# Scalar product of eigenstates of $\hat{S}_z$ and $\hat{S}_x$ operators

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