Consider the states $\left|j,m_x\right>_x$ and $\left|j,m_z\right>_z$ with total angular momentum $j$ and the angular momentum operators $\hat{S}_x$ and $\hat{S}_y$. In particular, assume that the states satisfy $\hat{S}_x\left|j,m_x\right>_x=m_x\left|j,m_x\right>_x$ and $\hat{S}_z\left|j,m_z\right>_z=m_z\left|j,m_z\right>_z$. I am wondering about the closed form of the overlap between the eigenstates shown, that is, what is the analytical expression for $_x\left<j,m_x\lvert j,m_z \right>_z$.

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    $\begingroup$ $S$, $L$, $J$ are usually spin, orbit, total angular momentum, with quantum numbers $|s, s_z\rangle$, $|l, m\rangle$, $|j, j_z\rangle$ or $3$ in place of $z$. $\endgroup$
    – JEB
    Commented Sep 2, 2023 at 0:48

1 Answer 1


The transformation that takes you from $S_z$ to $S_x$ is a rotation about $\hat y$ by $\pi/2$ so all eigenstates of $S_x$ are of the form $$ \vert jm\rangle_x=R^{-1}_y(\pi/2)\vert jm\rangle_z $$ so that $$ S_xR^{-1}_y(\pi/2)\vert jm\rangle_z= [R^{-1}_y(\pi/2) S_z R_y(\pi/2)]R^{-1}_y(\pi/2)\vert jm\rangle_z=m R^{-1}_y(\pi/2)\vert jm\rangle_z $$ Thus, the closed form is $$ _x\langle jm’\vert jm\rangle_z =_z\langle jm’\vert R_y(\pi/2)\vert jm\rangle_z=d^{j}_{m’m}(\pi/2) $$ where $d$ is a Wigner little-d function.

(I hope I didn’t mess up the sign between $\pi/2$ and $-\pi/2$ but aside from this the answer is correct.)

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    $\begingroup$ I think you are right in your signs; checks for j=1/2, unless I messed up myself... $\endgroup$ Commented Sep 1, 2023 at 21:39

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