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 Exercise [22.26] 
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Joined: 23 Aug 2010, 13:12
Posts: 33
Post Exercise [22.26]
Partial solution attached


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19 Sep 2010, 17:32

Joined: 12 Jul 2010, 07:44
Posts: 154
Post Re: Exercise [22.26]
Smith has shown above that any state \{a,b\} where |a|^2+|b|^2=1 is an eigenstate of some operator x\boldsymbol{L_1}+y\boldsymbol{L_2}+z\boldsymbol{L_3} where x^2+y^2+z^2=1, \; x,y,z\in\mathbb{R}. But since this also holds true for all complex multiples of the operator or the eigenstate, we can conclude that every state \{a,b\} is an eigenstate of the angular momentum operator x\boldsymbol{l_1}+y\boldsymbol{l_2}+z\boldsymbol{l_3} along the (not necessarily unit) vector \left(x,y,z\right) in real 3-dimensional space. (Recall \boldsymbol{L_1}=i\hbar\boldsymbol{l_1}, etc).

That completes part (i) of the problem.

For part (ii) we note that since \boldsymbol{H}^2 is a representation space of SO(3), we can rotate the spin-up state \left|\boldsymbol{\uparrow}\right> to point in any direction (by applying the appropriate rotation R from SO(3)). If \mathbb{P}\boldsymbol{H}^2 represents the set of physically distinguishable eigenstates (i.e. we consider complex multiples of \{a,b\} to represent the same thing), then there must be an element of \mathbb{P}\boldsymbol{H}^2 corresponding to each possible rotated spin state R\left|\boldsymbol{\uparrow}\right>. But \mathbb{P}\boldsymbol{H}^2 is isomorphic to the Riemann sphere, i.e. to exactly the set of all possible directions in 3-space; so a topology preserving mapping R\left|\boldsymbol{\uparrow}\right>\rightarrow\mathbb{P}\boldsymbol{H}^2, R\in SO(3) must therefore be onto \mathbb{P}\boldsymbol{H}^2; and hence every element of \mathbb{P}\boldsymbol{H}^2 (and hence every spin state \{a,b\}) corresponds to some rotation R\left|\boldsymbol{\uparrow}\right> of the spin-up state... i.e. to a spin eigenstate along some axis in 3-space.


\blacksquare


02 Dec 2012, 09:23
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