2
$\begingroup$

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

$$\mathscr{Y}\psi_{jm}=\lambda\psi_{jm}$$

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?

$\endgroup$

2 Answers 2

3
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ 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
0
$\begingroup$

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;

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.