← Part IV/Orbital Angular Momentum

3. Orbital Angular Momentum

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Angular momentum is the generator of rotations in quantum mechanics, with profound implications for atomic structure and symmetries.

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Video Lecture

Quantum Angular Momentum - Introduction

Comprehensive introduction to quantum angular momentum operators, commutation relations, and eigenvalues

💡 Tip: Watch at 1.25x or 1.5x speed for efficient learning. Use YouTube's subtitle feature if available.

📜

Development of Angular Momentum Theory

Quantum Commutation Relations

1925

Werner Heisenberg & Max Born

Discovered the non-commutative algebra of quantum observables, leading to the commutation relations for angular momentum components.

Spherical Harmonics as Eigenfunctions

1926

Erwin Schrödinger

Identified spherical harmonics Y_ℓ^m(θ,φ) as the eigenfunctions of L² and L_z operators in the wave formulation of quantum mechanics.

General Angular Momentum Theory

1927

Wolfgang Pauli & Paul Dirac

Developed the general theory of angular momentum using ladder operators, showing that ℓ can be half-integer (spin) or integer (orbital).

💡 These developments represent key milestones in the evolution of quantum mechanics.

Classical Definition

Classical angular momentum: measure of rotational motion

$$\vec{L} = \vec{r} \times \vec{p}$$

In components:

$$L_x = yp_z - zp_y$$
$$L_y = zp_x - xp_z$$
$$L_z = xp_y - yp_x$$

Key insight: Angular momentum is conserved when the Hamiltonian has rotational symmetry (Noether's theorem).

Quantum Angular Momentum Operators

Quantum version: $\hat{\vec{L}} = \hat{\vec{r}} \times \hat{\vec{p}}$

In Cartesian coordinates:

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

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

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

These are Hermitian operators corresponding to observable quantities

▶️

Video Lecture

Angular Momentum Operators - Derivation

Step-by-step derivation of angular momentum operators in quantum mechanics from classical definitions

💡 Tip: Watch at 1.25x or 1.5x speed for efficient learning. Use YouTube's subtitle feature if available.

Commutation Relations

Fundamental algebra of angular momentum:

$$[\hat{L}_i, \hat{L}_j] = i\hbar\epsilon_{ijk}\hat{L}_k$$

where $\epsilon_{ijk}$ is the Levi-Civita symbol (antisymmetric tensor)

Explicitly:

$$[\hat{L}_x, \hat{L}_y] = i\hbar\hat{L}_z$$
$$[\hat{L}_y, \hat{L}_z] = i\hbar\hat{L}_x$$
$$[\hat{L}_z, \hat{L}_x] = i\hbar\hat{L}_y$$

⚠️ Crucial Consequence: Since the components don't commute, we cannot measure all three components simultaneously with arbitrary precision. This is fundamentally different from classical mechanics!

📝 Worked Example: Verifying Commutator

Problem: Prove that $[\hat{L}_x, \hat{L}_y] = i\hbar\hat{L}_z$ using the Cartesian representations.

Solution:

Start with definitions:

$$\hat{L}_x = \hat{y}\hat{p}_z - \hat{z}\hat{p}_y, \quad \hat{L}_y = \hat{z}\hat{p}_x - \hat{x}\hat{p}_z$$

Calculate the commutator term by term:

$$[\hat{L}_x, \hat{L}_y] = [\hat{y}\hat{p}_z - \hat{z}\hat{p}_y, \hat{z}\hat{p}_x - \hat{x}\hat{p}_z]$$

Using $[\hat{x}, \hat{p}_x] = i\hbar$ and $[\hat{x}, \hat{p}_y] = 0$ for $x \neq y$, most terms vanish. The surviving terms are:

$$= \hat{y}[\hat{p}_z, \hat{x}]\hat{p}_z + \hat{z}\hat{p}_y[\hat{z}, \hat{p}_x] - \hat{z}[\hat{p}_y, \hat{x}\hat{p}_z]$$

After working through the algebra:

$$= i\hbar(\hat{x}\hat{p}_y - \hat{y}\hat{p}_x) = i\hbar\hat{L}_z \quad \checkmark$$

Total Angular Momentum Squared

Define the magnitude squared operator:

$$\hat{L}^2 = \hat{L}_x^2 + \hat{L}_y^2 + \hat{L}_z^2$$

Key property:

$$[\hat{L}^2, \hat{L}_i] = 0 \quad \text{for } i = x,y,z$$

$\hat{L}^2$ commutes with all components!

Standard Choice:

We conventionally diagonalize $\hat{L}^2$ and $\hat{L}_z$ simultaneously. This gives us the quantum numbers $\ell$ and $m$.

(Could choose any component instead of $L_z$, but z is conventional)

Eigenvalues and Quantum Numbers

Simultaneous eigenstates of $\hat{L}^2$ and $\hat{L}_z$:

$$\hat{L}^2|\ell,m\rangle = \hbar^2\ell(\ell+1)|\ell,m\rangle$$

$$\hat{L}_z|\ell,m\rangle = \hbar m|\ell,m\rangle$$

Allowed values:

Orbital angular momentum: $\ell = 0, 1, 2, 3, \ldots$ (non-negative integer)
Magnetic quantum number: $m = -\ell, -\ell+1, \ldots, \ell-1, \ell$ (integer)
Degeneracy: $2\ell + 1$ states for each $\ell$

Why $\ell(\ell+1)$ and not $\ell^2$?

The eigenvalue $\hbar^2\ell(\ell+1)$ (not $\hbar^2\ell^2$) comes from solving the eigenvalue equation using ladder operators. The $+1$ term ensures that $|\vec{L}| = \hbar\sqrt{\ell(\ell+1)} > |L_z|_{\text{max}} = \hbar\ell$, consistent with the uncertainty principle.

Ladder Operators

Define raising $(\hat{L}_+)$ and lowering $(\hat{L}_-)$ operators:

$$\hat{L}_\pm = \hat{L}_x \pm i\hat{L}_y$$

Commutation relations:

$$[\hat{L}_z, \hat{L}_\pm] = \pm\hbar\hat{L}_\pm$$
$$[\hat{L}^2, \hat{L}_\pm] = 0$$
$$[\hat{L}_+, \hat{L}_-] = 2\hbar\hat{L}_z$$

Action on eigenstates:

$$\hat{L}_+|\ell,m\rangle = \hbar\sqrt{\ell(\ell+1) - m(m+1)}|\ell,m+1\rangle$$

$$\hat{L}_-|\ell,m\rangle = \hbar\sqrt{\ell(\ell+1) - m(m-1)}|\ell,m-1\rangle$$

Ladder analogy: $\hat{L}_+$ "climbs up" the ladder (increases $m$), $\hat{L}_-$ "climbs down" (decreases $m$). At the top/bottom rungs ($m = \pm\ell$), the coefficient vanishes and you can't go further.

📝 Worked Example: Using Ladder Operators

Problem: Starting from $|2, 2\rangle$, find $\hat{L}_-|2,2\rangle$ and $\hat{L}_-^2|2,2\rangle$.

Step 1: Apply $\hat{L}_-$ once

$$\hat{L}_-|2,2\rangle = \hbar\sqrt{2(2+1) - 2(2-1)}|2,1\rangle = \hbar\sqrt{6-2}|2,1\rangle = 2\hbar|2,1\rangle$$

Step 2: Apply $\hat{L}_-$ again to $|2,1\rangle$

$$\hat{L}_-|2,1\rangle = \hbar\sqrt{6 - 1(0)}|2,0\rangle = \hbar\sqrt{6}|2,0\rangle$$

Step 3: Combine the results

$$\hat{L}_-^2|2,2\rangle = \hat{L}_-(2\hbar|2,1\rangle) = 2\hbar^2\sqrt{6}|2,0\rangle$$

Note: $\ell$ doesn't change, only $m$ decreases by 1 per application.

▶️

Video Lecture

Ladder Operators for Angular Momentum

Detailed explanation of raising and lowering operators and their application to angular momentum states

💡 Tip: Watch at 1.25x or 1.5x speed for efficient learning. Use YouTube's subtitle feature if available.

Spherical Coordinates Representation

In spherical coordinates $(r, \theta, \phi)$, angular momentum operators simplify:

$$\hat{L}_z = -i\hbar\frac{\partial}{\partial\phi}$$

$$\hat{L}^2 = -\hbar^2\left[\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial}{\partial\theta}\right) + \frac{1}{\sin^2\theta}\frac{\partial^2}{\partial\phi^2}\right]$$

Note: $\hat{L}_z$ depends only on $\phi$, while $\hat{L}^2$ depends on both $\theta$ and $\phi$ (but not $r$!)

Physical interpretation: Angular momentum operators act only on the angular variables $(\theta, \phi)$, not on the radial distance $r$. This reflects that angular momentum describes rotation, not radial motion.

Spectroscopic Notation

Traditional letter code for $\ell$ values (from atomic spectroscopy):

$\ell$	Symbol	Name	Example (H atom)
0	s	sharp	1s, 2s, 3s...
1	p	principal	2p, 3p, 4p...
2	d	diffuse	3d, 4d, 5d...
3	f	fundamental	4f, 5f, 6f...
4, 5, 6...	g, h, i...	(alphabetical)	5g, 6h...

Notation: "3d" means $n=3, \ell=2$ (principal quantum number n, angular momentum quantum number $\ell$)

🤔 Self-Check Question

How many degenerate states (different $m$ values) are there for a d orbital ($\ell = 2$)?

Show Answer

For $\ell = 2$, we have $m = -2, -1, 0, +1, +2$, giving us 5 states. In general, there are $2\ell + 1$ states for each $\ell$. This is why d orbitals can hold 10 electrons (5 spatial states × 2 spin states).

Uncertainty Relations

Since components don't commute, they obey uncertainty relations:

$$\Delta L_x \cdot \Delta L_y \geq \frac{\hbar}{2}|\langle L_z \rangle|$$

and cyclic permutations

Physical meaning: We cannot precisely know the direction of the angular momentum vector $\vec{L}$. We can know its magnitude ($|\vec{L}| = \hbar\sqrt{\ell(\ell+1)}$) and one component (say $L_z = m\hbar$), but the other two components remain uncertain.

Vector Model (Semiclassical Picture)

Semiclassical visualization:

Angular momentum vector $\vec{L}$ has fixed length $\sqrt{\ell(\ell+1)}\hbar$
z-component is fixed at $m\hbar$
Vector precesses around z-axis forming a cone
x and y components are indeterminate (time-averaged to zero)
Cone angle: $\cos\theta = \frac{m}{\sqrt{\ell(\ell+1)}}$

Caveat: This is only a semiclassical picture! The true quantum state doesn't have a well-defined direction for $\vec{L}$. The precession represents the uncertainty in $L_x$ and $L_y$.

Relation to Rotations

Angular momentum generates rotations:

$$\hat{R}_z(\phi) = e^{-i\hat{L}_z\phi/\hbar}$$

Rotation by angle $\phi$ around z-axis

General rotation:

$$\hat{R}_{\hat{n}}(\theta) = e^{-i\vec{L}\cdot\hat{n}\theta/\hbar}$$

Rotation by angle $\theta$ around axis $\hat{n}$

Connection to Noether's theorem: Conservation of angular momentum follows from rotational symmetry of the Hamiltonian. If $[\hat{H}, \hat{R}] = 0$, then $[\hat{H}, \hat{L}] = 0$ and angular momentum is conserved.

💡 Application: Atomic Structure and the Periodic Table

Angular momentum quantum numbers $\ell$ and $m$ determine the shapes and orientations of atomic orbitals. The (2ℓ+1)-fold degeneracy for each $\ell$ explains why:

s orbitals ($\ell=0$) hold 2 electrons (1 state × 2 spins)
p orbitals ($\ell=1$) hold 6 electrons (3 states × 2 spins)
d orbitals ($\ell=2$) hold 10 electrons (5 states × 2 spins)
f orbitals ($\ell=3$) hold 14 electrons (7 states × 2 spins)

This structure underlies the entire periodic table!

Summary: Orbital Angular Momentum

• Components don't commute: $[\hat{L}_i, \hat{L}_j] = i\hbar\epsilon_{ijk}\hat{L}_k$
• Can simultaneously measure $\hat{L}^2$ and one component (conventionally $\hat{L}_z$)
• Eigenvalues: $L^2 = \hbar^2\ell(\ell+1)$ and $L_z = m\hbar$ where $m = -\ell, ..., +\ell$
• Ladder operators $\hat{L}_\pm$ raise/lower $m$ while preserving $\ell$
• (2ℓ+1)-fold degeneracy for each $\ell$ value
• Angular momentum generates rotations via $\hat{R} = e^{-i\vec{L}\cdot\hat{n}\theta/\hbar}$
• Conservation of $\vec{L}$ follows from rotational symmetry (Noether's theorem)

🔗 Related Topics: Spherical Harmonics - The explicit wave functions that are eigenfunctions of $\hat{L}^2$ and $\hat{L}_z$

🔗 Related Topics: Spin Angular Momentum - Intrinsic angular momentum that obeys the same algebra but with half-integer values allowed

Runnable Simulations

Angular Momentum Ladder Operators and Matrix Representation

Python

script.py50 lines

import numpy as np

# Angular momentum algebra: [J_x, J_y] = i*J_z (cyclic)
# J_+|j,m> = sqrt(j(j+1)-m(m+1)) |j,m+1>
# J_-|j,m> = sqrt(j(j+1)-m(m-1)) |j,m-1>

print("Angular Momentum Operators: Matrix Representations")
print("[J_x, J_y] = i*J_z, J^2|j,m> = j(j+1)|j,m>")
print("=" * 55)

for j_val in [0.5, 1.0, 1.5, 2.0]:
    j = j_val
    dim = int(2*j + 1)
    m_vals = np.arange(j, -j-1, -1)

# Build J_+, J_-, J_z matrices
    Jp = np.zeros((dim, dim))
    Jm = np.zeros((dim, dim))
    Jz = np.diag(m_vals)

for i in range(dim-1):
        m = m_vals[i+1]
        Jp[i, i+1] = np.sqrt(j*(j+1) - m*(m+1))
        Jm[i+1, i] = np.sqrt(j*(j+1) - m*(m+1))

Jx = (Jp + Jm) / 2
    Jy = (Jp - Jm) / (2j)
    Jy_correct = (Jp - Jm) / (2*1j)

J2 = Jx @ Jx + Jy_correct @ Jy_correct + Jz @ Jz

# Verify commutation relations
    comm_xy = Jx @ Jy_correct - Jy_correct @ Jx
    expected_z = 1j * Jz
    check = np.allclose(comm_xy, expected_z)

# J^2 eigenvalue
    j2_eigenval = np.diag(J2.real)[0]

print(f"\nj = {j}: dim = {dim}, J^2 eigenvalue = {j2_eigenval:.4f} (expected {j*(j+1):.4f})")
    print(f"  m values: {m_vals}")
    print(f"  [Jx, Jy] = i*Jz: {'PASS' if check else 'FAIL'}")

# Ladder operator actions
    if dim <= 5:
        print(f"  J_+ matrix:")
        for row in Jp:
            print(f"    [{', '.join(f'{v:6.3f}' for v in row)}]")

Click Run to execute the Python code

Code will be executed with Python 3 on the server

Orbital Angular Momentum Eigenvalues and Counting

Fortran

program.f9054 lines

program angular_momentum
  implicit none
  integer :: l, m, n, total_states, l_max
  double precision :: L2, Lz, E_n
  double precision, parameter :: hbar = 1.0d0

print *, "Orbital Angular Momentum Eigenvalues"
  print *, "L^2|l,m> = l(l+1)*hbar^2 |l,m>"
  print *, "L_z|l,m> = m*hbar |l,m>, m = -l,...,+l"
  print *, "========================================"
  print *, ""

print '(A6, A12, A10, A20)', "l", "L^2/hbar^2", "# states", "m values"
  print *, "----------------------------------------------------"

do l = 0, 6
    L2 = dble(l) * dble(l + 1)
    print '(I6, F12.1, I10, A, I3, A, I3, A)', &
      l, L2, 2*l+1, "  m = ", -l, " to ", l, ""
  end do

! Degeneracy counting for hydrogen-like atoms
  print *, ""
  print *, "Hydrogen atom state counting:"
  print '(A4, A10, A10, A10, A10)', &
    "n", "l range", "states", "total", "E_n (eV)"
  print *, "--------------------------------------------"

total_states = 0
  do n = 1, 6
    l_max = n - 1
    E_n = -13.6d0 / dble(n)**2

! Count states for this n
    total_states = 0
    do l = 0, l_max
      total_states = total_states + 2*(2*l + 1)  ! factor 2 for spin
    end do

print '(I4, A, I2, A, I2, I10, I10, F10.4)', &
      n, "  0-", l_max, "", 2*n**2, total_states, E_n
  end do

print *, ""
  print *, "Spectroscopic notation:"
  print *, "  l = 0: s (sharp)"
  print *, "  l = 1: p (principal)"
  print *, "  l = 2: d (diffuse)"
  print *, "  l = 3: f (fundamental)"
  print *, "  l = 4: g, l = 5: h, ..."

end program angular_momentum

Click Run to execute the Fortran code

Code will be compiled with gfortran and executed on the server

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