[QM] Total angular momentum rotation operator

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Homework Help Overview

The discussion revolves around proving a property of the rotation operator in quantum mechanics, specifically how the state \( e^{-i{\pi}J_x/\hbar}|j,m\rangle \) relates to \( |j,-m\rangle \). The context involves angular momentum operators and their eigenstates.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the Taylor series expansion of the rotation operator and its implications. There is a reference to transformation laws of rotation generators and eigenvalue equations. Questions arise regarding the nature of eigenstates and the potential for phase factors in the states.

Discussion Status

Participants are actively engaging with the mathematical framework and questioning the assumptions about eigenstates. Some guidance has been offered regarding the nature of eigenstates and the role of phase factors, but no consensus has been reached on the exact relationship between the states.

Contextual Notes

There is an ongoing discussion about the definitions of eigenstates and the implications of phase factors in quantum mechanics. The original poster's inquiry is situated within the constraints of quantum mechanics principles and the properties of angular momentum operators.

Rovello
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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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