-1
$\begingroup$

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 $m_\mathrm{max}$

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

$\endgroup$
1
$\begingroup$

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 $m=-l$.

$\endgroup$
1
$\begingroup$

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 $\hat J_-$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.