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 Exercise [22.23] 
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Joined: 23 Aug 2010, 13:12
Posts: 33
Post Exercise [22.23]
Partial Solution attached

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18 Sep 2010, 20:24

Joined: 12 Jul 2010, 07:44
Posts: 154
Post Re: Exercise [22.23]
Smith, what makes you think that L^+ and L^- are adjoint operators?

The adjoint of L^+ would be:

\left(L^+\right)^* = \left(L_1+i L_2\right)^* = L_1^*-i L_2^*

This is only equal to L^- if L_1 and L_2 are both self-adjoint (Hermitian). It isn't obvious to me that this is the case.

Also, your solution is indeed only (very) partial, because you haven't established that all your eigenstates |\psi> are also eigenstates of J^2, nor that they all share the same eigenvalue j(j+1), nor why values of m below -j-1 or above j+1 can't be counted, nor why *all* values from -j to j should be, nor that there is only one eigenstate (and hence one dimension) per eigenvalue...

Furthermore, your solution is misleading, because it suggests that the dimension is 2j, corresponding to m values from -j+1 to j. It is indeed true that these are the values for which L^-|\psi> is nonzero, but by comparison the values of m for which L^+|\psi> is nonzero are -j to j-1; as the main text says, m actually runs from -j to j. The 2j dimension in the exercise statement is actually a typo - it ought to read:

"...then the dimension is an integer 2j+1, where..."

...as the representation space is in fact (2j+1)-dimensional.

11 Nov 2012, 21:30
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