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[QM] Total angular momentum rotation operator

  • Thread starter Rovello
  • Start date
  • #1
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Homework Statement


How to prove that for any representation of the spin, the state [tex] e^{-i{\pi}J_x/\hbar}|j,m\rangle[/tex]
is proportional to [tex]|j,-m\rangle[/tex]
The exponential term is the rotation operator where [itex] J_x [/itex] is the x-component of the total angular momentum operator,
and [itex]|j,m\rangle[/itex] is an eigenket.

Homework Equations



[itex] J_x=\frac{1}{2}(J_+ + J_-)[/itex] where [itex]J_+ [/itex] and [itex] J_- [/itex] are the ladder operators.
[itex] J_±|j,m\rangle=\sqrt{(j{\mp}m)(j±m+1)}|j,m±1> [/itex]

The Attempt at a Solution


Taylor series expansion of the exponential term?
[itex]e^{-i{\pi}J_x/\hbar}=1-i\frac{{\pi}J_x}{\hbar} - \frac{1}{2}(\frac{{\pi}J_x}{\hbar})^2 +... [/itex]
 

Answers and Replies

  • #2
123
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You have to use the equation:

[itex]U\left[ {{R}_{1}}\left( \pi \right) \right]{{J}_{3}}{{U}^{-1}}\left[ {{R}_{1}}\left( \pi \right) \right]=-{{J}_{3}}[/itex]

which comes from the transformation law of the rotation generators:

[itex]U\left( R \right){{J}^{ij}}{{U}^{-1}}\left( R \right)={{R}_{k}}^{i}{{R}_{\ell }}^{j}{{J}^{k\ell }}[/itex]

Now consider the eigen-value equation:

[itex]{{J}_{3}}\left| jm \right\rangle =m\left| jm \right\rangle [/itex]

and make the [itex]U\left[ {{R}_{1}}\left( \pi \right) \right]{{J}_{3}}{{U}^{-1}}\left[ {{R}_{1}}\left( \pi \right) \right] [/itex] appear in it, like that:

[itex]{{J}_{3}}\left| jm \right\rangle =m\left| jm \right\rangle \Rightarrow U{{J}_{3}}{{U}^{-1}}U\left| jm \right\rangle =mU\left| jm \right\rangle \Rightarrow -{{J}_{3}}U\left| jm \right\rangle =mU\left| jm \right\rangle [/itex]

This shows that [itex]U\left| jm \right\rangle \equiv \exp \left( -i\pi {{J}_{1}} \right)\left| jm \right\rangle [/itex] is an eigen-state of [itex]{{J}_{3}}[/itex] , which corresponds to the eigen-value [itex]-m[/itex] .
 
  • #3
3
0
Thank you, cosmic dust!

I have one question (doubt),

if [itex] U|j,m\rangle[/itex] is an eigen-state of [itex]J_z[/itex], therefore [itex] U|j,m\rangle=|j,-m\rangle[/itex], because [itex] J_z(U|j,m\rangle)=J_z|j,-m\rangle=-m|j,-m\rangle[/itex].

Isn't it?
 
  • #4
123
0
Almost... To show that a state is an eigenstate of some operator, all you have to do is to show that when that operator acts on that state, gives the same state multiplied by some constant (like the last of the equalities I presented). In general, U|jm> does not have to be equal to |j,-m>, since it can be any state of the form z|j,-m>, where z is a phase factor. But, without loss of generality, you can always redifine the eigenstates of J3 or J3 its self, in such a way that the phase factor gets absorbed by the new definitions.
 
  • #5
3
0
Ok! Thank you so much!
 

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