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My aim is to prove the restriction for the orbital momentum quantum number $-\ell \leq m \leq \ell$. My professor was giving me the hint, that I should use the norm of the state $|| L_+ |\ell,m\rangle ||$ with the state vector $|\ell, m\rangle$.

So, I start with the norm

$$|| L_+ |\ell,m\rangle ||^2 = \langle\ell,m| L_-L_+ |\ell,m\rangle \geq 0.$$

First of all, I calculate the operator product:

$$L_-L_+ = (L_x - iL_y)(L_x + i L_y) = L_x^2 + L_y^2 + i[L_x, L_y] = \vec{L}^2 - L_z^2 - \hbar L_z.$$

Using the eigenvalues of the operator, I get

$$|| L_+ |\ell,m\rangle || ^2 = \underbrace{\langle\ell, m|\ell,m\rangle}_{=1} \, \hbar^2 \big( \ell(\ell+1) - m(m+1) \big) \geq 0.$$

Finally I have

$$\ell(\ell + 1) \geq m(m+1).$$

Calculating now the norm $|| L_- |\ell, m\rangle ||$ leads me to a similar inequality:

$$\ell(\ell + 1) \geq m(m-1).$$

My question now is, how do get from these two inequalities to the restriction $-\ell \leq m \leq \ell$?

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    $\begingroup$ There is a standard notation for kets and bras. Why not use it? $\endgroup$
    – G. Smith
    Oct 14 '20 at 20:16
  • $\begingroup$ I tried \bra{.} and \ket{.} they didnt seem to work, I don't know another command. $\endgroup$
    – Octavius
    Oct 14 '20 at 20:16
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    $\begingroup$ You can use \langle, \mid, and \rangle. $\endgroup$
    – G. Smith
    Oct 14 '20 at 20:17
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    $\begingroup$ Using | for \mid works as well. You can also do \right > or \left < as well. So some options for ket of $\psi$. \mid\psi\rangle gives $\mid\psi\rangle$, \left|\psi\right> also gives $\left|\psi\right>$. I usually just do |\psi\rangle : $|\psi\rangle$ $\endgroup$ Oct 14 '20 at 21:09
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Your inequality can be read as

$$f(\ell)\geq f(m)$$ where $f(x)=x(x+1)$. This can be reduced to $\ell\geq m$ by applying $f^{-1}$ to both sides. But there's a caveat. Our function $f$ is not an injection since multiple x-values can map to the same number. This means we can't define the inverse easily and we have to do more work.

Our function is a parabola with a minimum at $x=-1/2$. We can make this inversion a little easier by shifting to this minimum using $\tilde \ell=\ell+1/2,\ \tilde m=m+1/2$. \begin{align} \ell(\ell+1)&\ge m(m+1)\\ (\tilde \ell-1/2)(\tilde\ell+1/2)&\ge(m-1/2)(m+1/2)\\ \tilde\ell^2-1/4&\geq \tilde m^2-1/4\\ \tilde \ell^2&\geq \tilde m^2\\ |\tilde \ell|&\ge|\tilde m| \end{align} This last line means $-\tilde \ell\le\tilde m\leq\tilde l\implies-\ell\le m\le \ell$ where I assumed $\ell$ to be positive.

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    $\begingroup$ Clear and complete. Thank you very much for your answer. $\endgroup$
    – Octavius
    Oct 14 '20 at 23:47

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