First-Order Energy and State Corrections

The previous article built the tower; this one climbs its first floor — and the first floor is where most of practical quantum mechanics lives. Two formulas are derived, interpreted, and spent here. The energy: $E_n^{(1)}=\langle n^{(0)}|\hat{H}'|n^{(0)}\rangle$ — the perturbation averaged over the unperturbed state, a one-integral answer requiring no knowledge of any other level. The state: $|n^{(1)}\rangle=\sum_{m\ne n}\frac{H'_{mn}}{E_n^{(0)}-E_m^{(0)}}|m^{(0)}\rangle$ — the leading lean of the eigenstate toward its neighbors, weighted by coupling over gap. The energy formula's slogan — first do no mixing — is the article's conceptual spine: to leading order the system does not rearrange itself; it pays (or collects) the perturbation's bill as it stands. That is why symmetry can silence the formula: a perturbation odd where the density is even averages to zero, and the calculation announces its own inadequacy, handing the stage to second order. The mixing formula, meanwhile, explains who participates when rearrangement does begin: near levels and strongly coupled levels dominate, in exact proportion. Everything is exercised on computable systems and machine-verified. The slanted-floor well shifts every level by exactly $\varepsilon/2$ (and a numerical diagonalization confirms the residual error scales as $\varepsilon^2$ , ratio $4.00$ on doubling). The delta-bumped well shifts odd levels by $2\alpha/L$ and even levels — which hold a node exactly on the bump — by strictly zero. The oscillator in a uniform field gives $E^{(1)}=0$ by parity: the null result that hands the next serial its cleanest customer. The quartic oscillator yields $E^{(1)}_n=\tfrac{3\gamma}{4}(\hbar/m\omega)^2(2n^2+2n+1)$ , verified against exact diagonalization so precisely that the leftover is the known second-order coefficient, term for term. And the Hellmann–Feynman theorem — first-order perturbation theory's differential alter ego, and a 21-year-old's undergraduate thesis — is proven and verified exactly on the two-level model.

Key takeaways

THE energy formula: $E_n^{(1)}=\langle n^{(0)}|\hat{H}'|n^{(0)}\rangle$ — the perturbation's mean over the unperturbed density; one integral, no other levels needed.
THE state formula: $|n^{(1)}\rangle=\sum_{m\ne n}\dfrac{H'_{mn}}{E_n^{(0)}-E_m^{(0)}}|m^{(0)}\rangle$ — mixing weights = coupling ÷ gap; near neighbors and strong couplers dominate; lower levels push up, upper push down.
Symmetry zeros are features: parity-odd $\hat{H}'$ on a parity-even density ⇒ $E^{(1)}=0$ — the calculation tells you the story starts at second order (owner named).
Slanted-floor well ( $H'=\varepsilon x/L$ ): $E^{(1)}_n=\varepsilon/2$ for every $n$ (symbolic ×3) — uniform shifts move spectra, not spectroscopy; grid check: residual $\propto\varepsilon^2$ (ratio 4.00 on doubling $\varepsilon$ ).
Delta bump at center: $E^{(1)}_n=2\alpha/L$ (odd $n$ ), $0$ (even $n$ — node on the bump); mixing $c_3=\alpha/(4LE_1)$ , $c_5=-\alpha/(12LE_1)$ (computed).
Quartic oscillator: $\langle n|\hat{x}^4|n\rangle=\tfrac34(2n^2{+}2n{+}1)(\hbar/m\omega)^2$ (symbolic $n\le3$ ); vs exact diagonalization the residual equals the known $E^{(2)}$ coefficient to 10% at $\gamma=0.002$ — first order verified through its own error term.
Hellmann–Feynman: $\dfrac{dE}{d\lambda}=\langle\psi_\lambda|\dfrac{\partial\hat{H}}{\partial\lambda}|\psi_\lambda\rangle$ — proven (normalization derivative cancels), verified exactly on the two-level model; at $\lambda=0$ it is the first-order formula.
Key idea: at first order the state stands still and the energy takes the perturbation's average — mixing begins at the same order in the wavefunction, one floor below where it shows in the energy.

The core insight

First order charges each level the unperturbed-state average of the perturbation ( $E^{(1)}=\langle\hat{H}'\rangle_0$ ) while leaning the state toward other levels with weights $H'_{mn}/(E_n^{(0)}-E_m^{(0)})$ — energy shifts without rearrangement, rearrangement without yet costing energy, and symmetry zeros in the average as the signal that second order owns the physics.

Where this lives today

First-order perturbation theory is the default first calculation of working physics: every "estimate the shift" in atomic clocks (blackbody-radiation shifts on optical transitions are first-order averages over the clock states), every nuclear-structure single-particle estimate, every solid-state band-shift-under-strain calculation begins at equation (2). Hellmann–Feynman forces power the multi-billion-dollar ab-initio simulation industry — molecular dynamics with quantum forces, from drug binding to battery electrolytes, differentiates one wavefunction instead of finite-differencing many. And the sensor-placement physics of the delta-bump example is engineering practice in cavity QED and nanomechanics: to make a qubit or a mechanical mode insensitive to a known disturbance, park its node on the disturbance — first-order immunity by geometry, exactly the even states' trick.

The physical picture

Picture the unperturbed probability cloud $|\psi_n^{(0)}|^2$ frozen in place, and the perturbation as a new landscape laid under it. First order is the rent audit: charge each patch of the cloud the local value of $\hat{H}'$ and total the bill. No motion, no adaptation — the cloud pays as it lies. That is the entire content of $E^{(1)}=\langle\hat{H}'\rangle_0$ , and it explains the formula's two signature behaviors: a perturbation localized where the cloud is thin costs little (the delta bump under a node costs nothing), and a perturbation antisymmetric across a symmetric cloud nets to zero.

Figure

First order as a rent audit: the frozen cloud pays the new landscape as it lies

The state's first-order story is different, and keeping the two separate is half the pedagogy of this chapter. While the energy needs no rearrangement at first order, the wavefunction is already leaning: $|n^{(1)}\rangle$ mixes in other levels with weight coupling-over-gap. The lean is invisible in the energy at this order (that is a theorem — the cross terms cancel), but perfectly visible in everything else: densities shift, transition amplitudes open, selection rules soften. Energy at order $k$ , state at order $k-1$ : the tower's staggered bookkeeping, met here in its first instance.

The two first-order formulas

Take the tower's order-one equation from Non-Degenerate Time-Independent Perturbation Theory (its hierarchy, consumed — never re-derived):

(\hat{H}_0-E_n^{(0)})|n^{(1)}\rangle=(E_n^{(1)}-\hat{H}')|n^{(0)}\rangle \qquad (1)

where: the left side lives entirely in the orthogonal complement of $|n^{(0)}\rangle$ (intermediate normalization, consumed), so the equation splits cleanly under projection: the component along $\langle n^{(0)}|$ fixes the energy, the components along every $\langle m^{(0)}|$ fix the state. Two projections, two formulas — the whole derivation is bookkeeping honesty.

Derivation AProjecting out the energy and the mixing coefficients

We extract both first-order results from equation (1) by projection, using hermiticity where it earns its keep, and prove the reading that gives the energy formula its power: no property of any other level enters $E^{(1)}$ .

Step 1. The energy. Apply $\langle n^{(0)}|$ to (1). The left side is $\langle n^{(0)}|(\hat{H}_0-E_n^{(0)})|n^{(1)}\rangle$ — and here hermiticity acts (Hermitian Operators and Real Eigenvalues' grammar, one line): let $\hat{H}_0$ act leftward on its own eigenbra, giving $(E_n^{(0)}-E_n^{(0)})\langle n^{(0)}|n^{(1)}\rangle=0$ identically. The right side survives as $E_n^{(1)}-\langle n^{(0)}|\hat{H}'|n^{(0)}\rangle$ , whence

E_n^{(1)}=\langle n^{(0)}|\hat{H}'|n^{(0)}\rangle \qquad (2)

Why this is remarkable and not routine: the formula contains no sum, no gap, no other state — the leading energy response of level $n$ is computable knowing only level $n$ . It is also exactly the expectation value a variational evaluation with the frozen state would give, which is why $E^{(0)}+E^{(1)}$ always sits above the true ground energy (the variational reading, one line; the bound's own serial is ahead).

Step 2. The state. Apply $\langle m^{(0)}|$ , $m\ne n$ , to (1): the left side gives $(E_m^{(0)}-E_n^{(0)})\langle m^{(0)}|n^{(1)}\rangle$ — no vanishing this time, and invertible precisely because the level is non-degenerate (the divorce clause, cashed). The right side gives $-H'_{mn}$ . Solve and assemble:

|n^{(1)}\rangle=\sum_{m\ne n}\frac{H'_{mn}}{E_n^{(0)}-E_m^{(0)}}\,|m^{(0)}\rangle \qquad (3)

Why the anatomy is worth memorizing: numerator = how strongly the perturbation connects the two states; denominator = how much energy separates them. Distant levels resist mixing in exact proportion to their distance; levels below $n$ (positive denominator) mix in with the sign of $H'_{mn}$ , levels above with the opposite — the directional lean that becomes level repulsion when the energy bill arrives at second order (owner named). And the formula's failure mode is visible on its face: a vanishing denominator with nonvanishing numerator is the degenerate catastrophe, deferred BY NAME to Degenerate Perturbation Theory.

Step 3. The staggered theorem, made explicit: insert $|n\rangle\approx|n^{(0)}\rangle+\lambda|n^{(1)}\rangle$ into $\langle\hat{H}\rangle$ and expand — the terms linear in $|n^{(1)}\rangle$ cancel by exactly the Step-1 mechanism. The mixing (3) costs energy only at $\mathcal{O}(\lambda^2)$ : the wavefunction leans for free at first order. This is the $k$ -versus- $k{-}1$ bookkeeping of the tower, and its general form (" $2n+1$ rule") is stated with its owner in the next serial. Verified: (2) and (3) evaluated on the worked systems below — slanted floor, delta bump (coefficients $c_3=\alpha/4LE_1$ , $c_5=-\alpha/12LE_1$ computed), quartic and linear oscillator — each against symbolic integrals and/or exact numeric diagonalization (article verifier).

Go deeper · Derivation B · 1 minThe Hellmann–Feynman theorem: first order as a derivative

Derivation BThe Hellmann–Feynman theorem: first order as a derivative

We prove the theorem that repackages equation (2) as an exact statement about any parameter of any Hamiltonian, verify it symbolically on the two-level model, and place it honestly: an exact identity whose $\lambda\to0$ instance is first-order perturbation theory. (The theorem is a gap-queue topic in its own right — Feynman's undergraduate thesis made it the working tool of molecular forces — so this card proves and uses it without claiming its full applications story.)

Step 1. Let $\hat{H}(\lambda)$ depend smoothly on a parameter, with normalized eigenstate $|\psi_\lambda\rangle$ and eigenvalue $E(\lambda)$ . Differentiate $E=\langle\psi_\lambda|\hat{H}(\lambda)|\psi_\lambda\rangle$ :

\frac{dE}{d\lambda}=\Big\langle\frac{\partial\psi}{\partial\lambda}\Big|\hat{H}\Big|\psi\Big\rangle+\Big\langle\psi\Big|\hat{H}\Big|\frac{\partial\psi}{\partial\lambda}\Big\rangle+\Big\langle\psi\Big|\frac{\partial\hat{H}}{\partial\lambda}\Big|\psi\Big\rangle \qquad (4)

Step 2. The first two terms are $E(\lambda)\frac{d}{d\lambda}\langle\psi|\psi\rangle$ (let $\hat{H}$ act on its own eigenvector in each) — and normalization is constant, so its derivative vanishes. What survives is the theorem:

\frac{dE}{d\lambda}=\Big\langle\psi_\lambda\Big|\frac{\partial\hat{H}}{\partial\lambda}\Big|\psi_\lambda\Big\rangle \qquad (5)

Why it is first-order PT in disguise — and more: at $\lambda=0$ , with $\partial\hat{H}/\partial\lambda=\hat{H}'$ , equation (5) reads $dE/d\lambda|_0=\langle n^{(0)}|\hat{H}'|n^{(0)}\rangle$ — formula (2) verbatim. But (5) holds at every $\lambda$ , evaluated on the exact state: it is a nonperturbative identity, the reason one can extract forces (derivatives of energy with respect to nuclear positions) from a single electronic wavefunction without re-solving anything. The subtlety worth one honest line: for approximate wavefunctions the cancellation in Step 2 fails unless the approximation is variationally optimal — the practical fine print of every quantum-chemistry force computation.

Step 3. Machine check where everything is exact: on the two-level model, differentiate the exact $E_-(\lambda)$ and separately compute $\langle\psi_\lambda|\partial_\lambda\hat{H}|\psi_\lambda\rangle$ with the exact normalized eigenvector — equal, symbolically, at all $\lambda$ (computed). Not just at the origin: the theorem's full nonperturbative strength, exhibited on one matrix. Verified: (5) on the two-level model, exact and symbolic (article verifier).

Figure

Coupling over gap: who leans into whom, and with what sign

Limiting cases

case	result	meaning
$\hat{H}'=$ const	$E^{(1)}=$ const, $\|n^{(1)}\rangle=0$	pure gauge: energies shift, physics doesn't
$\hat{H}'$ diagonal in the $\hat{H}_0$ basis	(2) exact, (3) zero, series terminates	shared eigenbasis: perturbation theory gratuitous
symmetry-odd $\hat{H}'$	$E^{(1)}=0$	the null that delegates to second order
node on the perturbation's support	that level immune at first order	the bump's even states; sensor-placement physics
gap $\to\infty$ for all partners	$\|n^{(1)}\rangle\to0$ , $E^{(0)}+E^{(1)}$ nearly exact	isolated levels are perturbatively rigid
$\lambda$ -derivative at finite $\lambda$	Hellmann–Feynman (5)	first order's exact, nonperturbative extension

Common trapA zero first-order shift is information, not failure — and the state correction is not optional

Two chronic misreadings. First: when symmetry forces $E^{(1)}=0$ (oscillator in a uniform field; any parity-odd perturbation on a parity-eigenstate), students report "no effect." Wrong — the energy effect merely starts one order later, and the state is already mixed at first order: the field-displaced oscillator polarizes immediately even though its energy responds quadratically. Read $E^{(1)}=0$ as a referral note to the next serial, never as a verdict of immunity. Second: truncating at "energies to first order" while silently using unperturbed wavefunctions for everything else is inconsistent bookkeeping — transition rates, dipole moments, and densities all change at the same order as the energy does, through (3). The staggered rule is precise: energy to order $k$ requires the state to order $k-1$ , so first-order energies pair with zeroth-order states only for the energy itself — any other observable computed at "first order" needs $|n^{(1)}\rangle$ .

Worked examples

Example 1 — the slanted floor, symbolic and numeric. Tilt the infinite well: $H'=\varepsilon x/L$ on $0<x<L$ .

Step 1. The average: $\langle n|x|n\rangle=L/2$ for every ISW state (each density is symmetric about the center — symbolic for $n=1,2,3$ ), so $E^{(1)}_n=\varepsilon/2$ uniformly: the whole ladder rises by the tilt's mean, and every gap — everything spectroscopy measures — is untouched at first order.

Step 2. Audit against truth: exact grid diagonalization ( $N=2000$ ) versus $E_n^{(0)}+\varepsilon/2$ . Residuals at $\varepsilon=0.5$ : $2.7\times10^{-4}$ , $8.2\times10^{-5}$ , $4.9\times10^{-5}$ for $n=1,2,3$ (computed) — and doubling $\varepsilon$ multiplies the ground-state residual by $4.00$ (computed): the leftover is second order, cleanly.

Step 3. Read the trend in $n$ : residuals shrink up the ladder — higher levels have larger gaps and mix less, formula (3)'s denominators in action. Answer: $E^{(1)}_n=\varepsilon/2$ exactly, all $n$ ; the neglected piece measured and scaling as $\varepsilon^2$ — first order behaving precisely as advertised.

Example 2 — the quartic oscillator, verified through its own error. Stiffen the oscillator: $H'=\gamma\hat{x}^4$ .

Step 1. The matrix element, by ladder machinery (Ladder Operators and the Algebraic Oscillator's $\hat{x}=(\hat{a}+\hat{a}^\dagger)\sqrt{\hbar/2m\omega}$ , consumed): expanding $\hat{x}^4$ and keeping number-conserving strings gives $\langle n|\hat{x}^4|n\rangle=\tfrac34(2n^2+2n+1)(\hbar/m\omega)^2$ — verified symbolically by explicit Hermite integrals for $n\le3$ (values $\tfrac34,\tfrac{15}4,\tfrac{39}4,\tfrac{75}4$ in natural units). So $E^{(1)}_n=\tfrac{3\gamma}{4}\big(\tfrac{\hbar}{m\omega}\big)^2(2n^2+2n+1)$ : the stiffening grows quadratically up the ladder — anharmonicity bites excited states hardest, the structural fact behind every vibrational-spectroscopy anharmonic constant.

Step 2. Audit at $\gamma=0.002$ (natural units) against exact diagonalization in a 60-state basis: first-order predictions miss by $1.0\times10^{-5}$ , $8.1\times10^{-5}$ , $3.0\times10^{-4}$ for $n=0,1,2$ (computed).

Step 3. Identify the miss: the known second-order coefficient for the quartic problem gives $|E^{(2)}_n|=\tfrac{\gamma^2}{8}(34n^3+51n^2+59n+21)$ — numerically $1.05\times10^{-5}$ , $8.25\times10^{-5}$ , $3.07\times10^{-4}$ : the residuals match the next order's prediction to within 10% (computed). Answer: $E^{(1)}_n=\tfrac{3\gamma}{4}(\hbar/m\omega)^2(2n^2+2n+1)$ , verified so tightly that its error is recognizably the second-order term — the tower auditing itself floor by floor. (For the record: the quartic series is the textbook divergent asymptotic series — the previous article's honesty card in its most famous costume; low orders remain superbly accurate at small $\gamma$ , as just measured.)

Origins · 1 minRichard Feynman and how this was discovered

Richard Feynman

1918–1988 · MIT, then Princeton

Before the path integrals (that revolution has its own portrait at this track's time-dependent Schrödinger article — nothing of it here), Feynman's first publication was this article's third derivation. As an MIT senior in 1939, assigned a thesis by John Slater, the 21-year-old asked a question of almost insolent practicality: molecular physicists computed energies by heroic integrals, then re-computed them at displaced nuclei to get forces — why not differentiate once and be done? His thesis, "Forces in Molecules," proved equation (5): the force on a nucleus is just the classical electrostatic pull of the quantum charge cloud, evaluated on one wavefunction — no second calculation, no perturbative apparatus, provided the wavefunction is exact (or variationally honest — he flagged the fine print too). Priority is shared, and the name says so: Hans Hellmann, a German quantum chemist exiled to Moscow, had published the theorem in his 1937 textbook — and was arrested and executed in Stalin's purges in 1938, at 34, months before Feynman's thesis; the double name is a small act of historical justice. The theorem became the engine of computational chemistry (every molecular-dynamics force in a quantum simulation is a Hellmann–Feynman evaluation) — and its author went on to Los Alamos, Cornell, Caltech, and the other portrait. The habit visible already at 21: refuse the standard hard way if an exact identity offers a shortcut.

History

1926

— Schrödinger's third paper computes first-order shifts by the projection formula

1937

— Hellmann's Einführung in die Quantenchemie states the derivative theorem

1939

— Feynman's MIT senior thesis proves it independently: Forces in Molecules

1951

— Ehrenfest-style force computations mature into quantum chemistry's standard machinery

1985

— Car and Parrinello build ab-initio molecular dynamics on Hellmann–Feynman forces

Going further

Griffiths & Schroeter Ch. 7 — the first-order formulas at the source register, with the well examples.
Sakurai & Napolitano Ch. 5 — the projection derivation and the staggered bookkeeping.
Shankar Ch. 17 — careful state-correction treatment.
Feynman, Phys. Rev. 56, 340 (1939) — the undergraduate thesis; still readable in an afternoon.

Provenance note: All formulas are the standard non-degenerate first-order results derivable from $H=H_0+\lambda H'$ ; a verifier should recompute the order- $\lambda^1$ projection (Griffiths Ch. 7, Sakurai Ch. 5, Shankar Ch. 17). Numbers cited: the oscillator matrix element $\langle 0|x^4|0\rangle=3(\hbar/2m\omega)^2$ (verify via ladder operators) and the weak-field Zeeman splitting $\Delta E=g_J\mu_B B m_J$ (Cohen-Tannoudji Vol. II). The Hellmann–Feynman cross-check follows Feynman (Phys. Rev. 56, 340, 1939). — The projection derivation is executed against exactly solvable systems in this article's verify script (slanted floor symbolic ×3 + grid diagonalization with $\varepsilon^2$ residual scaling; delta bump shifts and mixing coefficients; quartic $\langle x^4\rangle$ symbolic $n\le3$ — the quoted $\langle0|x^4|0\rangle=3(\hbar/2m\omega)^2$ is the $n=0$ row — plus diagonalization residuals matching the second-order coefficient to 10%; oscillator parity nulls; Hellmann–Feynman exact on the two-level model). The quoted weak-field Zeeman splitting $g_J\mu_BBm_J$ is an application owned BY NAME by the Zeeman serials (Weak-Field (Anomalous) Zeeman Effect, with the Landé machinery at its own serial) — stated for the record, neither derived nor verified here per the chapter's fine-structure bank. Edition note: the Hellmann–Feynman theorem as a standalone topic (molecular forces, electrostatic theorem) is a gap-queue candidate; this article proves the identity and its perturbative reading only.

Why does the energy formula need no other levels when the state formula needs all of them?

Because the two questions probe different components of equation (1). The energy is the equation's component along the unperturbed state — and there the operator $(\hat{H}_0-E^{(0)})$ dies identically, leaving pure bookkeeping. The state correction is the component perpendicular — where the operator is invertible and its inverse manufactures the gap denominators. One equation, two projections, two very different information requirements. (At second order the energy inherits the state's sum — the staggered rule.)

Is the first-order energy an upper or lower bound on anything?

$E^{(0)}+\lambda E^{(1)}$ is the expectation of the full $\hat{H}$ in the unperturbed state — a variational evaluation — so for the ground state it is a rigorous upper bound on the true energy. The second-order correction (always negative for the ground state, the next serial's theorem) is the first installment of the repair. For excited states no general bound holds without extra structure (orthogonality constraints — the variational chapter's business).

What happens to formula (3) if the spectrum has a continuum?

The sum becomes a sum-plus-integral: $\sum_m\to\sum_{\text{discrete}}+\int dE\,\rho(E)$ over continuum states with the same matrix-element-over-gap weight (δ-normalized continuum states, the free-particle chapter's grammar). Nothing conceptual changes at first order; the integrals converge whenever the coupling to the continuum falls off. Second order is where continua bite harder (level pushed by an entire integral of partners) — flagged there.

Does E⁽¹⁾ = 0 by symmetry survive to all odd orders?

Often, and the two-level model shows the mechanism: with a purely off-diagonal perturbation, every odd-order energy vanished (verified through fifth order in the previous article). The general statement: if a symmetry operator commutes with $\hat{H}_0$ , anticommutes-in-effect with $\hat{H}'$ (odd parity), and the state is a symmetry eigenstate, then odd powers of $\hat{H}'$ average to zero between like-parity states. Selection rules organize entire series, not just first terms — one of the quiet payoffs of the symmetry chapter.

How accurate is first order in practice — when do I need more?

Audit the ratios: the neglected relative error is of order $|H'_{mn}/(E_n-E_m)|$ for the dominant partner. In this article's examples: slanted floor at $\varepsilon=0.5$ — residual $10^{-3}$ -level; quartic at $\gamma=0.002$ — residual $10^{-5}$ -level, both measured. Real atomic applications (fine structure, Zeeman at laboratory fields) sit at ratios $10^{-3}$ – $10^{-6}$ : first order is often beyond experimental resolution. Need more when: ratios exceed a few percent, when $E^{(1)}=0$ hands you to second order, or when a gap closes — degeneracy's territory.

The mixing coefficients have signs — do they mean anything measurable?

Yes: interference. The perturbed density $|\psi^{(0)}+\lambda\psi^{(1)}|^2$ contains cross terms linear in each mixing coefficient, so the sign determines whether density piles up on the perturbation or flees it. For the delta bump, $c_3>0$ mixes the third state in-phase near the center in just the way that reduces density at the bump — the cloud dodging the toll, at order $\lambda$ . Dipole moments induced by fields (polarization) are exactly such sign-carrying first-order interference terms; the next serial's exact oscillator case makes the dodge quantitative.