# Eigenvalue of Orbital Operator

I read a literature, which discusses the orbital operator.

$$𝐿^2$$ and $$𝐿_𝑧$$ are square of the orbital operator and the $$z$$-axis component of orbital operator respectively. $$|l\rangle$$ and $$|m\rangle$$ are the eigenvectors for the $$𝐿^2$$ and $$𝐿_𝑧$$ operators respectively. $$𝐿_+$$ is defined as $$𝐿_+=𝐿_𝑥+𝑖𝐿_𝑦$$.

The literature says that $$(𝐿^2+2ℏ𝐿𝑧−2ℏ𝐿_+)|𝑙𝑙\rangle=(𝑙^2+3𝑙)ℏ^2|𝑙𝑙\rangle \tag{1}$$ and $$𝐿_𝑧|𝑙𝑙\rangle=𝑙ℏ|𝑙𝑙\rangle \tag{2}$$

I do not understand why it is like that. I can understand that $$𝐿^2|𝑙𝑙\rangle=𝑙(𝑙+1)ℏ^2|𝑙𝑙\rangle$$ because $$|l\rangle$$ is the eigenvector of the square of the orbital operator ($$L^2$$). If $$|l\rangle$$ is replaced by $$|m\rangle$$, I can understand that $$𝐿_𝑧|𝑚\rangle=𝑚ℏ|𝑚\rangle$$.

However, $$|l\rangle$$ is not the eigenvector of the $$z$$-axis component of orbital operator ($$𝐿_𝑧$$). How does the author obtain that $$(2ℏ𝐿_𝑧−2ℏ𝐿_+)|𝑙𝑙\rangle=2𝑙ℏ^2|𝑙𝑙\rangle$$ and $$𝐿_𝑧|𝑙𝑙\rangle=(𝑙+1)ℏ|𝑙𝑙\rangle$$

Could anyone give me some hint on these equations (1) and (2)? Thank you very much in advance.

• Are you sure about $(2)$? That should be $L_z|l,l\rangle=l\hbar|l,l\rangle$. Oct 27 '20 at 18:08
• "The literature says that" -- which literature? Your second equation is definitely wrong, but it's hard to tell if your source is wrong, or whether you've misinterpreted it, without a precise reference. Oct 27 '20 at 18:24

$$L^2$$ and $$L_z$$ commute, so it's possible to find a simultaneous eigenbasis of both operators. We label the states in this basis $$|l,m\rangle$$, where $$L^2|l,m\rangle = l(l+1)\hbar^2|l,m\rangle$$ $$L_z|l,m\rangle = m\hbar^2|l,m\rangle$$
The state $$|l,l\rangle$$ is such an eigenstate with $$m=l$$, so $$L_z|l,l\rangle=l\hbar|l,l\rangle$$ and $$L^2|l,l\rangle=l(l+1)\hbar^2|l,l\rangle$$
Note also that $$L_+|l,l\rangle=0$$.
• @JMurray@EmilioPisanty Thank you very much for your comments and suggestions. I wrote equation 2 in the wrong way. It should be $L_z|ll\rangle=l\hbar |ll\rangle$. I have already modified my post. On the other hand, the second $l$ in the simultaneous eigenbasis $|ll\rangle$ is still the eigen vector $m$ for the orbital operator $L$. This $m$ just takes the maximum state and becomes $l$ when the first $l$ takes the state $l$. This is why the $m$ is written as $l$ in the second simultaneous eigenbasis $|ll\rangle$. Is my understanding correct? Thank you again. Oct 28 '20 at 2:12