First-Order Perturbation Theory

Purpose

Time-independent perturbation theory follows an eigenvalue and eigenstate of

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

away from a solved Hamiltonian $H_0$. For an isolated, normalized, nondegenerate state

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

the first-order energy coefficient is

$$E_n^{(1)} = \langle n^{(0)}\rvert V\lvert n^{(0)}\rangle.$$

The physical energy shift at this order is $\lambda E_n^{(1)}$. If the level is degenerate, this one-state formula is not the starting point: diagonalize $V$ inside the entire degenerate subspace first.

The order-by-order derivation is at Nondegenerate Perturbation Theory. This card combines the formulas with validity tests and the degenerate branch needed in calculations.

Expansion and normalization

Write

$$E_n(\lambda) = E_n^{(0)} + \lambda E_n^{(1)} + \lambda^2E_n^{(2)} + \cdots$$

and

$$\lvert n(\lambda)\rangle = \lvert n^{(0)}\rangle + \lambda\lvert n^{(1)}\rangle + \lambda^2\lvert n^{(2)}\rangle + \cdots.$$

Intermediate normalization fixes the phase and longitudinal component by

$$\langle n^{(0)}\vert n(\lambda)\rangle=1,$$

so

$$\langle n^{(0)}\vert n^{(r)}\rangle=0, \qquad r\ge1.$$

The truncated vector $\lvert n^{(0)}\rangle+\lambda\lvert n^{(1)}\rangle$ is normalized through first order, not exactly. Its norm differs from one at order $\lambda^2$. Energy corrections do not depend on this normalization choice, while explicit state-correction formulas do.

At a glance

Task	Formula
First-order energy coefficient	$E_n^{(1)}=V_{nn}$
Physical first-order shift	$\Delta E_n=\lambda V_{nn}$
State correction	$\lvert n^{(1)}\rangle=\sum_{m\ne n}V_{mn}\lvert m^{(0)}\rangle/(E_n^{(0)}-E_m^{(0)})$
Mixing diagnostic	$\eta_{mn}=\lvert\lambda V_{mn}/(E_n^{(0)}-E_m^{(0)})\rvert$
Degenerate first order	diagonalize $PVP$
Fixed-observable response	$\delta\langle A\rangle^{(1)}=2\operatorname{Re}\sum_{m\ne n}A_{nm}V_{mn}/(E_n^{(0)}-E_m^{(0)})$

Here

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

Nondegenerate energy correction

For an isolated level,

$$E_n(\lambda) = E_n^{(0)} + \lambda V_{nn} + O(\lambda^2).$$

The correction is real when $V$ is self-adjoint. It is the expectation value of the perturbing operator in the unperturbed state, not in the corrected state.

Several immediate checks follow:

• if $V=cI$, every level shifts by exactly $\lambda c$ and no state changes;
• if a symmetry makes $V_{nn}=0$, the energy has no first-order shift, but the state may still change at first order;
• if $V$ is diagonal in the $H_0$ eigenbasis, the state correction vanishes and the diagonal energy shift is exact for the linear family $H_0+\lambda V$;
• if $\lambda$ has been absorbed into $V$, do not multiply by it a second time.

The focused physical interpretation and symmetry examples are at First-Order Energy Corrections.

Nondegenerate state correction

With intermediate normalization,

$$\lvert n^{(1)}\rangle = \sum_{m\ne n} \frac{V_{mn}} {E_n^{(0)}-E_m^{(0)}} \lvert m^{(0)}\rangle.$$

The numerator says which states the perturbation connects. The denominator penalizes distant levels and exposes dangerous near-degeneracies. The full first-order state is

$$\lvert n(\lambda)\rangle = \lvert n^{(0)}\rangle + \lambda \sum_{m\ne n} \frac{V_{mn}} {E_n^{(0)}-E_m^{(0)}} \lvert m^{(0)}\rangle + O(\lambda^2).$$

Let

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

The reduced resolvent is

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

where the inverse is taken only on the orthogonal complement of the target state. Then

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

This form remains useful when the unperturbed spectrum contains both discrete and continuum sectors. In a spectral expansion, the sum must then be augmented by continuum integrals with the chosen normalization.

Use First-Order State Corrections for phase freedom, normalization beyond first order, and geometric interpretation.

Observable response

For an operator $A$ with no explicit $\lambda$ dependence,

$$\begin{aligned} \langle A\rangle_n(\lambda) &= A_{nn} + \lambda \Bigl[ \langle n^{(1)}\rvert A\lvert n^{(0)}\rangle \\ &\qquad+ \langle n^{(0)}\rvert A\lvert n^{(1)}\rangle \Bigr] + O(\lambda^2). \end{aligned}$$

Therefore

$$\delta\langle A\rangle_n^{(1)} = 2\operatorname{Re} \sum_{m\ne n} \frac{A_{nm}V_{mn}} {E_n^{(0)}-E_m^{(0)}}.$$

If $A=A(\lambda)$ also changes explicitly, add

$$\left\langle n^{(0)}\left\rvert \frac{\partial A}{\partial\lambda} \right\lvert n^{(0)}\right\rangle_{\lambda=0}$$

to the first derivative. A vanishing first-order energy shift therefore does not imply vanishing first-order response of other observables.

Hellmann–Feynman check

For an exact normalized eigenstate of a differentiable Hamiltonian,

$$\frac{dE_n}{d\lambda} = \left\langle n(\lambda)\left\rvert \frac{\partial H}{\partial\lambda} \right\lvert n(\lambda)\right\rangle.$$

At $\lambda=0$ for $H=H_0+\lambda V$,

$$\left.\frac{dE_n}{d\lambda}\right|_{0} = \langle n^{(0)}\rvert V\lvert n^{(0)}\rangle = E_n^{(1)}.$$

This is an independent interpretation of the first-order energy coefficient as a parameter derivative. Degeneracies require choosing differentiable energy branches after diagonalizing the projected derivative in the degenerate subspace.

Degenerate first order

Suppose $H_0$ has a $d$-dimensional eigenspace $\mathcal D$ at energy $E^{(0)}$. Let

$$P = \sum_{a=1}^{d} \lvert a\rangle\langle a\rvert$$

project onto $\mathcal D$. Construct

$$W = PVP, \qquad W_{ab}=\langle a\rvert V\lvert b\rangle.$$

Diagonalize $W$:

$$W\lvert\alpha^{(0)}\rangle = w_\alpha\lvert\alpha^{(0)}\rangle.$$

Then the first-order branches are

$$E_\alpha(\lambda) = E^{(0)} + \lambda w_\alpha + O(\lambda^2).$$

The eigenvectors of $PVP$ are the good zeroth-order combinations. The perturbation can rotate the original basis by an order-one angle even as $\lambda\to0$, which is why the nondegenerate denominator formula cannot be repaired inside the degenerate subspace.

After this diagonalization, mixing with states outside $\mathcal D$ is

$$\lvert\alpha_\perp^{(1)}\rangle = \sum_{r\notin\mathcal D} \frac{ \langle r\rvert V\lvert\alpha^{(0)}\rangle }{ E^{(0)}-E_r^{(0)} } \lvert r\rangle.$$

If $PVP$ is proportional to the identity, the degeneracy is not split at first order. A higher-order effective Hamiltonian or additional symmetry analysis may still be required.

Use Degenerate Perturbation Theory for the canonical subspace treatment.

Near-degenerate levels

For an isolated-level expansion, inspect

$$\eta_{mn} = \left\lvert \frac{\lambda V_{mn}} {E_n^{(0)}-E_m^{(0)}} \right\rvert.$$

Relevant values should be much smaller than one. This is a diagnostic, not a universal theorem: cumulative couplings, unbounded operators, and large state spaces can demand stronger analysis.

If one or more $\eta_{mn}$ is not small, enlarge the model subspace and diagonalize the resulting effective or quasi-degenerate Hamiltonian. The appropriate canonical method is Quasi-Degenerate Perturbation Theory.

For a two-level block,

$$H = \begin{pmatrix} E_1^{(0)}+\lambda V_{11}&\lambda V_{12}\\ \lambda V_{21}&E_2^{(0)}+\lambda V_{22} \end{pmatrix},$$

direct diagonalization is usually simpler and more reliable than expanding a small denominator.

Symmetry diagnostics

If $\lvert n^{(0)}\rangle$ is a parity eigenstate and $V$ is parity odd, then

$$V_{nn}=0.$$

The first-order energy shift vanishes, while $V$ can mix the state with opposite-parity levels through $V_{mn}$. More generally, selection rules can make entire sums sparse and identify the symmetry sector of the state correction.

If a symmetry protects a degenerate irreducible multiplet and $V$ respects that symmetry, Schur’s lemma may force $PVP$ to be proportional to the identity. If $V$ breaks the symmetry, the block structure under the remaining symmetry often predicts the splitting pattern before diagonalization.

Validity and error interpretation

The formulas assume the target eigenvalue or chosen eigenspace is isolated from states omitted from the perturbative model. For sufficiently regular operator families and a finite spectral gap, eigenvalues and spectral projectors can be analytic near $\lambda=0$. In many physical problems the series is only asymptotic, and the best truncation order depends on the coupling.

First order means

$$E_n = E_n^{(0)}+\lambda E_n^{(1)}+O(\lambda^2),$$

not that the neglected error is numerically small for arbitrary $\lambda$. Identify the dimensionless expansion ratio before setting a bookkeeping parameter to one.

An isolated bound state approaching a continuum threshold, an embedded state, a level crossing, or a resonance may require resolvent, effective-Hamiltonian, or scattering methods rather than ordinary bound-state perturbation theory.

Calculation workflow

1. Write $H=H_0+\lambda V$ and identify the physical dimensionless small parameter.
2. Determine whether the target energy is nondegenerate, degenerate, or nearly degenerate.
3. Use symmetry to find zero matrix elements and invariant blocks.
4. For an isolated state, compute $V_{nn}$ and the relevant ratios $\eta_{mn}$.
5. For a degenerate subspace, diagonalize $PVP$ before using outside-state denominators.
6. Compute only the state correction needed for the requested observable.
7. Check Hermiticity, units, limiting cases, and comparison with an exactly solvable small block when available.

Common mistakes

• Applying $E_n^{(1)}=V_{nn}$ to an arbitrary vector inside a degenerate subspace.
• Dividing by zero or a small gap instead of enlarging and diagonalizing the model space.
• Calling $E_n^{(1)}$ the physical shift while forgetting the factor $\lambda$.
• Setting $\lambda=1$ before identifying a dimensionless control parameter.
• Assuming $V_{nn}=0$ means the perturbation has no first-order effect on the state or other observables.
• Treating the first-order corrected ket as exactly normalized.
• Omitting continuum integrals when the spectral resolution is not purely discrete.
• Using state-vector corrections as if they were basis-independent observables.
• Ignoring symmetry-protected blocks or accidental near-degeneracies.

Canonical links

• Small Parameters and Error Estimates explains control parameters and truncation logic.
• Higher-Order Structure develops recursion, convergence, and asymptotic behavior.
• Rayleigh–Schrodinger Perturbation Theory gives the historical and all-order framework.
• Anharmonic Oscillator supplies a standard bound-state application.
• Stark Effect Example shows symmetry, degeneracy, and field response.

References

• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• 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, Wiley, 1977.
• A. Messiah, Quantum Mechanics, Dover, 1999.
• T. Kato, Perturbation Theory for Linear Operators, 2nd ed., Springer, 1976.

Exercises

1. Let $H_0$ have parity eigenstate $\lvert n^{(0)}\rangle$ and let $V$ be parity odd. What vanishes at first order, and what need not vanish?

Solution

Parity gives

$$\langle n^{(0)}\rvert V\lvert n^{(0)}\rangle=0,$$

so

$$E_n^{(1)}=0.$$

However, matrix elements between opposite-parity states can be nonzero. Thus

$$\lvert n^{(1)}\rangle = \sum_{m\ne n} \frac{V_{mn}}{E_n^{(0)}-E_m^{(0)}} \lvert m^{(0)}\rangle$$

need not vanish and has the parity opposite to the original state when the selection rule is exact.

2. Consider

$$H_0 = \begin{pmatrix}E_1&0\\0&E_2\end{pmatrix}, \qquad V = \begin{pmatrix}a&v\\v^*&c\end{pmatrix},$$

with $E_1\ne E_2$. Find the first-order energy and state corrections for the lower basis state $\lvert1\rangle$.

Solution

The energy coefficient is

$$E_1^{(1)}=a.$$

The only other state is $\lvert2\rangle$, with $V_{21}=v^*$. Therefore

$$\lvert1^{(1)}\rangle = \frac{v^*}{E_1-E_2} \lvert2\rangle.$$

The expansion is controlled when

$$\left\lvert \frac{\lambda v}{E_1-E_2} \right\rvert \ll1.$$

3. For the two-level system in Exercise 2, let $A=\lvert1\rangle\langle2\rvert+\lvert2\rangle\langle1\rvert$. Find the first-order change of $\langle A\rangle$ in the branch starting from $\lvert1\rangle$.

Solution

Here $A_{12}=A_{21}=1$ and $A_{11}=A_{22}=0$. The response formula gives

$$\delta\langle A\rangle_1^{(1)} = 2\operatorname{Re} \frac{A_{12}V_{21}}{E_1-E_2} = 2\operatorname{Re} \frac{v^*}{E_1-E_2}.$$

Thus

$$\langle A\rangle_1(\lambda) = 2\lambda \operatorname{Re} \frac{v^*}{E_1-E_2} + O(\lambda^2).$$

4. A two-dimensional degenerate subspace has

$$PVP = \begin{pmatrix} a&b\\ b^*&a \end{pmatrix}.$$

Find the first-order shifts and normalized good zeroth-order states.

Solution

Write $b=\lvert b\rvert e^{i\phi}$. The eigenvalues are

$$w_\pm=a\pm\lvert b\rvert.$$

One phase convention for normalized eigenvectors is

$$\lvert+\rangle = \frac{1}{\sqrt2} \left( e^{i\phi}\lvert1\rangle+\lvert2\rangle \right),$$ $$\lvert-\rangle = \frac{1}{\sqrt2} \left( -e^{i\phi}\lvert1\rangle+\lvert2\rangle \right).$$

The first-order energies are

$$E_\pm = E^{(0)} + \lambda \left( a\pm\lvert b\rvert \right) + O(\lambda^2).$$