Vector calculus: angular momentum operator in spherical coordinates

• MisterX
In summary, the gradient operator is defined in spherical coordinates (except along the polar axis, where spherical coordinates are singular). To evaluate \hat{\vec{L}}^2, it's easier to go back to Cartesian coordinates first.
MisterX
Note: physics conventions, $\theta$ is measured from z-axis
We have a vector operator

$\vec{L} = -i \vec{r} \times \vec{\nabla} = -i\left(\hat{\phi} \frac{\partial}{\partial \theta} - \hat{\theta} \frac{1}{\sin\theta} \frac{\partial}{\partial \phi} \right)$
And apparently
$\vec{L}\cdot\vec{L}= -\left(\frac{\partial^2}{\partial \theta^2} + \cot \theta\frac{\partial}{\partial \theta}+ \frac{1}{\sin^2\theta} \frac{\partial^2}{\partial \phi^2} \right)$
I am wondering about a way to obtain the second expression (the $\cot \theta$ term in particular) from the first expression without taking the circuitous route followed in my references. I realize the unit vectors aren't constant.
$\frac{\partial \boldsymbol{\hat{\theta}}} {\partial \phi} = \cos \theta\boldsymbol{\hat \phi}$
$\frac{\partial \boldsymbol{\hat{\phi}}} {\partial \theta} = 0$

The basis vectors are orthonormal however, and for the norm-squared of a regular vector in spherical coordinates, we can just square each of the components and add.

while I'm not quite sure what you're asking for (not a physicist), i can say the del operator is not explicitly defined in spherical coordinates. thus, if you are looking for a way to literally take $\vec{L}\cdot \vec{L}$ perhaps the only way is the "circuitous" route from your references.

maybe this helps a little?

while I'm not quite sure what you're asking for (not a physicist), i can say the del operator is not explicitly defined in spherical coordinates. thus, if you are looking for a way to literally take $\vec{L}\cdot \vec{L}$ perhaps the only way is the "circuitous" route from your references.

maybe this helps a little?

while I'm not quite sure what you're asking for (not a physicist), i can say the del operator is not explicitly defined in spherical coordinates. thus, if you are looking for a way to literally take $\vec{L}\cdot \vec{L}$ perhaps the only way is the "circuitous" route from your references.

maybe this helps a little?

The gradient operator is defined in spherical coordinates (except along the polar axis, where spherical coordinates are singular).

To evaluate $\hat{\vec{L}}^2$, it's easier to go back to Cartesian coordinates first.
$$\hat{\vec{L}}^2 \psi =-\epsilon_{ijk}x_j \partial_k (\epsilon_{ilm} x_l \partial_m) \psi.$$
First we use the product rule and then to get
$$\epsilon_{ijk} \epsilon_{ilm}=\delta_{jl} \delta_{km}-\delta_{jm} \delta_{kl}$$
$$\hat{\vec{L}}^2 \psi =-(delta_{jl} \delta_{km}-\delta_{jm} \delta_{kl}) x_j (\delta_{kl} \partial_m+x_j x_l \partial_k \partial_m)=-(r^2 \Delta - 2 (\vec{x} \cdot \vec{\nabla}) \psi - x_j (\vec{x}\cdot \vec{\nabla}) \partial_j \psi).$$
The latter expression is not nice, but we can use
$$(\vec{x} \cdot \vec{\nabla})(\vec{x} \cdot \vec{\nabla})=x_j \partial_j (x_k \partial_k \psi) = (\vec{x} \cdot \vec{\nabla}) \psi + x_j (\vec{x} \cdot \vec{\nabla}) \partial_j \psi$$
to finally write
$$\hat{\vec{L}}^2 \psi = - (r^2 \Delta - (\vec{x} \cdot \vec{\nabla}) \psi - (\vec{x} \cdot \vec{\nabla})(\vec{x} \cdot \vec{\nabla}) \psi.$$
Now you use the Laplacian in spherical coordinates and
$$\vec{x} \cdot \vec{ \nabla} \psi = r \hat{r} \cdot \vec{\nabla} \psi = r \partial_r \psi$$
to finally get
$$\hat{\vec{L}}^2 \psi=-\frac{1}{\sin \vartheta} \left [\frac{\partial}{\partial \vartheta} \left (\sin \vartheta \frac{\partial \psi}{\partial \vartheta} \right ) + \frac{1}{\sin \vartheta}\frac{\partial^2 \psi}{\partial \varphi^2} \right ].$$

vanhees71 said:
To evaluate $\hat{\vec{L}}^2$, it's easier to go back to Cartesian coordinates first.

I guess what I'm wondering is if there is a way without going back to Cartesian coordinates.

It's a bit tricky in non-Cartesian coordinates as soon as vector (or higher-rank tensor) fields are involved, and $\hat{\vec{L}} \psi$ already is a vector field. That's why it is much simpler to first derive a scalar expression in Cartesian coordinates, which you already now how to express in spherical coordinates.

1. What is the angular momentum operator in spherical coordinates?

The angular momentum operator in spherical coordinates is a mathematical operator that represents the angular momentum of a particle or system in three-dimensional space. It is denoted as L and is defined as the cross product of the position vector and the linear momentum vector.

2. How is the angular momentum operator calculated in spherical coordinates?

In spherical coordinates, the angular momentum operator is calculated using the formula L = -iħ(sinθ∂/∂φ + cotθcosφ∂/∂θ + cotθsinφ∂/∂r), where ħ is the reduced Planck's constant, θ is the polar angle, and φ is the azimuthal angle.

3. What is the physical significance of the angular momentum operator in spherical coordinates?

The angular momentum operator in spherical coordinates is a fundamental quantity in quantum mechanics and has a physical significance that describes the rotational motion of a particle or system in three-dimensional space. It is a conserved quantity that plays a crucial role in understanding the behavior of atoms, molecules, and other quantum systems.

4. How does the angular momentum operator behave under rotations in spherical coordinates?

The angular momentum operator in spherical coordinates behaves like a vector under rotations. This means that it follows the same transformation rules as a vector, and its components change accordingly. This property is important in the study of symmetries and the conservation of angular momentum in quantum systems.

5. Can the angular momentum operator be used to find the quantum states of a particle in spherical coordinates?

Yes, the angular momentum operator can be used to find the quantum states of a particle in spherical coordinates. The eigenvalues and eigenvectors of the operator correspond to the allowed values of the angular momentum, and the associated quantum states, respectively. This allows for the prediction and analysis of the behavior of quantum systems in spherical coordinates.

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