Degenerate Perturbation Theory

A hydrogen atom placed in a weak electric field suddenly splits its spectral lines apart. This phenomenon is called the Stark effect. It occurs even though the ordinary formula for a first-order energy shift, $E^{(1)} = \langle \psi^{(0)} | \hat{H}' | \psi^{(0)} \rangle$ , seems to give a single unambiguous number. The catch is that hydrogen's $n=2$ level is four-fold degenerate. There are four equally valid unperturbed wavefunctions ( $2s$ , $2p_0$ , $2p_{+1}$ , $2p_{-1}$ ). The formula gives a different answer depending on which one, or which combination, you plug in. Ordinary First-Order Perturbation Theory silently assumes the zeroth-order state is unique. When it isn't, the theory doesn't just get sloppy. It becomes internally contradictory unless repaired. This matters far beyond hydrogen. Degeneracy is the rule, not the exception, in chemistry. Examples include the three $2p$ orbitals of any atom, the five $d$ orbitals of a transition metal ion. Also included are the doubly degenerate $e_g$ and $t_{2g}$ sets that ligand fields split apart. The pairs of $\pi$ and $\pi^*$ molecular orbitals in symmetric aromatic systems belong here too. All of these are degenerate manifolds waiting to be split by some perturbation. That perturbation may be an electric field, a crystal field, spin–orbit coupling, or a Jahn–Teller distortion. Crystal field theory, ligand field theory, the Zeeman and Stark effects, and vibronic coupling all rest on the same repaired machine: degenerate perturbation theory. In this approach, you diagonalize the perturbing operator within the degenerate subspace before ever computing an energy correction. The fix is conceptually small but structurally deep. Instead of asking "what is $\langle \psi^{(0)}|\hat H'|\psi^{(0)}\rangle$ for this particular degenerate state," you take a different approach. You build the full matrix of $\hat H'$ between all the degenerate basis functions. You diagonalize that finite matrix. Only then do you have "good" zeroth-order states for which non-degenerate perturbation theory can safely proceed. The eigenvalues of this small matrix are the first-order energy corrections. Its eigenvectors tell you the correct linear combinations that the perturbation itself selects out of the degenerate manifold. These combinations are often dictated by symmetry. This article builds the degenerate perturbation machinery from the ground up. It explains why the naive formula fails. It shows how diagonalizing a secular determinant inside the degenerate block repairs it. It also shows how this single idea explains the linear Stark effect and orbital splitting patterns. Finally, it covers the general strategy chemists use whenever "extra" quantum numbers or symmetry labels must be assigned before any energy can be computed.

Key takeaways

When $n$ states share energy $E_n^{(0)}$ , $\langle\psi^{(0)}|\hat H'|\psi^{(0)}\rangle$ is basis-dependent and meaningless until you pick the right basis.
The right basis is found by diagonalizing $\hat H'$ restricted to the degenerate subspace — a small secular determinant, not the full Hamiltonian.
The eigenvalues of this restricted matrix are the correct first-order energy corrections; degeneracy is lifted (or partly lifted) according to how many eigenvalues coincide.
The eigenvectors are the "good" zeroth-order wavefunctions — frequently forced by symmetry, e.g. real $p$ or $d$ combinations diagonalizing a crystal field.
The linear Stark effect in hydrogen, $d$ -orbital splitting in ligand fields, and near-degenerate perturbation theory (avoided crossings) are all instances of this one procedure.
If $\hat H'$ is diagonal already in the symmetry-adapted basis (often guaranteed by group theory), no extra work is needed — the naive formula was secretly fine all along.
Key idea: diagonalize the perturbation inside the degenerate subspace first; only the resulting eigenbasis gives a well-defined first-order energy.

Where this lives today

Ligand field theory, the working language of inorganic and bioinorganic chemistry for describing transition-metal complexes, is built entirely on diagonalizing an electrostatic (or more refined) perturbation within the $d$ -orbital manifold — the octahedral, tetrahedral, and square-planar splitting diagrams used to rationalize colors, magnetism, and reactivity of coordination compounds are direct outputs of this procedure. The Jahn–Teller effect, responsible for the characteristic tetragonal distortion of Cu(II) complexes and for structural distortions that gate electron transfer in many metalloenzymes, is a dynamical extension in which the perturbation itself (nuclear displacement) is diagonalized jointly with vibrational degrees of freedom inside an orbitally degenerate electronic manifold. In molecular spectroscopy, the linear Zeeman and Stark effects in atoms and diatomic molecules with degenerate angular-momentum states (electronic, rotational, or nuclear spin degeneracies) are computed by exactly the secular-determinant procedure developed here; modern precision spectroscopy of Rydberg atoms, used in quantum sensing and atomic clocks, routinely relies on diagonalizing electric or magnetic perturbations within near-degenerate manifolds of high- $n$ hydrogenic states. In materials science, crystal-field splitting of $f$ -orbitals in lanthanide-doped phosphors and laser gain media (Nd:YAG, Er-doped fiber amplifiers) determines emission wavelengths through precisely this kind of degenerate-subspace diagonalization, now typically automated inside quantum chemistry packages that build and diagonalize the relevant perturbation matrix as a matter of course whenever a calculation detects (numerically or by symmetry) a degenerate reference state — for instance, multireference and near-degenerate perturbation methods (like CASPT2 and NEVPT2) used throughout computational chemistry are direct descendants of the degenerate-perturbation-theory logic, generalized to handle near- rather than exactly-degenerate configurations.

Ask the tutor: How does this same diagonalization strategy change when two levels are only nearly (not exactly) degenerate, and why does that regime dominate modern multireference quantum chemistry?

The physical picture

Suppose an unperturbed Hamiltonian $\hat H^{(0)}$ has an eigenvalue $E_n^{(0)}$ shared by $g$ linearly independent eigenfunctions $\psi_1^{(0)}, \psi_2^{(0)}, \dots, \psi_g^{(0)}$ . Any linear combination

\phi^{(0)} = \sum_{i=1}^g c_i \, \psi_i^{(0)}

is equally an eigenfunction of $\hat H^{(0)}$ with the same eigenvalue $E_n^{(0)}$ — the degenerate subspace is a $g$ -dimensional vector space, and $\hat H^{(0)}$ alone gives no reason to prefer any particular basis within it. This is precisely the situation for the $2s, 2p_0, 2p_{+1}, 2p_{-1}$ hydrogenic orbitals: all four solve $\hat H^{(0)}\psi = -\tfrac{1}{4}\,\text{Ry}\,\psi$ , and $\hat H^{(0)}$ itself is blind to how you mix them. Now switch on a perturbation $\hat H'$ , such as a static electric field $\hat H' = e\mathcal{E}z$ (the Stark perturbation). The true perturbed Hamiltonian $\hat H = \hat H^{(0)} + \hat H'$ generically breaks whatever symmetry made the states degenerate, so it typically has no reason to leave $E_n^{(0)}$ degenerate at all. The correct zeroth-order states — the ones that the true eigenstates approach continuously as $\hat H' \to 0$ — are a specific combination of $\psi_1^{(0)}, \dots, \psi_g^{(0)}$ , not an arbitrary one. Non-degenerate perturbation theory applied blindly to an arbitrary member of the degenerate set gives an answer that is simply wrong, because it silently assumes that $\psi_n^{(0)}$ itself survives smoothly as the "parent" of the perturbed state — an assumption that fails whenever off-diagonal matrix elements of $\hat H'$ connect the degenerate states to one another.

Figure

Degenerate subspace before and after diagonalizing the perturbation

The core physical claim of degenerate perturbation theory is that the correct linear combinations are exactly the eigenvectors of the $g \times g$ matrix formed by the matrix elements $H'_{ij} = \langle \psi_i^{(0)} | \hat H' | \psi_j^{(0)} \rangle$ taken only among the degenerate set. You never need to touch states outside the degenerate manifold at this order — their effect only enters at second order, exactly as in the non-degenerate theory. This restriction to a small block is what makes the method tractable: diagonalizing a $4\times4$ or $5\times5$ matrix analytically (often using symmetry to block it further) replaces an intractable infinite-dimensional problem. A second, complementary picture is symmetry. If $\hat H^{(0)}$ commutes with some symmetry operations (rotations, reflections) that connect the degenerate states, the degeneracy is often required by that symmetry — this is Wigner's theorem territory. A perturbation that breaks the symmetry (an external field, distortion of a ligand cage) generally does connect the degenerate basis functions through nonzero off-diagonal matrix elements, and it is exactly this connectivity that both demands diagonalization and, at the same time, usually fixes what the correct linear combinations must be — often you can identify them from symmetry alone (this is the practical shortcut used constantly in ligand-field and Hückel problems).

Figure

d-orbital splitting in an octahedral field: the diagonalized picture

The core ideas

Write the perturbed Hamiltonian in the usual way,

\hat H = \hat H^{(0)} + \lambda \hat H'

where: $\hat H^{(0)}$ is exactly solvable, $\hat H'$ is the perturbation, and $\lambda$ is a bookkeeping parameter set to 1 at the end. Suppose $E_n^{(0)}$ is $g$ -fold degenerate with orthonormal eigenfunctions $\psi_1^{(0)}, \dots, \psi_g^{(0)}$ . Zeroth-order "good" states are combinations

\phi_k^{(0)} = \sum_{i=1}^g c_{ki}\,\psi_i^{(0)}, \qquad k = 1, \dots, g

where: $c_{ki}$ are coefficients to be determined — not assumed known in advance. Substituting the Rayleigh–Schrödinger expansion $\phi_k = \phi_k^{(0)} + \lambda \phi_k^{(1)} + \dots$ and $E_k = E_n^{(0)} + \lambda E_k^{(1)} + \dots$ into $\hat H \phi_k = E_k \phi_k$ and collecting first-order terms gives the same starting identity as in the non-degenerate case,

\hat H^{(0)} \phi_k^{(1)} + \hat H' \phi_k^{(0)} = E_n^{(0)} \phi_k^{(1)} + E_k^{(1)} \phi_k^{(0)}

where: this is an operator equation valid to first order in $\lambda$ . Projecting onto each $\psi_i^{(0)}$ (for $i = 1, \dots, g$ , inside the degenerate manifold) rather than an arbitrary outside state is what produces the eigenvalue problem central to this theory,

\sum_{j=1}^g \left( H'_{ij} - E_k^{(1)} \delta_{ij} \right) c_{kj} = 0, \qquad H'_{ij} \equiv \langle \psi_i^{(0)} | \hat H' | \psi_j^{(0)} \rangle

where: $H'_{ij}$ is the matrix representation of $\hat H'$ restricted to the degenerate subspace, and $E_k^{(1)}$ , $c_{kj}$ are the unknowns. Nontrivial solutions require the secular determinant to vanish,

\det\left( H'_{ij} - E^{(1)} \delta_{ij} \right) = 0

where: this is a $g \times g$ polynomial equation in $E^{(1)}$ , structurally identical to a Solving the Secular Determinant problem, giving $g$ roots (not necessarily distinct). Each root $E_k^{(1)}$ is a valid first-order energy correction, and its eigenvector $\{c_{ki}\}$ specifies the correct zeroth-order wavefunction $\phi_k^{(0)}$ for which ordinary non-degenerate perturbation theory becomes valid at all higher orders (with the caveat that if some $E_k^{(1)}$ remain equal, degeneracy persists to this order and one must go to second order, or use additional symmetry, to separate them). The reading of this result: within the degenerate subspace, the perturbation behaves exactly like a finite-dimensional quantum-mechanical problem in its own right — the $g$ -dimensional degenerate manifold is treated the way you would treat a small isolated Hilbert space, and $\hat H'$ is diagonalized inside it precisely as one diagonalizes any Hermitian operator in a finite basis.

Second-order and residual degeneracy

If diagonalizing $H'_{ij}$ leaves two or more eigenvalues equal (a "residual" degeneracy that the first-order perturbation does not lift), the standard prescription is:

E_k^{(2)} = \sum_{m \neq n} \frac{\left| \langle \psi_m^{(0)} | \hat H' | \phi_k^{(0)} \rangle \right|^2}{E_n^{(0)} - E_m^{(0)}}

where: the sum runs only over states $\psi_m^{(0)}$ outside the degenerate manifold (all states with $m \ne n$ ), and $\phi_k^{(0)}$ is the already-diagonalized zeroth-order combination. If the residual degeneracy is not lifted even by this second-order sum, one must construct and diagonalize a further effective matrix among the still-degenerate combinations using contributions from outside states (the technique behind, e.g., Jahn–Teller vibronic coupling and near-degenerate configuration mixing) — but for a huge fraction of chemically relevant problems (linear Stark effect, crystal-field splitting, spin–orbit splitting of orbital triplets) the first-order diagonalization already lifts the degeneracy completely, and this refinement is not needed.

Derivation ALinear Stark effect in the hydrogen n = 2 manifold

Goal. Find the first-order energy shifts of the four degenerate $n=2$ hydrogen states, $2s$ , $2p_0$ , $2p_{+1}$ , $2p_{-1}$ , in a uniform static electric field $\vec{\mathcal E} = \mathcal{E}\hat z$ , using $\hat H' = e\mathcal{E}\hat z$ .

Step 1 — Set up the basis and note the symmetry. The four degenerate hydrogenic $n=2$ states are $\psi_1 = \psi_{200}$ (the $2s$ orbital), $\psi_2 = \psi_{210}$ ( $2p_0$ , i.e. $2p_z$ ), $\psi_3 = \psi_{21,+1}$ , $\psi_4 = \psi_{21,-1}$ . The operator $\hat z$ has odd parity under inversion, while spherical harmonics carry definite parity $(-1)^\ell$ ; $\hat H' = e\mathcal E z$ therefore connects only states differing by $\Delta\ell = \pm1$ (Laporte-type selection rule) and, since $z \propto Y_1^0$ , requires $\Delta m_\ell = 0$ . This immediately kills most of the $4\times4$ matrix: only the $2s$ – $2p_0$ block can be nonzero, and the $2p_{\pm1}$ states are decoupled from everything (including each other, since $z$ carries $\Delta m_\ell = 0$ but $2p_{+1}$ and $2p_{-1}$ differ by $\Delta m_\ell = 2$ ).

Step 2 — Evaluate the surviving matrix element. Direct integration over the hydrogenic radial and angular parts (a standard but lengthy integral) gives the well-known result

\langle \psi_{200} | \hat z | \psi_{210} \rangle = -3 a_0

where: $a_0$ is the Bohr radius. All diagonal elements $\langle \psi_i | \hat z | \psi_i \rangle$ vanish by parity (each $|\psi_i|^2$ is an even function of $z$ when $\psi_i$ has definite parity), so $\langle \psi_i|\hat H'|\psi_i\rangle = 0$ for every one of the four basis states individually — this is exactly the naive, wrong-looking result that motivates the whole calculation: applying non-degenerate theory blindly to any single hydrogenic state predicts zero linear Stark shift, which is experimentally false.

Step 3 — Build and diagonalize the 4×4 matrix. Ordering the basis as $(2s, 2p_0, 2p_{+1}, 2p_{-1})$ , the matrix of $\hat H' = e\mathcal E \hat z$ restricted to the degenerate subspace is

H' = e\mathcal E \begin{pmatrix} 0 & -3a_0 & 0 & 0 \\ -3a_0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}

where: the top-left $2\times2$ block couples $2s$ and $2p_0$ , and the bottom-right $2\times2$ block is already diagonal (zero) for $2p_{\pm1}$ . The secular determinant factors into the $2\times2$ block (giving two nontrivial roots) times the trivial $2\times2$ zero block (giving two zero roots). For the $2\times2$ block,

\det\begin{pmatrix} -E^{(1)} & -3e\mathcal E a_0 \\ -3e\mathcal E a_0 & -E^{(1)} \end{pmatrix} = 0 \;\;\Longrightarrow\;\; (E^{(1)})^2 - (3e\mathcal E a_0)^2 = 0

E^{(1)} = \pm\,3e\mathcal E a_0

where: the two roots are equal and opposite, and the other two roots (from the $2p_{\pm1}$ block) are exactly zero.

Step 4 — Read off eigenvectors (the "good" states) and the full splitting pattern. Solving $(H' - E^{(1)}\mathbb{1})\mathbf{c}=0$ for each root gives, for $E^{(1)} = +3e\mathcal{E}a_0$ , the eigenvector $c=(1,-1,0,0)/\sqrt2$ , and for $E^{(1)} = -3e\mathcal E a_0$ , $c=(1,1,0,0)/\sqrt2$ . So the correct zeroth-order states are

\psi_+ = \frac{1}{\sqrt2}(2s - 2p_0), \qquad \psi_- = \frac{1}{\sqrt2}(2s + 2p_0)

with energies shifted by $+3e\mathcal{E}a_0$ and $-3e\mathcal{E}a_0$ respectively, while $2p_{+1}$ and $2p_{-1}$ remain unshifted (and degenerate with each other) at this order. The original four-fold degenerate level thus splits into three first-order levels: one raised by $3e\mathcal{E}a_0$ , one lowered by $3e\mathcal E a_0$ , and one (doubly degenerate) unshifted — this is exactly the observed linear Stark pattern for hydrogen's $n=2$ level, and it could never have been obtained by plugging a single hydrogenic orbital into the naive first-order formula.

Go deeper · Derivation B · 2 minCrystal-field splitting of a d-orbital manifold — a symmetry shortcut

Derivation BCrystal-field splitting of a d-orbital manifold — a symmetry shortcut

Goal. Show how symmetry alone identifies the diagonalizing basis for a perturbation acting on a 5-fold degenerate $d$ -orbital set placed in an octahedral ligand field, without brute-force computing all 25 matrix elements.

Step 1 — Identify the perturbation and the symmetry it must respect. The perturbation is the electrostatic potential $\hat H' = V_{\text{oct}}(\vec r)$ due to six point-charge-like ligands at $(\pm a,0,0), (0,\pm a,0), (0,0,\pm a)$ along the Cartesian axes. This potential, expanded in spherical harmonics, is invariant under the full octahedral point group $O_h$ — any operation of $O_h$ (90° rotations about $x,y,z$ , diagonal $C_3$ axes, reflections) leaves $V_{\text{oct}}$ unchanged, because it merely permutes the six ligand positions among themselves.

Step 2 — Use group theory to pre-block-diagonalize the problem. The five real $d$ -orbitals ( $d_{z^2}, d_{x^2-y^2}, d_{xy}, d_{xz}, d_{yz}$ ) transform under $O_h$ as the reducible representation $E_g \oplus T_{2g}$ : the pair $(d_{z^2}, d_{x^2-y^2})$ spans the two-dimensional irreducible representation $E_g$ , and the trio $(d_{xy}, d_{xz}, d_{yz})$ spans the three-dimensional irreducible representation $T_{2g}$ . A foundational theorem of group theory (a close relative of the Wigner–Eckart theorem) guarantees that a totally symmetric operator like $V_{\text{oct}}$ cannot connect states belonging to different irreducible representations. This means, without evaluating a single integral, that all matrix elements between an $E_g$ -type orbital and a $T_{2g}$ -type orbital vanish identically — the $5\times5$ matrix is guaranteed block-diagonal, at worst a $2\times2$ block and a $3\times3$ block, in the real $d$ -orbital basis.

Step 2 also fixes the shape within each block. By the same theorem, within the $T_{2g}$ block, the three orbitals must be degenerate with each other (any operator that is totally symmetric under $O_h$ , when restricted to an irreducible representation, is a multiple of the identity on that subspace — this is a form of Schur's lemma). The same applies within the $E_g$ block. So symmetry alone tells you the matrix must be

H'_{\text{restricted}} = \begin{pmatrix} E_{e_g} & 0 & & & \\ 0 & E_{e_g} & & & \\ & & E_{t_{2g}} & 0 & 0\\ & & 0 & E_{t_{2g}} & 0\\ & & 0 & 0 & E_{t_{2g}} \end{pmatrix}

where: $E_{e_g}$ and $E_{t_{2g}}$ are two numbers still to be computed by explicit integration, but the structure — which orbitals mix and which don't — is already fixed by symmetry with zero computation of matrix elements. This is the general chemist's shortcut: symmetry frequently supplies the diagonalizing basis before any integral is touched.

Step 3 — Fix the two energies using the barycenter (trace) rule. Actually evaluating $\langle d_{z^2}|V_{\text{oct}}|d_{z^2}\rangle$ requires integrating the crystal field potential (expanded through $\ell = 4$ spherical harmonics) against the $d$ -orbital angular parts; the standard result, conventionally reported relative to the spherically-averaged ("barycenter") energy and expressed in units of $\Delta_o$ (the octahedral splitting parameter), is

E_{e_g} = +0.6\,\Delta_o, \qquad E_{t_{2g}} = -0.4\,\Delta_o

where: these satisfy the barycenter (center-of-gravity) rule required by trace invariance of the perturbation matrix,

2(+0.6\,\Delta_o) + 3(-0.4\,\Delta_o) = 1.2\Delta_o - 1.2\Delta_o = 0

confirming that the crystal field only redistributes energy among the five orbitals without changing their average — a useful, exactly-checkable identity for any crystal- or ligand-field splitting pattern, since $\mathrm{Tr}(H'_{ij})$ computed in any orthonormal basis of the degenerate subspace must equal the sum of eigenvalues, and for a purely electrostatic multipolar perturbation acting on a full, symmetric orbital set this trace vanishes by the orthogonality properties of spherical harmonics.

Step 4 — State the result. The five-fold degenerate free-ion $d$ level splits, under an octahedral field, into a doubly degenerate $e_g$ level at $+0.6\,\Delta_o$ and a triply degenerate $t_{2g}$ level at $-0.4\,\Delta_o$ — the textbook crystal-field diagram, derived here purely as an instance of diagonalizing a perturbation within a degenerate subspace, with group theory supplying the block structure and a short explicit integral supplying only the two numbers.

Worked examples

Example 1 — Foundational: an explicit 2×2 degenerate block with asymmetric coupling. Consider two degenerate zeroth-order states $\psi_1^{(0)}, \psi_2^{(0)}$ (orthonormal, energy $E_n^{(0)} = 4.0$ eV) perturbed by an operator with matrix elements, in this basis,

H'_{11} = 0.30\text{ eV}, \quad H'_{22} = -0.10\text{ eV}, \quad H'_{12} = H'_{21} = 0.45\text{ eV}

Deliberately asymmetric and non-round numbers are used so that no accidental symmetry can mask an error. The secular determinant is

\det\begin{pmatrix} 0.30 - E^{(1)} & 0.45 \\ 0.45 & -0.10 - E^{(1)} \end{pmatrix} = 0

Expanding,

(0.30 - E^{(1)})(-0.10 - E^{(1)}) - (0.45)^2 = 0

E^{(1)2} - 0.20\,E^{(1)} - 0.030 - 0.2025 = 0

E^{(1)2} - 0.20\,E^{(1)} - 0.2325 = 0

Using the quadratic formula,

E^{(1)} = \frac{0.20 \pm \sqrt{0.0400 + 0.9300}}{2} = \frac{0.20 \pm \sqrt{0.9700}}{2} = \frac{0.20 \pm 0.98489}{2}

E^{(1)}_+ = 0.59244\text{ eV}, \qquad E^{(1)}_- = -0.39244\text{ eV}

As a check, the sum of the roots must equal the trace, $H'_{11}+H'_{22} = 0.30 - 0.10 = 0.20$ eV, and indeed $0.59244 + (-0.39244) = 0.20000$ eV — confirmed. The product of the roots must equal the determinant of $H'$ : $\det H' = (0.30)(-0.10) - (0.45)^2 = -0.030 - 0.2025 = -0.2325$ , and indeed $(0.59244)(-0.39244) = -0.23252$ — matching within rounding. The two perturbed levels are $E_n^{(0)} + 0.59244$ eV and $E_n^{(0)} - 0.39244$ eV, i.e. $4.59244$ eV and $3.60756$ eV: the two-fold degeneracy is completely lifted. Solving for the eigenvector at $E^{(1)}_+ = 0.59244$ : from row 1, $(0.30 - 0.59244)c_1 + 0.45\,c_2 = 0 \Rightarrow c_2 = 0.6499\,c_1$ , giving (after normalization) $c_1 = 0.8375, c_2=0.5443$ ; this fixes the "good" zeroth-order combination $\phi_+^{(0)} = 0.8375\,\psi_1^{(0)} + 0.5443\,\psi_2^{(0)}$ .

Limiting case check. If instead $H'_{12} \to 0$ (no coupling between the degenerate states), the quadratic becomes $E^{(1)2} - 0.20E^{(1)} - 0.030 = 0$ , whose roots are exactly $E^{(1)}=H'_{11}=0.30$ and $E^{(1)}=H'_{22}=-0.10$ — the diagonal elements themselves, recovering the naive (non-degenerate-style) formula exactly, as it must when there is no actual mixing to worry about.

Example 2 — Applied: linear Stark splitting magnitude for real hydrogen. Using the result of the first mechanism, $E^{(1)} = \pm 3e\mathcal E a_0$ , evaluate the splitting for a laboratory field of $\mathcal{E} = 1.00\times10^5$ V/m (a strong but experimentally accessible static field) with $a_0 = 5.292\times10^{-11}$ m and $e = 1.602\times10^{-19}$ C.

3e\mathcal{E}a_0 = 3 \times (1.602\times10^{-19}\text{ C}) \times (1.00\times10^5\text{ V/m}) \times (5.292\times10^{-11}\text{ m})

= 3 \times 1.602\times10^{-19} \times 1.00\times10^5 \times 5.292\times10^{-11}

= 3 \times (8.478\times10^{-25})\ \text{J} = 2.543\times10^{-24}\text{ J}

Converting to more familiar spectroscopic units (dividing by $hc = 1.986\times10^{-25}$ J·m to get wavenumbers, or simply converting to eV using $1\text{ eV} = 1.602\times10^{-19}$ J):

2.543\times10^{-24}\text{ J} \times \frac{1\text{ eV}}{1.602\times10^{-19}\text{ J}} = 1.588\times10^{-5}\text{ eV}

So the total splitting between the two shifted sublevels ( $\pm 3e\mathcal E a_0$ ) is $2\times1.588\times10^{-5}\text{ eV} \approx 3.18\times10^{-5}\text{ eV}$ , corresponding to roughly $0.256\ \text{cm}^{-1}$ — small but very much resolvable in high-resolution spectroscopy, and crucially it is linear in $\mathcal E$ (unlike the ordinary quadratic Stark effect seen in non-degenerate levels), a direct experimental fingerprint of degenerate perturbation theory at work: only levels with accidental or symmetry-required degeneracy that is directly connected by the perturbing operator show a linear, rather than quadratic, field dependence.

Limiting case check. As $\mathcal E \to 0$ , the splitting $3e\mathcal{E}a_0 \to 0$ linearly and continuously, and all four $n=2$ states smoothly reconverge to their common unperturbed energy $E_2^{(0)} = -3.4$ eV, exactly as required.

Example 3 — Exam/competition level: a 3×3 degenerate block with an accidental residual degeneracy. Three orthonormal degenerate states $\psi_1^{(0)}, \psi_2^{(0)}, \psi_3^{(0)}$ share energy $E_n^{(0)} = 10.0$ eV. A perturbation has matrix (in eV, in this basis)

H' = \begin{pmatrix} 2.0 & 1.5 & 0 \\ 1.5 & 2.0 & 0 \\ 0 & 0 & 4.0 \end{pmatrix}

Find all first-order corrections and identify which combinations of the original states are the correct zeroth-order eigenstates. Because the matrix is already block-diagonal (a $2\times2$ block plus a $1\times1$ block — an intentional design so the student must notice the block structure rather than expand a full $3\times3$ determinant blindly), the $1\times1$ block gives immediately $E^{(1)}_3 = 4.0$ eV with eigenvector $\phi_3^{(0)} = \psi_3^{(0)}$ unchanged. For the $2\times2$ block,

\det\begin{pmatrix} 2.0-E^{(1)} & 1.5 \\ 1.5 & 2.0-E^{(1)} \end{pmatrix}=0

(2.0-E^{(1)})^2 - (1.5)^2 = 0 \;\;\Longrightarrow\;\; 2.0 - E^{(1)} = \pm1.5

E^{(1)}_1 = 0.5\text{ eV}, \qquad E^{(1)}_2 = 3.5\text{ eV}

with eigenvectors $\phi_1^{(0)} = (\psi_1^{(0)}-\psi_2^{(0)})/\sqrt2$ (eigenvalue $0.5$ ) and $\phi_2^{(0)} = (\psi_1^{(0)}+\psi_2^{(0)})/\sqrt2$ (eigenvalue $3.5$ ), obtained the same way as in Example 1's symmetric-coupling special case (here $H'_{11}=H'_{22}$ exactly, so the eigenvectors are the simple symmetric/antisymmetric combinations rather than needing the general formula). The full first-order picture: the original three-fold degenerate level at $10.0$ eV splits into three distinct levels at $10.5, 13.5, 14.0$ eV — fully lifted, with no residual degeneracy, because all three eigenvalues ( $0.5, 3.5, 4.0$ ) are numerically distinct even though the perturbation matrix looked "mostly zero." A student who instead mistakenly diagonalized only the diagonal entries (treating $H'_{12}=1.5$ as negligible) would have wrongly reported $2.0, 2.0, 4.0$ eV — an error of exactly $\pm1.5$ eV on two of the three levels, illustrating why off-diagonal elements inside the degenerate block can never be dropped, however small they might look compared to the diagonal.

Limiting cases, scope & common traps

Situation	What happens	Why
$H'_{ij}=0$ for all $i\ne j$ in the chosen basis	Naive formula $E^{(1)}=H'_{ii}$ is already correct	The degenerate-subspace matrix is already diagonal; no mixing needed
Some, but not all, off-diagonal $H'_{ij}\ne 0$	Must diagonalize the relevant sub-block only	Symmetry or accidental structure can block-diagonalize part of the matrix (Example 3)
All eigenvalues of $H'_{ij}$ coincide	Degeneracy survives to first order	Go to second order, forming an effective matrix using outside states
Perturbation shares full symmetry of $\hat H^{(0)}$	Degeneracy often persists exactly (not just to some order)	Wigner's theorem: symmetry-protected degeneracy cannot be lifted by a symmetric perturbation
$\lambda \to 0$	All $\phi_k^{(0)}$ smoothly recombine into arbitrary degenerate combinations again	The distinguishing information (which combination is "good") only exists because of $H'$
$g=1$ (no degeneracy)	Reduces exactly to ordinary non-degenerate perturbation theory	The secular determinant becomes the trivial $1\times1$ statement $E^{(1)}=H'_{11}$

The chemist behind it

The formalism traces to the joint development of perturbation theory for degenerate quantum systems by Erwin Schrödinger in his 1926 papers establishing wave mechanics, refined immediately afterward in the group-theoretic language that made symmetry-based diagonalization systematic. It was Eugene Wigner who, in the years following, supplied the deep connection between degeneracy, symmetry, and the block-diagonal structure exploited throughout this article — his theorem that a symmetric perturbation cannot connect states of different irreducible representations is the rigorous foundation behind every crystal-field and ligand-field splitting diagram used in inorganic chemistry today.

Origins · 1 minEugene Paul Wigner and how this was discovered

Eugene Paul Wigner

1902–1995 · Central Europe to the United States, mathematical physics and quantum chemistry

Wigner trained as a chemical engineer in Berlin before turning to theoretical physics, a trajectory that left a permanent imprint on how he thought about molecules and symmetry. Working alongside John von Neumann and in constant dialogue with the Göttingen school (Born, Heisenberg, Jordan), Wigner recognized that the representation theory of finite groups — developed decades earlier by pure mathematicians with no physical application in mind — was exactly the right language for organizing degeneracies in atoms and molecules. His 1931 book Gruppentheorie und ihre Anwendung auf die Quantenmechanik der Atomspektren systematized how symmetry dictates which matrix elements of a perturbation must vanish, turning degenerate perturbation theory from a case-by-case algebraic exercise into a predictive, symmetry-first discipline. He shared the 1963 Nobel Prize in Physics for work on the structure of the atomic nucleus and elementary particles, particularly through the discovery and application of fundamental symmetry principles.

History

- 1926 — Schrödinger's wave mechanics papers introduce perturbation theory, including remarks on degenerate cases.

- 1927 — Early applications to the linear Stark effect in hydrogen, showing the

n=2

level splits linearly in a field.

- 1931 — Wigner's Gruppentheorie systematizes the symmetry-based treatment of degenerate perturbations.

- 1929–1935 — Bethe and Van Vleck apply degenerate perturbation methods to build crystal-field and ligand-field theory for transition-metal complexes.

- 1937 — Jahn and Teller show that any nonlinear molecule with an orbitally degenerate electronic state is unstable to symmetry-breaking distortion, a striking dynamical consequence of degenerate perturbation theory.

Why doesn't ordinary first-order perturbation theory just work for degenerate states?

The standard formula $E^{(1)} = \langle\psi_n^{(0)}|\hat H'|\psi_n^{(0)}\rangle$ is derived by projecting the first-order Schrödinger equation onto a unique zeroth-order state $\psi_n^{(0)}$ . When several linearly independent states share the same energy, that projection step becomes ambiguous — you get a different number for every different combination of degenerate states you happen to plug in. Worse, the derivation implicitly assumes $\hat H'\psi_n^{(0)}$ has no component along other states of the same energy, which is exactly what fails when off-diagonal matrix elements inside the degenerate subspace are nonzero. Diagonalizing the perturbation restores a well-defined, basis-independent set of energies and identifies which particular combinations behave like ordinary non-degenerate states.

Do you ever need to diagonalize the full Hamiltonian, or just the perturbation, within the degenerate subspace?

Just the perturbation. Within the degenerate subspace, $\hat H^{(0)}$ is already a multiple of the identity matrix (every basis state has the same eigenvalue $E_n^{(0)}$ ), so diagonalizing $\hat H^{(0)}+\hat H'$ restricted to that subspace gives exactly the same eigenvectors as diagonalizing $\hat H'$ alone — the eigenvalues merely shift by the constant $E_n^{(0)}$ . This is why the secular determinant in degenerate perturbation theory is built purely from $H'_{ij}$ , never from the full Hamiltonian matrix elements.

How large can the degenerate subspace be before this method becomes impractical?

In principle the method works for any finite $g$ , since diagonalizing a $g\times g$ Hermitian matrix is always possible in principle; in practice, symmetry (as in the $d$ -orbital example) frequently block-diagonalizes a large degenerate manifold into much smaller pieces before any numerical diagonalization is needed. For genuinely large or symmetry-poor degenerate spaces, numerical diagonalization of the $g \times g$ matrix on a computer replaces hand algebra, but the underlying logic — diagonalize the restricted perturbation first — is unchanged regardless of $g$ .

What happens if the perturbation matrix within the degenerate subspace has repeated eigenvalues?

Some or all of the original degeneracy survives to first order — the perturbation simply doesn't distinguish those states at this order of approximation. You then either go to second order (building an effective matrix using virtual transitions through outside, non-degenerate states) or recognize, often from symmetry, that the residual degeneracy is protected and will persist to all orders unless a symmetry-breaking perturbation of a different type is introduced (this is the essence of the Jahn–Teller theorem for orbitally degenerate electronic states in nonlinear molecules).

Is the diagonalized zeroth-order basis always dictated by symmetry?

Very often, yes — this is the practical shortcut chemists rely on constantly. If the degenerate states can be organized by irreducible representations of the symmetry group shared by $\hat H^{(0)}$ , and if the perturbation itself is invariant under a subgroup of that symmetry, group theory alone frequently identifies the correct linear combinations (or at least reduces a large secular determinant to much smaller blocks) before a single integral is evaluated. When the perturbation has no useful residual symmetry (e.g., a completely generic distortion), explicit diagonalization of numerical matrix elements is unavoidable.

How does this connect to configuration interaction and multireference quantum chemistry?

Configuration interaction and multireference methods (CASSCF, CASPT2, NEVPT2, MRCI) generalize exactly this logic to electronic structure: whenever multiple electronic configurations lie close in energy (exactly or nearly degenerate), a naive single-reference perturbative treatment fails for the same reason ordinary perturbation theory fails for degenerate atomic levels. The remedy is the same in spirit — diagonalize the relevant electronic Hamiltonian (or an approximation to it) within the near-degenerate subspace of configurations first, producing "good" reference states, before treating the remaining dynamic correlation perturbatively.

Can degenerate perturbation theory predict which combination of orbitals is observed experimentally?

Yes, and this is one of its most chemically useful outputs. The eigenvectors of the diagonalized perturbation matrix are not a mathematical afterthought — they are literally the physically correct zeroth-order states that the true perturbed eigenstates reduce to as the perturbation strength goes to zero. In the hydrogen Stark example, this identifies $(2s\pm 2p_0)/\sqrt2$ , not $2s$ or $2p_0$ individually, as the physically meaningful states in a field — these combinations carry a permanent linear Stark dipole and are directly observable in the spectrum's field-dependence.