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Dimension of Hilbert space (quantum mechanics)
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[QUOTE="Haye, post: 4498177, member: 363552"] s should be 1/2, sorry for that. Thanks for pointing me in the right direction, I'm a lot closer to the answer now (I think so, atleast). If I am correct, the 'spin-angle functions' would give me the number of bases vector, and thus the dimension of the Hilbert-space. This is dependent on the quantum numbers s and m[SUB]s[/SUB], l and m[SUB][/SUB] and j and m[SUB]j[/SUB]. s=1/2, so m[SUB]s[/SUB]=-1/2 , +1/2 l=1, so m[SUB]l[/SUB] = -1, 0, +1 j= 1/2 or 3/2, so m[SUB]j[/SUB] = -1/2 , 1/2 OR -3/2, -1/2, +1/2, +3/2 For j=1/2 you get 2*3*2=12 different combinations, and for j=3/2 it's 2*3*4=24 different combinations. This gives 36 different base vectors, and therefore the dimension of Hilbert space would be 36? I am completely unsure if my logic is correct, and I would greatly appreciate it if someone could tell me if I am wrong or not. vela, thank you so much for pointing me in the right direction at least. I feel a lot less lost than I was. [/QUOTE]
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Dimension of Hilbert space (quantum mechanics)
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