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.


1 Answer 1


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.


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