[QM] Total angular momentum rotation operator

Click For Summary
SUMMARY

The discussion centers on proving that the state e^{-i{\pi}J_x/\hbar}|j,m\rangle is proportional to |j,-m\rangle for any representation of spin. The rotation operator is defined by J_x, the x-component of the total angular momentum operator. Key equations include J_x=\frac{1}{2}(J_+ + J_-) and the ladder operators J_±|j,m\rangle. The transformation law of rotation generators is utilized to demonstrate that U|j,m\rangle is an eigenstate of J_z, leading to the conclusion that U|j,m\rangle can be expressed as |j,-m\rangle, subject to the absorption of a phase factor.

PREREQUISITES
  • Understanding of angular momentum operators in quantum mechanics
  • Familiarity with the exponential form of operators
  • Knowledge of ladder operators and their action on eigenstates
  • Concept of eigenstates and eigenvalues in quantum mechanics
NEXT STEPS
  • Study the properties of rotation operators in quantum mechanics
  • Learn about the implications of the transformation law for angular momentum operators
  • Explore the significance of phase factors in quantum state representations
  • Investigate the role of eigenstates in quantum mechanics, particularly in relation to symmetry operations
USEFUL FOR

Quantum mechanics students, physicists specializing in angular momentum, and researchers exploring spin representations will benefit from this discussion.

Rovello
Messages
3
Reaction score
0

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 +...
 
Physics news on Phys.org
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!
 

Similar threads

Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
0
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
17
Views
3K
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K