Restriction on orbital angular momentum quantum numbers

Question

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 $-\ell \leq m \leq \ell$?

Answer 1

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.