# Proving that rotations of the angular momentum eigenstates are corresponding eigenstates

If I apply a rotation operator about an arbitrary axis to a typical $$\mathbf{J},J_z$$ angular momentum eigenstate $$|j,m \rangle$$ then my sense from the development in Ballentine is that I also obtain an eigenstate (of the same $$j,m$$?) for the angular momentum along this axis. I can't seem to mathematize this, as a statement like $$(\mathbf{J} \cdot \hat{\mathbf{n}})e^{-i \mathbf{J} \cdot \hat{\mathbf{n}} \theta /\hbar}|j,m \rangle = m\hbar e^{-i \mathbf{J} \cdot \hat{\mathbf{n}} \theta /\hbar}|j,m \rangle$$ has one free parameter ($$\theta$$) and is therefore surely wrong.

Can someone set me on the right track? How do I obtain angular momentum eigenstates about a different axis by applying a rotation?

Indeed, your statement is wrong, but you almost have the right idea. Consider $$J_z|j,m \rangle = m\hbar |j,m \rangle~~~\leadsto \\ \left (e^{-i \mathbf{J} \cdot \hat{\mathbf{n}} \theta /\hbar} J_z e^{i \mathbf{J} \cdot \hat{\mathbf{n}} \theta /\hbar}\right ) \left( e^{-i \mathbf{J} \cdot \hat{\mathbf{n}} \theta /\hbar} |j,m \rangle \right ) = m\hbar \left (e^{-i \mathbf{J} \cdot \hat{\mathbf{n}} \theta /\hbar}|j,m \rangle\right ),$$ which means you already have the eigenstates of the rotated generator, $$J_{z'}\equiv e^{-i \mathbf{J} \cdot \hat{\mathbf{n}} \theta /\hbar} J_z e^{i \mathbf{J} \cdot \hat{\mathbf{n}} \theta /\hbar} ,$$ namely $$J_{z'} \left( e^{-i \mathbf{J} \cdot \hat{\mathbf{n}} \theta /\hbar} |j,m \rangle \right ) = m\hbar \left (e^{-i \mathbf{J} \cdot \hat{\mathbf{n}} \theta /\hbar}|j,m \rangle\right ).$$

In particular, you might take $$\theta=\pi/2$$ around $$\hat x$$, so $$z'=y$$, and marvel... For example, in the doublet representation where these finite rotations are explicitly computable, $$\exp (-i\pi \sigma_x /4) ~~\sigma_z ~\exp (i\pi \sigma_x /4)\\ = (\cos\pi/4 -i\sigma_x \sin\pi/4)\sigma_z (\cos\pi/4 +i\sigma_x \sin\pi/4)\\ =\sigma_y.$$

• Thank you for your answer! I am confused though. Does not your $J_{z'}$ equal my $\mathbf{J} \cdot \hat{\mathbf{n}}$?
– EE18
Commented Jul 27, 2023 at 1:47
• Not if you rotate z around the n axis! Think of rotating z around the x axis by a right angle: you get y, not x….. Commented Jul 27, 2023 at 2:06
• Ah, so you are saying in that case I gave that $\mathbf{J} \cdot \hat{\mathbf{n}}= J_x \neq J_y$, where the RHS is what I would get if I used the transformation law you gave?
– EE18
Commented Jul 27, 2023 at 2:17
• Yes. I fleshed out the example. Commented Jul 27, 2023 at 14:01
• Got it, thank you (again)!
– EE18
Commented Jul 28, 2023 at 18:31