# Connection between half and whole integer eigenvalues for orbital angular momentum [duplicate]

I have been trying to follow this derivation from Sakurai and Shankar, pulling from both. I would like to see how the following derivation can be extended to orbital angular momentum, and thus find that $\ell$ is an integer. The details are omitted but the core of the proof is here.

Consider the following

\begin{align*} [\hat{J}^2,\hat{J}_z] &= 0 \\ [\hat{J}_z,\hat{J}_\pm] &= \pm\hbar\hat{J}_\pm \\ \hat{J}_\pm &\equiv \hat{J}_x \pm \mathrm{i}\hat{J}_y. \end{align*}

Now let \begin{align*} \hat{J}^2 \lvert \alpha, \beta \rangle &= \alpha \lvert \alpha, \beta \rangle \\ % \hat{J}_z \lvert \alpha, \beta \rangle &= \beta \lvert \alpha, \beta \rangle \end{align*}

From the above we can see that $$\hat{J}_z \hat{J}_\pm \lvert \alpha, \beta \rangle = (\beta \pm \hbar) \hat{J}_\pm \lvert \alpha, \beta \rangle$$

and $$\hat{J}^2 \hat{J}_\pm \lvert \alpha, \beta \rangle = \hat{J}_\pm \hat{J}^2 \lvert \alpha, \beta \rangle = \alpha \hat{J}_\pm \lvert \alpha, \beta \rangle$$

From this we see that $$\hat{J}_\pm \lvert \alpha, \beta \rangle \propto \lvert \alpha, (\beta + \hbar) \rangle \\ % \implies \hat{J}_\pm \lvert \alpha, \beta \rangle = C(\alpha,\beta) \lvert \alpha, (\beta + \hbar) \rangle.$$

Now we see that there is an upper limit on $\beta$ $$\langle \alpha \beta \rvert \hat{J}^2 - \hat{J}_z^2 \lvert \alpha \beta \rangle = \langle \alpha \beta \rvert \hat{J}_x^2 + \hat{J}_y^2 \lvert \alpha \beta \rangle \\ \implies \alpha \geq \beta^2$$

So $$\hat{J}_- \hat{J}_+ \lvert \alpha, \beta_{max} \rangle = (\hat{J}^2 - \hat{J}_z^2 - \hbar\hat{J}_z) \lvert \alpha, \beta_{max} \rangle = 0 \\ \implies \alpha = \beta_{max}(\beta_{max} +\hbar).$$

Similarly $$\alpha = \beta_{min}(\beta_{min} +\hbar)$$

From this we can show $$\beta_{min} = -\beta_{max}.$$

So $$\beta_{max} = \frac{\hbar n}{2} \\ \implies \frac{\beta_{max}}{\hbar} = \frac{n}{2} = j.$$

Thus the eigenvalues are $$\alpha = \hbar^2 j(j + 1).$$

Now we define $$\beta \equiv m \hbar$$

and we have the eigen kets $$\lvert j, m \rangle$$

where $j$ increments in half integer steps.

Question: So the only added restriction to the derivation of the integer values of the orbital angular momentum quantum number $\ell$ is $$\vec{L} = \vec{r} \times \vec{p}.$$ How does this added restriction require that the orbital angular momentum quantum number $\ell$ be an integer and more importantly how can I show this using the proof above?

## marked as duplicate by Kyle Kanos, Neuneck, Qmechanic♦Feb 23 '15 at 14:29

This may not be exactly what you want, but it's something. The definition $$\vec{L}=\vec{r}\times\vec{p}$$ combined with $\vec{p}=-i\hbar\vec{\nabla}$ implies that the the $z$ projection may be written as $$L_{z} = -i\hbar\frac{\partial}{\partial \phi}.$$ Where $\phi$ is the azimuthal angle. Since $L_{z}|\ell m\rangle = m\hbar |\ell m\rangle$, the $\phi$ dependence of the wave function can be written as $e^{im\phi}$. In order for the wave function to be single-valued, we require $$e^{im\phi} = e^{im(\phi+2\pi)}$$ and so $m$ must be an integer, and thus so must $\ell$.