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

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

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 $m_\mathrm{max} = -m_\mathrm{min} = \lambda$.