Hydrogen in Static External Fields

Author

Daniel Fischer

Hydrogen in External Fields

This chapter collects commonly used formulae and order-of-magnitude estimates for how a hydrogenic atom responds to static external fields. We concentrate on the Zeeman effect (magnetic fields) and the leading quadratic Stark effect (electric fields), emphasising the interplay with fine and hyperfine structure and which quantum numbers remain “good” in the different parameter regimes.

The presentation is perturbative: choose the largest interaction as the unperturbed Hamiltonian and treat smaller interactions as perturbations. Where a derivation is somewhat lengthy it is placed in a collapsible note.

1. Zeeman effect: Magnetic interaction Hamiltonian

The magnetic moment operator of an atom has contributions from electronic orbital motion, electronic spin, and the nuclear magnetic moment. In SI units we write

\[ \hat{\vec{\mu}} = -\mu_B\,\hat{\vec{L}} - g_s \mu_B\,\hat{\vec{S}} + g_I \mu_N\,\hat{\vec{I}}, \]

where

$\mu_B = \dfrac{e\hbar}{2m_e}$ is the Bohr magneton,
$g_s\approx 2.002319\ldots$ is the electron spin $g$-factor (often approximated as $2$),
$\mu_N = \dfrac{e\hbar}{2m_p}$ is the nuclear magneton,
$g_I$ is the nuclear $g$-factor (proton: $g_p\approx 5.5857$),
$\hat{\vec{L}},\hat{\vec{S}},\hat{\vec{I}}$ are the orbital, spin and nuclear spin operators respectively.

An external (static) magnetic field $\vec{B}=B_z \vec{e}_z$ couples via

\[ \hat{H}_{\rm ZE} \;=\; -\hat{\vec{\mu}}\cdot\vec{B} \;=\; \frac{\mu_B}{\hbar} B_z\big(\hat{L}_z + g_s \hat{S}_z\big) - g_I\frac{\mu_N}{\hbar} B_z \hat{I}_z. \]

Because $\mu_N/\mu_B \approx m_e/m_p \approx 1/1836$, the direct nuclear Zeeman term is usually negligible for first-order electronic Zeeman shifts (but it does matter for very high precision hyperfine spectroscopy):

\[ \hat{H}_{\rm ZE} \;\approx\; \frac{\mu_B}{\hbar} B_z\big(\hat{L}_z + g_s \hat{S}_z\big). \]

Useful numerical values

$\mu_B \approx 9.2740100783\times10^{-24}\ \mathrm{J/T} \approx 5.7883818\times10^{-5}\ \mathrm{eV/T}.$
In frequency units (divide energy by Planck’s constant $h$): \[\frac{\mu_B}{h} \approx 13.9962449\ \mathrm{GHz/T}.\]

These conversions are handy when comparing Zeeman splittings (in GHz) to hyperfine frequencies (also usually quoted in MHz/GHz).

1.2 Why eigenstates change: non-commuting parts of the Hamiltonian

Let us write down the interaction Hamiltonians explicitly (schematically, suppressing constants):

Spin–orbit (fine structure): \[ \hat H_{\rm SO} \;\propto\; \hat{\vec L}\cdot\hat{\vec S}. \]
Hyperfine: \[ \hat H_{\rm HF} \;\propto\; \hat{\vec I}\cdot\hat{\vec J},\qquad \hat{\vec J}=\hat{\vec L}+\hat{\vec S}. \]
Zeeman (magnetic field $\vec B \parallel z$): \[ \hat H_{\rm ZE} \;\propto\; \hat L_z + g_s \hat S_z. \]

Step 1. What would be the ideal basis?

If all three Hamiltonians commuted, there would be a common eigenbasis diagonalising them simultaneously. The obvious candidate would be the coupled basis \[ |L,S,J,I;F,M_F\rangle, \] which diagonalises $\hat L^2,\hat S^2,\hat J^2,\hat I^2,\hat F^2,\hat F_z$.

Step 2. Commutation relations

$[\hat H_{\rm SO}, \hat H_{\rm HF}] = 0$
Reason: Both are rotational scalars, i.e. invariant under rotations and independent of the choice of coordinate system.
⇒ They share common eigenstates $|L,S,J,I;F,M_F\rangle$.
$[\hat H_{\rm ZE}, \hat H_{\rm SO}] \neq 0$
Reason: $\hat H_{\rm ZE}$ singles out the $z$-axis, breaking rotational symmetry, while $\hat H_{\rm SO}$ is rotationally invariant.
⇒ States $|L,S,J,M_J\rangle$ are no longer exact eigenstates once $\hat H_{\rm ZE}$ is significant.
$[\hat H_{\rm ZE}, \hat H_{\rm HF}] \neq 0$
Reason: $\hat H_{\rm HF}$ is a rotational scalar, but $\hat H_{\rm ZE}$ is not.
⇒ Hyperfine eigenstates $|F,M_F\rangle$ cease to be eigenstates in the Zeeman regime.

Step 3. Consequence for basis choice

If fine + hyperfine dominate (Zeeman weak): good basis is $|L,S,J,I;F,M_F\rangle$.
If Zeeman dominates over hyperfine but not over spin–orbit: good basis is $|L,S,J,M_J; I,M_I\rangle$.
If Zeeman dominates completely (hyperfine Paschen–Back): good basis is closer to the uncoupled product $|L,M_L\rangle|S,M_S\rangle|I,M_I\rangle$.

Rule of thumb. Always pick the basis corresponding to the largest rotationally invariant interaction (SO or HF), unless the Zeeman term overwhelms it and breaks the symmetry.

2. Practical regimes (and convenient bases)

Depending on the relative sizes of the Zeeman splitting $\Delta E_{\rm ZE}$, fine-structure splitting $\Delta E_{\rm FS}$, and hyperfine splitting $\Delta E_{\rm HF}$, different bases are useful:

Weak fields relative to fine structure ($\Delta E_{\rm ZE} \ll \Delta E_{\rm FS}$, assume $I=0$).
Natural basis: $|L,S,J,M_J\rangle$.
Spin–orbit coupling fixes $\hat{\vec J}$, and the Zeeman term is treated as a perturbation via $\hat{\vec J}\cdot\vec B$.
Energy shifts are proportional to $g_J \mu_B B M_J$.
Fields weak compared to hyperfine splitting ($\Delta E_{\rm ZE} \ll \Delta E_{\rm HF}$, while $\Delta E_{\rm FS}$ remains much larger).
Natural basis: $|L,S,J,I;F,M_F\rangle$.
Hyperfine coupling fixes $\hat{\vec F}$, and the Zeeman effect produces first-order shifts proportional to $g_F \mu_B B M_F$.
Strong fields (Paschen–Back / hyperfine Paschen–Back limit) ($\Delta E_{\rm ZE} \gg \Delta E_{\rm HF}$).
Electronic and nuclear angular momenta decouple.
A convenient product basis is $|J,M_J; I,M_I\rangle$, with dominant energy $E\approx g_J \mu_B B M_J$ plus small nuclear contributions and hyperfine corrections.

2.1 Case 1: Landé $g_J$ and the electronic (LS) Zeeman shift

If $\Delta E_{\rm ZE} \ll \Delta E_{\rm FS}$ and the nucleus carries no spin ($I=0$), then $M_J$ is a good quantum number and the convenient basis is $|L,S,J,M_J\rangle$.

The Zeeman interaction Hamiltonian (electronic part) is \[ \hat H_{\rm ZE} \;=\; -\hat{\vec\mu}_e\cdot\vec B \;=\; \frac{\mu_B}{\hbar}\,B\;\big(\hat L_z + g_s \hat S_z\big), \] where $\mu_B=\dfrac{e\hbar}{2m_e}$ is the Bohr magneton and $g_s\approx 2$ is the electron spin $g$-factor.

To first order in perturbation theory the energy shift of an eigenstate $|L,S,J,M_J\rangle$ is \[ \Delta E^{(1)} \;=\; \langle L,S,J,M_J|\hat H_{\rm ZE}|L,S,J,M_J\rangle \;=\; \frac{\mu_B}{\hbar}\,B\;\big(\langle \hat L_z\rangle + g_s \langle \hat S_z\rangle\big). \]

The expectation values $\langle \hat L_z\rangle$ and $\langle \hat S_z\rangle$ in the coupled state align with the total angular momentum direction. They can be written in the compact form \[ \langle \hat L_z\rangle = \frac{\langle \hat{\vec L}\cdot\hat{\vec J}\rangle}{J(J+1)\hbar^2}\; M_J\hbar, \qquad \langle \hat S_z\rangle = \frac{\langle \hat{\vec S}\cdot\hat{\vec J}\rangle}{J(J+1)\hbar^2}\; M_J\hbar, \] where the scalar products evaluate (on $|L,S,J\rangle$) to \[ \langle \hat{\vec L}\!\cdot\!\hat{\vec J}\rangle = \tfrac{\hbar^2}{2}\big[J(J+1)+L(L+1)-S(S+1)\big], \] \[ \langle \hat{\vec S}\!\cdot\!\hat{\vec J}\rangle = \tfrac{\hbar^2}{2}\big[J(J+1)-L(L+1)+S(S+1)\big]. \]

Inserting these expressions into the first-order shift leads to the familiar Landé formula in the LS-coupling limit:

Zeeman shift in the LS-coupling limit

\[ \Delta E^{(1)} = \mu_B\,B\,g_J\,M_J\, \] with \[ \begin{align} g_J =& \frac{1}{2J(J+1)}\Big[ \big(J(J+1)-S(S+1)+L(L+1)\big)\\ &+ g_s\big(J(J+1)+S(S+1)-L(L+1)\big)\Big]. \end{align} \]

Thus, qualitatively: when $\hat{\vec J}$ is fixed by strong spin–orbit coupling, the Zeeman interaction projects onto $\hat J_z$ and produces shifts proportional to $M_J$; the proportionality factor is the Landé $g_J$ which encodes how orbital and spin contributions combine.

Plausibility argument for the projection formulas (sketch)

We give a heuristic (geometric) justification for the projection results used above — it is intended to build intuition rather than to replace a full operator derivation.

In the LS coupling regime $\vec L$ and $\vec S$ are well defined in magnitude, and their vector sum is the total electronic angular momentum $\vec J=\vec L+\vec S$. Because of the coupling, the directions of $\vec L$ and $\vec S$ are not sharp: each precesses about the fixed direction of $\vec J$, so their instantaneous orientations fill cones around $\vec J$.
The expectation values $\langle\vec L\rangle$ and $\langle\vec S\rangle$ therefore point along $\vec J$ (by symmetry), i.e., they are proportional to $\vec J$: \[ \langle\vec L\rangle = \alpha_L\,\vec J,\qquad \langle\vec S\rangle = \alpha_S\,\vec J. \] The scalars $\alpha_L,\alpha_S$ are fixed by taking scalar products with $\vec J$: \[ \langle\vec L\cdot\vec J\rangle = \alpha_L\,|\vec J|^2,\qquad \langle\vec S\cdot\vec J\rangle = \alpha_S\,|\vec J|^2. \] Solving for $\alpha_{L,S}$ and replacing the eigenvalues of $\vec{J}^2$ gives \[ \alpha_L = \frac{\langle\vec L\cdot\vec J\rangle}{J(J+1)\hbar^2},\qquad \alpha_S = \frac{\langle\vec S\cdot\vec J\rangle}{J(J+1)\hbar^2}. \]
Projecting onto the laboratory $z$-axis (the direction of $\vec B$), where $\langle J_z\rangle = M_J\hbar$, yields the projection formulas used in the main text: \[ \langle L_z\rangle=\alpha_L\,M_J\hbar,\qquad \langle S_z\rangle=\alpha_S\,M_J\hbar. \]

This geometric picture is illustrated in the figure: the cones of allowed orientations for $\vec L$ and $\vec S$ average to vectors aligned with $\vec J$, and the $z$-component is the corresponding projection. The algebraic operator identities (used in the text) make this intuitive picture precise.

Vector average: $\langle\vec L\rangle$ and $\langle\vec S\rangle$ project onto $\vec J$.

The scalar product $\langle\vec L\cdot\vec J\rangle$ and $\langle\vec S\cdot\vec J\rangle$ can be rewritten using the identies \[ \begin{align} \vec{S}^2 &= (\vec J - \vec L)^2 =\vec{J}^2 + 2\, \vec{L}\cdot\vec{J} - \vec{L}^2\\ \vec{L}^2 &= (\vec J - \vec S)^2 =\vec{J}^2 + 2\, \vec{S}\cdot\vec{J} - \vec{S}^2 \end{align} \] yielding (in operator form): \[ \hat{\vec L}\cdot\hat{\vec J}=\tfrac{1}{2}(\hat J^2+\hat L^2-\hat S^2), \qquad \hat{\vec S}\cdot\hat{\vec J}=\tfrac{1}{2}(\hat J^2-\hat L^2+\hat S^2). \]

The Landé factor can the be calculated using the eigenvalues of $\hat J^2,\hat L^2,\hat S^2$.

Remarks / assumptions

We assumed (i) LS coupling so $L,S,J$ are good quantum numbers, and (ii) the applied field is weak enough that first-order perturbation theory applies (no level mixing beyond perturbative shifts).
The projection argument (expectation of an electronic vector operator is parallel to $\langle\hat{\vec J}\rangle$) follows from rotational symmetry and is a standard Wigner–Eckart result; a short justification is: the only vector available in the $|J,M_J\rangle$ subspace is $\langle\hat{\vec J}\rangle$, so any vector expectation must be proportional to it — the proportionality constant is fixed by taking the scalar product with $\hat{\vec J}$ (done above).
If $L,S$ are not good quantum numbers (e.g. strong spin–orbit mixing beyond LS coupling), $g_J$ must be obtained from the appropriate matrix elements in the actual eigenbasis.

2.2 Case 2: Hyperfine regime and $g_F$

If hyperfine coupling dominates ($\Delta E_{\rm ZE} \ll \Delta E_{\rm HF}$) and the $z$ component of $\hat{\vec F} = \hat{\vec J} + \hat{\vec I}$ is a good quantum number, the first-order Zeeman shift is

\[ \Delta E^{(1)} \;=\; \langle L,S,J,F,M_F|\hat H_{\rm ZE}|L,S,J,F,M_F\rangle. \]

From the previous section we may approximate the Zeeman Hamiltonian as

\[ \hat H_{\rm ZE} \;\approx\; g_J \frac{\mu_B}{\hbar}\,\hat{J}_z\,B. \]

Using the same projection argument as in the electronic case, and neglecting the nuclear contribution to the Zeeman interaction, we obtain

\[ \Delta E^{(1)} \;=\; \frac{\mu_B}{\hbar}\,B\,g_J \,\langle \hat J_z\rangle. \]

The expectation value $\langle \hat J_z \rangle$ can be expressed in terms of $\vec J$ projected onto $\vec F$, giving

\[ \langle \hat J_z\rangle \;=\; \frac{\langle \hat{\vec J}\cdot \hat{\vec F}\rangle}{F(F+1)\hbar^2}\;M_F \hbar. \]

Inserting this into the expression above yields

\[ \Delta E^{(1)} \;=\; \mu_B\,B \,\frac{\langle \hat{\vec J}\cdot \hat{\vec F}\rangle}{F(F+1)}\, M_F. \]

This can be written compactly as

Zeeman shift in the hyperfine regime

\[ \Delta E^{(1)} \;=\; \mu_B\,B\,g_F\,M_F, \]

with the standard result for the hyperfine Landé factor

\[ g_F \;=\; g_J \,\frac{F(F+1)-I(I+1)+J(J+1)}{2F(F+1)}. \]

2.3 Case 3: Strong fields — (Hyperfine) Paschen–Back limit

When the Zeeman splitting is much larger than the hyperfine interaction ($\Delta E_{\rm ZE} \gg \Delta E_{\rm HF}$), the electronic and nuclear moments effectively decouple. In this regime the product basis

\[ |J,M_J; I,M_I\rangle \]

is appropriate (formally identical to Case 1, but now with nuclear spin included explicitly). The dominant Zeeman energy shift is

\[ \Delta E_{\rm ZE} \;=\; \mu_B\,B\,g_J\,M_J, \]

where the electronic Landé factor $g_J$ governs the splitting of the electronic angular momentum states.

The hyperfine Hamiltonian can then be treated perturbatively. To first order we write

\[ \Delta E^{(1)}_{\rm HF} \;=\; \langle J,M_J; I,M_I|\hat H_{\rm HF}|J,M_J; I,M_I\rangle \;=\; A\,\langle \hat{\vec I}\cdot \hat{\vec J}\rangle, \]

where $A$ is the hyperfine constant.

Expanding the scalar product,

\[ \langle \hat{\vec I}\cdot \hat{\vec J}\rangle = \langle \hat I_x \hat J_x\rangle \;+\; \langle \hat I_y \hat J_y\rangle \;+\; \langle \hat I_z \hat J_z\rangle. \]

In the strong-field limit, the transverse terms $\langle \hat I_x \hat J_x\rangle$ and $\langle \hat I_y \hat J_y\rangle$ vanish on average. This is because both $\vec I$ and $\vec J$ precess rapidly around the external field axis ($z$), so only the $z$ components remain well defined in expectation values.

Thus, the leading hyperfine contribution reduces to

\[ \langle \hat{\vec I}\cdot \hat{\vec J}\rangle \;\approx\; \langle \hat I_z \hat J_z\rangle \;=\; M_I M_J. \]

Accordingly, the hyperfine correction produces only small, $M_I M_J$–dependent energy shifts:

Hyperfine shift in the LS-coupling limit

\[ \Delta E^{(1)}_{\rm HF} \;=\; A\,M_I M_J. \]

3. Zeeman effect: summary and some exact results

3.1 Exact diagonalization for hydrogen ground state — Breit–Rabi behaviour

For the hydrogen ground state ($n=1$, $J=\tfrac12$, $I=\tfrac12$) the hyperfine + Zeeman Hamiltonian can be diagonalized exactly within the two $2\times2$ subspaces (labelled by $m_F=0$ and $m_F=\pm1$). The resulting $E(B)$ dependence is commonly referred to as the Breit–Rabi curve. It captures the gradual transition from hyperfine-dominated eigenstates at low field to Zeeman-dominated eigenstates at high field.

Code

import numpy as np
import matplotlib.pyplot as plt

# Set the global font size
plt.rcParams['font.size'] = 14

# Enable LaTeX rendering
plt.rcParams['text.usetex'] = True

# Optionally, configure the font family to match your LaTeX document
plt.rcParams['font.family'] = 'serif'
plt.rcParams['font.serif'] = 'Computer Modern'

# Example LaTeX preambles can be set for more advanced customization
plt.rcParams['text.latex.preamble'] = r'\usepackage{amsmath} \usepackage{physics}'

# --- Physical Constants (in SI units) ---
h = 6.62607015e-34  # Planck's constant (J·s)
mu_B = 9.2740100783e-24  # Bohr magneton (J/T)
g_J = 2.00231930436256  # Electron g-factor
g_I_proton = 5.585694702  # Proton g-factor

# --- Hydrogen Ground State Parameters ---
delta_E_hz = 1420.405751768e6  # Hyperfine splitting frequency (Hz)
delta_E_joules = h * delta_E_hz  # Hyperfine splitting energy (Joules)
I = 0.5  # Nuclear spin for Hydrogen

# --- Magnetic Field Range (Tesla) ---
B_field = np.linspace(0, 0.2, 500)  # From 0 to 0.2 Tesla

# --- Breit-Rabi Parameter 'x' ---
# The parameter x is defined as: x = (g_J * mu_B - g_I * mu_N) * B / delta_E
# Often, the g_I*mu_N term is neglected as it's much smaller, but let's use the full expression
# x = (g_J * mu_B * B_field) / delta_E_joules
# A more common form of x uses the total magnetic moments to be more accurate
x = (g_J*mu_B + g_I_proton*mu_B/1836.15) * B_field / delta_E_joules

# --- Breit-Rabi Formula Calculation ---
# For I = 1/2 and S = 1/2, the states are F=1, F=0
# The formula is E = -delta_E / 4 +- delta_E/2 * sqrt(1 + 2*mF*x + x^2)
# The '+' is for F=1, the '-' is for F=0, but this is only for the mF=0 states.
# We must use the general formula from the beginning.
# The general Breit-Rabi formula is:
# E_mF = -(delta_E/2) * (1/(2I+1)) +- (delta_E/2)*sqrt(1+x^2 + 4mF*x) for I=1/2

# Let's use the full formula for the four energy levels
# F=1, m_F=+1 state:
E1_plus1 = +delta_E_joules/4 + (delta_E_joules/2)*x

# F=1, m_F=0 state:
E1_0 = -delta_E_joules/4 + (delta_E_joules/2)*np.sqrt(1 + x**2)

# F=0, m_F=0 state:
E0_0 = -delta_E_joules/4 - (delta_E_joules/2)*np.sqrt(1 + x**2)

# F=1, m_F=-1 state:
E1_minus1 = delta_E_joules/4 - (delta_E_joules/2)*x

# --- Plotting ---
plt.figure(figsize=(10, 7))

# Plot the energy levels, converting to GHz for the y-axis
plt.plot(B_field * 1000, E1_plus1 / h / 1e9, label=r'$F=1, m_F=+1$', color='red', linewidth=3)
plt.plot(B_field * 1000, E1_0 / h / 1e9, label=r'$F=1, m_F=0$', color='blue', linewidth=3)
plt.plot(B_field * 1000, E0_0 / h / 1e9, label=r'$F=0, m_F=0$', color='purple', linewidth=3)
plt.plot(B_field * 1000, E1_minus1 / h / 1e9, label=r'$F=1, m_F=-1$', color='green', linewidth=3)

plt.text(75, 1.2, r'$(M_J=\tfrac{1}{2},M_I=\tfrac{1}{2})$', fontsize=18, color='red', rotation=14)
plt.text(72, 0.68, r'$(M_J=\tfrac{1}{2},M_I=-\tfrac{1}{2})$', fontsize=18, color='blue', rotation=12)
plt.text(72, -0.88, r'$(M_J=-\tfrac{1}{2},M_I=\tfrac{1}{2})$', fontsize=18, color='green', rotation=-14)
plt.text(69, -1.74, r'$(M_J=-\tfrac{1}{2},M_I=-\tfrac{1}{2})$', fontsize=18, color='purple', rotation=-11.5)

plt.text(2, 0.5, r'$(F=1,M_F=1)$', fontsize=18, color='red', rotation=14)
plt.text(15, 0.2, r'$(F=1,M_F=0)$', fontsize=18, color='blue', rotation=0)
plt.text(2, -0.25, r'$(F=1,M_F=-1)$', fontsize=18, color='green', rotation=-14)
plt.text(2, -0.97, r'$(F=0,M_F=0)$', fontsize=18, color='purple', rotation=0)

plt.axvline(x=10 , color='gray', linestyle='--', linewidth=0.8,
            label=r'low field region')
plt.axvline(x=53 , color='gray', linestyle='--', linewidth=0.8,
            label=r'high field region')

plt.text(1, 1.2, r'low field', fontsize=18, rotation=90)
plt.text(5, 1.2, r'$\ket{F M_F}$', fontsize=18, rotation=90)

plt.text(15, 1.85, r'transition region', fontsize=18)
plt.text(15, 1.6, r'(mixed states)', fontsize=18)



plt.text(55, 1.85, r'high field (Paschen--Back)', fontsize=18)
plt.text(55, 1.6, r'$\ket{J M_J; I M_I}$', fontsize=18)



plt.xlabel('Magnetic Field B (mT)')
plt.ylabel('Energy shift (GHz)')
plt.grid(False)
#plt.legend()
plt.xlim(0, 100.0)
plt.ylim(-2.0, 2.0)
plt.show()

Figure 1: Hydrogen ground state energy shifts in a magnetic field.

(For completeness: the full Breit–Rabi formula and its derivation are standard and can be found in many textbooks — see the note below for a compact statement and derivation sketch.)

The Breit–Rabi formula (reference and sketch)

The Breit–Rabi formula gives closed expressions for the energy levels of an atom with $J=\tfrac12$ and arbitrary nuclear spin $I$ in an external magnetic field. For the special case $I=\tfrac12$ (hydrogen ground state) the formula simplifies considerably and yields the usual two linear Zeeman branches ($m_F=\pm1$) and two levels that mix and show the characteristic avoided-crossing behaviour for $m_F=0$.

Derivation sketch: 1. Write the Hamiltonian in the $\{|m_J,m_I\rangle\}$ product basis. 2. Use that $m_F=m_J+m_I$ is conserved, which decomposes the problem into $2\times2$ blocks. 3. Diagonalize each $2\times2$ block to get the analytic expressions.

For an authoritative source and compact formula see: C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics (Volume II) or the review by Steck (Rubidium data) where the Breit–Rabi formula is written out explicitly and applied to alkali ground states.

3.2 Order-of-magnitude numbers (useful estimates)

Hyperfine splitting of hydrogen ground state (21-cm line): \[\nu_{\rm hfs} \approx 1420.405751768\ \mathrm{MHz},\qquad \Delta E_{\rm hfs}=h\nu_{\rm hfs}\approx 5.87\times10^{-6}\ \mathrm{eV}.\]
Convert frequency to effective field: \[B_{\rm crossover}\approx \frac{\nu_{\rm hfs}}{\mu_B/h} \approx \frac{1.4204\ \mathrm{GHz}}{13.996245\ \mathrm{GHz/T}} \approx 0.1015\ \mathrm{T}.\]

Thus around $\sim 0.1\ \mathrm{T}$ the Zeeman splitting becomes comparable to the zero-field hyperfine splitting for hydrogen.
At $B=1\ \mathrm{T}$ the single-electron Zeeman frequency (in energy units) \[\Delta E_{\rm Z} = \mu_B B \approx 5.79\times10^{-5}\ \mathrm{eV} \quad\Longleftrightarrow\quad \frac{\Delta E_{\rm Z}}{h}\approx 13.996\ \mathrm{GHz}.\]

These numbers help decide which interactions dominate in a given laboratory situation.

Typical experimental magnitudes

Weak laboratory fields ($B\lesssim 1\ \mathrm{mT}$): Zeeman shifts are tiny; hyperfine structure typically dominates.
Modest fields ($B\sim 0.01\!-\!0.2\ \mathrm{T}$): interesting interplay — use the Breit–Rabi description.
Strong fields ($B\gtrsim 1\ \mathrm{T}$): Zeeman splitting dominates; $M_J,M_I$ are the convenient labels.

4. Atoms in electric fields — quadratic Stark effect

The interaction of an atom with a static electric field $\vec{\mathcal E}$ is given by

\[ \hat H_{\rm St} = -\hat{\vec d}\cdot\vec{\mathcal E}, \]

where $\hat{\vec d} = -e\,\hat{\vec r}$ is the electric dipole operator of the electron.
For simplicity we assume the field points along $z$:

\[ \hat H_{\rm St} = e\,\mathcal E\,\hat z. \]

4.1 Perturbation theory setup

We apply time-independent perturbation theory in the basis of hydrogen eigenstates $|n,\ell,m\rangle$. The first- and second-order energy corrections are

\[ \Delta E^{(1)}_{n\ell m} = \langle n\ell m| e\mathcal E z |n\ell m\rangle, \]

\[ \Delta E^{(2)}_{n\ell m} = \sum_{n'\ell' m' \neq n\ell m} \frac{|\langle n'\ell' m'| e\mathcal E z |n\ell m\rangle|^2}{E_{n\ell}-E_{n'\ell'}}. \]

4.2 Why the first order vanishes

For hydrogen eigenstates, the expectation value $\langle n\ell m| z |n\ell m\rangle$ vanishes.
This can be seen in two ways:

Symmetry argument. The $z$ operator is odd under inversion $(x,y,z)\mapsto (-x,-y,-z)$. Hydrogen eigenfunctions have definite parity $(-1)^\ell$. Thus the integrand in $\int \psi^*(r)\,z\,\psi(r)\,d^3r$ is odd, and the integral is zero.
Physical picture. Stationary hydrogen states have no permanent dipole moment. The probability distribution is symmetric around the nucleus, so the average $z$ coordinate vanishes.

Therefore, $\Delta E^{(1)}=0$ for all $|n,\ell,m\rangle$ eigenstates.

4.3 Second-order shift and dipole matrix elements

The leading Stark effect is therefore quadratic, arising from the second-order term. It depends on dipole matrix elements of the form $\langle n'\ell' m'| z |n\ell m\rangle$, which connect states of opposite parity ($\ell'=\ell\pm 1$). These terms encode the atom’s ability to be polarized by the field.

We will not evaluate these matrix elements here — they become central in the discussion of light–matter interaction, where the dipole operator governs optical transitions.

Worked example: $2s$–$2p$ mixing

The $2s$ state has $\ell=0$ (even parity). The operator $z$ connects it to $2p$ states ($\ell=1$), which have odd parity:

\[ \langle 2p,m|\;z\;|2s\rangle \neq 0. \]

Thus the second-order Stark shift of the $2s$ level includes terms like

\[ \Delta E^{(2)}_{2s} = \sum_{m=-1}^{1}\frac{|\langle 2p,m|\,e\mathcal E z\,|2s\rangle|^2}{E_{2s}-E_{2p}}. \]

The numerator is the squared dipole matrix element (strength of $2s\leftrightarrow 2p$ coupling).
The denominator is the energy difference between $2s$ and $2p$ (which is zero in pure Schrödinger theory, but shifted by Lamb shift in reality).

This shows explicitly why the Stark effect is quadratic in $\mathcal E$:
the energy correction involves $|\langle\cdot|z|\cdot\rangle|^2\mathcal E^2$.

4.4 Result for hydrogen ground state

For the hydrogen ground state ($1s$), the quadratic Stark shift can be summarized as

\[ \Delta E = -\tfrac{1}{2}\,\alpha\,\mathcal E^2, \]

where $\alpha$ is the static electric polarizability. For $1s$ hydrogen:

\[ \alpha(1s) \approx 4.5\,a_0^3, \]

with $a_0$ the Bohr radius. In SI units, $\alpha$ may be converted to J/(V/m)$^2$.

4.5 Order of magnitude estimate

For a field of $\mathcal E=10^6\ \mathrm{V/m}$ (a large laboratory field) the shift is tiny:

\[ \Delta E \sim -\tfrac{1}{2}\,(4.5\,a_0^3)\,(10^6\ \mathrm{V/m})^2, \]

which is negligible compared with hyperfine and Zeeman splittings under similar conditions.

Quick remark on polarizability and units

The quoted polarizability $\alpha(1s)\approx 4.5\,a_0^3$ is in atomic units. To convert to SI (J m$^2$/V$^2$ or C m$^2$/V), apply the standard atomic-unit conversion factors. For conceptual discussions, the $a_0^3$ form is typically sufficient.

Hydrogen in Static External Fields

Hydrogen in External Fields

1. Zeeman effect: Magnetic interaction Hamiltonian

1.2 Why eigenstates change: non-commuting parts of the Hamiltonian

Step 1. What would be the ideal basis?

Step 2. Commutation relations

Step 3. Consequence for basis choice

2. Practical regimes (and convenient bases)

2.1 Case 1: Landé \(g_J\) and the electronic (LS) Zeeman shift

2.2 Case 2: Hyperfine regime and \(g_F\)

2.3 Case 3: Strong fields — (Hyperfine) Paschen–Back limit

3. Zeeman effect: summary and some exact results

3.1 Exact diagonalization for hydrogen ground state — Breit–Rabi behaviour

3.2 Order-of-magnitude numbers (useful estimates)

Typical experimental magnitudes

4. Atoms in electric fields — quadratic Stark effect

4.1 Perturbation theory setup

4.2 Why the first order vanishes

4.3 Second-order shift and dipole matrix elements

Worked example: \(2s\)–\(2p\) mixing

4.4 Result for hydrogen ground state

4.5 Order of magnitude estimate