In the context of diatomic molecules I have seen it has come up that an eigenvalue $M_L\hbar$ of $L_z$ is doubly degenerate with respect to reflection about a plane containing the $z$ axis, whose corresponding operator I shall call $A_y$, since $A_yL_zA_y = -L_z$. Therefore one chooses the basis of $L_z^2,A_y$ instead of $L_z$ on its own since the hamiltonian also commutes with $A_y$.

My question is thus the following: how does one pass from the basis determined by $L_z^2,L_z$ to the basis determined by $L_z^2,A_y$?

$ |\lambda,m\rangle \to |\lambda,a\rangle ?$

where $\lambda = |m|$ and $a$ is the eigenvalue ($\pm1$) of $A_y$. This is a two dimensional space for which the two vectors $|\lambda,m\rangle,|\lambda,-m\rangle$ or $|\lambda,+1\rangle,|\lambda,-1\rangle$ form a basis.

Due to symmetry I expect something along the lines of:

$ |\lambda,a\rangle = \frac{1}{\sqrt{2}} (|\lambda\rangle \pm |-\lambda\rangle )$

where there is some link between $a,\lambda$ and the sign of $\pm$.

Since the real spherical harmonics $Y_{lm}$ are invariant under this mirror reflection I then thought that the answer would be like:

$ |\lambda,\pm1\rangle = |\lambda\rangle \pm (-1)^{\lambda}|-\lambda\rangle $

But I really don't know how to do this, I thought of using a similar method to the one used for spherical harmonics by means of the ladder operators but it doesn't apply (at least in an obvious way) in this case. I'm pretty sure the answer is easy but I can't figure it out.

Thanks for any help.


I believe the only difference is the labelling, since due to the degeneracy introduced by the reflection symmetry you need to use a different set of commuting observables, see (Axial Symmetry bellow):


For the situation you describe $L^2$ and $L_z$ are not a Complete Set of Commuting Observables (CSCO). To illustrate, let us imagine a Hamiltonian of the form $H=\alpha L_z^2$. It is invariant under the action of $A_y$, since $A_yL_z^2=L_z^2A_y$. The energy eigenbasis is formed simply by the eigenvectors of $L_z$, which we denote by $|m\rangle$ - thus $L_z|m\rangle=m|m\rangle$. Now, measurements of energy will give us $H|m\rangle=\alpha m^2|m\rangle$ and we cannot differentiate between, say, $|m=-1\rangle$ or $|m=1\rangle$. An extra quantum number is necessary and the value $(\pm 1)$ given by $A_y$ gives us this extra information. The basis is still the same, the labelling is different.

Perhaps another way of seeing this is to consider the wave functions in position representation $\psi(\mathbf{r})=\langle \mathbf{r}|m\rangle$. $|\lambda,+1\rangle$ are those $\psi(\mathbf{r})$ even under the reflection, while $|\lambda,+1\rangle$ are those wave functions that are odd under the reflection.

Hope this helps somehow.

  • $\begingroup$ I edited the question as to make it clearer what I was asking. I wasn't talking about the whole space, just the two dimensional subspace generated by $|\pm m\rangle$ (with $\lambda$ fixed), which you can imagine as fixing everything else and varying only the eigenvalues of $L_z$ or equivalently $A_y$. $\endgroup$ Jul 21 '20 at 9:19

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