Case Study 1: The Stark Effect — Electric Fields Meet Atoms

The Physical Situation

In 1913, the same year Niels Bohr published his model of the hydrogen atom, Johannes Stark placed hydrogen gas in a strong electric field and observed something remarkable: the spectral lines split. Where there had been a single emission line at a particular wavelength, there were now several closely spaced lines. This was the first direct observation of how external electric fields modify the internal quantum structure of atoms — the Stark effect.

The Stark effect remains one of the most important applications of perturbation theory, for several reasons. It provides a clean, quantitative test of quantum mechanics. It is the physical basis for understanding atomic and molecular polarizabilities. And it connects to modern technology: electric-field control of quantum states is essential in atomic clocks, quantum computing with trapped ions, and electro-optic devices.

Setting Up the Perturbation

When a hydrogen atom is placed in a uniform electric field $\mathbf{\mathcal{E}} = \mathcal{E}\hat{z}$, the Hamiltonian becomes:

$$\hat{H} = \hat{H}_0 + \hat{H}' = \left[-\frac{\hbar^2}{2m_e}\nabla^2 - \frac{e^2}{4\pi\epsilon_0 r}\right] + e\mathcal{E}r\cos\theta$$

The unperturbed problem is the hydrogen atom (Chapter 5), with energies $E_n^{(0)} = -13.6\,\text{eV}/n^2$ and eigenstates $|n,l,m\rangle$. The perturbation $\hat{H}' = e\mathcal{E}r\cos\theta$ represents the interaction between the electric field and the electron.

Key question: How does the electric field affect the energy levels?

The Ground State: Quadratic Stark Effect

Why First-Order Vanishes

The hydrogen ground state $|1,0,0\rangle$ is non-degenerate and has $l = 0$ (spherical symmetry). The perturbation $r\cos\theta$ has the angular symmetry of $Y_1^0$ — it is odd under parity ($\cos\theta \to -\cos\theta$ under inversion). Since $|1,0,0\rangle$ has even parity and the perturbation is odd:

$$E_1^{(1)} = \langle 1,0,0|e\mathcal{E}r\cos\theta|1,0,0\rangle = 0$$

Physically, the spherically symmetric ground state has no permanent electric dipole moment. A symmetric charge cloud, when averaged over all orientations, has no preferred direction.

Second-Order: The Induced Dipole

Although the atom has no permanent dipole, the field induces one by pulling the electron cloud slightly away from the nucleus. This distortion is a second-order effect:

$$E_1^{(2)} = e^2\mathcal{E}^2 \sum_{n,l,m \neq 1,0,0} \frac{|\langle n,l,m|r\cos\theta|1,0,0\rangle|^2}{E_1^{(0)} - E_n^{(0)}}$$

Selection rules ($\Delta l = \pm 1$, $\Delta m = 0$) restrict the sum to $|n,1,0\rangle$ states. The exact result (obtained by the Dalgarno-Lewis method or direct summation including the continuum) is:

$$E_1^{(2)} = -\frac{9}{4}a_0^3\mathcal{E}^2 \quad \text{(Gaussian units)}$$

This corresponds to a polarizability:

$$\alpha = \frac{9}{2}a_0^3 = 0.667\,\text{\AA}^3 \quad \text{(in SI: } \alpha/(4\pi\epsilon_0) = 4.5a_0^3\text{)}$$

Contributions from Individual States

The dominant contribution comes from the $n = 2$ states. The matrix element $\langle 2,1,0|r\cos\theta|1,0,0\rangle$ can be computed explicitly:

$$\langle 2,1,0|r\cos\theta|1,0,0\rangle = -\frac{2^7}{3^5}\sqrt{\frac{2}{3}}a_0 \approx -0.745\,a_0$$

This single state accounts for approximately $\frac{2^{15}}{3^{10}\cdot 3} \cdot \frac{4}{3} \approx 0.436$ of the total polarizability $4.5a_0^3$, or about 56%. The remaining 44% comes from $n = 3, 4, \ldots$ discrete states plus the continuum.

This table illustrates an important practical lesson: the infinite sum converges reasonably quickly, with the first few terms capturing most of the physics. But the continuum contribution (11%) is not negligible — a reminder that the sum in the perturbation formula must include all states, not just the discrete ones.

The $n = 2$ Level: Linear Stark Effect (Preview)

For the $n = 2$ level of hydrogen, the situation is fundamentally different. The states $|2,0,0\rangle$ and $|2,1,0\rangle$ are degenerate in the unperturbed atom ($E_2^{(0)} = -3.4$ eV for both). When we attempt non-degenerate perturbation theory, the formula for the wavefunction correction:

$$c_m^{(1)} = \frac{\langle m^{(0)}|\hat{H}'|n^{(0)}\rangle}{E_n^{(0)} - E_m^{(0)}}$$

has a zero denominator for $m = |2,0,0\rangle$ and $n = |2,1,0\rangle$ (or vice versa).

The resolution (Chapter 18) is to diagonalize $\hat{H}'$ within the degenerate subspace before applying perturbation theory. For the $n=2$ Stark effect, this gives a linear splitting:

$$E_2^{(\pm)} = E_2^{(0)} \pm 3ea_0\mathcal{E}$$

The energy shift is proportional to $\mathcal{E}$ (not $\mathcal{E}^2$), which is much larger for weak fields. This linear Stark effect was what Stark originally observed: the hydrogen Balmer lines (transitions to $n = 2$) show a clear splitting that grows linearly with field strength.

Experimental Verification

Historical Measurements

Stark's original 1913 experiment used discharge tubes to generate hydrogen emission and parallel-plate capacitors to create fields of $\sim 10^5$ V/m. He observed the splitting of the Balmer-$\alpha$ line (H$\alpha$, $656.3$ nm) into multiple components. The observed splittings agreed quantitatively with the predictions of quantum mechanics (once Schrödinger derived them in 1926), providing powerful evidence for the new theory.

Modern Precision

Today, the hydrogen Stark effect is measured using laser spectroscopy and atomic beams in controlled electric fields. The experimental value of the ground-state polarizability is:

$$\alpha/(4\pi\epsilon_0) = 4.50 \pm 0.01\,a_0^3$$

in agreement with theory at the $0.2\%$ level. The limitation is primarily experimental (controlling stray fields, measuring the applied field precisely), not theoretical.

Technological Applications

The Stark effect has numerous practical applications:

1. Stark deceleration of molecular beams: Molecules in specific quantum states experience electric-field-dependent forces. By rapidly switching inhomogeneous fields, molecules can be slowed to near rest — enabling cold molecule experiments.

2. Electric-field tuning of atomic transitions: In atomic clocks and precision spectroscopy, the Stark shift is a systematic error that must be carefully controlled. Understanding it at the perturbative level is essential.

3. Quantum computing with Rydberg atoms: Rydberg atoms (atoms in highly excited states with large $n$) have enormous polarizabilities (scaling as $n^7$). The Stark effect provides a mechanism for controlling and entangling Rydberg qubits.

4. Electro-optic materials: The Stark effect in bulk materials (the Pockels and Kerr effects) is the basis of electro-optic modulators used in telecommunications.

Quantitative Analysis

Critical Field Strengths

At what field strength does perturbation theory break down for hydrogen? The second-order energy shift equals the unperturbed level spacing when:

$$\frac{1}{2}\alpha\mathcal{E}^2 \sim E_2^{(0)} - E_1^{(0)} = \frac{3}{4} \times 13.6\,\text{eV} = 10.2\,\text{eV}$$

Solving: $\mathcal{E}_{\text{critical}} \sim 3 \times 10^{11}$ V/m. This is far beyond laboratory fields ($\sim 10^7$ V/m), confirming that perturbation theory is excellent for the ground state.

For excited states, the story is different. The $n = 2$ and $n = 3$ levels are separated by only $1.89$ eV, and the polarizability grows rapidly with $n$. For Rydberg states with $n \sim 30$, perturbation theory breaks down at relatively modest fields of $\sim 10^3$ V/m.

Comparison with Classical Model

A classical model of the atom as a uniformly charged sphere of radius $R$ gives a polarizability:

$$\alpha_{\text{classical}} = 4\pi\epsilon_0 R^3$$

Setting $R = a_0$: $\alpha_{\text{classical}} = 4\pi\epsilon_0 a_0^3$. The quantum result $\alpha = 4.5 \times 4\pi\epsilon_0 a_0^3$ is about 4.5 times larger, reflecting the diffuse nature of the quantum charge distribution (the exponential tail extends well beyond $a_0$).

Discussion Questions

1. The Stark effect for hydrogen excited states ($n \ge 2$) is qualitatively different from the ground state because of degeneracy. But alkali atoms (Li, Na, K) do not have degenerate excited states (the quantum defect lifts the $l$-degeneracy). Predict whether the Stark effect for the first excited state of sodium is linear or quadratic in $\mathcal{E}$, and explain your reasoning.

2. The polarizability of helium ($\alpha/(4\pi\epsilon_0) = 0.205\,\text{\AA}^3$) is much smaller than that of hydrogen ($0.667\,\text{\AA}^3$), despite helium having two electrons. Explain why in terms of energy level spacing and wavefunction size.

3. In strong electric fields, the hydrogen atom eventually ionizes — the electron tunnels through the potential barrier created by the combined Coulomb and electric field. This ionization rate is proportional to $e^{-c/\mathcal{E}}$ for some constant $c$. Is this effect perturbative or non-perturbative? Can any order of perturbation theory capture it?

Key Takeaways

• The Stark effect for the hydrogen ground state is quadratic ($\propto \mathcal{E}^2$) because parity symmetry forbids a first-order correction.
• The energy shift is characterized by the polarizability $\alpha$, which measures how easily the electron cloud is distorted.
• The perturbative result agrees with experiment to better than $0.2\%$, validating the method.
• For degenerate excited states, the Stark effect becomes linear ($\propto \mathcal{E}$) — a fundamentally different regime requiring degenerate perturbation theory (Chapter 18).
• Non-perturbative effects (field ionization) lie beyond the reach of any finite-order perturbation calculation.