I want to compute the square of the angular momentum operator in spherical coordinates. I already know how the cartesian components look like: \begin{align} L_x &= -i\hbar \left(-\sin\phi\,\partial_{\theta} - \cos\phi\,\cot\theta \,\partial_{\phi}\right)\\ L_y &= -i\hbar \left(\cos\phi\,\partial_{\theta} - \sin\phi\,\cot\theta \,\partial_{\phi}\right)\\ L_z &= -i\hbar\,\partial_{\phi} \end{align} The square should then be given by: $L^2 = L_x^2 + L_y^2 + L_z^2$

With that, what I would get is this:

$$L^2 = -\hbar^2 \left(\partial_{\theta}^2 + \cot^2\theta\,\partial_{\phi}^2 +\partial_{\phi}^2 \right) = -\hbar^2 \left(\partial_{\theta}^2 + (\cot^2\theta + 1)\partial_{\phi}^2\right) = -\hbar^2\left(\partial_{\theta}^2 + \frac{1}{\sin^2\theta}\partial_{\phi}^2\right)$$

From the 3 terms in the bracket, the first term comes from $\cos^2\phi + \sin^2\phi = 1$, the second term from the same rule, and the 3rd term is the square of the $L_z$ component. The mixed terms you get, when you square $L_x$ and $L_y$ should cancel:

$$2\,\sin\phi\,\partial_{\theta}\,\cos\phi\,\cot\theta\,\partial_{\phi} - 2\,\cos\phi\,\partial_{\theta}\,\sin\phi\,\cot\theta\,\partial_{\phi} = 0 $$

However, the solution I find everywhere is:

$$L^2 = \left(\partial_{\theta}^2 + \cot\theta\,\partial_{\theta} + \frac{1}{\sin^2\theta}\partial_{\phi}^2\right)$$

I can't find where the error is.


The error is that \begin{align} &L_x^2f(\theta,\phi)\\ &=\left( -i\hbar \left(-\sin\phi\,\partial_{\theta} - \cos\phi\,\cot\theta \,\partial_{\phi}\right)\right) \left( -i\hbar \left(-\sin\phi\,\partial_{\theta} - \cos\phi\,\cot\theta \,\partial_{\phi}\right)\right)f(\theta,\phi)\\ &\ne-\hbar^2\left(\sin^2\phi\partial^2_{\theta^2}+2\cos\phi\sin\phi\cos\theta \partial^{2}_{\phi,\theta} +\cos^2\phi\cot^2\theta\partial^2_{\phi^2}\right) f(\theta,\phi) \end{align} since the derivatives $\partial_\theta$ and $\partial_\phi$ do not commute with the functions $\sin\phi$ and $\cos\phi\cot\theta$ in the expressions of the operators. In other words, there are additional cross-terms in $L_x^2$ beyond $2\cos\phi\sin\phi\cos\theta \partial^{2}_{\phi,\theta}\partial_{\phi\theta}$; these extra cross-terms are linear in $\partial_\theta$ and $\partial_\phi$ and come - for instance - from the product rule applied to \begin{align} -\sin\phi\partial_\theta \left(-\cos\phi\cot\theta\partial_\phi f(\theta,\phi)\right)&= -\sin\phi\cos\phi\csc^2\theta\partial_\phi f(\theta,\phi)\\ &\qquad +\sin\phi\cot\theta\cos\phi\partial^2_{\theta\phi}f(\theta,\phi)\, . \end{align} The same argument applies to $L_y^2$. $L_z^2$ is easy since the coefficients in front of the derivative are constant.

| cite | improve this answer | |

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.