Quantum Mechanics Course

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← 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

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

πŸ“œ

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 SchrΓΆdinger

Identified spherical harmonics Y_β„“^m(ΞΈ,Ο†) as the eigenfunctions of LΒ² 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 β„“ can be half-integer (spin) or integer (orbital).

πŸ’‘ 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

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Video Lecture

Angular Momentum Operators - Derivation

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

πŸ’‘ 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!

πŸ“ 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.

πŸ“ 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

πŸ’‘ 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):

$\ell$SymbolNameExample (H atom)
0ssharp1s, 2s, 3s...
1pprincipal2p, 3p, 4p...
2ddiffuse3d, 4d, 5d...
3ffundamental4f, 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 $\ell$)

πŸ€” 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 Γ— 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.

πŸ’‘ Application: Atomic Structure and the Periodic Table

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

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

This structure underlies the entire periodic table!

Summary: Orbital Angular Momentum

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

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

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

← Central PotentialSpherical Harmonics β†’