Second-Order Perturbation Theory

Purpose

For an isolated nondegenerate eigenstate of

$$H(\lambda) = H_0+\lambda V,$$

the second-order energy coefficient is

$$E_n^{(2)} = \sum_{m\ne n} \frac{ \left\lvert \langle m^{(0)}\rvert V\lvert n^{(0)}\rangle \right\rvert^2 }{ E_n^{(0)}-E_m^{(0)} }.$$

The physical quadratic shift is $\lambda^2E_n^{(2)}$. This is the leading energy effect of off-diagonal state mixing: levels above the target contribute negatively, while levels below contribute positively.

The derivation, ground-state concavity argument, and extended examples are at Second-Order Energy Corrections. This card emphasizes how to evaluate, check, and generalize the formula.

Setup and explicit quadratic terms

For

$$H_0\lvert n^{(0)}\rangle = E_n^{(0)}\lvert n^{(0)}\rangle,$$

define

$$V_{mn} = \langle m^{(0)}\rvert V\lvert n^{(0)}\rangle.$$

If the Hamiltonian is linear in $\lambda$,

$$\begin{aligned} E_n(\lambda) &= E_n^{(0)} + \lambda V_{nn} \\ &\quad+ \lambda^2 \sum_{m\ne n} \frac{\lvert V_{mn}\rvert^2} {E_n^{(0)}-E_m^{(0)}} + O(\lambda^3). \end{aligned}$$

If instead

$$H(\lambda) = H_0+\lambda V+\lambda^2W+O(\lambda^3),$$

then

$$E_n^{(2)} = W_{nn} + \sum_{m\ne n} \frac{\lvert V_{mn}\rvert^2} {E_n^{(0)}-E_m^{(0)}}.$$

The first term is an explicit quadratic term in the Hamiltonian. The second is generated by two applications of the linear perturbation. Omitting $W_{nn}$ is a common error in field expansions, minimal-coupling Hamiltonians, and parameter-dependent effective models.

At a glance

Task	Formula or rule
Mixing correction	$E_n^{(2)}=\sum_{m\ne n}\lvert V_{mn}\rvert^2/(E_n^{(0)}-E_m^{(0)})$
Reduced-resolvent form	$E_n^{(2)}=\langle n^{(0)}\rvert VR_n'V\lvert n^{(0)}\rangle$
Ground-state sign	$E_0^{(2)}\le0$ for the mixing term
Physical curvature	$d^2E_n/d\lambda^2
Static ground-state polarizability	$\alpha=2\sum_{m\ne0}\lvert D_{m0}\rvert^2/(E_m-E_0)\ge0$
Degenerate model space	diagonalize the second-order effective matrix after $PVP$

Reduced-resolvent form

Define

$$P_n = \lvert n^{(0)}\rangle \langle n^{(0)}\rvert, \qquad Q_n=I-P_n.$$

On the orthogonal complement of the target state, define

$$R_n' = Q_n \frac{1}{E_n^{(0)}-H_0} Q_n.$$

For a discrete nondegenerate spectrum,

$$R_n' = \sum_{m\ne n} \frac{ \lvert m^{(0)}\rangle \langle m^{(0)}\rvert }{ E_n^{(0)}-E_m^{(0)} }.$$

Then

$$\lvert n^{(1)}\rangle = R_n'V\lvert n^{(0)}\rangle$$

and

$$E_n^{(2)} = \langle n^{(0)}\rvert VR_n'V \lvert n^{(0)}\rangle.$$

The $Q_n$ projectors are essential: the ordinary resolvent is singular at the eigenvalue $E_n^{(0)}$. The reduced inverse exists only after the target eigenspace is removed and only when the remaining spectrum is sufficiently separated.

Use Resolvent Operator for the canonical operator treatment.

Sign and level repulsion

Separate intermediate states below and above the target:

$$\begin{aligned} E_n^{(2)} &= \sum_{E_m^{(0)}<E_n^{(0)}} \frac{\lvert V_{mn}\rvert^2} {E_n^{(0)}-E_m^{(0)}} \\ &\quad- \sum_{E_m^{(0)}>E_n^{(0)}} \frac{\lvert V_{mn}\rvert^2} {E_m^{(0)}-E_n^{(0)}}. \end{aligned}$$

Thus:

Target	General sign information for the mixing term
Nondegenerate ground state	nonpositive
Highest level in a finite-dimensional model	nonnegative
Generic excited state	no fixed sign
State with $Q_nV\lvert n^{(0)}\rangle=0$	zero

For the ground state,

$$E_0^{(2)} = -\sum_{m\ne0} \frac{\lvert V_{m0}\rvert^2} {E_m^{(0)}-E_0^{(0)}} \le0.$$

Equality holds exactly when $V\lvert0^{(0)}\rangle$ has no component orthogonal to the ground state. A nonzero diagonal $V_{00}$ can still produce a first-order shift.

For an excited state, never infer a sign without identifying both lower and upper coupled levels. The statement “second order lowers the energy” is a ground-state result, not a general rule.

Exact two-level check

Consider

$$H(\lambda) = \begin{pmatrix} E_a&\lambda v\\ \lambda v^*&E_b \end{pmatrix}, \qquad \Delta=E_b-E_a>0.$$

The exact eigenvalues are

$$E_\pm = \frac{E_a+E_b}{2} \pm \frac12 \sqrt{ \Delta^2+4\lambda^2\lvert v\rvert^2 }.$$

For $\lvert\lambda v\rvert\ll\Delta$,

$$E_-(\lambda) = E_a - \lambda^2 \frac{\lvert v\rvert^2}{\Delta} + O(\lambda^4),$$ $$E_+(\lambda) = E_b + \lambda^2 \frac{\lvert v\rvert^2}{\Delta} + O(\lambda^4).$$

The exact square root displays both level repulsion and the failure criterion. When coupling and gap are comparable, diagonalize the block instead of retaining a large second-order term.

Continuum contributions

If $H_0$ has discrete states and continuum channels $c$, completeness gives

$$\begin{aligned} E_n^{(2)} &= \sum_{m\in\mathrm{disc},,m\ne n} \frac{\lvert V_{mn}\rvert^2} {E_n^{(0)}-E_m^{(0)}} \\ &\quad+ \sum_c \int_{E_{c,\mathrm{th}}}^{\infty} dE\, \frac{ \left\lvert \langle E,c\rvert V\lvert n^{(0)}\rangle \right\rvert^2 }{E_n^{(0)}-E}. \end{aligned}$$

The integration measure depends on whether continuum states are normalized in energy, momentum, or another channel variable. The matrix element and density factor must use the same convention.

For a bound target below threshold, denominators do not cross zero. For an embedded state or an energy above an open threshold, poles and outgoing boundary conditions lead to energy shifts and widths. Principal values, self-energies, or resonance theory then replace the naive bound-state formula.

Computing without an explicit sum

Solve the projected inhomogeneous equation

$$\left( H_0-E_n^{(0)} \right) \lvert\chi_n\rangle = -Q_nV\lvert n^{(0)}\rangle,$$

with

$$\langle n^{(0)}\vert\chi_n\rangle=0.$$

Then

$$\lvert\chi_n\rangle = R_n'V\lvert n^{(0)}\rangle$$

and

$$E_n^{(2)} = \langle n^{(0)}\rvert V\lvert\chi_n\rangle.$$

This Dalgarno–Lewis strategy can include an infinite discrete spectrum and continuum effects without constructing every intermediate state. In a finite numerical basis, remove or constrain the null direction along $\lvert n^{(0)}\rangle$ before solving.

Use Sum Rules and Completeness Tricks for closure identities and numerical checks.

Static polarizability

Let a static electric field couple through

$$H(\boldsymbol{\mathcal E}) = H_0 - \boldsymbol{\mathcal E}\cdot\mathbf D.$$

For a nondegenerate state with no linear Stark shift,

$$E_n(\boldsymbol{\mathcal E}) = E_n(0) - \frac12 \alpha_{ij}^{(n)} \mathcal E_i\mathcal E_j + O(\mathcal E^3).$$

The symmetric static-polarizability tensor is

$$\begin{aligned} \alpha_{ij}^{(n)} &= 2\operatorname{Re} \sum_{m\ne n} \frac{ \langle n\rvert D_i\lvert m\rangle \langle m\rvert D_j\lvert n\rangle }{E_m-E_n}. \end{aligned}$$

Along unit vector $\mathbf e$,

$$\alpha_{\mathbf e}^{(n)} = 2 \sum_{m\ne n} \frac{ \left\lvert \langle m\rvert \mathbf e\cdot\mathbf D \lvert n\rangle \right\rvert^2 }{E_m-E_n}.$$

For a nondegenerate ground state, $\alpha_{\mathbf e}^{(0)}\ge0$, so the quadratic Stark energy is nonpositive. Excited-state static coefficients need not have a fixed sign. Frequency- dependent driving requires dynamic response denominators and resonance prescriptions, not this static expression.

Degenerate second-order effective matrix

Let $P$ project onto an exactly degenerate eigenspace of $H_0$ with energy $E^{(0)}$, and let $Q=I-P$. For

$$H = H_0+\lambda V+\lambda^2W+O(\lambda^3),$$

the model-space effective Hamiltonian through second order is

$$\begin{aligned} H_{\mathrm{eff}} &= E^{(0)}P + \lambda PVP \\ &\quad+ \lambda^2 \left[ PWP + PVQ \frac{1}{E^{(0)}-QH_0Q} QVP \right] + O(\lambda^3). \end{aligned}$$

In a basis $\lvert a\rangle$ of the model space, the mixing part has matrix elements

$$\left(H_{\mathrm{eff}}^{(2)}\right)_{ab} = W_{ab} + \sum_{r\notin P} \frac{V_{ar}V_{rb}} {E^{(0)}-E_r^{(0)}}.$$

First diagonalize $PVP$. If its eigenvalues are distinct, evaluate second-order shifts in those good first-order states. If a residual degeneracy remains, diagonalize the second-order effective matrix within that residual subspace.

For quasi-degenerate levels whose unperturbed splittings are comparable to $\lambda V$, place all of them in $P$ and retain their $H_0$ splittings in the model-space matrix. Use Projection Methods for the canonical effective-Hamiltonian framework.

Convergence and numerical checks

The formal summand is not enough. Check:

1. Completeness: include every discrete state, continuum channel, and internal label reached by $V$.
2. Near-degeneracy: inspect $\lvert\lambda V_{mn}/(E_n^{(0)}-E_m^{(0)})\rvert$ before trusting the sum.
3. High-energy convergence: matrix-element decay must overcome the density of states and inverse-gap behavior.
4. Operator domains: unbounded or singular perturbations may require a common domain or quadratic-form treatment.
5. Cutoff convergence: report changes under basis or energy-cutoff enlargement.
6. Independent checks: compare a spectral sum with a projected linear solve or exact diagonalization at several small couplings.

For a ground-state sum over fixed exact intermediate states, every term is nonpositive. Increasing a literal energy cutoff can only leave the partial sum unchanged or make it more negative. This monotonicity need not hold when the approximate eigenstates themselves change with basis size.

Calculation workflow

1. Expand the Hamiltonian and identify any explicit $\lambda^2W$ term.
2. Decide whether the target is isolated or belongs in a degenerate or quasi-degenerate model space.
3. Apply symmetry and selection rules to the matrix elements.
4. Inspect every materially coupled energy gap before evaluating the sum.
5. Separate lower, upper, and continuum contributions so sign and completeness remain visible.
6. Use either the spectral sum or the projected inhomogeneous equation.
7. Restore the factor $\lambda^2$ in the reported physical energy.
8. Test dimensions, sign theorems, cutoff stability, and an exact small-block limit.

Common mistakes

• Including $m=n$ and creating a zero denominator.
• Dropping the square modulus for complex matrix elements.
• Using perturbed energies in a fixed-order Rayleigh–Schrodinger denominator.
• Forgetting an explicit quadratic Hamiltonian term $W_{nn}$.
• Claiming every second-order shift is negative.
• Treating a small denominator as a large valid result rather than a breakdown signal.
• Omitting continuum channels or mismatching their normalization and measure.
• Using static polarizability at finite frequency or resonance.
• Applying a one-state formula inside a degenerate model space.
• Reporting a numerical partial sum without a cutoff or basis-convergence study.

Canonical links

• First-Order Perturbation Theory fixes the state correction entering the derivation.
• Higher-Order Structure treats recursion, convergence radii, and asymptotic behavior.
• Degenerate Perturbation Theory establishes the model-space basis before second order.
• Stark Effect Example applies the polarizability structure.
• Anharmonic Oscillator Worked Problem provides an explicit sum-over-states benchmark.

References

• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• C. Cohen-Tannoudji, B. Diu, and F. Laloe, Quantum Mechanics, Volume II, Wiley, 1977.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• A. Messiah, Quantum Mechanics, Volume II, North-Holland, 1962.
• T. Kato, Perturbation Theory for Linear Operators, 2nd ed., Springer, 1976.
• A. Dalgarno and J. T. Lewis, “The exact calculation of long-range forces between atoms by perturbation theory,” Proceedings of the Royal Society A 233, 70-74 (1955).

Exercises

1. Prove the sign of the second-order mixing correction for a nondegenerate ground state.

Solution

For every excited unperturbed state,

$$E_m^{(0)}-E_0^{(0)}>0.$$

Therefore

$$E_0^{(2)} = -\sum_{m\ne0} \frac{\lvert V_{m0}\rvert^2} {E_m^{(0)}-E_0^{(0)}} \le0.$$

Equality requires every off-diagonal coupling $V_{m0}$ to vanish, equivalently $Q_0V\lvert0^{(0)}\rangle=0$.

2. Suppose $H=H_0+\lambda V+\lambda^2W$ and $[H_0,V]=0$. Find the second-order energy coefficient for a nondegenerate common eigenstate.

Solution

Because $H_0$ and $V$ share the nondegenerate eigenbasis,

$$V_{mn}=0 \qquad m\ne n.$$

The mixing sum vanishes. The remaining coefficient is

$$E_n^{(2)} = \langle n^{(0)}\rvert W\lvert n^{(0)}\rangle.$$

If $W=0$ as well, the energy branch is linear in $\lambda$ in this common eigenstate.

3. Expand the lower exact eigenvalue of the two-level model through fourth order in $\lambda$.

Solution

Write

$$E_-(\lambda) = \frac{E_a+E_b}{2} - \frac{\Delta}{2} \sqrt{ 1+\frac{4\lambda^2\lvert v\rvert^2}{\Delta^2} }.$$

Using

$$\sqrt{1+x} = 1+\frac{x}{2}-\frac{x^2}{8}+O(x^3)$$

gives

$$E_-(\lambda) = E_a - \lambda^2\frac{\lvert v\rvert^2}{\Delta} + \lambda^4\frac{\lvert v\rvert^4}{\Delta^3} + O(\lambda^6).$$

The quadratic coefficient matches perturbation theory, while the expansion parameter is $\lvert\lambda v\rvert/\Delta$.

4. A two-dimensional degenerate model space has $PVP=0$ and couples to one outside state $\lvert r\rangle$ through $\langle r\rvert V\lvert1\rangle=g_1$ and $\langle r\rvert V\lvert2\rangle=g_2$. Find the second-order effective matrix when $W=0$.

Solution

Let the model-space energy be $E^{(0)}$ and the outside-state energy be $E_r^{(0)}$. Then

$$H_{\mathrm{eff}}^{(2)} = \frac{1}{E^{(0)}-E_r^{(0)}} \begin{pmatrix} \lvert g_1\rvert^2&g_1^*g_2\\ g_2^*g_1&\lvert g_2\rvert^2 \end{pmatrix}.$$

The matrix is rank one. One linear combination proportional to $g_1^*\lvert1\rangle+g_2^*\lvert2\rangle$ couples to $\lvert r\rangle$ and shifts, while the orthogonal combination has zero second-order shift in this three-state model.