Proving that rotations of the angular momentum eigenstates are corresponding eigenstates

Question

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?

Answer 1

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, $$ \exp (-i\pi \sigma_x /4) ~~\sigma_z ~\exp (i\pi \sigma_x /4)\\ = (\cos\pi/4 -i\sigma_x \sin\pi/4)\sigma_z (\cos\pi/4 +i\sigma_x \sin\pi/4)\\ =\sigma_y. $$