# Failure or incomplete demostration about egeinvalues of $J^2$ and $J_z$ using lowering and raising operators

In many books show how find eigenvalues of $J^2$ and $J_z$

\begin{align} \hat{J}^2 |\ell,m\rangle & = \hbar^2 \ell(\ell+1) |\ell,m\rangle , \\ \hat{J_z} |\ell,m\rangle & = \hbar m |\ell,m \rangle . \end{align}

We use the rules of commutation and ladder operators to find the values of $\ell$ and $m$ and we arrive at the relation:

$$m_\mathrm{min} \leq m \leq m_\mathrm{max}.$$

Now, if we apply $J_{-}$ or $J_{+}$ $N$ times on eigenkets until reaching or exceeding $m_\mathrm{min}$ or $m_\mathrm{max}$, of course it follows that

$$\hat{J}^N_{-} |\ell,m\rangle = 0. \tag{1}$$

but how do you come to the conclusion that this alone is true?

$$\hat{J}_{-} |\ell,m_\mathrm{min} \rangle = 0$$

If in the interval $N-1$ and $N+1$ there are others kets that can fulfill the condition (1) and not only $m_\mathrm{min}$. Equal for $m_\mathrm{max}$

$$\hat{J_{-}} |\chi \rangle = 0 \\ \hat{J_{-}} |\phi \rangle = 0$$

First $J_+$ and $J_-$ are adjoint of each others, so
$$\|J_-|l,m\rangle\|^2=\langle l,m|J_+J_-|l,m\rangle=\langle l,m|J^2-J_z^2+\hbar J_z|l,m\rangle=\big(l(l+1)-m(m-1)\big)\hbar^2$$
This is 0 for $m=-l$.
Easy. If $\hat{J}_{-} |\ell,m_\mathrm{min} \rangle$ were nonzero, then it would be a nonzero common eigenstate of $\hat{J}^2$ and $\hat{J}_z$, with eigenvalue $\hbar(m_\mathrm{min}-1)$ for the latter, which is a contradiction with the definition of $m_\mathrm{min}$. Since we know that this $m_\mathrm{min}$ must exist (because $\hat{J}^2-\hat{J}_z$ is positive semidefinite) then it follows that it must vanish under $\hat J_-$.