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Diagonalizing the perturbation within degenerate sets, second-order level repulsion, the linear Stark effect of hydrogen, and the Landé factor of the Zeeman effect.
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By the end of this lesson you will be able to find first-order shifts of degenerate levels by diagonalizing the perturbation, compute second-order pushes between nearby levels, and predict Stark and Zeeman splittings.
You know first-order perturbation theory, $E_n^1 = \langle\psi_n^0|H'|\psi_n^0\rangle$, and the first-order state, whose coefficients have $E_n^0 - E_m^0$ in the denominator. You can diagonalize a $2 \times 2$ matrix, and you know that hydrogen's levels are degenerate and that spin and orbit combine into $j$. This lesson handles the case the previous one could not: perturbations of degenerate levels.
| Term | What it means |
|---|---|
| Degenerate level | An energy shared by two or more independent unperturbed states. |
| Good states | The combinations of degenerate states that the perturbation does not mix; the eigenvectors of $W$. |
| $W$ matrix | $W_{ij} = \langle\psi_i^0\vert \hat{H}'\vert \psi_j^0\rangle$ within the degenerate set. |
| Stark effect | The shift and splitting of atomic levels in an electric field. |
| Zeeman effect | The shift and splitting of atomic levels in a magnetic field. |
| Landé $g$ factor | $g_J$, which gives the weak-field Zeeman shift $g_J\mu_BBm_j$ of a level with total angular momentum $j$. |
| Level repulsion | The pushing apart of two coupled levels, by $V^2/\Delta$ at second order. |
If several unperturbed states $\psi_a^0, \psi_b^0, \ldots$ share an energy $E^0$, any combination of them is equally good as an unperturbed state, and the first-order formula $\psi_n^1 = \sum H'_{mn}\psi_m^0/(E_n^0 - E_m^0)$ divides by zero. The resolution is that the perturbation picks out particular combinations. Build the matrix of the perturbation within the degenerate set,
$$W_{ij} = \langle\psi_i^0|\hat{H}'|\psi_j^0\rangle,$$
and diagonalize it. Its eigenvalues are the first-order energy shifts, and its eigenvectors are the good states, the combinations that evolve smoothly into the true states as the perturbation is turned on. If the basis already makes $W$ diagonal — often because some operator commutes with both $\hat{H}^0$ and $\hat{H}'$ — ordinary perturbation theory works in that basis.
For levels that are close but not degenerate, the second-order correction matters:
$$E_n^2 = \sum_{m \ne n}\frac{|H'_{mn}|^2}{E_n^0 - E_m^0}.$$
Each coupled level pushes level $n$ away from itself by $|H'_{mn}|^2/|E_n^0 - E_m^0|$: levels repel. The ground state, with every other level above it, is always pushed down at second order.
Another way: picture
Picture two identical pendulums hanging side by side, each swinging at the same frequency. Any combination of their motions is a valid way to describe them until you connect them with a weak spring. Then two particular combinations — swinging together and swinging opposite — become the true modes, with slightly different frequencies. The spring is the perturbation; the two modes are the good states. It was the coupling, not the pendulums, that chose them.
Another way: steps
Checks. The sum of the first-order shifts equals the trace of $W$, often zero. The number of good states equals the degeneracy. A perturbation that commutes with an operator — such as $L_z$ for a field along $z$ — cannot mix states of different $m$, so $W$ is block diagonal. And a linear shift in the field is possible only when degenerate states of opposite parity exist, which is special to hydrogen.
Imagine turning the perturbation on slowly, $\lambda$ from zero to one. The true eigenstates change smoothly, and as $\lambda \to 0$ they approach particular combinations of the degenerate states — not arbitrary ones. Those limits are the good states. Writing the first-order equation and projecting onto each degenerate state shows that the good states' coefficients must satisfy $\sum_j W_{ij}c_j = E^1c_i$: an eigenvalue problem for $W$.
When $W$ has distinct eigenvalues, the degeneracy is lifted completely at first order and ordinary perturbation theory applies from then on. When some remain equal, the perturbation leaves a residual degeneracy — often a sign of a symmetry the perturbation still respects, such as rotation about the field direction, which leaves $\pm m$ degenerate in the Stark effect.
An electric field along $z$ adds $\hat{H}' = eE\hat{z}$. For a nondegenerate state of definite parity, $\langle z\rangle = 0$ and the first-order shift vanishes; the leading effect is second order, $-\tfrac{1}{2}\alpha_{\text{pol}}E^2$, set by the atom's polarizability. Hydrogen's ground state has $\alpha_{\text{pol}} = 4.5 \times 4\pi\varepsilon_0a^3$.
Hydrogen's $n = 2$ level is different, because $2s$ and $2p$ are degenerate and have opposite parity. The good states $(2s \pm 2p_0)/\sqrt{2}$ are lopsided, with their charge displaced along $\pm z$, and they carry a permanent dipole moment $3ea$. Their energies shift linearly, $\pm 3eaE$, a far larger effect in weak fields. In a field of $10^6$ V/m the shift is $0.16$ meV, and the line splits into three components. Johannes Stark observed the splitting of hydrogen's Balmer lines in 1913, and the linear effect was one of the early triumphs of the new quantum mechanics.
A magnetic field adds $\hat{H}_Z = \tfrac{\mu_B}{\hbar}(\hat{L}_z + 2\hat{S}_z)B$; the spin counts double because $g_s \approx 2$. When the field is weak compared with spin–orbit coupling, the unperturbed states are $|l, s, j, m_j\rangle$, and within each $j$ level $\hat{H}_Z$ is already diagonal in $m_j$. The first-order shift is $g_J\mu_BBm_j$ with the Landé factor
$$g_J = 1 + \frac{j(j + 1) + s(s + 1) - l(l + 1)}{2j(j + 1)},$$
obtained by projecting $\vec{L} + 2\vec{S}$ onto $\vec{J}$. For a pure spin state, $l = 0$, $g_J = 2$; for pure orbital motion, $s = 0$, $g_J = 1$; in between it varies. Different levels have different $g_J$, so each spectral line splits into a pattern that identifies the levels involved — the anomalous Zeeman effect that puzzled physicists until spin was discovered. In strong fields the spin–orbit coupling becomes the small perturbation, and the good states switch to $|m_l, m_s\rangle$: the Paschen–Back effect.
Second-order perturbation theory says that coupled levels push each other apart by $|V|^2/\Delta$. As two levels approach, the push grows, and when $\Delta$ becomes comparable to $V$ the second-order formula fails. The exact two-level result, $E_\pm = \bar{E} \pm \sqrt{(\Delta/2)^2 + V^2}$, shows what happens: the levels never cross. Their closest approach is $2|V|$, where they mix equally.
Such avoided crossings appear throughout physics. In molecules they govern which chemical reactions happen as atoms approach; in solids they open band gaps where the free-electron energies of two waves would otherwise cross; in neutrinos passing through the Sun, an avoided crossing between flavor states, the MSW effect, converts electron neutrinos into other flavors. The only levels that can cross are those with $V = 0$, typically because a symmetry forbids their coupling.
Degenerate perturbation theory rewards choosing the basis well. If an operator $\hat{A}$ commutes with both $\hat{H}^0$ and $\hat{H}'$, and its eigenvalues distinguish the degenerate states, then $W$ is already diagonal in the eigenbasis of $\hat{A}$ and the shifts are just the diagonal elements. For a magnetic field along $z$ acting on hydrogen, $\hat{J}_z$ commutes with everything relevant, so working in states of definite $m_j$ turns a potentially large matrix into small blocks.
This is why physicists spend so much effort finding conserved quantities. Every symmetry of the perturbation shrinks the matrix to diagonalize. For hydrogen's fine structure — relativity, spin–orbit and the Darwin term together — $\hat{J}^2$, $\hat{J}_z$, $\hat{L}^2$ and $\hat{S}^2$ all commute with the perturbation, and the entire calculation reduces to first-order averages in the $|n, l, j, m_j\rangle$ basis, giving $E_{nj}^1 = -\tfrac{(E_n^0)^2}{2mc^2}\left(\tfrac{4n}{j + 1/2} - 3\right)$.
The Helioseismic and Magnetic Imager on NASA's Solar Dynamics Observatory photographs the Sun's full disk every $45$ seconds in six narrow wavelength bands around an iron line at $617.33$ nm. That line was chosen because its Landé factor is large, $g = 2.5$, so it splits strongly in a magnetic field. From the splitting and the polarization of the components, the instrument maps the field's strength and direction over the whole visible Sun.
Over a sunspot, with a field of about $0.3$ T, the outer Zeeman components shift by only $13$ pm — two hundred-thousandths of the wavelength — yet the measurement is routine. These maps feed the space-weather forecasts of the National Oceanic and Atmospheric Administration, because tangled fields in active regions store the energy released in solar flares and coronal mass ejections, which can disrupt satellites, GPS and power grids. George Ellery Hale first detected sunspot fields this way at Mount Wilson in 1908.
The Stark effect is a practical control knob. In semiconductor quantum wells, an applied electric field shifts the absorption edge by the quantum-confined Stark effect; electro-absorption modulators built on it switch light on and off billions of times per second in the transmitters of fiber-optic networks and data centers.
In atomic physics, Rydberg atoms, whose high-$n$ levels are nearly degenerate like hydrogen's, show huge linear Stark shifts, which lets experimenters tune them into resonance with microwave photons or with each other. The same sensitivity makes Rydberg atoms excellent electric-field sensors: vapor cells of cesium or rubidium now serve as antennas that measure radio-frequency fields with traceable accuracy, a technology developed at the National Institute of Standards and Technology. Every one of these devices is designed with the diagonalization of a degenerate block in mind.
It is tempting to apply the nondegenerate formula to each degenerate state separately, taking $E^1 = \langle\psi_i^0|H'|\psi_i^0\rangle$ for each. That gives the right answer only if the basis happens to diagonalize the perturbation. For hydrogen's $n = 2$ in an electric field, the diagonal elements are all zero, and the naive answer — no first-order shift — is wrong: the true shifts are $\pm 3eaE$, and they belong to the mixtures $(2s \pm 2p_0)/\sqrt{2}$, not to $2s$ or $2p_0$ alone.
A related error is to think levels can cross as a parameter changes. Coupled levels repel; they approach within $2|V|$ and then veer apart, exchanging character. Only levels a symmetry prevents from coupling can cross.
Two degenerate states have $W = \begin{pmatrix} 3 & 4 \\ 4 & -3 \end{pmatrix}$ meV. Write the eigenvalue equation.
$\det\begin{pmatrix} 3 - \lambda & 4 \\ 4 & -3 - \lambda \end{pmatrix} = 0$
The shifts are the eigenvalues.
Expand the determinant.
$\lambda^2 - 9 - 16 = 0$
$(3 - \lambda)(-3 - \lambda) - 16$.
Solve for the shifts.
$\lambda = \pm 5\ \text{meV}$
The level splits by $10$ meV.
Find the good state for $+5$ meV.
$-2c_a + 4c_b = 0 \quad\Rightarrow\quad \tfrac{1}{\sqrt{5}}(2, 1)$
From the first row of $(W - 5I)c = 0$.
Find the good state for $-5$ meV.
$\tfrac{1}{\sqrt{5}}(1, -2)$
Orthogonal to the first, as eigenvectors of a Hermitian matrix must be.
List the degenerate $n = 2$ states.
$2s, \ 2p_{1}, \ 2p_{0}, \ 2p_{-1}$
Four orbital states at $-3.40$ eV.
Apply the selection rules for $eEz$.
$\Delta m = 0, \ \text{opposite parity}: \quad \text{only } \langle 2s|z|2p_0\rangle \ne 0$
$z$ is odd and does not change $m$.
Evaluate the matrix element.
$\langle 2s|eEz|2p_0\rangle = -3eaE$
From the radial integral of $R_{20}R_{21}r^3$.
Diagonalize the $2s$–$2p_0$ block.
$\begin{pmatrix} 0 & -3eaE \\ -3eaE & 0 \end{pmatrix}: \quad \lambda = \pm 3eaE$
Equal and opposite shifts.
Identify the unshifted states.
$2p_{\pm 1}: \quad E^1 = 0$
Not coupled to anything at first order.
Evaluate for $E = 10^6$ V/m.
$3eaE = 3 \times 0.0529 \times 10^{-9} \times 10^{6}\ \text{eV} = 1.59 \times 10^{-4}\ \text{eV}$
About $0.16$ meV.
Write the Landé factor for a $p_{3/2}$ level.
$g_J = 1 + \dfrac{\frac{15}{4} + \frac{3}{4} - 2}{2 \times \frac{15}{4}}$
$j = \tfrac{3}{2}$, $s = \tfrac{1}{2}$, $l = 1$.
Simplify the factor.
$g_J = 1 + \dfrac{2.5}{7.5} = \dfrac{4}{3}$
Between the orbital value $1$ and the spin value $2$.
Find the factor for $p_{1/2}$.
$g_J = 1 + \dfrac{\frac{3}{4} + \frac{3}{4} - 2}{2 \times \frac{3}{4}} = \dfrac{2}{3}$
The spin is mostly opposed to the orbit.
Find the factor for $s_{1/2}$.
$g_J = 1 + \dfrac{\frac{3}{4} + \frac{3}{4} - 0}{2 \times \frac{3}{4}} = 2$
Pure spin.
Find the shifts of the $p_{3/2}$ states in $1$ T.
$\tfrac{4}{3} \times 57.88 \times m_j = \pm 115.8, \ \pm 38.6\ \mu\text{eV}$
Four states, evenly spaced.
Find the shifts of the $s_{1/2}$ states.
$2 \times 57.88 \times (\pm\tfrac{1}{2}) = \pm 57.9\ \mu\text{eV}$
Two states.
Count the lines in the $p_{3/2} \to s_{1/2}$ transition.
$\Delta m_j = 0, \pm 1: \quad 6 \text{ lines}$
The anomalous Zeeman pattern of the sodium $D_2$ line.
Compare with the normal Zeeman effect.
$3 \text{ lines, spaced by } \mu_BB$
Only when spin plays no role, as in transitions between singlet states.
Write the formula with the values.
$g_J = 1 + \dfrac{\frac{35}{4} + \frac{3}{4} - 6}{2 \times \frac{35}{4}}$
$j(j + 1) = \tfrac{35}{4}$ and $l(l + 1) = 6$.
Simplify the fraction.
$g_J = 1 + \dfrac{3.5}{17.5}$
Numerator $3.5$, denominator $17.5$.
Evaluate the factor.
Two degenerate states have diagonal perturbation elements $W_{aa} = W_{bb} = 0$ and off-diagonal element $W_{ab} = 10$ meV. What are the first-order energy shifts?
Complete the worked solution: hydrogen's $n = 2$ level, at $-3400$ meV, is placed in an electric field of $13 \times 10^{6}$ V/m. With $a = 0.0529$ nm, find the linear Stark shift $3eaE$, the splitting between the shifted states, and the upper state's energy, all in meV.
Evaluate the linear Stark shift.
$3eaE = 3 \times 0.0529 \times 10^{-9} \times 13 \times 10^{6}\ \text{eV} =$ s meV
The field times the $2s$–$2p_0$ dipole matrix element.
Double it for the splitting.
$\Delta E = 6eaE =$ d meV
One state shifts up, one down.
Add the shift to the unperturbed energy.
$E_+ = -3400 + 3eaE =$ u meV
The upper good state.
Match each part of degenerate perturbation theory to its expression.
| $\langle\psi_i^0|H'|\psi_j^0\rangle$ in the degenerate set | the eigenvalues of $W$ | $g_J\mu_BBm_j$ | $\pm 3eaE$ and $0$ | |
|---|---|---|---|---|
| the matrix to diagonalize | ||||
| the first-order shifts | ||||
| the weak-field Zeeman shift | ||||
| hydrogen's $n = 2$ Stark shifts |
A doubly degenerate level at $E^0 = 56$ meV is perturbed, with $W = \begin{pmatrix} 3 & 4 \\ 4 & 3 \end{pmatrix}$ meV in the degenerate pair. Fill in the two perturbed energies and their splitting, in meV.
| meV | |
|---|---|
| upper energy | |
| lower energy | |
| splitting |
Two levels are $\Delta = 10$ meV apart and are coupled by a matrix element $V = 3$ meV. By how much does the coupling push the lower level down, to second order, in meV?
Answer: meV
A one-electron atom is in a $d$ level with $l = 2$, $s = \tfrac{1}{2}$ and $j = \tfrac{5}{2}$, and in the state $m_j = \tfrac{5}{2}$. By how much does a weak field of $1$ T shift its energy, in μeV? Use $\mu_B = 57.88$ μeV/T.
Answer: μeV
NASA's Solar Dynamics Observatory measures magnetic fields on the Sun from the Zeeman splitting of an iron line at $\lambda = 617.33$ nm with effective Landé factor $g = 2.5$. Over a sunspot, each outer component is shifted by $13.34$ pm. What is the field, in tesla? Use $hc = 1239.84$ eV nm and $\mu_B = 5.788 \times 10^{-5}$ eV/T.
Answer: T
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A doubly degenerate level at $E^0 = 140$ meV is perturbed, with $W = \begin{pmatrix} 5 & 7 \\ 7 & 5 \end{pmatrix}$ meV in the degenerate pair. Fill in the two perturbed energies and their splitting, in meV.
| meV | |
|---|---|
| upper energy | |
| lower energy | |
| splitting |
You can handle perturbations of degenerate levels. Explain to someone why hydrogen's $n = 2$ level shifts linearly in an electric field while its ground state does not.
17. Your turn: find the Landé factor of a $d_{5/2}$ level ($l = 2$, $j = \tfrac{5}{2}$)., step 3
$g_J = 1.2$
Spin aligned with a larger orbit.