Linear Stark Effect in Hydrogen

Derivation

Linear Stark Effect in Hydrogen

D-303 Home PU-306 Threads energy · fields · matter · symmetry Depends on Bound-State Spectrum of the Hydrogen Atom, Degenerate Perturbation Theory

Statement

For the hydrogen atom in a uniform external electric field \(\vec{\mathcal{E}} = \mathcal{E}\,\hat{z}\), degenerate perturbation theory applied to the \(n=2\) manifold yields a first-order energy shift that is linear in \(\mathcal{E}\). The four-fold degenerate \(n=2\) level (states \(|2s\rangle,\ |2p_0\rangle,\ |2p_{+1}\rangle,\ |2p_{-1}\rangle\)) splits into three levels with shifts \(\{+3ea_0\mathcal{E},\ 0,\ 0,\ -3ea_0\mathcal{E}\}\), the outer two being non-degenerate and the central one two-fold degenerate. This linear response exists because the hydrogenic degeneracy places states of opposite parity (\(2s\) and \(2p_0\)) at the same energy; in non-hydrogenic (many-electron) atoms that degeneracy is lifted and the leading Stark shift is quadratic.

Why it matters

The linear Stark effect is the cleanest laboratory demonstration that the Coulomb potential has an "accidental" \(SO(4)\) symmetry beyond the geometric \(SO(3)\) rotations. An atom with a genuine permanent-dipole-like response in first order is otherwise forbidden by parity — hydrogen evades the theorem only because its exact degeneracy lets the field mix states of opposite parity at zero energy cost. Historically it explained the field-induced splitting of the Balmer lines seen by Stark (1913) and was one of the first quantitative triumphs of the new quantum mechanics.

Practically, the effect governs field ionization of Rydberg atoms, the design of electrostatic lenses and Stark decelerators for cold molecules, and the interpretation of spectral line broadening in plasmas (the Holtsmark/Stark broadening of hydrogen lines is a direct consequence of the linear term). Understanding when the shift is linear versus quadratic is the difference between a first-order and a second-order sensitivity to stray fields in precision spectroscopy.

Assumptions

The field is weak enough that \(e\mathcal{E}a_0 \ll\) the \(n=2\)-to-\(n=1,3\) energy spacing.If dropped, the perturbation mixes different principal quantum numbers and the manifold cannot be treated in isolation; one must diagonalise on a larger basis or use parabolic coordinates for the exact treatment.

Fine structure, spin, and the Lamb shift are neglected (pure Coulomb degeneracy).If dropped, the exact \(2s\)–\(2p\) degeneracy is broken by \(\sim 10^{-5}\,\text{eV}\) (Lamb shift, spin-orbit), and the linear regime only survives for fields large compared to that splitting; below it the response reverts to quadratic. Tag as the subtle one.

The nucleus is a fixed point charge and the field is static and spatially uniform over the atom.If dropped, field gradients couple to the quadrupole moment and time dependence introduces the AC (dynamic) Stark effect with a frequency-dependent polarisability.

The potential energy of the electron in the field is \(H' = e\mathcal{E}z\) with the electron charge \(-e\) (\(e>0\)).If the sign convention is dropped, the labelling of which mixed state rises and which falls in energy is reversed, though the magnitude of the splitting is unchanged.

Derivation

\[ H = H_0 + H', \qquad H' = -q\,\vec{\mathcal{E}}\cdot\vec{r} = -(-e)\,\mathcal{E}\,z = e\,\mathcal{E}\,z \]

Write the total Hamiltonian as the Coulomb term \(H_0\) plus the interaction of the electron (charge \(-e\)) with the external field. Potential energy of a charge in a uniform field. A

\[ E_n^{(0)} = -\frac{13.6\ \text{eV}}{n^2}, \qquad \text{for } n=2:\ \{\,|2\ell m\rangle\,\} = \{|2s\rangle,\,|2p_{0}\rangle,\,|2p_{+1}\rangle,\,|2p_{-1}\rangle\} \]

The \(n=2\) level of \(H_0\) is four-fold degenerate; ordinary (non-degenerate) perturbation theory diverges here, so we must diagonalise \(H'\) within the degenerate subspace. Prior result: hydrogen bound-state spectrum. A

\[ [\,\hat{L}_z,\ z\,] = 0 \ \Rightarrow\ \langle 2\ell m' | z | 2\ell m\rangle \propto \delta_{m'm} \]

\(z = r\cos\theta\) carries \(m=0\), so the perturbation cannot change the magnetic quantum number: only states with equal \(m\) couple. This block-diagonalises the \(4\times4\) matrix by \(m\). B

\[ \langle 2\ell m | z | 2\ell m\rangle = 0 \quad (\text{parity}), \qquad \langle 2p_{\pm1}|z|2p_{\pm1}\rangle = 0 \]

\(z\) is odd under parity while each \(|n\ell m\rangle\) has definite parity \((-1)^\ell\); diagonal elements vanish. The \(m=\pm1\) states have no partner of equal \(m\) and opposite parity in the manifold, so they are unshifted. B

\[ \text{Only } \langle 2s | z | 2p_0\rangle \text{ survives: } m=0 \text{ both, opposite parity } (\ell=0 \leftrightarrow \ell=1). \]

Selection rules \(\Delta m = 0\) and \(\Delta\ell = \pm1\) (parity) leave a single independent nonzero matrix element. The problem reduces to a \(2\times2\) in the \(\{|2s\rangle, |2p_0\rangle\}\) subspace. C

\[ \langle 2s|z|2p_0\rangle = \int \psi_{200}^{*}\,(r\cos\theta)\,\psi_{210}\,d^3r = -3a_0 \]

Insert the hydrogen wavefunctions \(\psi_{200}\) and \(\psi_{210}\) and carry out the radial and angular integrals (the angular part gives \(\langle Y_0^0|\cos\theta|Y_1^0\rangle = 1/\sqrt3\); the radial part yields \(-3\sqrt3\,a_0\)). This is the one number the whole result depends on. C

\[ W = e\mathcal{E}\begin{pmatrix} \langle 2s|z|2s\rangle & \langle 2s|z|2p_0\rangle \\ \langle 2p_0|z|2s\rangle & \langle 2p_0|z|2p_0\rangle \end{pmatrix} = e\mathcal{E}\begin{pmatrix} 0 & -3a_0 \\ -3a_0 & 0 \end{pmatrix} \]

Assemble the perturbation matrix on the coupled subspace using steps 4 and 6, with \(\langle 2p_0|z|2s\rangle = \langle 2s|z|2p_0\rangle^{*}\) real by Hermiticity. B

\[ \det(W - E^{(1)}\mathbb{1}) = 0 \ \Rightarrow\ (E^{(1)})^2 - (3e a_0 \mathcal{E})^2 = 0 \]

The first-order shifts are the eigenvalues of \(W\); solve the secular (characteristic) equation. This is the defining equation of degenerate perturbation theory. B

\[ E^{(1)}_{\pm} = \pm 3 e a_0 \mathcal{E}, \qquad |{\pm}\rangle = \frac{1}{\sqrt2}\big(|2s\rangle \mp |2p_0\rangle\big) \]

Solve the quadratic and read off the eigenvectors: symmetric/antisymmetric hybrids of \(2s\) and \(2p_0\). These "sp-hybrids" carry a permanent electric dipole \(\mp 3ea_0\hat z\). A

Result

\[ \Delta E^{(1)}_{n=2} = \{\,+3ea_0\mathcal{E},\ \ 0,\ \ 0,\ \ -3ea_0\mathcal{E}\,\} \]

Reading. The applied field splits the four-fold degenerate \(n=2\) level into three: two shifted states (the \(sp\)-hybrids, one raised and one lowered by \(3ea_0\mathcal{E}\)) and a doubly-degenerate unshifted pair (the \(2p_{\pm1}\) states). The splitting grows linearly with field strength — the signature of the effect — and its scale \(3ea_0\) is an induced-independent permanent dipole moment of the hybrid states, magnitude three Bohr radii times the elementary charge.

Units check. \([e][a_0][\mathcal{E}] = \text{C}\cdot\text{m}\cdot(\text{V/m}) = \text{C}\cdot\text{V} = \text{J}\), an energy. Equivalently \(ea_0 = 8.478\times10^{-30}\,\text{C·m}\) is a dipole moment, and (dipole)×(field) = energy. Correct.

Limiting cases

\(\mathcal{E}\to 0\): all shifts vanish and the four states collapse back to the unperturbed degenerate \(n=2\) level — no zero-field permanent dipole exists.
Field along a general direction \(\hat n\): by rotational covariance the result is unchanged with \(z\to \hat n\cdot\vec r\); the quantisation axis is simply chosen along \(\vec{\mathcal{E}}\), so \(3ea_0\mathcal{E}\) is direction-independent.
Higher manifolds \(n\): the linear splitting generalises to \(\Delta E^{(1)} = \tfrac{3}{2}n(n_1-n_2)ea_0\mathcal{E}\) in parabolic quantum numbers; for \(n=2\), \(n_1-n_2=\pm1\) recovers \(\pm3ea_0\mathcal{E}\).
\(\mathcal{E}\) comparable to the fine-structure/Lamb splitting: the linear law is recovered only well above that scale; below it the two-level model is invalid and the shift is quadratic.

Breaks when

Non-hydrogenic atoms. In any many-electron atom the quantum defect lifts the \(\ell\)-degeneracy, so \(s\) and \(p\) states differ in zeroth-order energy. The off-diagonal coupling then produces a second-order (quadratic) shift \(\Delta E \propto -\tfrac12\alpha\mathcal{E}^2\); the linear term is strictly absent.
Strong fields \(e\mathcal{E}a_0 \gtrsim\) inter-\(n\) spacing. The perturbation mixes \(n=2\) with \(n=1,3,\dots\); the isolated-manifold assumption fails, higher-order and continuum (field-ionisation) effects dominate, and the atom eventually ionises by tunnelling through the tilted Coulomb barrier.
Fine structure / Lamb shift regime. When \(e\mathcal{E}a_0\) is below the \(\sim4.4\times10^{-6}\,\text{eV}\) Lamb shift, the \(2s_{1/2}\)–\(2p_{1/2}\) states are no longer degenerate and the response is quadratic in \(\mathcal{E}\), not linear.

Failure modes

Using non-degenerate perturbation theory. Applying \(E^{(1)} = \langle n\ell m|H'|n\ell m\rangle\) gives zero for every state (parity), missing the effect entirely. The degeneracy requires diagonalisation.
Keeping the diagonal elements. Writing \(\langle 2s|z|2s\rangle \neq 0\) or \(\langle 2p_0|z|2p_0\rangle \neq 0\); these vanish by parity and their inclusion is a sign the wavefunction parities were mishandled.
Coupling \(m=\pm1\) states. Forgetting the \(\Delta m=0\) selection rule and building a full \(4\times4\) with spurious entries linking \(2p_{+1}\) to \(2s\).
Dropping the \(-3\) coefficient. Quoting \(ea_0\mathcal{E}\) instead of \(3ea_0\mathcal{E}\) by using \(|\langle 2s|z|2p_0\rangle| = a_0\); the radial integral genuinely gives \(3a_0\).
Sign/charge slip. Taking the electron charge as \(+e\) so \(H'=-e\mathcal{E}z\), which flips which hybrid rises — harmless for the splitting magnitude but wrong for state labelling.
Claiming a permanent dipole at zero field. The hybrids only exist as eigenstates because the field selects them; there is no dipole for \(\mathcal{E}=0\).

Discussion

The deep reason hydrogen shows a linear Stark effect is symmetry. Parity forbids a first-order energy shift in any non-degenerate, parity-definite state, because \(\langle\psi|z|\psi\rangle = 0\). Hydrogen escapes this only because its "accidental" degeneracy — traceable to the conserved Runge–Lenz vector and the hidden \(SO(4)\) symmetry of the \(1/r\) potential — places the opposite-parity states \(2s\) and \(2p_0\) at exactly the same energy. The field can then mix them at no zeroth-order cost, and the true eigenstates are parity-mixed hybrids carrying a permanent dipole. The linear Stark effect is thus a direct fingerprint of the special \(1/r\) form of the Coulomb law.

The eigenstates \(\tfrac1{\sqrt2}(|2s\rangle \mp |2p_0\rangle)\) are the \(sp\)-hybrids familiar from chemistry: electron density is displaced toward \(-z\) or \(+z\), giving a static dipole \(\mp3ea_0\hat z\) that couples to the field like a classical dipole \(U = -\vec p\cdot\vec{\mathcal{E}}\). This is why the exact treatment in parabolic coordinates is so natural: the parabolic states are already dipole eigenstates, and the linear shift \(\tfrac32 n(n_1-n_2)ea_0\mathcal{E}\) follows without any \(2\times2\) diagonalisation.

More rigorously, the Stark Hamiltonian \(H_0 + e\mathcal{E}z\) has no true bound states: the potential \(-e^2/4\pi\varepsilon_0 r + e\mathcal{E}z\) is unbounded below as \(z\to-\infty\), so every "level" is really a resonance with a finite tunnelling width \(\Gamma\). The perturbation series in \(\mathcal{E}\) is asymptotic (Borel-summable) rather than convergent, and the linear shift is only the leading real part of a complex quasi-energy \(E - i\Gamma/2\). For \(n=2\) at laboratory fields \(\Gamma\) is astronomically small, so the resonance is razor-sharp and the perturbative shift is physical, but the conceptual point — that field ionisation is always present — is exact.

Common misconceptions. The linear Stark effect does not mean hydrogen has a permanent dipole moment in its ground state — the \(n=1\) level is non-degenerate and shows only the ordinary quadratic effect. The linear term is exclusive to degenerate excited manifolds. Nor does the splitting imply broken rotational symmetry of the atom itself; it is the external field that defines the axis, and the total system (atom + field) respects axial symmetry about \(\hat z\), which is exactly why \(m\) remains a good quantum number.

Worked examples

Compute the \(n=2\) Stark splitting for a laboratory field \(\mathcal{E} = 1.0\times10^{5}\ \text{V/m}\).

Set up: the extreme states are shifted by \(\pm3ea_0\mathcal{E}\); the full splitting between them is \(\Delta = 6ea_0\mathcal{E}\). A

\[ 3ea_0 = 3(1.602\times10^{-19}\,\text{C})(5.29\times10^{-11}\,\text{m}) = 2.54\times10^{-29}\,\text{C·m} \]

Evaluate the dipole scale (symbols before numbers). A

\[ E^{(1)}_{\pm} = \pm(2.54\times10^{-29}\,\text{C·m})(1.0\times10^{5}\,\text{V/m}) = \pm2.54\times10^{-24}\,\text{J} \]

Multiply by the field. A

\[ E^{(1)}_{\pm} = \frac{\pm2.54\times10^{-24}\,\text{J}}{1.602\times10^{-19}\,\text{J/eV}} = \pm1.59\times10^{-5}\ \text{eV} \]

Convert to eV. A

\[ E^{(1)}_{\pm} = \pm1.6\times10^{-5}\ \text{eV}, \qquad \Delta_{\text{total}} = 3.2\times10^{-5}\ \text{eV} \]

Reading. A modest \(10^5\,\text{V/m}\) field splits the extreme \(n=2\) sublevels by \(\sim32\ \mu\text{eV}\) — comparable to the Lamb shift, which is exactly the field scale at which the linear regime sets in.

What field strength makes the linear Stark shift equal the \(n=2\) Lamb shift \(E_L = 4.37\times10^{-6}\ \text{eV}\)? (This marks the onset of the linear regime.)

Set up: require the single-state shift \(3ea_0\mathcal{E} = E_L\) and solve for \(\mathcal{E}\). B

\[ \mathcal{E} = \frac{E_L}{3ea_0} = \frac{E_L}{2.54\times10^{-29}\,\text{C·m}} \]

Rearrange symbolically before inserting numbers. B

\[ E_L = 4.37\times10^{-6}\,\text{eV} \times 1.602\times10^{-19}\,\text{J/eV} = 7.00\times10^{-25}\ \text{J} \]

Convert the Lamb shift to joules. A

\[ \mathcal{E} = \frac{7.00\times10^{-25}\,\text{J}}{2.54\times10^{-29}\,\text{C·m}} = 2.76\times10^{4}\ \text{V/m} \]

Divide. A

\[ \mathcal{E}_{\text{onset}} \approx 2.8\times10^{4}\ \text{V/m} \]

Reading. Below \(\sim3\times10^4\,\text{V/m}\) the Lamb shift dominates and the response is quadratic; above it the \(2s\)–\(2p\) states behave as effectively degenerate and the shift becomes linear. This crossover field is why precision hydrogen spectroscopy must control stray fields to well below \(10^4\,\text{V/m}\).

Problems

Verify the units-consistency of the atomic-unit expression: show that \(3ea_0\mathcal{E}\) in SI equals \(3\mathcal{E}\) (in atomic units of \(e a_0\)). State the atomic field unit.
Solution
In atomic units \(e=a_0=1\), so \(3ea_0\mathcal{E}\to 3\mathcal{E}\) with energy measured in Hartree and field in the atomic unit \(\mathcal{E}_{\text{au}} = E_h/(ea_0) = 5.14\times10^{11}\,\text{V/m}\). Check: \(E_h = 27.2\,\text{eV} = 4.36\times10^{-18}\,\text{J}\); \(E_h/(ea_0) = 4.36\times10^{-18}/(1.602\times10^{-19}\times5.29\times10^{-11}) = 5.14\times10^{11}\,\text{V/m}\). The atomic field unit is that at which the field energy over \(a_0\) equals one Hartree.
Show explicitly that \(\langle 2p_{+1}|z|2s\rangle = 0\) using the \(\phi\)-integral only, without evaluating the radial part.
Solution
\(z = r\cos\theta\) is independent of \(\phi\). The states carry \(e^{im\phi}\): \(2s\) has \(m=0\), \(2p_{+1}\) has \(m=+1\). The \(\phi\)-integral is \(\int_0^{2\pi} e^{-i(+1)\phi}e^{i(0)\phi}\,d\phi = \int_0^{2\pi} e^{-i\phi}d\phi = 0\). Hence the matrix element vanishes regardless of the radial and \(\theta\) integrals — this is the \(\Delta m=0\) selection rule.
Compute the total splitting \(\Delta = 6ea_0\mathcal{E}\) for \(\mathcal{E}=2.5\times10^{6}\,\text{V/m}\) and express it as an equivalent frequency \(\Delta/h\).
Solution
\(6ea_0 = 2\times(2.54\times10^{-29}) = 5.08\times10^{-29}\,\text{C·m}\). Then \(\Delta = 5.08\times10^{-29}\times2.5\times10^{6} = 1.27\times10^{-22}\,\text{J}\). As frequency: \(\Delta/h = 1.27\times10^{-22}/6.626\times10^{-34} = 1.92\times10^{11}\,\text{Hz} \approx 192\,\text{GHz}\). Equivalently \(0.79\,\text{meV}\).
For sodium (a non-hydrogenic atom) the \(3s\)–\(3p\) quantum-defect splitting is \(\approx 2.1\,\text{eV}\). Estimate the second-order (quadratic) Stark shift of the \(3s\) state at \(\mathcal{E}=1.0\times10^{6}\,\text{V/m}\), taking the dipole matrix element \(\langle 3s|ez|3p\rangle \approx 5ea_0\), and explain why it is not linear.
Solution
Second-order shift \(\Delta E^{(2)} = -\dfrac{|\langle 3s|H'|3p\rangle|^2}{E_{3p}-E_{3s}} = -\dfrac{(5ea_0\mathcal{E})^2}{2.1\,\text{eV}}\). With \(5ea_0 = 4.24\times10^{-29}\,\text{C·m}\), \(5ea_0\mathcal{E} = 4.24\times10^{-23}\,\text{J} = 2.65\times10^{-4}\,\text{eV}\). Then \(\Delta E^{(2)} = -(2.65\times10^{-4})^2/2.1 = -3.3\times10^{-8}\,\text{eV}\). It is quadratic because \(3s\) and \(3p\) are non-degenerate (quantum defect lifts the \(\ell\)-degeneracy), so degenerate perturbation theory does not apply and the leading term is the second-order energy denominator, \(\propto\mathcal{E}^2\).
Using the parabolic-coordinate formula \(\Delta E^{(1)} = \tfrac32 n(n_1-n_2)ea_0\mathcal{E}\), list the distinct first-order shifts of the \(n=3\) manifold and count the resulting levels.
Solution
For \(n=3\), \(n_1+n_2+|m|+1 = 3\), so \((n_1-n_2)\) ranges over integer values \(\{-2,-1,0,+1,+2\}\). The shifts are \(\Delta E^{(1)} = \tfrac32(3)(n_1-n_2)ea_0\mathcal{E} = \tfrac92(n_1-n_2)ea_0\mathcal{E}\), giving \(\{-9,-\tfrac92,0,+\tfrac92,+9\}\,ea_0\mathcal{E}\). That is five distinct, equally-spaced levels separated by \(\tfrac92 ea_0\mathcal{E}\) (the \(n=3\) manifold has \(n^2=9\) states distributed among these five shifts, with the central \(0\) shift most degenerate). The equal spacing \(\propto n\) is the hallmark of the linear effect.