First-Order Energy Corrections

The first-order energy correction is the linear response of an isolated energy level to a controlled change in the Hamiltonian. It is simple to evaluate,

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

but using it responsibly requires more than taking an expectation value. One must identify the parameter being varied, verify that the reference level is isolated, exploit symmetry before integrating, distinguish the coefficient from the physical shift, and estimate the terms that the linear approximation omits.

This page owns those interpretations and tests. The full order-by-order derivation is on Nondegenerate Perturbation Theory, while the formula card provides a compact lookup.

Result at a Glance

Consider a differentiable Hamiltonian family

$$\begin{aligned} H(\lambda) &= H_0+\lambda V+O(\lambda^2), \\ V &= \left. \frac{\partial H}{\partial\lambda} \right\rvert_{\lambda=0}. \end{aligned}$$

and an isolated normalized eigenstate

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

If the corresponding eigenvalue branch is differentiable at $\lambda=0$, then

$$\begin{aligned} E_n(\lambda) &= E_n^{(0)} + \lambda E_n^{(1)} + O(\lambda^2), \\ E_n^{(1)} &= \langle n^{(0)}\rvert V \lvert n^{(0)}\rangle. \end{aligned}$$

The actual linear shift is therefore

$$\Delta E_n^{\mathrm{linear}} = \lambda E_n^{(1)},$$

not $E_n^{(1)}$ by itself. If the physical Hamiltonian is written as $H=H_0+W$ with no explicit parameter, introduce $H(\lambda)=H_0+\lambda W$ for order counting and set $\lambda=1$ only after checking that the perturbative ratios are small.

Where the Formula Comes From

The first-order part of the eigenvalue equation is

$$\bigl(H_0-E_n^{(0)}\bigr) \lvert n^{(1)}\rangle = \bigl(E_n^{(1)}-V\bigr) \lvert n^{(0)}\rangle.$$

Projecting onto $\langle n^{(0)}\rvert$ annihilates the left side and leaves the diagonal matrix element of $V$. This short projection is independent of the normalization convention used for $\lvert n^{(1)}\rangle$.

The result is also the parameter derivative

$$\left. \frac{dE_n}{d\lambda} \right\rvert_{\lambda=0} = \left\langle n^{(0)} \left\rvert \left. \frac{\partial H}{\partial\lambda} \right\rvert_{\lambda=0} \right\lvert n^{(0)} \right\rangle.$$

This is the local form of the Hellmann–Feynman Theorem. The full theorem requires care with degeneracy, parameter-dependent bases, domains, and approximate states.

Physical Meaning

$E_n^{(1)}$ is the expectation value of the infinitesimal change in the Hamiltonian, evaluated in the unperturbed state. It answers a local question:

If the Hamiltonian begins to change in the direction $V$, what is the initial slope of this energy branch?

For a local potential perturbation in $d$ spatial dimensions,

$$V = V(\mathbf x),$$

the coefficient becomes

$$E_n^{(1)} = \int_{\mathbb R^d} d^d x\, \lvert\psi_n^{(0)}(\mathbf x)\rvert^2 V(\mathbf x).$$

It is therefore a probability-weighted spatial average of the perturbing potential. This interpretation does not extend unchanged to momentum-dependent, differential, spin-dependent, or nonlocal operators; for those, the operator matrix element is the primary object.

The formula is local in parameter space. It need not predict the energy accurately at finite $\lambda$, and it says nothing by itself about the convergence radius of the perturbation series.

Assumptions

Assumption	Why it matters	Warning sign
$\lvert n^{(0)}\rangle$ is normalizable	The expectation value must be defined	Continuum normalization or a resonance
$E_n^{(0)}$ is isolated	A unique eigenvalue branch must emerge from the reference level	Exact or near degeneracy
$H(\lambda)$ is differentiable at $\lambda=0$	$V$ must represent the tangent to the Hamiltonian family	A cusp, abrupt domain change, or singular limit
The relevant matrix elements exist	Unbounded operators require domain control	Divergent integrals or regulator dependence
Coupling to the complement is weak compared with gaps	The omitted state mixing must remain perturbative	$\lvert\lambda V_{mn}\rvert$ comparable to a level spacing
The same spectral branch is followed	Numerical or experimental comparisons must match the state continuously	Avoided crossings or reordered eigenvalues

The coefficient itself contains no energy denominator. The validity of truncating after it does. A small value of $E_n^{(1)}$ is not evidence that the entire perturbation is small.

What Diagonal Means

Introduce the unperturbed-basis matrix elements

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

Then

$$E_n^{(1)}=V_{nn}.$$

The word diagonal refers to the eigenbasis of $H_0$, not to whichever basis happens to be convenient. For a nondegenerate eigenvalue, the ray of $\lvert n^{(0)}\rangle$ is fixed by $H_0$, and a phase change leaves $V_{nn}$ invariant.

Several immediate checks follow.

• If $V=cI$, then every level shifts by the same coefficient $c$ and no state mixes.
• If $V$ is positive semidefinite, then $E_n^{(1)}\ge0$.
• If $V$ is negative semidefinite, then $E_n^{(1)}\le0$.
• If $v_{\min}I\le V\le v_{\max}I$ for a bounded self-adjoint $V$, then $v_{\min}\le E_n^{(1)}\le v_{\max}$.
• If $H_0$ and $V$ possess a common eigenbasis, their eigenvectors do not mix in that basis. For a strictly linear family, the corresponding energies are linear in $\lambda$ as long as the branch and common-domain assumptions remain valid.

These are structural statements. They often catch a sign, unit, or basis error before any detailed calculation.

Symmetry Tests

Suppose a unitary symmetry $U$ leaves the reference ray invariant,

$$U\lvert n^{(0)}\rangle = e^{i\phi_n} \lvert n^{(0)}\rangle,$$

and transforms the perturbation as

$$U^\dagger VU = \chi V.$$

Then

$$E_n^{(1)} = \chi E_n^{(1)}.$$

If $\chi\ne1$, the diagonal matrix element must vanish. This is the diagonal version of a selection rule.

Parity

For a nondegenerate parity eigenstate,

$$\Pi\lvert n^{(0)}\rangle = \pi_n\lvert n^{(0)}\rangle, \qquad \pi_n=\pm1,$$

and a parity-odd perturbation,

$$\Pi^\dagger V\Pi=-V,$$

one obtains

$$\begin{aligned} \langle n^{(0)}\rvert V \lvert n^{(0)}\rangle &= \langle n^{(0)}\rvert \Pi^\dagger V\Pi \lvert n^{(0)}\rangle \\ &= -\langle n^{(0)}\rvert V \lvert n^{(0)}\rangle =0. \end{aligned}$$

No integral is needed.

Symmetry breaking

A perturbation may transform nontrivially under a symmetry of $H_0$. Its expectation value then measures whether the chosen state can support the corresponding symmetry character. In a degenerate multiplet, however, individual basis vectors need not define physical branches. One must first diagonalize the perturbation within the entire multiplet.

Use Symmetry Constraints on Hamiltonians for the general symmetry framework.

A Vanishing Linear Shift Is Not No Effect

The conditions

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

and

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

are very different. The first removes only the component of $V\lvert n^{(0)}\rangle$ parallel to the reference state. If

$$Q_nV\lvert n^{(0)}\rangle \ne0, \qquad Q_n = I-\lvert n^{(0)}\rangle \langle n^{(0)}\rvert,$$

then the state generally changes at order $\lambda$, observables can change at order $\lambda$, and the energy can begin changing at order $\lambda^2$.

This distinction is central in parity-odd perturbations: the diagonal matrix element vanishes, while off-diagonal matrix elements between opposite-parity states can be nonzero.

The induced mixing, its normalization conventions, and the resulting linear changes in other observables are developed on First-Order State Corrections.

Example: A Localized Probe

Consider a one-dimensional weak point perturbation

$$V(x)=g\,\delta(x-x_0),$$

where $g$ has units of energy times length. For a normalized bound state,

$$E_n^{(1)} = g\, \lvert\psi_n^{(0)}(x_0)\rvert^2.$$

The linear shift samples the unperturbed probability density at the probe location.

For an infinite well on $0\lt x\lt L$,

$$\psi_n^{(0)}(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{n\pi x}{L}\right),$$

so

$$E_n^{(1)} = \frac{2g}{L} \sin^2\left( \frac{n\pi x_0}{L} \right).$$

A repulsive point perturbation, $g\gt0$, gives a nonnegative linear shift. If the probe sits at a node, the first-order shift vanishes. Whether the state is unaffected beyond first order must be checked from the full operator problem rather than inferred from that one number.

Example: Oscillator Perturbations

Let

$$H_0 = \frac{p^2}{2m} + \frac{1}{2}m\omega^2x^2$$

and take

$$V=\alpha x^4, \qquad \alpha\gt0.$$

Using oscillator ladder operators,

$$\langle n\rvert x^4\lvert n\rangle = 3 \left( \frac{\hbar}{2m\omega} \right)^2 \bigl(2n^2+2n+1\bigr).$$

Therefore

$$E_n^{(1)} = \frac{3\alpha\hbar^2} {4m^2\omega^2} \bigl(2n^2+2n+1\bigr).$$

The coefficient is positive, grows quadratically with $n$, and has the dimensions of energy. These facts provide three independent checks on the calculation.

For an odd perturbation such as $V=Fx$, parity instead gives

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

The oscillator is displaced at first order in its state even though its energy begins at second order. The complete model comparison is on Anharmonic Oscillator, and the linear-force benchmark appears in the chapter guide.

Example: A Simple Stark Test

An electric field couples to the electric dipole operator through

$$V = -\mathbf d\cdot\boldsymbol{\mathcal E}.$$

The dipole operator is odd under parity. A nondegenerate atomic state of definite parity therefore has

$$E_n^{(1)} = -\boldsymbol{\mathcal E} \cdot \langle n^{(0)}\rvert \mathbf d \lvert n^{(0)}\rangle =0.$$

For example, the hydrogen $1s$ state has no linear Stark shift; its leading weak-field energy response is quadratic.

This argument cannot be applied separately to arbitrary basis states in the degenerate hydrogen $n=2$ manifold. There the electric field mixes opposite-parity states within the degenerate subspace, and diagonalizing the projected perturbation produces a linear splitting. The apparent contradiction is exactly the degenerate warning, not a failure of parity.

Example: Spin in a Weak Additional Field

Take

$$H_0 = -\gamma B_0S_z, \qquad B_0\gt0,$$

and add

$$V = -\gamma\mathbf B_1\cdot\mathbf S.$$

For the unperturbed state $\lvert s,m\rangle$,

$$\begin{aligned} E_{s,m}^{(1)} &= -\gamma \langle s,m\rvert \mathbf B_1\cdot\mathbf S \lvert s,m\rangle \\ &= -\gamma\hbar m B_{1z}. \end{aligned}$$

Only the component parallel to the reference field contributes to the linear energy shift. The transverse field has zero diagonal matrix element, although it mixes magnetic sublevels.

For spin one-half, introduce the field magnitude

$$\begin{aligned} B(\lambda) &\equiv \left\lvert B_0\hat{\mathbf z} + \lambda\mathbf B_1 \right\rvert, \\ E_m(\lambda) &= -\gamma\hbar m B(\lambda). \end{aligned}$$

Its small-$\lambda$ expansion is

$$\begin{aligned} B(\lambda) &= B_0+\lambda B_{1z} \\ &\quad +\lambda^2\frac{B_{1\perp}^2}{2B_0} +O(\lambda^3). \end{aligned}$$

Here $m=\pm1/2$. The exact expansion confirms the first-order result and shows that a purely transverse field first changes the energy at second order.

If $B_0=0$, the unperturbed spin levels are degenerate and the nondegenerate formula is not the correct starting point. Use Spin in Magnetic Fields for the full magnetic dynamics.

Degenerate Warning

Let a two-dimensional degenerate eigenspace of $H_0$ have energy $E_0$, and suppose the perturbation restricted to it is

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

The first-order coefficients are not generally the two diagonal entries in an arbitrarily chosen basis. They are the eigenvalues of $W$:

$$w_\pm = \frac{a+d}{2} \pm \sqrt{ \left( \frac{a-d}{2} \right)^2 + \lvert b\rvert^2 }.$$

The perturbed energies begin as

$$E_\pm(\lambda) = E_0 + \lambda w_\pm + O(\lambda^2).$$

This result is basis independent. The diagonal entries $a$ and $d$ are not.

For a nearly degenerate pair, the same subspace treatment is needed whenever the perturbative coupling is comparable to the small internal splitting. See Degenerate Perturbation Theory.

Estimating What Was Omitted

Let

$$\Delta_n = \inf_{m\ne n} \left\lvert E_n^{(0)}-E_m^{(0)} \right\rvert$$

denote the spectral gap from the target level in a discrete problem. A useful mixing diagnostic is

$$\eta_n = \frac{ \lvert\lambda\rvert \left\lVert Q_nV\lvert n^{(0)}\rangle \right\rVert } {\Delta_n}.$$

When $\eta_n\ll1$, the first-order state deformation is plausibly small. This is a diagnostic, not a universal error theorem.

When an unperturbed basis is available, the scale

$$\mathcal R_n^{(2)} = \lambda^2 \sum_{m\ne n} \frac{ \lvert V_{mn}\rvert^2 }{ \left\lvert E_n^{(0)}-E_m^{(0)} \right\rvert }$$

indicates the possible size of the leading omitted energy contribution. Cancellations may make the signed second-order correction smaller, while dense spectra, continuum states, or unbounded operators require more careful estimates.

The key comparison is not merely

$$\lvert\lambda E_n^{(1)}\rvert \ll \lvert E_n^{(0)}\rvert.$$

Energy zeros are conventional, and a small diagonal element can coexist with strong off-diagonal mixing. Compare couplings with gaps and test the result against a second order, exact limit, variational bound, or converged numerical calculation.

Practical Workflow

1. Specify the family $H(\lambda)$ and identify $V=\partial_\lambda H\rvert_0$.
2. Confirm that the target level is isolated, or enlarge the model subspace.
3. Normalize the reference state and verify that the matrix element exists.
4. Apply symmetry, positivity, dimensional, and sign checks before integrating.
5. Evaluate $E_n^{(1)}=\langle n^{(0)}\rvert V\lvert n^{(0)}\rangle$.
6. Restore the factor of $\lambda$ when reporting the physical shift.
7. Estimate coupling-to-gap ratios and the leading omitted scale.
8. Track the same eigenvalue branch in any exact, numerical, or experimental comparison.

Common Mistakes

• Reporting $E_n^{(1)}$ as the physical shift while omitting the factor of $\lambda$.
• Evaluating a diagonal entry in a convenient basis rather than in an isolated eigenstate of $H_0$.
• Applying the nondegenerate formula separately to states inside a degenerate multiplet.
• Concluding that $E_n^{(1)}=0$ means the state and all observables are unchanged.
• Using parity without checking that both the state and perturbation have the required transformation properties.
• Judging validity from the diagonal matrix element while ignoring off-diagonal coupling and small gaps.
• Treating a singular or domain-changing perturbation as an ordinary bounded operator without justification.
• Comparing sorted numerical eigenvalues across an avoided crossing instead of following eigenvector overlap.

Exercises

Positivity and a uniform offset

Let

$$V=cI+A^\dagger A.$$

Show that $E_n^{(1)}\ge c$ for every normalized reference state. When is equality attained?

Solution

The correction is

$$\begin{aligned} E_n^{(1)} &= c + \langle n^{(0)}\rvert A^\dagger A \lvert n^{(0)}\rangle \\ &= c + \left\lVert A\lvert n^{(0)}\rangle \right\rVert^2 \ge c. \end{aligned}$$

Equality holds precisely when $A\lvert n^{(0)}\rangle=0$.

Point probe in an infinite well

For

$$V(x)=g\,\delta(x-x_0)$$

in an infinite well on $0\lt x\lt L$, derive $E_n^{(1)}$ and find the interior probe positions at which it vanishes.

Solution

Using

$$\psi_n(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{n\pi x}{L}\right),$$

gives

$$E_n^{(1)} = \frac{2g}{L} \sin^2\left( \frac{n\pi x_0}{L} \right).$$

It vanishes at the interior nodes

$$x_0 = \frac{jL}{n}, \qquad j=1,\ldots,n-1.$$

The ground state has no interior node, so a nonzero interior point perturbation has a nonzero first-order shift for $n=1$.

Odd oscillator perturbation

For a harmonic oscillator with $V=Fx$, show that $E_n^{(1)}=0$ and identify which unperturbed states can mix at first order.

Solution

Every oscillator eigenstate has definite parity, while $x$ is parity odd. Therefore

$$\langle n\rvert x\lvert n\rangle=0.$$

Using

$$x = \sqrt{\frac{\hbar}{2m\omega}} \bigl(a+a^\dagger\bigr)$$

shows that $x\lvert n\rangle$ contains only $\lvert n-1\rangle$ and $\lvert n+1\rangle$. Thus the diagonal energy response vanishes while the state mixes with its nearest opposite-parity neighbors.

Tilted magnetic field

For spin one-half, let

$$H(\lambda) = -\gamma \left[ B_0\hat{\mathbf z} + \lambda \bigl( B_\parallel\hat{\mathbf z} + B_\perp\hat{\mathbf x} \bigr) \right]\cdot\mathbf S.$$

Find the first-order energy correction and compare it with the expansion of the exact eigenvalues.

Solution

In an $S_z$ eigenstate,

$$\langle m\rvert S_x\lvert m\rangle=0, \qquad \langle m\rvert S_z\lvert m\rangle=\hbar m.$$

Hence

$$E_m^{(1)} = -\gamma\hbar m B_\parallel.$$

The exact eigenvalues are

$$\begin{aligned} \mathcal B(\lambda) &= \sqrt{ \bigl(B_0+\lambda B_\parallel\bigr)^2 +\lambda^2B_\perp^2 }, \\ E_m(\lambda) &= -\gamma\hbar m\mathcal B(\lambda). \end{aligned}$$

For $B_0\gt0$,

$$\begin{aligned} E_m(\lambda) &= -\gamma\hbar m \Bigl[ B_0+\lambda B_\parallel \\ &\qquad +\lambda^2\frac{B_\perp^2}{2B_0} +O(\lambda^3) \Bigr]. \end{aligned}$$

The linear terms agree. The transverse field first enters the energy at second order.

Basis dependence in a degenerate subspace

Suppose

$$H_0=E_0I, \qquad V = \begin{pmatrix} 0 & v \\ v^* & 0 \end{pmatrix}$$

on a two-dimensional subspace. Explain why the two zero diagonal entries are not the first-order shifts and find the correct shifts.

Solution

The reference energy is degenerate, so its basis vectors do not define unique eigenvalue branches. The eigenvalues of the projected perturbation are

$$w_\pm=\pm\lvert v\rvert.$$

Therefore

$$E_\pm(\lambda) = E_0 \pm \lambda\lvert v\rvert + O(\lambda^2).$$

The zero diagonal entries are basis dependent; the eigenvalues of the full projected matrix are invariant.

Cross-Links

• Time-Independent Perturbation Theory
• Nondegenerate Perturbation Theory
• First-Order State Corrections
• Second-Order Energy Corrections
• Hellmann–Feynman Theorem
• Degenerate Perturbation Theory
• Small Parameters and Error Estimates
• Common Failure Modes
• Anharmonic Oscillator
• Symmetry Constraints on Hamiltonians
• Spin in Magnetic Fields
• First-Order Perturbation Theory Formula Card

References

• T. Kato, Perturbation Theory for Linear Operators, 2nd ed., Springer, 1976.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Vol. 2, Wiley, 1977.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• A. Messiah, Quantum Mechanics, Vol. 2, North-Holland, 1962.
• B. Zwiebach, Quantum Physics III, MIT OpenCourseWare 8.06, Chapter 1, 2018.