4.4 Angular Momentum and Hydrogen Atom

Quantum Mechanical Angular Momentum

Angular Momentum in Quantum Mechanics

In classical mechanics, angular momentum can take any value. In quantum mechanics, it's quantized: only specific discrete values are allowed. This distinction is central to understanding atomic structure and spectra.

The angular momentum operators $\hat{L}_x$, $\hat{L}_y$, and $\hat{L}_z$ are built from the position and momentum operators. The total angular momentum operator is defined as:

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

Two key eigenvalue results follow:

• The eigenvalues of $\hat{L}^2$ are $l(l+1)\hbar^2$, where $l = 0, 1, 2, \ldots$ is the angular momentum quantum number.
• The eigenvalues of $\hat{L}_z$ are $m_l\hbar$, where $m_l = -l, -l+1, \ldots, l-1, l$ is the magnetic quantum number.

For a given $l$, there are $2l+1$ allowed values of $m_l$. This means the z-component of angular momentum is quantized, but the x- and y-components remain indeterminate (you cannot simultaneously know all three components).

Spherical Harmonics as Eigenfunctions

The simultaneous eigenfunctions of $\hat{L}^2$ and $\hat{L}_z$ are the spherical harmonics, $Y_l^{m_l}(\theta, \phi)$. These functions solve the angular part of the Schrödinger equation in spherical coordinates and describe how the wave function is distributed over angles.

Familiar examples and their connection to orbital shapes:

• $Y_0^0$: spherically symmetric, corresponds to s orbitals
• $Y_1^0$: has a $\cos\theta$ dependence, corresponds to the $p_z$ orbital
• $Y_1^{\pm1}$: linear combinations give the $p_x$ and $p_y$ orbitals

Spherical harmonics are orthonormal and form a complete basis set, meaning any angular function on a sphere can be expanded in terms of them.

Angular Momentum Operators and Eigenfunctions

Commutation Relations and the Uncertainty Principle

The angular momentum operators obey cyclic commutation relations:

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

Because $\hat{L}_x$, $\hat{L}_y$, and $\hat{L}_z$ don't commute with each other, you can't simultaneously determine all three components. However, $\hat{L}^2$ commutes with each individual component ($[\hat{L}^2, \hat{L}_z] = 0$), so you can simultaneously know the total angular momentum magnitude and one component (conventionally $\hat{L}_z$). That's why $l$ and $m_l$ are both good quantum numbers.

Raising and Lowering Operators

The ladder operators $\hat{L}_+ = \hat{L}_x + i\hat{L}_y$ and $\hat{L}_- = \hat{L}_x - i\hat{L}_y$ shift the magnetic quantum number:

• $\hat{L}_+$ acting on $Y_l^{m_l}$ produces a state with $m_l + 1$
• $\hat{L}_-$ acting on $Y_l^{m_l}$ produces a state with $m_l - 1$

These operators hit zero when they reach the boundaries ($m_l = +l$ or $m_l = -l$), which is how the allowed range of $m_l$ values is established.

Coupling of Angular Momenta

When a system has more than one source of angular momentum, you combine them using Clebsch-Gordan coefficients. For two angular momenta $l_1$ and $l_2$, the total angular momentum quantum number $L$ ranges from $|l_1 - l_2|$ to $l_1 + l_2$.

A classic example: coupling two spin-$\frac{1}{2}$ particles gives total spin $S = 0$ (singlet, one state) and $S = 1$ (triplet, three states). The singlet is antisymmetric under particle exchange, while the triplet is symmetric.

Hydrogen Atom Schrödinger Equation

Hydrogen Atom Model

The hydrogen atom (one proton, one electron) is the only atom whose Schrödinger equation can be solved exactly. The Hamiltonian includes the electron's kinetic energy and the Coulomb potential:

$\hat{H} = -\frac{\hbar^2}{2\mu}\nabla^2 - \frac{e^2}{4\pi\epsilon_0 r}$

Here $\mu$ is the reduced mass of the electron-proton system (very close to the electron mass, but the distinction matters for precision spectroscopy). Working in spherical coordinates, the equation separates into a radial part and an angular part.

Solving the Schrödinger Equation

The separation of variables yields three quantum numbers, each from a different part of the equation:

1. The angular part gives the spherical harmonics $Y_l^{m_l}(\theta, \phi)$, introducing $l$ and $m_l$.
2. The radial part gives the associated Laguerre polynomials $R_{n,l}(r)$, introducing the principal quantum number $n$.

The full wave function is:

$\psi_{n,l,m_l}(r,\theta,\phi) = R_{n,l}(r) \cdot Y_l^{m_l}(\theta,\phi)$

The energy eigenvalues depend only on $n$:

$E_n = -\frac{13.6 \text{ eV}}{n^2}, \quad n = 1, 2, 3, \ldots$

This $n$-only dependence is a special feature of the Coulomb potential. It means that for hydrogen, orbitals with the same $n$ but different $l$ (e.g., 2s and 2p) are degenerate. This degeneracy is broken in multi-electron atoms.

The ground state (1s) wave function is:

$\psi_{1,0,0} = \frac{1}{\sqrt{\pi}} \left(\frac{1}{a_0}\right)^{3/2} e^{-r/a_0}$

where $a_0 \approx 0.529 \text{ Å}$ is the Bohr radius. Notice this function has no angular dependence (spherically symmetric) and decays exponentially with distance from the nucleus.

Quantum Numbers and Electronic Structure of Hydrogen

Quantum Numbers and Orbital Shapes

Each hydrogen orbital is specified by three quantum numbers, and the electron carries a fourth:

Electronic Transitions and Spectral Lines

The Pauli exclusion principle requires that no two electrons share the same set of all four quantum numbers. For hydrogen (one electron), this simply means the electron occupies one orbital at a time. The ground state is $n = 1, l = 0, m_l = 0$.

When the electron transitions between energy levels, it emits or absorbs a photon whose energy matches the gap:

$\Delta E = 13.6 \text{ eV} \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$

Different series of spectral lines correspond to different final states:

• Lyman series: transitions to $n_f = 1$ (ultraviolet)
• Balmer series: transitions to $n_f = 2$ (visible)
• Paschen series: transitions to $n_f = 3$ (infrared)

For example, the $H_\alpha$ line of the Balmer series results from the $n = 3 \to n = 2$ transition. Plugging in: $\Delta E = 13.6(1/4 - 1/9) = 1.89 \text{ eV}$, corresponding to a wavelength of 656.3 nm (red light). These spectral series provided some of the earliest experimental evidence for quantized energy levels.