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Angular Momentum Operator in Spherical Coordinates 📂Quantum Mechanics

Angular Momentum Operator in Spherical Coordinates

Formula

The angular momentum operator is expressed in spherical coordinates as follows.

$$ \begin{align*} L_{x} &= \i\hbar \left(\sin\phi\dfrac{\partial }{\partial \theta} + \cos\phi \cot\theta \dfrac{\partial }{\partial \phi}\right) \\ L_{y} &= -\i\hbar \left( \cos\phi \dfrac{\partial }{\partial \theta} - \sin\phi \cot\theta \dfrac{\partial }{\partial \phi}\right) \\ L_{z} &= -\i\hbar \dfrac{\partial }{\partial \phi} \end{align*} $$

Derivation

The definition of the angular momentum operator is as follows.

$$ L = \mathbf{r} \times P = - \i\hbar (\mathbf{r} \times \nabla) $$

At this time, $\mathbf{r} = (X, Y, Z)$ is the position operator, $P$ is the momentum operator, $\nabla$ is the del operator. The del operator in spherical coordinates is as follows.

$$ \nabla = \widehat{\mathbf{r}}\dfrac{\partial}{\partial r} + \widehat{\boldsymbol{\theta}}\dfrac{1}{r}\dfrac{\partial}{\partial \theta} + \widehat{\boldsymbol{\phi}}\dfrac{1}{r\sin\theta}\dfrac{\partial}{\partial \phi} $$

Since $\mathbf{r} = r \widehat{\mathbf{r}}$, $L$ is as follows.

$$ \begin{align*} L &= -\i\hbar (r \widehat{\mathbf{r}} \times \nabla) \\ &= -\i\hbar \left( \widehat{\mathbf{r}} \times \widehat{\mathbf{r}} r \dfrac{\partial}{\partial r} + \widehat{\mathbf{r}} \times \widehat{\boldsymbol{\theta}}\dfrac{\partial}{\partial \theta} + \widehat{\mathbf{r}} \times \widehat{\boldsymbol{\phi}} \dfrac{1}{\sin\theta}\dfrac{\partial}{\partial \phi} \right) \\ &= -\i\hbar \left( \widehat{\boldsymbol{\phi}} \dfrac{\partial}{\partial \theta} - \widehat{\boldsymbol{\theta}} \dfrac{1}{\sin\theta}\dfrac{\partial}{\partial \phi} \right) \\ \end{align*} $$

At this time, the two unit vectors can be represented as follows in Cartesian coordinates.

$$ \begin{align*} \widehat{\boldsymbol{\theta}} &= \cos\phi \cos\theta \widehat{\mathbf{x}} + \sin\phi \cos\theta \widehat{\mathbf{y}} - \sin\theta\widehat{\mathbf{z}} \\ \widehat{\boldsymbol{\phi}} &= -\sin\phi \widehat{\mathbf{x}} + \cos\phi \widehat{\mathbf{y}} \end{align*} $$

Therefore, $L$ is as follows.

$$ \begin{align*} L &= -\i\hbar \left[ \left( -\sin\phi \widehat{\mathbf{x}} + \cos\phi \widehat{\mathbf{y}} \right) \dfrac{\partial}{\partial \theta} - \left( \cos\phi \cos\theta \widehat{\mathbf{x}} + \sin\phi \cos\theta \widehat{\mathbf{y}} - \sin\theta\widehat{\mathbf{z}} \right) \dfrac{1}{\sin\theta}\dfrac{\partial}{\partial \phi} \right] \\ &= -\i\hbar \left[ \widehat{\mathbf{x}}\left(-\sin\phi\dfrac{\partial }{\partial \theta} - \cos\phi \cot\theta \dfrac{\partial }{\partial \phi}\right) + \widehat{\mathbf{y}}\left( \cos\phi \dfrac{\partial }{\partial \theta} - \sin\phi \cot\theta \dfrac{\partial }{\partial \phi}\right) + \widehat{\mathbf{z}}\left( \dfrac{\partial }{\partial \phi} \right) \right]\\ \end{align*} $$

Therefore, each component is as follows.

$$ \begin{align*} L_{x} &= \i\hbar \left(\sin\phi\dfrac{\partial }{\partial \theta} + \cos\phi \cot\theta \dfrac{\partial }{\partial \phi}\right) \\ L_{y} &= -\i\hbar \left( \cos\phi \dfrac{\partial }{\partial \theta} - \sin\phi \cot\theta \dfrac{\partial }{\partial \phi}\right) \\ L_{z} &= -\i\hbar \dfrac{\partial }{\partial \phi} \end{align*} $$