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.

  • $\begingroup$ $\hat{L_{+}}|\lambda,m_{min}\rangle=|0\rangle$ is not 0, it should be sth like $\hat{L_{+}}|\lambda,m_{min}\rangle=|1\rangle$, the next level (of ang. momentum) L+ and L- operators raise or lower the levels of ang. momentum, also cannot raise above max and cannot lower below min. $\endgroup$ – Nikos M. Jun 1 '14 at 3:57
  • $\begingroup$ I have already mentioned this in my question. $\endgroup$ – Thiago Jun 1 '14 at 5:05
  • $\begingroup$ The result is zero because this level has no states of the system that are compatible, i.e zero number of states of the system. Does this answer your question? Since L+ raised mmin to one (from zero), applying L- will lower the state back to zero $\endgroup$ – Nikos M. Jun 1 '14 at 5:08
  • $\begingroup$ No, it does not. The result is a ket vector $|0\rangle$ not the number 0. And my question was on how to compute $\hat{L^2}|\lambda,m_{min}\rangle$ $\endgroup$ – Thiago Jun 1 '14 at 5:11
  • $\begingroup$ Ket vector representing the number of states in this level, correct? It is a quantized harmonic oscillator-type creation-annihilation model, if i am not mistaken $\endgroup$ – Nikos M. Jun 1 '14 at 5:13

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

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    $\begingroup$ How does an operator acting on a (quantum) state have a simple number (in this case $0$) as a result? $\endgroup$ – Nikos M. Jul 2 '14 at 22:19
  • $\begingroup$ @NikosM. Maybe a more precise way to put it is to use the properly-normalized version, $L_+\left|jm\right> = \sqrt{(j-m)(j+m+1)}\left|j,m+1\right>$. In that case whatever state vector is produced by $L_+\left|jj\right>$ is multiplied by a numerical constant which happens to be zero. That makes the state act like an additive identity. Compare to the state with zero angular momentum $\left|0,0\right>$, whose eigenvalue under $L^2$ is $\hbar^2$, or to a vacuum state $\left|0\right>$, which doesn't vanish under particle creation operators. $\endgroup$ – rob Jul 2 '14 at 22:56
  • $\begingroup$ yeap this makes it more clear. my answer has same problem with this notation. $\endgroup$ – Nikos M. Jul 2 '14 at 23:43

$\hat{L_{-}}\hat{L_{+}}|\lambda,m_{min}\rangle=|0\rangle$, since the first operator raises to next state and the next lowers back to "zero state"

From there you can proceed as previously.

  • $\begingroup$ ket |0> definitely not same as the number 0... $\endgroup$ – ZeroTheHero Dec 29 '16 at 5:58

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