CSU East Bay logo

Chemistry 352

Angular Momentum Operators

Orbital Angular Momentum Operators

In quantum mechanics, angular momentum is represented by a set of operators rather than by classical vectors. For a particle moving in three dimensions, the orbital angular momentum operator \( \hat{\mathbf{L}} \) is defined as

\[ \hat{\mathbf{L}} = \mathbf{r} \times \hat{\mathbf{p}}, \]

where \( \mathbf{r} = (x,y,z) \) is the position operator and \( \hat{\mathbf{p}} = -i\hbar\nabla \) is the linear momentum operator. Because \( \hat{\mathbf{p}} \) involves derivatives, the components of \( \hat{\mathbf{L}} \) are differential operators acting on wavefunctions.

Cartesian Components

Expanding the cross product gives the three Cartesian components of the orbital angular momentum operator:

\[ \hat{L}_x = -i\hbar\!\left( y\frac{\partial}{\partial z} - z\frac{\partial}{\partial y} \right), \] \[ \hat{L}_y = -i\hbar\!\left( z\frac{\partial}{\partial x} - x\frac{\partial}{\partial z} \right), \] \[ \hat{L}_z = -i\hbar\!\left( x\frac{\partial}{\partial y} - y\frac{\partial}{\partial x} \right). \]

Each component generates an infinitesimal rotation of the wavefunction about the corresponding Cartesian axis. For example, \( \hat{L}_z \) generates rotations about the \( z \)-axis.

Commutation Relations

Unlike the components of a classical angular momentum vector, the operators \( \hat{L}_x, \hat{L}_y, \hat{L}_z \) do not commute with one another. Their fundamental commutation relations are

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

These relations imply that it is impossible to simultaneously measure more than one Cartesian component of orbital angular momentum with arbitrary precision. If one component is known exactly, the other two are fundamentally uncertain.

Derivation: Explicitly showing cancellation in \( [\hat L_x,\hat L_y] \)

Recall \( \hat L_x=-i\hbar\!\left(y\frac{\partial}{\partial z}-z\frac{\partial}{\partial y}\right) \) and \( \hat L_y=-i\hbar\!\left(z\frac{\partial}{\partial x}-x\frac{\partial}{\partial z}\right) \). We apply each product to an arbitrary function \( \psi(x,y,z) \).

Step 2: Evaluate both products (color shows terms that cancel)

\[ \hat L_x(\hat L_y\psi) = -\hbar^2\Big( y\,\frac{\partial\psi}{\partial x} + yz\,\frac{\partial^2\psi}{\partial x\,\partial z} - xy\,\frac{\partial^2\psi}{\partial z^2} - z^2\,\frac{\partial^2\psi}{\partial x\,\partial y} + xz\,\frac{\partial^2\psi}{\partial y\,\partial z} \Big) \]

\[ \hat L_y(\hat L_x\psi) = -\hbar^2\Big( x\,\frac{\partial\psi}{\partial y} + yz\,\frac{\partial^2\psi}{\partial x\,\partial z} - xy\,\frac{\partial^2\psi}{\partial z^2} - z^2\,\frac{\partial^2\psi}{\partial x\,\partial y} + xz\,\frac{\partial^2\psi}{\partial y\,\partial z} \Big) \]

The second-derivative terms appear in both expressions and therefore cancel when the difference is taken to derive the commutator.

Step 3: Form the commutator

\[ [\hat L_x,\hat L_y]\psi = -\hbar^2\!\left( y\,\frac{\partial\psi}{\partial x} - x\,\frac{\partial\psi}{\partial y} \right) \]

\[ = \hbar^2\!\left( x\,\frac{\partial\psi}{\partial y} - y\,\frac{\partial\psi}{\partial x} \right) = i\hbar\,\hat L_z\psi, \]

\[ \boxed{[\hat L_x,\hat L_y]=i\hbar\,\hat L_z} \]

Similarly, cyclic permutation of the Cartesian components gives

\[ [\hat L_y,\hat L_z] = i\hbar\,\hat L_x, \qquad [\hat L_z,\hat L_x] = i\hbar\,\hat L_y. \]

Total Angular Momentum

The square of the total orbital angular momentum operator is defined as

\[ \hat{L}^2 = \hat{L}_x^2 + \hat{L}_y^2 + \hat{L}_z^2. \]

Importantly, \( \hat{L}^2 \) commutes with each Cartesian component:

\[ [\hat{L}^2, \hat{L}_x] = [\hat{L}^2, \hat{L}_y] = [\hat{L}^2, \hat{L}_z] = 0. \]

As a result, a quantum state can be chosen to have definite values of \( \hat{L}^2 \) and one component of angular momentum (by convention, \( \hat{L}_z \)). This structure underlies the quantization of angular momentum and leads directly to the appearance of the spherical harmonics as angular eigenfunctions.

Key idea: The non-commuting nature of \( \hat{L}_x, \hat{L}_y, \hat{L}_z \) is not a mathematical curiosity—it encodes the rotational symmetry of three-dimensional space and determines the allowed angular momentum quantum numbers used throughout atomic and molecular spectroscopy.

Angular Momentum in Spherical Polar Coordinates

For the rigid rotor (and for central-force problems more generally), it is most natural to use spherical polar coordinates \( (r,\theta,\phi) \). In these coordinates, the orbital angular momentum operators simplify dramatically. In particular, the operator \( \hat L_z \) becomes a single derivative with respect to the azimuthal angle \( \phi \).

The \( \hat L_z \) operator

\[ \hat L_z = -\,i\hbar\,\frac{\partial}{\partial \phi}. \]

This form reflects the physical meaning of \( \phi \): changing \( \phi \) corresponds to a rotation about the \( z \)-axis. Thus \( \hat L_z \) is the generator of rotations about \( z \).

The \( \hat L^2 \) operator

The total orbital angular momentum operator \( \hat L^2 \) contains only angular derivatives (no radial derivatives). In spherical polar coordinates,

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

This operator depends only on the angles \( \theta \) and \( \phi \), which is why the rigid rotor problem separates cleanly into an angular equation.

Spherical harmonics as eigenfunctions

The spherical harmonics \( Y_\ell^{m}(\theta,\phi) \) are special functions that arise naturally as simultaneous eigenfunctions of \( \hat L^2 \) and \( \hat L_z \). They satisfy the eigenvalue equations:

\[ \hat L^2\,Y_\ell^{m}(\theta,\phi) = \hbar^2\,\ell(\ell+1)\,Y_\ell^{m}(\theta,\phi), \] \[ \hat L_z\,Y_\ell^{m}(\theta,\phi) = \hbar\,m\,Y_\ell^{m}(\theta,\phi). \]

Here, \( \ell = 0,1,2,\ldots \) is the orbital angular momentum quantum number and \( m = -\ell, -\ell+1, \ldots, +\ell \) is the magnetic quantum number.

Why \( [\hat L^2,\hat L_z]=0 \)

Because the spherical harmonics form a set of functions that can be chosen to have definite values of both \( \hat L^2 \) and \( \hat L_z \) simultaneously, the operators must be compatible (they can be diagonalized at the same time). This is expressed by the commutator:

\[ \boxed{[\hat L^2,\hat L_z]=0.} \]

Key idea: In rigid-rotor language, the quantum number \( \ell \) fixes the total rotational angular momentum, while \( m \) fixes its projection on the laboratory \( z \)-axis. The spherical harmonics provide the angular wavefunctions that carry these quantum numbers.

Practice: Angular Momentum Operators and Spherical Harmonics

Problem 1
Which statement best explains why the operators \( \hat L_x \), \( \hat L_y \), and \( \hat L_z \) cannot all be simultaneously diagonalized?
A. Because angular momentum is not conserved in quantum mechanics
B. Because the operators involve second derivatives
C. Because the Cartesian components do not commute with one another
D. Because only one component of angular momentum exists
Problem 2
The operator \( \hat L_z = -i\hbar\,\partial/\partial\phi \) acts on a spherical harmonic \( Y_\ell^{m}(\theta,\phi) \). What property of the spherical harmonic guarantees that it is an eigenfunction of \( \hat L_z \)?
A. Its dependence on \( \theta \) through Legendre polynomials
B. Its azimuthal dependence of the form \( e^{im\phi} \)
C. The fact that it contains no radial dependence
D. The orthogonality of different spherical harmonics
Problem 3
Spherical harmonics \( Y_\ell^{m}(\theta,\phi) \) are eigenfunctions of both \( \hat L^2 \) and \( \hat L_z \). What does this imply about the commutator \( [\hat L^2,\hat L_z] \)?
A. It is proportional to \( \hat L_x \)
B. It is proportional to \( \hat L_y \)
C. It depends on the value of \( m \)
D. It is zero

Further Considerations: Ladder Operators

Although not essential for solving the rigid rotor problem in this course, the ladder operators provide a powerful and elegant way to understand the structure of angular momentum states. They are especially useful in more advanced treatments of atomic and molecular spectroscopy.

Definition

The angular momentum ladder operators are defined as

\[ \hat L_{\pm} = \hat L_x \pm i\,\hat L_y. \]

These operators do not correspond to physical components of angular momentum. Instead, they are algebraic combinations chosen to simplify the action of angular momentum on its eigenstates.

Action on Spherical Harmonics

When acting on a spherical harmonic \( Y_\ell^{m}(\theta,\phi) \), the ladder operators change the magnetic quantum number \( m \) without changing \( \ell \):

\[ \hat L_{\pm}\,Y_\ell^{m} = \hbar\,\sqrt{\ell(\ell+1)-m(m\pm1)}\;Y_\ell^{\,m\pm1}. \]

Repeated application of \( \hat L_{+} \) or \( \hat L_{-} \) therefore generates the entire set of \( 2\ell+1 \) states with \( m=-\ell,\ldots,+\ell \) starting from a single known state.

Why Ladder Operators Are Useful

Perspective: Ladder operators are introduced here as a preview of the algebraic methods used extensively in upper-level quantum mechanics. While the rigid rotor can be solved without them, ladder operators provide a unifying framework that connects angular momentum across atomic, molecular, and spectroscopic problems.

Key points (one glance)

Big picture: Angular momentum operators encode the rotational symmetry of space. Their commutation relations determine which quantities can be known simultaneously, while their eigenfunctions—the spherical harmonics—provide the angular wavefunctions used throughout atomic and molecular spectroscopy. This framework underlies the rigid rotor, atomic orbitals, and the angular momentum coupling schemes encountered later in the course.