | kg__ | CC BY-SA 3.0 |
grammer changed. meaning is same.
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| Nov 14, 2017 at 19:40 | review | Suggested edits | |||
| S Nov 14, 2017 at 21:07 | |||||
| Nov 14, 2017 at 19:06 | history | edited | kbg | CC BY-SA 3.0 |
added 404 characters in body
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| Nov 14, 2017 at 18:50 | history | asked | kbg | CC BY-SA 3.0 |