[QM] Total angular momentum rotation operator

In summary, the homework statement is 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.
  • #1
Rovello
3
0

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]
 
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  • #2
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
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
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
Ok! Thank you so much!
 

1. What is the total angular momentum rotation operator in quantum mechanics?

The total angular momentum rotation operator in quantum mechanics is a mathematical operator that describes the rotational motion of a quantum system. It is used to calculate the changes in the quantum state of a system when it is rotated in space.

2. How is the total angular momentum rotation operator calculated?

The total angular momentum rotation operator is calculated using the angular momentum operators for each direction (x, y, z) and the Euler angles (α, β, γ) that describe the rotation. The operator can be written as a combination of these operators and angles.

3. What is the significance of the total angular momentum rotation operator in quantum mechanics?

The total angular momentum rotation operator is significant in quantum mechanics because it allows us to study the rotational properties of quantum systems. It helps us understand how the quantum state of a system changes when it is rotated and can be used to make predictions about the behavior of quantum particles.

4. How does the total angular momentum rotation operator relate to the conservation of angular momentum?

The total angular momentum rotation operator is related to the conservation of angular momentum in quantum mechanics. This is because the operator is a representation of the total angular momentum of a system, which is conserved in a closed system. This means that the total angular momentum of a system will remain constant even if the system is rotated.

5. Can the total angular momentum rotation operator be used for all quantum systems?

Yes, the total angular momentum rotation operator can be used for all quantum systems. It is a fundamental operator in quantum mechanics and can be applied to any quantum system, regardless of its size or complexity. However, the calculations may become more complex for larger or more complex systems.

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