Angular momentum - maximum and minimum values for $m_{\ell}$

Question

I want to work out the maximum and minimum values for $m_{\ell}$. I know that $\lambda \geq m_{\ell}$, therefore $m_{\ell}$ is bounded. In the lectures notes there is the following assumption: $$ \hat{L_{+}}|\lambda,m_{max}\rangle=|0\rangle \\ \hat{L_{-}}|\lambda,m_{min}\rangle=|0\rangle $$ I think I understand this. Since the action of the ladder operators is the keep the value of $\lambda$ and raise (or lower) $m_{\ell}$, you cannot "go up" from $m_{max}$ or down from $m_{min}$. However, I do not understand why the result of the operation should be $|0\rangle$.

It turns out we can write the produt of $\hat{L_{-}}\hat{L_{+}}$ as: $$ \hat{L_{-}}\hat{L_{+}}= \hat{L^2}-\hat{L_{z}^2}-\hbar\hat{L_{z}} $$ Then we we evalute the following expression: $$ \hat{L^2} |\lambda,m_{max}\rangle = (\hat{L_{-}}\hat{L_{+}}+\hat{L_{z}^2}+\hbar\hat{L_{z}})|\lambda,m_{max}\rangle $$ Since $\hat{L_{+}}|\lambda,m_{max}\rangle=|0\rangle $, then $\hat{L_{-}}\hat{L_{+}}|\lambda,m_{max}\rangle=\hat{L_{-}}|0\rangle =|0\rangle $. And $\hat{L_z}|\lambda,m_{max}\rangle = \hbar m_{\ell}|\lambda,m_{max}\rangle $. These two relations imply: $$ \hat{L^2} |\lambda,m_{max}\rangle =\hbar^2 m_{max}(m_{max}+1)|\lambda,m_{max}\rangle $$ Now I want to know how to compute $\hat{L^2} |\lambda,m_{min}\rangle $ since my lecture notes only state the result. My problem is that I will have $\hat{L_{-}}\hat{L_{+}}|\lambda,m_{min}\rangle$, but I can no longer say that $\hat{L_{+}}|\lambda,m_{min}\rangle=|0\rangle$. I tried to compute $\hat{L_{+}}\hat{L_{-}}$ and try to plug in the expression, but I had no success. How can I solve this?

PS. This is not homework, I'm just trying to derive the expression stated in the lecture notes.

Answer 1

Your lecture notes, or your transcription of them, are in error. You should have $$ \hat{L_{+}}|\lambda,m_\mathrm{max}\rangle= 0 \\ \hat{L_{-}}|\lambda,m_\mathrm{min}\rangle= 0 $$ That is, raising the maximum-projected state doesn't give you $\left|\lambda,m\right> = \left|0,0\right>$, a state with no angular momentum, and it doesn't give you a vacuum state $\left|0\right>$ with no particles in it. It gives you the number zero. This means, among other things, that the overlap of any state $\left|x\right>$ with $\hat{L_{+}}|\lambda,m_\mathrm{max}\rangle$ is zero.

As for your question about computing $ \hat {L^2} \left| \lambda,m \right>, $ your lecture notes should contain enough information for you to prove that the commutator between $L^2$ and $L_z$ is zero: \begin{align} L^2 L_z \left|x\right> = L_z L^2 \left|x\right>, \quad\quad \text{ for any state $\left|x\right>$} \end{align} which means that the eigenvalue of $L^2$ cannot depend on the eigenvalue $m$ of $L_z$. In fact the eigenvalue of $L^2$ on a state $\left|\lambda,m\right>$ is always $\hbar^2\lambda(\lambda+1)$, which is the same as your result since $m_\mathrm{max} = -m_\mathrm{min} = \lambda$.