Angular Momentum

Property	Linear Momentum	Angular Momentum
Physical nature	Motion along a straight line	Rotation about an axis
Vector form	$\vec{p} = m\vec{v}$	$\vec{L} = \vec{r} \times \vec{p}$
Effective mass	$m$	$I = m r^2$
Velocity	$v$	$\omega=\frac{v}{r}$
Magnitude (scalar form)	$\mid\vec{p}\mid = m v$	$\mid\vec{L}\mid = I \omega$
Kinetic energy	$E_k = \dfrac{p^2}{2m}$	$E_\text{rot} = \dfrac{L^2}{2I}$
Quantum operator	$\hat{p}_x = -i\hbar\dfrac{\partial}{\partial x}$	$\hat{L}_x = \hat{y}\hat{p}_z - \hat{z}\hat{p}_y$
Conservation law	Conserved if no net external force acts	Conserved if no net external torque acts
Conservation condition	$\sum \vec{F}_\text{ext} = 0$	$\sum \vec{\tau}_\text{ext} = 0$

velocity vs angular velocity

In circular motion, angular velocity ( $\omega$ ) measures how fast the angle changes, while linear velocity ( $v$ ) measures how fast the object moves along the circular path.

For a full rotation, the arc length traveled is related to the angle by

s = r\theta.

(1)

Differentiating with respect to time gives

\frac{ds}{dt} = r\frac{d\theta}{dt},

(2)

or equivalently,

v = r\omega \quad \Rightarrow \quad \omega = \frac{v}{r}.

(3)

Thus, angular velocity is the rate of rotation per unit radius: objects farther from the axis move faster linearly for the same angular speed.

Moment of inertia

Consider a rigid body rotating about a fixed axis with angular velocity $\omega$ . Each particle $i$ of mass $m_i$ at distance $r_i$ from the axis moves with linear velocity

v_i = r_i \omega.

(4)

The kinetic energy of that particle is

E_i = \tfrac{1}{2} m_i v_i^2 = \tfrac{1}{2} m_i (r_i \omega)^2 = \tfrac{1}{2} m_i r_i^2 \omega^2.

(5)

For the whole body, total rotational kinetic energy is the sum over all particles:

E_\text{rot} = \sum_i E_i = \tfrac{1}{2} \omega^2 \sum_i m_i r_i^2.

(6)

We define the term in parentheses as the moment of inertia:

I = \sum_i m_i r_i^2,

(7)

E_\text{rot} = \tfrac{1}{2} I \omega^2.

(8)

Interpretation

The $r_i^2$ factor naturally appears from the kinetic energy of rotational motion: it weights each mass element by how far it lies from the axis, determining its contribution to rotational inertia.

Conservation of momentum is due to symmetry¶

DeD0 — Figure 1:Emmy Noether, a German mathematician whose first and second theorems are fundamental to mathematical physics. See more about her here.

Conservation of momenta

Linear momentum is defined as $\vec{p} = m\vec{v}$

From Newton’s second law,

\frac{d\vec{p}}{dt} = \sum \vec{F}_\text{ext}.

(9)

Thus, when no external force acts on a system,

\sum \vec{F}_\text{ext} = 0 ;\Rightarrow; \frac{d\vec{p}}{dt} = 0,

(10)

meaning linear momentum is conserved.

Angular momentum is defined as $\vec{L} = \vec{r} \times \vec{p}$ .
Taking the time derivative gives
$\frac{d\vec{L}}{dt} = \vec{r} \times \frac{d\vec{p}}{dt} = \sum \vec{\tau}_\text{ext}.$
(11)

Therefore, when no external torque acts on the system,

\sum \vec{\tau}_\text{ext} = 0 ;\Rightarrow; \frac{d\vec{L}}{dt} = 0,

(12)

so angular momentum is conserved.

Summary:

No external force → linear momentum conserved.
No external torque → angular momentum conserved.

Classical angular momentum¶

In classical mechanics, the angular momentum is defined via a cross product of position and linear momentum. The cross product is convenient to write using a determinant:

{\vec{L} = \vec{r}\times\vec{p} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\ x & y & z\\ p_x & p_y & p_z\\ \end{vmatrix}= \left(yp_z - zp_y\right)\vec{i} + \left(zp_x - xp_z\right)\vec{j} + \left(xp_y - yp_x\right)\vec{k}}

(13)

where $\vec{i}, \vec{j}$ and $\vec{k}$ denote unit vectors along the $x, y$ and $z$ axes. $p_x$ and $L_x$ are components of linear and angular momentum, respectively.

Classical picture: Rotating dumbbell¶

The rigid rotor is a model of a rotating dumbbell: two unequal masses held together via a rigid stick. The system is not acted upon by any external potential; hence the only energy is the kinetic energy of rotation:

K=\frac{m_1 v_1^2}{2}+\frac{m_2 v_2^2}{2}=\frac{m_1 r_1^2+m_2 r^2_2}{2}\omega^2

(18)

where we have plugged in $v_1=\omega r_1$ and $v_2=\omega_2 r$ , the rotational velocities of the two masses rotating with frequency $\omega$ . The classical mechanical problem of two masses is once again reducible to a single reduced mass $\mu$ rotating at constant radius $r=r_1+r_2$ around the center of mass $m_1 r_1=m_2 r_2$ .

K=\frac{I \omega^2}{2}=\frac{L^2}{2I}

(19)

here $L=I \omega$ is the angular momentum, $I=\mu r^2$ is the moment of inertia, and $\mu=\frac{m_1 m_2}{m_1+m_2}$ .

Spherical coordinates¶

Spherical coordinates are more convenient for rotational problems where we replace $(x,y,z)$ by $(r, \phi, \theta)$ .
For instance when considering rotation with fixed orbit $r=const$ we are able to eliminate one degree of freedom associated with radial direction.

Laplacian

Cartesian

Polar

\nabla^2 = \frac{{\partial^2}}{{\partial x^2}} + \frac{{\partial^2}}{{\partial y^2}} + \frac{{\partial^2}}{{\partial z^2}}

(27)

Solution

\int^a_0\int^b_0\int^c_0 dxdydz=a\cdot b\cdot c

(29)

\int^r_0\int^{2\pi}_0\int^\pi_0 r^2 \sin \theta \, dr \, d\theta \, d\phi=\frac{r^3}{3}\Big|^r_0 \cdot 2\pi \cdot (-cos\theta) \Big|^\pi_0\cdot 2\pi=\frac{4\pi r^3}{3}

(30)

Solution

\nabla^2 = {\color{blue}\frac{1}{{ r^2}}}\Big[ {\color{green}\frac{1}{\sin(\theta)}}{\color{green}\frac{{\partial}}{{\partial \theta}} \left(\sin(\theta) \frac{{\partial}}{{\partial \theta}}\right)} + {{\color{green}\frac{1}{{ \sin^2(\theta)}}} {\color{orange} \frac{{\partial^2}}{{\partial \phi^2}}}}\Big] = {\color{blue}\frac{1}{r^2}}\nabla^2_{\theta, \phi}

(31)

Quantum angular momentum¶

In quantum mechanics, the classical angular momentum is replaced by the corresponding quantum mechanical operator.
In spherical coordinates, the angular momentum operators can be written in the following form. The derivations are quite tedious, involving multiple applications of the chain rule, but they are just a straightforward mathematical procedure. Note that the choice of $z$ -axis here was arbitrary. Sometimes the physical system implies such an axis naturally (for example, the direction of an external magnetic field).

Cartesian

Polar

{\hat{L}_x = -i\hbar\left(y\frac{\partial}{\partial z} - z\frac{\partial}{\partial y}\right)}

(32)

{\hat{L}_y = -i\hbar\left(z\frac{\partial}{\partial x} - x\frac{\partial}{\partial z}\right)}

(33)

{\hat{L}_z = -i\hbar\left(x\frac{\partial}{\partial y} - y\frac{\partial}{\partial x}\right)}

(34)

\hat{L}^2=\hat{L_x}^2+\hat{L_y}^2+\hat{L_z}^2

(35)

Solution

The eigenvalue equation is:

\hat{L}_z f(\phi) = a f(\phi)

(41)

Substituting the operator gives

-i\hbar\frac{df}{d\phi} = a f(\phi)

(42)

f(\phi) = e^{im\phi}, \quad a = \hbar m

(43)

Thus, the eigenfunctions of $\hat{L}_z$ are $e^{im\phi}$ and the corresponding eigenvalues are $\hbar m$ , where $m$ is an integer quantum number.

Components of angular momentum do not commute!¶

Unlike linear momentum, the components of angular momentum do not commute!
This implies that it is not possible to measure any of the Cartesian angular momentum pairs simultaneously with an infinite precision (the Heisenberg uncertainty relation).

Note the cyclical nature of the commutators between components.

Eigenfunctions and eigenvalues of $L$ and $L_z$ ¶

Since the operators for a component and the angular momentum square commute, it is possible to find functions that are eigenfunctions of both $\vec{\hat{L}}^2$ and $\hat{L}_z$ .

Note that here $m$ has nothing to do with magnetism but the name originates from the fact that (electron or nuclear) spins follow the same laws of angular momentum.
$m$ describes the projections of $L$ on the z axis. E.g., for l=2 there are five projections $(-2, -1, 0, 1, 2)$ .
Functions $Y_l^m$ are called spherical harmonics. Examples of spherical harmonics with various values of $l$ and $m$ are given below (with the Condon-Shortley phase convention).

Spherical harmonics¶

Spherical harmonics, denoted as $Y_{lm}(\theta, \phi)$ or $|l, m\rangle$ are important in many theoretical and practical applications, e.g., the representation of multipole electrostatic and electromagnetic fields, computation of atomic orbital electron configurations, representation of gravitational fields, MRI imaging for streamline tractography, and the magnetic fields of planetary bodies and stars.
Spherical harmonics emerge from the angular part of the solutions to the Laplace equation $\nabla^2 f=0$ in spherical coordinates, which is the type of equation we obtained for the rigid rotor problem and will see once more in the hydrogen atom problem.

$Y_{lm}(\theta, \phi)$	Expression	Colatitudinal Nodes	Azimuthal Nodes
$Y_{00}$	$\frac{1}{\sqrt{4\pi}}$	0	0
$Y_{10}$	$\sqrt{\frac{3}{4\pi}} \cos(\theta)$	1 (equatorial)	0
$Y_{1-1}$	$\sqrt{\frac{3}{4\pi}} \sin(\theta) e^{-i\phi}$	0	1
$Y_{20}$	$\sqrt{\frac{5}{16\pi}} (3\cos^2(\theta) - 1)$	2 (colatitudinal rings)	0
$Y_{2-1}$	$\sqrt{\frac{15}{4\pi}} \sin(\theta) \cos(\theta) e^{-i\phi}$	1 (equatorial)	1
$Y_{2-2}$	$\sqrt{\frac{15}{4\pi}} \sin^2(\theta) e^{-2i\phi}$	0	2

Nodes of Spherical Harmonics¶

The nodal structure of spherical harmonics is influenced by both the degree $l$ and the order $m$ , defining the number and type of nodes as follows:

Colatitudinal Nodes (Polar Nodes): The variable $\theta$ defines “polar” nodes, which appear as circular bands around the sphere. The expression in $\cos(\theta)$ or $\sin(\theta)$ creates colatitudinal nodes:
- For example, $Y_{10}$ with $\cos(\theta)$ has a single node at $\theta = \pi/2$ , creating an equatorial node.
- $Y_{20}$ , with $(3\cos^2(\theta) - 1)$ , introduces two polar nodes, dividing the sphere into three regions along the colatitude.
Azimuthal Nodes (Longitudinal Nodes): The variable $\phi$ defines “azimuthal” nodes due to the terms $e^{im\phi}$ , which create lines of longitude where the function changes phase. The number of azimuthal nodes is determined by $|m|$ :
- If $m = 0$ , there are no azimuthal nodes, as seen in $Y_{00}$ and $Y_{10}$ .
- For $m = \pm 1$ , a single azimuthal node occurs (e.g., $Y_{1-1}$ , $Y_{2-1}$ ), and for $m = \pm 2$ , two azimuthal nodes appear (e.g., $Y_{2-2}$ ).
Total Nodes: equal to the sum of the polar and longitudinal nodes, and exactly equal to $l$ . For instance, $l=0$ has no nodes and $l=4$ has four nodes.

Orthogonality of Spherical Harmonics¶

The orthogonality of spherical harmonics is expressed as:

\langle l', m'| l, m \rangle = \delta_{l,l'}\delta_{m,m'}

(46)

This orthogonality condition can be explicitly written as:

\int_0^{\pi} \int_0^{2\pi} Y_{l'}^{m'*}(\theta, \phi) Y_l^m(\theta, \phi) \sin(\theta) \, d\theta \, d\phi = \delta_{l,l'}\delta_{m,m'}

(47)

import numpy as np
from scipy.special import sph_harm_y
from scipy.integrate import dblquad

def orthogonality_test(l1, m1, l2, m2):
    """
    Computes the orthogonality integral of spherical harmonics Y(l1, m1) and Y(l2, m2).
    
    Parameters:
    l1, m1 (int): Degree and order of the first spherical harmonic.
    l2, m2 (int): Degree and order of the second spherical harmonic.
    
    Returns:
    float: The result of the orthogonality integral.
    """
    # Define the integrand for the orthogonality condition
    def integrand(theta, phi):
        Y_lm1 = sph_harm_y(l1, m1, theta, phi)
        Y_lm2 = sph_harm_y(l2, m2, theta, phi)
        return np.real(Y_lm1 * np.conj(Y_lm2)) * np.sin(theta)
    
    # Perform the integration over theta (0 to pi) and phi (0 to 2*pi)
    integral, error = dblquad(integrand, 0, 2 * np.pi, lambda _: 0, lambda _: np.pi)
    return integral

# Test orthogonality between different spherical harmonics
print("Orthogonality Tests:")
print(f"∫ Y(1,0) * Y(1,0)* = {orthogonality_test(1, 0, 1, 0):.5f} (should be close to 1)")
print(f"∫ Y(1,0) * Y(1,1)* = {orthogonality_test(1, 0, 1, 1):.5f} (should be close to 0)")
print(f"∫ Y(1,1) * Y(2,1)* = {orthogonality_test(1, 1, 2, 1):.5f} (should be close to 0)")
print(f"∫ Y(2,1) * Y(2,1)* = {orthogonality_test(2, 1, 2, 1):.5f} (should be close to 1)")
print(f"∫ Y(2,0) * Y(2,-1)* = {orthogonality_test(2, 0, 2, -1):.5f} (should be close to 0)")

Orthogonality Tests:
∫ Y(1,0) * Y(1,0)* = 1.00000 (should be close to 1)
∫ Y(1,0) * Y(1,1)* = 0.00000 (should be close to 0)
∫ Y(1,1) * Y(2,1)* = 0.00000 (should be close to 0)
∫ Y(2,1) * Y(2,1)* = 1.00000 (should be close to 1)
∫ Y(2,0) * Y(2,-1)* = -0.00000 (should be close to 0)

Plotting Spherical Harmonics¶

Here we will visualize a spherical harmonic on a 3D sphere of radius 1. The radius is fixed, so there are only two variables that dictate the nodal features:
Colatitudinal nodes depend on $l$ and represent bands around the sphere, arising from terms in $\cos(\theta)$ or $\sin(\theta)$ .
Azimuthal nodes depend on $|m|$ and create longitudinal nodes based on $e^{im\phi}$ .

import matplotlib.pyplot as plt
from matplotlib import cm
import numpy as np
from scipy.special import sph_harm_y
from mpl_toolkits.mplot3d import Axes3D

def generate_spherical_harmonic_data(l, m):
    """
    Generates the data needed to plot the spherical harmonic for given l and m values.
    
    Parameters:
    l (int): Degree of the spherical harmonic.
    m (int): Order of the spherical harmonic.

    Returns:
    tuple: (x, y, z, fcolors_normalized) where:
        - x, y, z are the Cartesian coordinates of the spherical surface,
        - fcolors_normalized is the color data for plotting.
    """
    
    # Define theta and phi grids
    phi = np.linspace(0, np.pi, 100)          # colatitude
    theta = np.linspace(0, 2 * np.pi, 100)    # azimuth
    phi, theta = np.meshgrid(phi, theta)

    # Cartesian coordinates for the unit sphere
    x = np.sin(phi) * np.cos(theta)
    y = np.sin(phi) * np.sin(theta)
    z = np.cos(phi)

    # Calculate the spherical harmonic Y(l, m) and normalize it to [-1, 1] for color mapping
    fcolors = sph_harm_y(l, m, phi, theta).real
    fcolors_normalized = (fcolors - fcolors.min()) / (fcolors.max() - fcolors.min())
    
    return x, y, z, fcolors_normalized

$l=1$ harmonics¶

# Define the range of m and l values for the first three spherical harmonics
harmonics = [(0, 1), (1, 1), (-1, 1)]  # List of (m, l) pairs to plot

# Create a figure to hold three subplots in one row
fig = plt.figure(figsize=(18, 6))
fig.suptitle("First Three Spherical Harmonics", fontsize=16)

for i, (m, l) in enumerate(harmonics, start=1):

    x, y, z, fcolors_normalized = generate_spherical_harmonic_data(l, m)

    # Create a subplot for each (m, l) pair in one row
    ax = fig.add_subplot(1, 3, i, projection='3d')
    ax.plot_surface(x, y, z, rstride=1, cstride=1, facecolors=cm.seismic(fcolors_normalized), 
                    linewidth=0, antialiased=False)

    # Customize each subplot
    ax.set_title(f"$Y_{{{l},{m}}}$", fontsize=14)
    ax.set_box_aspect([1, 1, 1])  # Ensures spherical aspect ratio
    ax.set_axis_off()             # Turn off axes for clarity

# Adjust layout for a clear view of all subplots in one row
plt.tight_layout(rect=[0, 0, 1, 0.95])
plt.show()

$l=2$ harmonics¶

import matplotlib.pyplot as plt
from matplotlib import cm
import numpy as np
from scipy.special import sph_harm_y
from mpl_toolkits.mplot3d import Axes3D

# Define l and the range of m values for spherical harmonics with l=2
l = 2
m_values = [-2, -1, 0, 1, 2]

# Create a figure to hold five subplots in a 2x3 layout
fig = plt.figure(figsize=(15, 10))
fig.suptitle("Spherical Harmonics with l=2", fontsize=18)

for i, m in enumerate(m_values, start=1):

    # Calculate the spherical harmonic Y(l, m) and normalize it to [-1, 1] for color mapping
    x, y, z, fcolors_normalized = generate_spherical_harmonic_data(l, m) 

    # Create a subplot for each (m, l) pair in a 2x3 grid
    ax = fig.add_subplot(2, 3, i, projection='3d')
    ax.plot_surface(x, y, z, rstride=1, cstride=1, facecolors=cm.seismic(fcolors_normalized),
                    linewidth=0, antialiased=False)

    # Customize each subplot
    ax.set_title(f"$Y_{{2,{m}}}$", fontsize=16)
    ax.set_box_aspect([1, 1, 1])  # Ensures spherical aspect ratio
    ax.set_axis_off()             # Turn off axes for clarity

# Adjust layout for a clear view of all subplots in a 2x3 grid
plt.tight_layout(rect=[0, 0, 1, 0.95])
plt.show()

Total number of nodes = $l$ ¶

import plotly.io as pio
pio.renderers.default = "plotly_mimetype"   # interactive in Jupyter-Book
import numpy as np
import plotly.graph_objects as go
from scipy.special import sph_harm_y

# Define the range of m and l values for the first three spherical harmonics
harmonics = [(1, 1), (1, 2), (2, 3)]  # List of (m, l) pairs to plot

# Initialize the figure
fig = go.Figure()


# Loop over each harmonic and create a separate subplot
for i, (m, l) in enumerate(harmonics, start=1):
    
    # Calculate the spherical harmonic Y(l, m) and normalize it to [0, 1]
    x, y, z, fcolors_normalized = generate_spherical_harmonic_data(l, m) 
    
    # Add the spherical harmonic to the figure as a surface in a new scene
    fig.add_trace(go.Surface(
        x=x, y=y, z=z,
        surfacecolor=fcolors_normalized,
        colorscale='rdbu',
        showscale=False,
        name=f"Y_{{{l},{m}}}",
        scene=f'scene{i}'  # Assign each plot to a different scene
    ))

# Update layout to arrange scenes in a row and add titles as annotations
fig.update_layout(
    title="Three Spherical Harmonics: m=1, l=1, 2, 3",
    scene=dict(domain=dict(x=[0, 0.33]), xaxis=dict(visible=False), yaxis=dict(visible=False), zaxis=dict(visible=False), aspectmode='cube'),
    scene2=dict(domain=dict(x=[0.33, 0.66]), xaxis=dict(visible=False), yaxis=dict(visible=False), zaxis=dict(visible=False), aspectmode='cube'),
    scene3=dict(domain=dict(x=[0.66, 1]), xaxis=dict(visible=False), yaxis=dict(visible=False), zaxis=dict(visible=False), aspectmode='cube'),
    annotations=[
        dict(text="Y_{2,0}", x=0.16, y=1.05, showarrow=False, xref="paper", yref="paper", font=dict(size=14)),
        dict(text="Y_{2,1}", x=0.5, y=1.05, showarrow=False, xref="paper", yref="paper", font=dict(size=14)),
        dict(text="Y_{2,-1}", x=0.83, y=1.05, showarrow=False, xref="paper", yref="paper", font=dict(size=14))
    ]
)

# Display the figure
fig.show()

Alternative visualizations of Spherical Harmonics¶

Before, we visualized spherical harmonics on the unit sphere $|r|=1$ . However, there are times when we want to visualize how the magnitude of a spherical harmonic changes in space.
For instance, to visualize $Y_{1,-1}\sim cos(\theta)$ we could either plot its values on a unit sphere, or we could plot how $\cos(\theta)$ changes as we move around in space.

import numpy as np
import plotly.graph_objects as go
from plotly.subplots import make_subplots
from scipy.special import sph_harm_y

# Set up the subplots with two side-by-side 3D plots
fig = make_subplots(rows=1, cols=2, specs=[[{'is_3d': True}, {'is_3d': True}]])

# Define theta and phi grids for spherical coordinates
theta = np.linspace(0, np.pi, 100)
phi = np.linspace(0, 2 * np.pi, 100)
theta, phi = np.meshgrid(theta, phi)

# Convert spherical to Cartesian coordinates for the surface plots
x = np.sin(theta) * np.cos(phi)
y = np.sin(theta) * np.sin(phi)
z = np.cos(theta)

# Compute the spherical harmonic Y_{1,-1}
m, l = -1, 1
r0a = np.sqrt(3 / (4 * np.pi)) * y  # Y_{1,-1} harmonic on y-axis
r0 = np.abs(r0a)  # Absolute value for magnitude scaling

# First plot: Scaled by the spherical harmonic (bead-like structure)
fig.add_trace(
    go.Surface(
        x=r0 * x, y=r0 * y, z=r0 * z,  # Scale coordinates by |Y_{1,-1}|
        surfacecolor=r0a,  # Color by Y_{1,-1} values
        colorscale='RdBu'
    ),
    row=1, col=1
)

# Second plot: Standard unit sphere with grayscale color
fig.add_trace(
    go.Surface(
        x=x, y=y, z=z,  # Unit sphere without scaling
        surfacecolor=z,  # Color by z-coordinate for simplicity
        colorscale='RdBu'
    ),
    row=1, col=2
)

# Update layout for both plots
fig.update_layout(
    font_family="JuliaMono",
    showlegend=False,
    margin=dict(l=0, r=0, b=0, t=0),
    paper_bgcolor='rgba(0,0,0,0)',
)

# Set scene properties with adjusted axis limits for each plot
fig.update_scenes(
    dict(
        xaxis=dict(nticks=4, range=[-0.5, 0.5]),  # Limit range to reduce empty space
        yaxis=dict(nticks=4, range=[-0.5, 0.5]),
        zaxis=dict(nticks=4, range=[-0.5, 0.5]),
        aspectratio=dict(x=1, y=1, z=1)
    ),
    row=1, col=1
)
fig.update_scenes(
    dict(
        xaxis=dict(nticks=4, range=[-1.8, 1.8]),
        yaxis=dict(nticks=4, range=[-1.8, 1.8]),
        zaxis=dict(nticks=4, range=[-1.8, 1.8]),
        aspectratio=dict(x=1, y=1, z=1)
    ),
    row=1, col=2
)

# Hide color scales for a cleaner look
fig.update_traces(showscale=False)

# Show the plot
fig.show()

import numpy as np
import plotly.graph_objects as go
from plotly.subplots import make_subplots
from matplotlib import cm
import matplotlib
import matplotlib.pyplot as plt

# Define theta (angle) values
theta = np.linspace(0, 2 * np.pi, 100)

# Compute spherical harmonic values (scaled sin(theta) for two-lobe structure)
Y_1m1_values = np.sin(theta)  # Y_{1,-1} varies with sin(theta) for the two-lobe shape
r_values = np.abs(Y_1m1_values)  # Radial scaling based on |Y_{1,-1}|

# Set up the color map
cmap = matplotlib.colormaps["RdBu"]
norm = plt.Normalize(vmin=-1, vmax=1)

# Create subplots for two polar plots
fig = make_subplots(rows=1, cols=2, specs=[[{'type': 'polar'}, {'type': 'polar'}]])

# First polar plot: Two-lobed structure with color gradient
for i in range(len(theta) - 1):
    fig.add_trace(
        go.Scatterpolar(
            r=[r_values[i], r_values[i + 1]],
            theta=[np.degrees(theta[i]), np.degrees(theta[i + 1])],
            mode="lines",
            line=dict(color=matplotlib.colors.to_hex(cmap(norm(Y_1m1_values[i]))), width=5),
            showlegend=False
        ),
        row=1, col=1
    )

# Second polar plot: Unit circle with consistent color gradient
for i in range(len(theta) - 1):
    fig.add_trace(
        go.Scatterpolar(
            r=[1, 1],  # Fixed radius for unit circle
            theta=[np.degrees(theta[i]), np.degrees(theta[i + 1])],
            mode="lines",
            line=dict(color=matplotlib.colors.to_hex(cmap(norm(Y_1m1_values[i]))), width=5),
            showlegend=False
        ),
        row=1, col=2
    )

# Update layout for both polar plots with fixed radius range and equal appearance
fig.update_layout(
    font_family="JuliaMono",
    showlegend=False,
    margin=dict(l=0, r=0, b=0, t=0),
    polar=dict(radialaxis=dict(range=[0, 1.2])),  # First plot axis range
    polar2=dict(radialaxis=dict(range=[0, 1.2]))  # Second plot axis range
)

# Show the plot
fig.show()

Conservation of momentum is due to symmetry¶

Classical angular momentum¶

Classical picture: Rotating dumbbell¶

Spherical coordinates¶

Quantum angular momentum¶

Components of angular momentum do not commute!¶

Eigenfunctions and eigenvalues of LLL and LzL_zLz​¶

Spherical harmonics¶

Nodes of Spherical Harmonics¶

Orthogonality of Spherical Harmonics¶

Plotting Spherical Harmonics¶

l=1l=1l=1 harmonics¶

l=2l=2l=2 harmonics¶

Total number of nodes = lll¶

Alternative visualizations of Spherical Harmonics¶

Learn More about Spherical Harmonics¶

Eigenfunctions and eigenvalues of $L$ and $L_z$ ¶

$l=1$ harmonics¶

$l=2$ harmonics¶

Total number of nodes = $l$ ¶