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Feb 25, 2023 at 16:56 comment added Cosmas Zachos I fear your getting the 12 in that question is pure luck. I get it as 3x(1+3)=12, but your 6x2=12 glosses over the fact that there are two separate channels for total antisymmetry, not impeding each other; so, accidentally, space and spin parts don't block each other. It is a fine shortcut, as long as you can justify it in your head, but I fear it won't always work. Unless you stated a general argument...
Feb 25, 2023 at 7:59 comment added curious_mind @CosmasZachos please have a look at this approach, whether is it consistent with symmetry constrained counting problem like you described or not ? Thanks.physics.stackexchange.com/questions/751784/…
Mar 31, 2018 at 16:15 vote accept Jhonny
Mar 31, 2018 at 2:05 history edited Cosmas Zachos CC BY-SA 3.0
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Mar 31, 2018 at 1:52 comment added Cosmas Zachos Right you are. (000)(011) and its 2 permutants is there, contributing 12 when multiplied by spin states. In fact, this 12 comes out of the overcounted 6, which should have been 3. I'll tweak the answer.
Mar 30, 2018 at 23:01 comment added Jhonny Sorry for the late comment, but what about the (000) and (011) combination for the second excited state of fermions? Why is that not valid/does contribute to the degeneracy?
Mar 30, 2018 at 9:37 vote accept Jhonny
Mar 30, 2018 at 23:00
Mar 24, 2018 at 16:45 vote accept Jhonny
Mar 30, 2018 at 9:34
Mar 17, 2018 at 13:21 history edited Cosmas Zachos CC BY-SA 3.0
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Mar 16, 2018 at 10:55 history edited Cosmas Zachos CC BY-SA 3.0
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Mar 16, 2018 at 0:38 history edited Cosmas Zachos CC BY-SA 3.0
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Mar 15, 2018 at 22:43 history edited Cosmas Zachos CC BY-SA 3.0
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Mar 15, 2018 at 22:37 history answered Cosmas Zachos CC BY-SA 3.0