A scalar operator $\mathscr{Y}$ is defined as an operator where the inner product $\langle\phi|\mathscr{Y}|\phi\rangle$ is unaltered by a rotation of the coordinate system. Then for an operator to be scalar it must commute with every element of the total angular momentum operator $\textbf{J}$:

$$[\mathscr{Y},\textbf{J}]=0 \tag{1}$$

from which it follows that if $\psi_{jm}$ is a simultaneus eigenfunction of $\textbf{J}^2$ and $J_z$ belonging to the quantum numbers $j$ and $m$, then $\langle\psi_{jm}|\mathscr{Y}|\psi_{j'm'}\rangle$ vanishes unless $j=j'$ and $m=m'$, and


Whereas the last condition is obvious; the state $\psi_{jm}$ must be an eigenstate of the operator $\mathscr{Y}$, I am not sure why $j=j'$ and $m=m'$. Why must we have the same eigenvector on each side of the inner product? I know this has something to do with orthogonal states, but why does the commutation relation in eq. (1) imply that the inner product can only yield a value if the two eigenstates are equal?


2 Answers 2


From Eq(1), $$ \begin{align} [ \mathscr Y,J_z]&=0,\\ \mathscr YJ_z &=J_z\mathscr Y,\\\tag{a} \langle \psi_{jm}|\mathscr YJ_z|\psi_{j'm'} \rangle &= \langle \psi_{jm}|\mathscr YJ_z|\psi_{j'm'} \rangle ,\\\tag{b} m'\langle \psi_{jm}|\mathscr Y|\psi_{j'm'} \rangle &= m\langle \psi_{jm}|\mathscr Y|\psi_{j'm'} \rangle ,\\ (m'-m)\langle \psi_{jm}|\mathscr Y|\psi_{j'm'} \rangle &= 0.\\ \end{align} $$ What goes from Eq(a) to (b): $$ \begin{align} J_z|\psi_{j'm'}\rangle&=m',\\ \langle\psi _{jm}|J_z&=\langle\psi _{jm}|J_z^\dagger \\&=\left( J_z|\psi_{jm}\rangle \right)^\dagger \\&= (m|\psi_{jm}\rangle)^\dagger \\&=m^*\langle \psi_{jm}| \\&=m\langle \psi_{jm}|. \end{align} $$

Similar to $J^2$.

$|\psi_{jm}\rangle$ doesn't neet to be non-degenerate. Suppose $|\psi_{jmi}\rangle$ are degenerate states with eigenvalue $j(j+1)$ for $J^2$ and $m$ for $J_z$. Any state $\psi_{jm}$ in the subspace spanned by these states can be expressed as their linear combination, and all above equations still hold.

  • $\begingroup$ How did you go from eq. 3 to eq. 4? Why are you allowed to multiply by different integers on each side? $\endgroup$ Jul 31, 2023 at 16:36
  • $\begingroup$ @RasmusAndersen because the two operators commute, on the right side you can switch them and apply Jz to the left state, which gives the m eigenvalue. $\endgroup$
    – pmal
    Jul 31, 2023 at 16:43
  • $\begingroup$ Thank you. Perhaps you can answer one more question; if an eigenvalue is non-degenerate it has only one eigenvector. If $\psi$ is the eigenvector of eigenvalue $\lambda$, then $\lambda$ is the eigenvalue of an integer multiple of $\psi$. Is this always the case? If so why? $\endgroup$ Jul 31, 2023 at 16:58
  • $\begingroup$ Also if what you say is true then $J_z\langle \psi_{jm} |=m\langle\psi_{jm} |$. Of course the angular momentum operator is hermitian, so we can move it to the dual space, but why does the dual space yield the eigenvalue m? $\endgroup$ Jul 31, 2023 at 17:14
  • $\begingroup$ @RasmusAndersen See my edits. Your proposition about eigenvalues is not true. $\endgroup$
    – Luessiaw
    Jul 31, 2023 at 19:47

Since the scalar operator $\mathscr{Y}$ commutes with every element of the total angular momentum operator $\textbf{J}$ these operators share the same eigenstates. From the angular momentum formalism we know that the eigenstates of the total angular momentum are orthonormal by definition. Therefore the eigenstates of $\mathscr{Y}$ must be orthonormal as well which leads to;

$$\langle\psi_{jm}|\mathscr{Y}|\psi_{j'm'}\rangle=\lambda \delta_{jj'}\delta_{mm'}$$


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