[QM] Total angular momentum rotation operator

AI Thread Summary
The discussion focuses on proving that the state e^{-i{\pi}J_x/\hbar}|j,m\rangle is proportional to |j,-m\rangle, where J_x is the x-component of the total angular momentum operator. The approach involves using the Taylor series expansion of the rotation operator and the transformation law of rotation generators. It is established that applying the rotation operator U to the eigenstate |j,m\rangle results in an eigenstate of J_z with eigenvalue -m. However, it is clarified that U|j,m\rangle does not necessarily equal |j,-m\rangle but can be expressed as a phase factor times |j,-m\rangle. The discussion concludes that redefining the eigenstates can simplify the relationship, allowing for the absorption of the phase factor.
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Homework Statement


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

Homework Equations



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

The Attempt at a Solution


Taylor series expansion of the exponential term?
e^{-i{\pi}J_x/\hbar}=1-i\frac{{\pi}J_x}{\hbar} - \frac{1}{2}(\frac{{\pi}J_x}{\hbar})^2 +...
 
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You have to use the equation:

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

which comes from the transformation law of the rotation generators:

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

Now consider the eigen-value equation:

{{J}_{3}}\left| jm \right\rangle =m\left| jm \right\rangle

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

{{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

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

I have one question (doubt),

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

Isn't it?
 
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.
 
Ok! Thank you so much!
 
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