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Quasi-Degenerate Perturbation Theory

Quasi-degenerate perturbation theory treats several nearby states as one model space. Their small internal splittings and mutual couplings are diagonalized together, while coupling to well-separated states is expanded perturbatively.

The method is needed when levels are not exactly degenerate, yet their detuning is comparable to the perturbation that mixes them. A one-state formula then contains small energy denominators and predicts large state corrections. Exact diagonalization of the nearby cluster removes those artificial singularities.

The central rule is:

Retain every strongly mixed nearby state in PP, diagonalize within PP, and perturb only the coupling between PP and QQ.

Here PP projects onto the retained model space and Q=I−PQ=I-P onto its complement. The method interpolates between Nondegenerate Perturbation Theory when PP is one-dimensional and Degenerate Perturbation Theory when the unperturbed levels inside PP coincide exactly.

Required background. Use degenerate perturbation theory for matrix-valued level shifts and projectors to split the Hilbert space into PP and QQ.

Suppose two unperturbed states satisfy

H0∣a⟩=Ea(0)∣a⟩,H0∣b⟩=Eb(0)∣b⟩,H_0\lvert a\rangle =E_a^{(0)}\lvert a\rangle, \qquad H_0\lvert b\rangle =E_b^{(0)}\lvert b\rangle,

and are coupled by Vab=⟨a∣V∣b⟩V_{ab}=\langle a\rvert V\lvert b\rangle. Nondegenerate perturbation theory tries to assign the mixing amplitude

cb←a(1)=λVbaEa(0)−Eb(0).c_{b\leftarrow a}^{(1)} = \lambda \frac{V_{ba}} {E_a^{(0)}-E_b^{(0)}}.

This is small only if

∣λVba∣≪∣Ea(0)−Eb(0)∣.\lvert\lambda V_{ba}\rvert \ll \left|E_a^{(0)}-E_b^{(0)}\right|.

When the two scales are comparable, the exact eigenvectors are order-11 mixtures of ∣a⟩\lvert a\rangle and ∣b⟩\lvert b\rangle. No higher-order correction to a state that remains “mostly aa” can repair a zeroth-order labeling that has already failed.

Near-degeneracy is therefore a power-counting statement, not merely visual closeness on an energy plot. A small spacing matters when an allowed matrix element is large enough to compete with it. Symmetry can make Vab=0V_{ab}=0, in which case nearby levels may cross without mixing.

Retain both states and diagonalize their projected Hamiltonian. Write

HP(λ)=(ε1(λ)w(λ)w∗(λ)ε2(λ)),H_P(\lambda) = \begin{pmatrix} \varepsilon_1(\lambda) & w(\lambda) \\ w^*(\lambda) & \varepsilon_2(\lambda) \end{pmatrix},

where the diagonal entries may already include perturbative shifts from remote states. Define

εˉ=ε1+ε22,d=ε1−ε22.\begin{aligned} \bar\varepsilon &= \frac{\varepsilon_1+\varepsilon_2}{2}, \\ d &= \frac{\varepsilon_1-\varepsilon_2}{2}. \end{aligned}

The two eigenvalues are

E±=εˉ±d2+∣w∣2.E_\pm = \bar\varepsilon \pm \sqrt{d^2+\lvert w\rvert^2}.

At zero detuning, d=0d=0, the gap is

E+−E−=2∣w∣.E_+-E_- = 2\lvert w\rvert.

The eigenstates are equal-weight mixtures there. Away from the crossing, write w=∣w∣eiϕw=\lvert w\rvert e^{i\phi} and choose a mixing angle satisfying

tan⁡(2θ)=2∣w∣ε1−ε2.\tan(2\theta) = \frac{2\lvert w\rvert} {\varepsilon_1-\varepsilon_2}.

The phase ϕ\phi fixes the relative phase of the two basis components. For eigenvector conventions and the Bloch-sphere interpretation, see Two-State Hamiltonians.

If ε1>ε2\varepsilon_1\gt\varepsilon_2 and ∣w∣\lvert w\rvert is small compared with their difference, define Δε=ε1−ε2\Delta_\varepsilon=\varepsilon_1-\varepsilon_2. Expansion of the exact roots gives

E+=ε1+∣w∣2Δε+O ⁣(∣w∣4Δε3),E−=ε2−∣w∣2Δε+O ⁣(∣w∣4Δε3).\begin{aligned} E_+ &= \varepsilon_1 + \frac{\lvert w\rvert^2} {\Delta_\varepsilon} + O\!\left( \frac{\lvert w\rvert^4}{\Delta_\varepsilon^3} \right),\\ E_- &= \varepsilon_2 - \frac{\lvert w\rvert^2} {\Delta_\varepsilon} + O\!\left( \frac{\lvert w\rvert^4}{\Delta_\varepsilon^3} \right). \end{aligned}

Thus ordinary second-order level repulsion is the far-detuned expansion of an avoided crossing. The exact square root, rather than a divergent inverse detuning, is the uniform local description.

Let the model space be spanned by dd orthonormal unperturbed states,

P=∑a=1d∣a⟩⟨a∣,Q=I−P.P = \sum_{a=1}^{d} \lvert a\rangle\langle a\rvert, \qquad Q=I-P.

Block the Hamiltonian as

H=(PHPPHQQHPQHQ).H = \begin{pmatrix} PHP & PHQ \\ QHP & QHQ \end{pmatrix}.

The internal block PHPPHP is not expanded into separate one-state answers. It is diagonalized as a matrix. Remote states influence it through induced diagonal shifts and induced off-diagonal couplings.

A nearby level cluster retained in a model space while remote levels are eliminated perturbatively

The model space PP contains the close levels and all strong internal mixing. Remote QQ states are eliminated perturbatively, producing an effective Hamiltonian on PP. The final diagonalization is exact within that retained space.

For a cluster with unperturbed width

WP=max⁡a∈PEa(0)−min⁡a∈PEa(0),W_P = \max_{a\in P}E_a^{(0)} - \min_{a\in P}E_a^{(0)},

and distance to the complementary spectrum

GPQ=dist⁡(spec⁡(PH0P),spec⁡(QH0Q)),G_{PQ} = \operatorname{dist} \left( \operatorname{spec}(PH_0P), \operatorname{spec}(QH_0Q) \right),

the useful regime is schematically

WP≲∣λ∣∥PVP∥,ηPQ≡∣λ∣∥PVQ∥GPQ≪1.\begin{aligned} W_P &\lesssim \lvert\lambda\rvert\lVert PVP\rVert, \\ \eta_{PQ} &\equiv \frac{ \lvert\lambda\rvert\lVert PVQ\rVert }{G_{PQ}} \ll1. \end{aligned}

The first relation is only a rough motivation for retaining a cluster: a common diagonal shift makes ∥PVP∥\|PVP\| large without mixing anything. Compare actual off-diagonal couplings with the corresponding internal detunings. The second relation is a leading diagnostic, not a guarantee: diagonal perturbations can close the PP–QQ gap even when PVQPVQ is small. Monitor the separation of the perturbed diagonal blocks as well. For unbounded operators, these norm expressions are only a finite-model diagnostic; domain and relative-boundedness hypotheses require separate analysis.

Project an exact eigenstate as

∣Ψ⟩=∣ΨP⟩+∣ΨQ⟩.\lvert\Psi\rangle = \lvert\Psi_P\rangle + \lvert\Psi_Q\rangle.

If E−QHQE-QHQ is invertible, the QQ equation gives

∣ΨQ⟩=1E−QHQQHP∣ΨP⟩.\lvert\Psi_Q\rangle = \frac{1}{E-QHQ} QHP\lvert\Psi_P\rangle.

Substitution into the PP equation yields

Heff(E)∣ΨP⟩=E∣ΨP⟩,H_{\mathrm{eff}}(E) \lvert\Psi_P\rangle = E\lvert\Psi_P\rangle,

with

Heff(E)=PHP+ΣP(E),ΣP(E)=PHQ1E−QHQQHP.\begin{aligned} H_{\mathrm{eff}}(E) &= PHP+\Sigma_P(E), \\ \Sigma_P(E) &= PHQ \frac{1}{E-QHQ} QHP. \end{aligned}

This identity is exact where the resolvent exists. Its energy dependence makes the eigenvalue problem nonlinear, and poles signal that a supposedly eliminated state is dynamically relevant. Projection methods derive this Schur complement and the reconstruction of the full state; Feshbach projection extends it to continua and resonances.

Quasi-degenerate perturbation theory expands this exact structure in the coupling to QQ while preserving matrix diagonalization inside PP.

Hermitian Effective Hamiltonian Through Second Order

Section titled “Hermitian Effective Hamiltonian Through Second Order”

Let

H=H0+λV,H=H_0+\lambda V,

and choose PP from eigenstates ∣a⟩\lvert a\rangle of H0H_0 with nearby, not necessarily equal, energies Ea(0)E_a^{(0)}. Let Greek indices μ\mu label QQ states, and define

Daμ≡1Ea(0)−Eμ(0).D_{a\mu} \equiv \frac{1}{E_a^{(0)}-E_\mu^{(0)}}.

A standard Hermitian energy-independent effective Hamiltonian through second order has matrix elements

(Heff)ab=Ea(0)δab+λVab+λ2(Heff(2))ab+O(λ3).\bigl(H_{\mathrm{eff}}\bigr)_{ab} ={} E_a^{(0)}\delta_{ab} +\lambda V_{ab} + \lambda^2 \bigl(H_{\mathrm{eff}}^{(2)}\bigr)_{ab} +O(\lambda^3).

Here

(Heff(2))ab=12∑μ∈QVaμVμb(Daμ+Dbμ).\bigl(H_{\mathrm{eff}}^{(2)}\bigr)_{ab} = \frac12 \sum_{\mu\in Q} V_{a\mu}V_{\mu b} \bigl(D_{a\mu}+D_{b\mu}\bigr).

This is often called a canonical Van Vleck or Löwdin effective Hamiltonian. It is Hermitian because the two denominators are symmetrized between the bra and ket states.

For a=ba=b, the second-order term reduces to

(Heff(2))aa=∑μ∈Q∣Vaμ∣2Ea(0)−Eμ(0).\bigl(H_{\mathrm{eff}}^{(2)}\bigr)_{aa} = \sum_{\mu\in Q} \frac{\lvert V_{a\mu}\rvert^2} {E_a^{(0)}-E_\mu^{(0)}}.

This is the usual second-order energy correction, but the sum excludes the other model-space states. Their effects are generated by diagonalizing HeffH_{\mathrm{eff}} and must not be added again with small denominators.

For a≠ba\ne b, this same matrix element can be nonzero even when Vab=0V_{ab}=0. Virtual excursions into the eliminated space can therefore shift levels, open or modify avoided-crossing gaps, and generate interactions absent in the directly projected block.

Different effective-Hamiltonian conventions can differ in their off-shell matrix elements. Bloch, canonical Van Vleck, Löwdin, and Schrieffer–Wolff constructions impose different normalization or block-diagonalization choices. When carried out consistently, they reproduce the target exact eigenvalues through the claimed order; some forms are manifestly Hermitian and others are not. Do not combine matrix elements from different conventions term by term.

If all model-space states have the same unperturbed energy E0E_0, the second-order matrix simplifies to

(Heff)ab=E0δab+λVab+λ2∑μ∈QVaμVμbE0−Eμ(0)+O(λ3).\bigl(H_{\mathrm{eff}}\bigr)_{ab} ={} E_0\delta_{ab} +\lambda V_{ab} + \lambda^2 \sum_{\mu\in Q} \frac{V_{a\mu}V_{\mu b}} {E_0-E_\mu^{(0)}} +O(\lambda^3).

At first order, this is precisely the instruction to diagonalize PVPPVP. At second order, virtual QQ states add another Hermitian matrix that may split any degeneracy left by PVPPVP.

Quasi-degenerate theory therefore does not compete with degenerate perturbation theory. It extends the same model-space logic to clusters whose internal splittings are small but nonzero.

A useful PP space is large enough to remove every dangerous denominator and small enough that eliminating QQ is worthwhile.

Use the following tests:

  1. Energy-window test. Include levels whose unperturbed detuning from the target cluster is comparable to retained direct or induced couplings.
  2. Matrix-element test. A nearby state with a symmetry-forbidden coupling need not be included for that perturbation, though another allowed interaction may change the conclusion.
  3. Stability test. Enlarge PP and verify that target eigenvalues and observables change only beyond the claimed order.
  4. Symmetry test. Build PP from complete symmetry multiplets when the perturbation preserves that symmetry. Cutting a multiplet can create basis-dependent results.
  5. Intruder-state test. Monitor the smallest PP–QQ denominator. If a QQ level approaches the cluster, promote it into PP.
  6. State-character test. Track eigenvectors by overlaps or symmetry labels rather than sorting eigenvalues independently at each parameter value.

There is no universal energy cutoff. The correct model space depends on both detunings and couplings.

An eigenvector ∣u⟩=∑aca∣a⟩|u\rangle=\sum_a c_a|a\rangle of the Hermitian effective Hamiltonian represents the full state in a transformed model space. It is not exactly the bare projection P∣Ψ⟩P|\Psi\rangle: the two already differ in normalization at second order. Undoing the block-diagonalizing transformation gives the leading complementary component,

∣ΨQ⟩≈λ∑μ∈Q∑a∈PVμacaEa(0)−Eμ(0)∣μ⟩.\lvert\Psi_Q\rangle \approx \lambda\sum_{\mu\in Q}\sum_{a\in P} \frac{V_{\mu a}c_a}{E_a^{(0)}-E_\mu^{(0)}}|\mu\rangle.

The leakage probability is therefore schematically

pQ=⟨ΨQ∣ΨQ⟩=O(ηPQ2).p_Q = \langle\Psi_Q\vert\Psi_Q\rangle = O(\eta_{PQ}^{2}).

Normalization, transition matrix elements, and expectation values require this dressing. An effective observable generally differs from the bare projection PAPPAP because the same transformation that removes PP–QQ coupling also dresses operators. Using a corrected Hamiltonian with undressed observables can lose terms of the very order one intended to retain.

For an energy-dependent Feshbach Hamiltonian, the derivative of the self-energy also enters normalization. For an energy-independent unitary construction, transform both states and operators with the same unitary map.

Localized states ∣L⟩\lvert L\rangle and ∣R⟩\lvert R\rangle form a two-dimensional model space. Bias supplies the detuning and tunneling supplies the off-diagonal coupling. Diagonalizing the pair gives the avoided crossing. For the spatial construction, see Coupled Wells and Avoided Crossings.

The tunneling amplitude itself may be exponentially small in a semiclassical parameter. Finite-order power-series perturbation theory need not generate that nonperturbative scale; quasi-degenerate theory organizes the mixing once the amplitude has been calculated or matched.

External electric or magnetic fields can tune two levels close together. Selection rules determine whether the crossing is exact or avoided. Remote levels contribute Stark, Zeeman, and spin–orbit shifts to the effective matrix, while the nearby pair or multiplet is diagonalized together. This is the setting in which smooth adiabatic potential curves and reliable state tracking matter more than assigning a fixed unperturbed label.

Two-level atoms and dressed-state truncations

Section titled “Two-level atoms and dressed-state truncations”

For a static two-level approximation, the retained pair must be separated from all other atomic levels by gaps large compared with the relevant couplings. Near resonant driving, the quasi-degenerate pair often lives in a rotating or Floquet representation rather than the bare atomic spectrum. The rotating-wave approximation explains that change of frame and its limits.

Near a crossing momentum k0\mathbf k_0, retain the few bands whose energies coincide or nearly coincide and eliminate remote bands. A two-band effective Hamiltonian often has the form

Heff(q)=d0(q)I+d(q)⋅σ,q=k−k0.\begin{aligned} H_{\mathrm{eff}}(\mathbf q) &= d_0(\mathbf q)I + \mathbf d(\mathbf q)\cdot\boldsymbol\sigma, \\ \mathbf q &= \mathbf k-\mathbf k_0. \end{aligned}

Symmetry may force components of d\mathbf d to vanish and protect a crossing. An allowed constant component can open a gap. The two-band form alone does not establish topological protection; that requires the momentum dependence and symmetry constraints on d\mathbf d.

Let

H=(δ/2gg∗−δ/2).H = \begin{pmatrix} \delta/2 & g \\ g^* & -\delta/2 \end{pmatrix}.

Find the eigenvalues, the minimum gap as δ\delta is varied, and the far-detuned expansion.

Solution

The characteristic equation is

E2=δ24+∣g∣2,E^2 = \frac{\delta^2}{4} +\lvert g\rvert^2,

so

E±=±δ24+∣g∣2.E_\pm = \pm \sqrt{ \frac{\delta^2}{4} +\lvert g\rvert^2 }.

The gap is

E+−E−=2δ24+∣g∣2,E_+-E_- = 2\sqrt{ \frac{\delta^2}{4} +\lvert g\rvert^2 },

whose minimum at δ=0\delta=0 is 2∣g∣2\lvert g\rvert. For δ>0\delta\gt0 and ∣g∣≪δ\lvert g\rvert\ll\delta,

E+=δ2+∣g∣2δ+O ⁣(∣g∣4δ3),E−=−δ2−∣g∣2δ+O ⁣(∣g∣4δ3).\begin{aligned} E_+ &= \frac{\delta}{2} + \frac{\lvert g\rvert^2}{\delta} + O\!\left( \frac{\lvert g\rvert^4}{\delta^3} \right), \\ E_- &= -\frac{\delta}{2} - \frac{\lvert g\rvert^2}{\delta} + O\!\left( \frac{\lvert g\rvert^4}{\delta^3} \right). \end{aligned}

The inverse detuning is valid only in the far-detuned regime; the square root remains finite through the crossing.

Let PP contain ∣1⟩\lvert1\rangle and ∣2⟩\lvert2\rangle, and let QQ contain one state ∣3⟩\lvert3\rangle. Assume V12=0V_{12}=0, but V13=g1V_{13}=g_1 and V32=g2V_{32}=g_2. Find the induced off-diagonal matrix element through second order.

Solution

The Hermitian second-order formula gives

(Heff(2))12=g1g22[1E1(0)−E3(0)+1E2(0)−E3(0)].\bigl(H_{\mathrm{eff}}^{(2)}\bigr)_{12} = \frac{g_1g_2}{2} \Biggl[ \frac{1}{E_1^{(0)}-E_3^{(0)}} + \frac{1}{E_2^{(0)}-E_3^{(0)}} \Biggr].

The physical contribution to the Hamiltonian is λ2(Heff(2))12\lambda^2(H_{\mathrm{eff}}^{(2)})_{12}. Even though there is no direct coupling, the path 1→3→21\to3\to2 mediates mixing. Hermiticity gives the reverse element as its complex conjugate.

Take E1(0)=E2(0)=E0E_1^{(0)}=E_2^{(0)}=E_0 in Exercise 2. Show how the symmetrized denominator simplifies.

Solution

Both denominators become E0−E3(0)E_0-E_3^{(0)}, so

(Heff(2))12=g1g2E0−E3(0).\bigl(H_{\mathrm{eff}}^{(2)}\bigr)_{12} = \frac{g_1g_2} {E_0-E_3^{(0)}}.

The factor 1/21/2 is canceled by the two equal terms. This is the usual second-order effective matrix within an exactly degenerate subspace.

Three unperturbed levels have energies 00, 0.03Δ0.03\Delta, and Δ\Delta. At the physical coupling, the first two are connected by a matrix element of magnitude 0.05Δ0.05\Delta, while their couplings to the third have magnitude at most 0.02Δ0.02\Delta. Which states should be in the first model-space trial?

Solution

The detuning between the first two states is 0.03Δ0.03\Delta, smaller than their coupling scale 0.05Δ0.05\Delta. They must be retained and diagonalized together. The third state is separated by a gap of order Δ\Delta, while its coupling is at most 0.02Δ0.02\Delta, giving a mixing ratio of order 0.020.02. It is a reasonable first QQ-space state.

This is a starting choice, not a proof. One should add the third state to PP as a stability check and verify that target quantities change only beyond the intended order.

Let ∣u⟩|u\rangle be the normalized eigenvector of the Hermitian effective Hamiltonian. Write the leading QQ-space probability after reconstructing the full state.

Solution

Let ca=⟨a∣u⟩c_a=\langle a\vert u\rangle and let ∣μ⟩\lvert\mu\rangle label QQ states. To first order,

⟨μ∣ΨQ⟩≈λ∑a∈PVμacaEa(0)−Eμ(0).\langle\mu\vert\Psi_Q\rangle \approx \lambda \sum_{a\in P}\frac{V_{\mu a}c_a}{E_a^{(0)}-E_\mu^{(0)}}.

Therefore

pQ≈λ2∑μ∈Q∣∑a∈PVμacaEa(0)−Eμ(0)∣2.p_Q \approx \lambda^2 \sum_{\mu\in Q} \left| \sum_{a\in P}\frac{V_{\mu a}c_a}{E_a^{(0)}-E_\mu^{(0)}} \right|^2.

Interference between amplitudes from different model-space components is physical and can enhance or suppress leakage.

For

HLR=(ϵ−K−K−ϵ),H_{LR} = \begin{pmatrix} \epsilon & -K \\ -K & -\epsilon \end{pmatrix},

find the energies and state character in the limits ∣ϵ∣≫K\lvert\epsilon\rvert\gg K and ϵ=0\epsilon=0.

Solution

The energies are

E±=±ϵ2+K2.E_\pm = \pm\sqrt{\epsilon^2+K^2}.

For ∣ϵ∣≫K\lvert\epsilon\rvert\gg K, each eigenstate is predominantly localized in one well and receives a small admixture of the other. At ϵ=0\epsilon=0, the eigenstates are the symmetric and antisymmetric combinations

∣L⟩±∣R⟩2,\frac{ \lvert L\rangle\pm\lvert R\rangle }{\sqrt2},

with a splitting 2∣K∣2|K|. The exact two-state result is uniform across the bias value where a one-state expansion would fail.

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