The Angular Momentum Operator in Spherical Coordinates
Matthew Louis / October 2025 (514 Words, 3 Minutes)
In quantum mechanics, the angular momentum operator is defined via
\[\hat{\boldsymbol{L}} = \hat{\boldsymbol{r}}\times \hat{\boldsymbol{p}} = -i\hbar\hat{\boldsymbol{r}}\times \boldsymbol{\nabla}\]There is a handy identity that lets us decompose the total kinetic energy operator into a radial part and an angular part (in spherical coordinates)
\[\hat{\boldsymbol{L}}^2 = \hat{\boldsymbol{r}}^2\hat{\boldsymbol{p}}^2 - (\hat{\boldsymbol{r}}\cdot \hat{\boldsymbol{p}})^2 + i\hbar \hat{\boldsymbol{r}}\cdot\hat{\boldsymbol{p}}\](Townsend, 2000) which can then be used to find expressions for the radial angular momentum operator \(\hat{p}_r\) and \(\hat{\boldsymbol{L}}\) (in spherical coordinates). This is the basis of 3D quantum mechanics for spherically symmetric potentials.There is, however, an even more direct way to derive the position space representation of \(\hat{\boldsymbol{L}}\) in spherical coordinates, using its covariant definition (see my last blog post). The expression for the \(\mu\)th contravariant component of the operator acting on a scalar field \(\phi\) is
\[(\hat{\boldsymbol{L}}\phi)^\mu = -i\hbar(\boldsymbol{r}\times \nabla\phi)^\mu = -i\hbar \varepsilon^{\mu \nu \lambda}x_\nu \partial_\lambda \phi\]Recall that for spherical coordinates, the Lame coefficients are \(h_r = 1, h_\theta = r, h_\varphi = r\sin\theta\), and thus
\[g_{\mu\nu} = \begin{pmatrix} 1 & 0 & 0\\ 0 & r^2 & 0\\ 0 & 0 & r^2\sin^2\theta \end{pmatrix}\]The physical position vector is given by \(\boldsymbol{r}=r\hat{\textbf{e}}_r\), and its “ordinary” components are \((\tilde{x}_r, \tilde{x}_\theta, \tilde{x}_\varphi)= (r, 0, 0)\). Its contravariant components may be obtained by using the relation \(A^\mu =\tilde{A}_\mu /h_\mu\) no sum, which gives $x^r = r, x^\theta = 0, x^\varphi = 0$. Now, the Levi-Civita Tensor $\varepsilon^{\mu \nu \lambda}$ is related to the Levi-Civita symbol $\epsilon^{\mu \nu \lambda}$ by $\varepsilon^{\mu \nu \lambda} = \frac{1}{\sqrt{g}}\epsilon^{\mu \nu \lambda}$, where $\sqrt{g} = r^2\sin\theta$. Now, the sum over the index $\nu$ in $x_\nu$ is only nonzero when $\nu=r$, where $x_r = r$, so we may write
\[(\hat{\boldsymbol{L}}\phi)^r = -i\hbar r \varepsilon^{\mu r \lambda} \partial_\lambda \phi\]Then, we may write each individual component
\[(\hat{\boldsymbol{L}}\phi)^r = -i\hbar r \varepsilon^{r r \lambda} \partial_\lambda \phi=0\]because the Levi-Civita symbol is zero when any index is repeated.
\[\nonumber \begin{equation} \begin{split} (\hat{\boldsymbol{L}}\phi)^\theta &= -i\hbar r \varepsilon^{\theta r \lambda} \partial_\lambda \phi \\ &= -i\hbar r \varepsilon^{\theta r \varphi} \partial_\varphi \phi\hspace{5mm} \text{Sum only nonzero when $\lambda = \varphi$}\\ &= \frac{i\hbar}{r\sin\theta}\frac{\partial \phi}{\partial \varphi} \end{split} \end{equation}\] \[\nonumber \begin{equation} \begin{split} (\hat{\boldsymbol{L}}\phi)^\varphi &= -i\hbar r \varepsilon^{\varphi r \lambda} \partial_\lambda \phi \\ &= -i\hbar r \varepsilon^{\varphi r \theta} \partial_\theta \phi\hspace{5mm} \text{Sum only nonzero when $\lambda = \theta$}\\ &= -\frac{i\hbar}{r\sin\theta}\frac{\partial \phi}{\partial \varphi} \end{split} \end{equation}\]These are the contravariant components. Now we just need to relate them to the ordinary physical components $\tilde{L}_\mu$.
\[\nonumber \begin{equation} \begin{split} \tilde{L}_r &= h_r L^r = 1 \cdot 0 = 0\\ \tilde{L}_\theta &= h_\theta L^\theta = \frac{i\hbar}{\sin\theta}\frac{\partial}{\partial \theta}\\ \tilde{L}_\varphi &= h_\varphi L^\varphi = -i\hbar \frac{\partial}{\partial \theta} \end{split} \end{equation}\]so in sum
\[\hat{\boldsymbol{L}} = -i\hbar\left( - \frac{1}{\sin\theta} \hat{\textbf{e}}_\theta\partial_\varphi + \frac{1}{r\sin\theta}\hat{\textbf{e}}_\varphi \partial_\varphi\right)\]If we use the fact that
\[\hat{\textbf{e}}_z = \cos\theta \hat{\textbf{e}}_r - \sin\theta \hat{\textbf{e}}_\theta\]then we may compute the dot product and find that
\[\hat{L}_z = \hat{\boldsymbol{L}}\cdot \hat{\textbf{e}}_z = -i\hbar\partial_\varphi\]which is the linear angular momentum in the $\varphi$ direction. We may use the same formalism to find an expression for $L^2$, the squared magnitude of the vector operator $\hat{\boldsymbol{L}}$. This is computed by taking the scalar product of the operator with itself, for which the covariant expression is the following contraction
\[L^2 = \hat{L}^\mu \hat{L}_\mu\]which we note is different than \(\hat{L}_\mu \hat{L}^\mu\), because $\hat{L}^\mu $ is a differential operator, and the metric components used to lower the index depend on the coordinates, so the order of operations matters. The ordering \(\hat{L}^\mu \hat{L}_\mu\) is the one that correctly corresponds to the geometric definition of the dot product for vector operators, analogous to the Laplacian $\nabla^2 = \nabla^\mu \nabla_\mu $. Now, we write out the components of this contraction as before. For the $\theta$ term:
\[\nonumber \begin{equation} \begin{split} \hat{L}^\theta(\hat{L}_\theta\phi) &= \left( \frac{i\hbar}{r\sin\theta}\frac{\partial}{\partial \varphi} \right) \left( \frac{i\hbar r}{\sin\theta} \frac{\partial\phi}{\partial \varphi} \right) \\ &= \frac{(i\hbar)^2 r}{r\sin^2\theta} \frac{\partial^2\phi}{\partial\varphi^2} \\ &= -\frac{\hbar^2}{\sin^2\theta} \frac{\partial^2\phi}{\partial\varphi^2} \end{split} \end{equation}\]For the $\varphi$ term, we must use the product rule:
\[\nonumber \begin{equation} \begin{split} \hat{L}^\varphi(\hat{L}_\varphi\phi) &= \left( -\frac{i\hbar}{r\sin\theta}\frac{\partial}{\partial \theta} \right) \left( -i\hbar r \sin\theta \frac{\partial\phi}{\partial \theta} \right) \\ &= -\frac{\hbar^2 r}{r\sin\theta} \frac{\partial}{\partial\theta} \left( \sin\theta \frac{\partial\phi}{\partial \theta} \right) \end{split} \end{equation}\]Combining the terms gives the final operator: \(\hat{\boldsymbol{L}}^2 = -\hbar^2 \left[ \frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial}{\partial\theta}\right) + \frac{1}{\sin^2\theta}\frac{\partial^2}{\partial\varphi^2} \right]\) Another example illustrating the convenience of covariant definitions of operators.
References
Townsend, J. S. (2000). A modern approach to quantum mechanics. University Science Books.
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