This question already has an answer here:

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 question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


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 $\ell$.

  • $\begingroup$ This is the reason. A Details worth adding might be that this shows up in the coordinate space representation, so in real space. Which is the part differentiating it from spin with half integers. $\endgroup$ – pindakaas Feb 23 '15 at 7:37

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