Smith, what makes you think that

and

are adjoint operators?

The adjoint of

would be:

This is only equal to

if

and

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

are also eigenstates of

, 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 2

j, corresponding to

m values from -

j+1 to

j. It is indeed true that these are the values for which

is nonzero, but by comparison the values of

m for which

is nonzero are -

j to

j-1; as the main text says,

m actually runs from -

j to

j. The 2

j dimension in the exercise statement is actually a typo - it ought to read:

Quote:

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

...as the representation space is in fact (2

j+1)-dimensional.