Suppose that $|s_z\rangle$ and $|s_x\rangle$ are eigenvectors of operators $\hat{S}_z$ and $\hat{S}_x$ correspondingly, with $s_z$ and $s_x$ being eigenvalues. Is there a known formula for the scalar product $\langle s_x|s_z\rangle$? Spin $1/2$ case is trivial. Spin $1$ case is simple enough to express $|s_x\rangle$ vectors as linear combinations of $|s_z\rangle$ vectors and to find this scalar product directly. What about the general case?


The general case is given by rotating the z-axis to the x-axis by $\pi/2$ around the y-axis. The matrices which achieve that in the spherical basis are Wigner's celebrated d rotation matrices where you plug in $\theta=-\pi /2$ (but beware of my signs, not fully guaranteed!).

Thus, for any spin s, $$ \langle s_x|= \langle s_z | e ^{-i(\pi/2) \hat S_y}~, \\ \langle s_x|s_z\rangle= d^s_{s_x ~ s_z} (-\pi/2) ~. $$

So, e.g., in Feynman vIII, 7-12 , you get $\langle +|+\rangle=\langle + | -\rangle = 1/\sqrt 2$ for spin 1/2, and you may work out for spin 2 that $\langle 2| 1\rangle = 1/2\sqrt 2 $ , etc... Note the interchange identities for d at the end of that section.

  • $\begingroup$ Thank you very much! For my work, I need these formulas for spin $2$ and spin $5/2$. So your answer solves the problem. $\endgroup$ – Gec Jul 10 '19 at 16:31

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