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 $𝐿_+=𝐿_π‘₯+𝑖𝐿_𝑦$.

The literature says that $$(𝐿^2+2β„πΏπ‘§βˆ’2ℏ𝐿_+)|𝑙𝑙\rangle=(𝑙^2+3𝑙)ℏ^2|𝑙𝑙\rangle \tag{1}$$ 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 $𝐿_𝑧|π‘š\rangle=π‘šβ„|π‘š\rangle$.

However, $|l\rangle$ is not the eigenvector of the $z$-axis component of orbital operator ($𝐿_𝑧$). How does the author obtain that $(2ℏ𝐿_π‘§βˆ’2ℏ𝐿_+)|𝑙𝑙\rangle=2𝑙ℏ^2|𝑙𝑙\rangle$ 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.

  • 2
    $\begingroup$ Are you sure about $(2)$? That should be $L_z|l,l\rangle=l\hbar|l,l\rangle$. $\endgroup$
    – J. Murray
    Oct 27 '20 at 18:08
  • 3
    $\begingroup$ "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. $\endgroup$ 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 $L_+|l,l\rangle=0$.

  • $\begingroup$ @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. $\endgroup$
    – Kieran
    Oct 28 '20 at 2:12

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