Hellmann–Feynman Theorem

The Hellmann–Feynman theorem converts the derivative of an exact energy eigenvalue into the expectation value of the Hamiltonian’s parameter derivative. It is simultaneously a spectral theorem, the differential form of first-order perturbation theory, and the foundation for interpreting energy gradients as generalized forces.

Its familiar one-line formula is easy to remember:

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

Using it correctly is less automatic. One must identify a differentiable eigenvalue branch, control degeneracy, distinguish exact from approximate states, include parameter dependence in bases and domains, and keep the operator derivative separate from the total state response.

The theorem is commonly associated with Hellmann’s 1937 quantum-chemistry text and Feynman’s 1939 analysis of molecular forces. The compound name records the standard modern attribution; the mathematical idea also has earlier perturbative antecedents.

Statement

Let $H(\lambda)$ be a differentiable family of self-adjoint operators on a Hilbert space. Suppose that, on an interval of $\lambda$, there is a normalized differentiable eigenbranch

$$H(\lambda)\lvert n(\lambda)\rangle = E_n(\lambda)\lvert n(\lambda)\rangle,$$

with

$$\langle n(\lambda)\vert n(\lambda)\rangle=1.$$

If the operator derivative is meaningful on the branch, then

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

For several real parameters $\mathbf R=(R_1,\ldots,R_k)$, the componentwise form is

$$\frac{\partial E_n}{\partial R_i} = \left\langle n\left\vert \frac{\partial H}{\partial R_i} \right\vert n\right\rangle.$$

Equivalently,

$$\nabla_{\mathbf R}E_n = \left\langle n\left\vert \nabla_{\mathbf R}H \right\vert n\right\rangle,$$

provided the same differentiability and domain assumptions hold in each parameter direction.

Assumptions

The textbook statement packages several hypotheses.

Exact eigenstate

The state must satisfy the eigenvalue equation for the Hamiltonian being differentiated. For a generic normalized trial state, an additional residual-response term appears.

Differentiable branch

$E_n(\lambda)$ and a representative of $\lvert n(\lambda)\rangle$ must be differentiable locally. A sorted list of eigenvalues can develop cusps at crossings even when smooth branch labels exist.

Spectral isolation or resolved degeneracy

A simple isolated eigenvalue has a locally well-defined branch under standard perturbative conditions. At a degeneracy, the derivative is obtained only after resolving the perturbation inside the degenerate subspace.

Fixed Hilbert-space identification

States at nearby parameter values must be compared in one fixed Hilbert space, or by an explicitly specified unitary identification. If a boundary, integration measure, or operator domain moves with $\lambda$, its contribution cannot be silently discarded.

Differentiable operator or quadratic form

For bounded matrices, entrywise differentiability is enough. For unbounded Hamiltonians, one commonly assumes a common dense domain or works with differentiable quadratic forms. Kato’s analytic perturbation theory gives rigorous frameworks for isolated spectral branches.

Derivation from the Eigenvalue Equation

Suppress the branch label and write a prime for $d/d\lambda$. Differentiate

$$H\lvert n\rangle=E_n\lvert n\rangle$$

to obtain

$$H'\lvert n\rangle + H\lvert n'\rangle = E_n'\lvert n\rangle + E_n\lvert n'\rangle.$$

Left-multiply by $\langle n\rvert$:

$$\begin{aligned} \langle n\rvert H'\lvert n\rangle &+ \langle n\rvert H\lvert n'\rangle \\ &= E_n'\langle n\vert n\rangle + E_n\langle n\vert n'\rangle. \end{aligned}$$

Self-adjointness and the eigenvalue equation imply

$$\langle n\rvert H = E_n\langle n\rvert.$$

The terms involving $\lvert n'\rangle$ therefore cancel. Since the state is normalized,

$$E_n' = \langle n\rvert H'\lvert n\rangle.$$

The cancellation is the theorem’s essential content. The state generally changes with $\lambda$, but its derivative is unnecessary for the first derivative of an exact eigenvalue.

Derivation from the Rayleigh Quotient

For a normalized exact eigenstate,

$$E_n(\lambda) = \langle n(\lambda)\rvert H(\lambda) \lvert n(\lambda)\rangle.$$

Differentiation gives

$$\begin{aligned} E_n' ={}& \langle n'\rvert H\lvert n\rangle + \langle n\rvert H'\lvert n\rangle \\ &+ \langle n\rvert H\lvert n'\rangle. \end{aligned}$$

Using $H\lvert n\rangle=E_n\lvert n\rangle$ on both state-response terms,

$$\begin{aligned} E_n' ={}& \langle n\rvert H'\lvert n\rangle \\ &+ E_n \frac{d}{d\lambda} \langle n\vert n\rangle. \end{aligned}$$

The norm derivative vanishes, leaving the theorem. This proof also shows why normalization and exact stationarity matter when the state is approximate.

Phase Choice and State Derivatives

The eigenstate may be rephased:

$$\lvert n(\lambda)\rangle \longrightarrow e^{i\alpha(\lambda)} \lvert n(\lambda)\rangle.$$

This changes $\lvert n'\rangle$ by a component parallel to $\lvert n\rangle$, but it does not change $\langle n\vert H'\vert n\rangle$. The energy derivative is gauge invariant.

For a distinct nondegenerate eigenstate $\lvert m\rangle$, project the differentiated eigenvalue equation with $\langle m\rvert$:

$$\langle m\vert H'\vert n\rangle + E_m\langle m\vert n'\rangle = E_n\langle m\vert n'\rangle.$$

Thus

$$\langle m\vert n'\rangle = \frac{ \langle m\vert H'\vert n\rangle }{ E_n-E_m }, \qquad m\ne n.$$

The diagonal theorem determines an energy derivative; this off-diagonal identity determines the physically relevant complementary part of the state derivative. The parallel part remains a phase convention and underlies geometric connections such as the Berry connection.

Generalized Forces

If $\lambda$ is a generalized coordinate, define the conjugate force by

$$\mathcal F_\lambda \equiv -\frac{dE_n}{d\lambda}.$$

The theorem gives

$$\mathcal F_\lambda = -\left\langle \frac{\partial H}{\partial\lambda} \right\rangle_n.$$

The minus sign follows the convention that force is minus the energy gradient. Examples include:

• mechanical force for a spatial coordinate or nuclear position;
• electric polarization or dipole response for an applied electric field;
• magnetization for an applied magnetic field;
• pressure-like response for a volume or strain parameter;
• expectation value of an interaction when $\lambda$ is a coupling constant.

The interpretation depends on how the Hamiltonian is parameterized. If

$$H(\mathcal E) = H_0-\mathcal E D,$$

then

$$\frac{dE_n}{d\mathcal E} = -\langle D\rangle_n.$$

The dipole moment is therefore $-\partial E_n/\partial\mathcal E$ with this sign convention.

Coupling-Constant Integration

For

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

the theorem applies at every regular point of the branch:

$$\frac{dE_n(\lambda)}{d\lambda} = \langle n(\lambda)\rvert V \lvert n(\lambda)\rangle.$$

Integrating from $\lambda=0$ to $\lambda=1$ gives

$$\begin{aligned} E_n(1)-E_n(0) = \int_0^1 d\lambda\, \langle n(\lambda)\rvert V \lvert n(\lambda)\rangle. \end{aligned}$$

This exact coupling-constant integral replaces an energy difference by expectation values along a path. It does not remove the need to know the interacting branch, but it is useful for formal identities, numerical thermodynamic integration, and adiabatic-connection constructions. “Adiabatic” here describes a parameter path, not necessarily real-time evolution under the adiabatic theorem.

Connection to First-Order Perturbation Theory

For the Rayleigh–Schrödinger expansion

$$\begin{aligned} E_n(\lambda) ={}& E_n^{(0)} +\lambda E_n^{(1)} \\ &+ \lambda^2E_n^{(2)} +\cdots, \end{aligned}$$

the coefficient of $\lambda$ is

$$E_n^{(1)} = \left. \frac{dE_n}{d\lambda} \right|_{\lambda=0}.$$

Since $\partial_\lambda H=V$,

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

First-order perturbation theory is therefore the Hellmann–Feynman theorem evaluated at the reference point. The theorem is stronger in one sense: when the exact eigenstate at a finite $\lambda$ is known, it gives the exact local slope there, not merely a coefficient about $\lambda=0$.

First-Order Energy Corrections owns the interpretation and examples of the linear coefficient. Rayleigh–Schrödinger Perturbation Theory owns the full order-by-order construction.

Curvature and Second-Order Response

Differentiate the theorem once more:

$$\begin{aligned} E_n'' ={}& \langle n\rvert H''\lvert n\rangle \\ &+ 2\operatorname{Re} \langle n'\rvert H'\lvert n\rangle. \end{aligned}$$

Insert a complete set of discrete eigenstates and use the off-diagonal state-derivative identity. For a simple discrete level,

$$\begin{aligned} E_n'' ={}& \langle n\rvert H''\lvert n\rangle \\ &+ 2\sum_{m\ne n} \frac{ \lvert\langle m\rvert H'\lvert n\rangle\rvert^2 }{ E_n-E_m }. \end{aligned}$$

A continuum contributes the corresponding spectral integral. If $H(\lambda)=H_0+\lambda V$, then $H''=0$ and, at $\lambda=0$,

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

This makes the hierarchy transparent:

• the first derivative is a diagonal matrix element;
• the second derivative requires the first-order change of the state;
• higher derivatives require progressively more response information.

See Second-Order Energy Corrections for signs, polarizabilities, continuum terms, and sum-over-states convergence.

Degenerate Eigenvalues

At a $d$-fold degeneracy, there is no unique eigenvector branch before the perturbation direction is specified. Let $P$ project onto the degenerate eigenspace at $\lambda_0$, and choose any orthonormal basis $\{\lvert a\rangle\}_{a=1}^d$. Form

$$W_{ab} = \left\langle a\left\vert H'(\lambda_0) \right\vert b\right\rangle.$$

The eigenvalues $w_\alpha$ of $W$ are the first derivatives of the branches that split linearly:

$$\begin{aligned} E_\alpha(\lambda) ={}& E_0+ (\lambda-\lambda_0)w_\alpha \\ &+ O\!\left( (\lambda-\lambda_0)^2 \right). \end{aligned}$$

The corresponding eigenvectors of $W$ identify the combinations to which the ordinary theorem applies on either side of the degeneracy. Taking the expectation of $H'$ in an arbitrary basis vector of the degenerate subspace generally gives a basis-dependent number, not a branch derivative.

There is another labeling subtlety. For

$$H(\lambda)=\lambda\sigma_z,$$

the analytic branches are $E_\pm(\lambda)=\pm\lambda$. If instead the levels are sorted by size, the ground-state energy is

$$E_{\mathrm g}(\lambda)=-\lvert\lambda\rvert,$$

which is not differentiable at $\lambda=0$. The theorem has not failed; the sorted ground-state label does not define a differentiable branch at the crossing.

Degenerate Perturbation Theory is the canonical treatment of $PH'P$ and higher-order corrections.

Approximate States and the Residual Term

Let $\lvert\phi(\lambda)\rangle$ be normalized but not necessarily an eigenstate, and define its energy expectation

$$\mathcal E(\lambda) = \langle\phi(\lambda)\rvert H(\lambda) \lvert\phi(\lambda)\rangle.$$

Differentiating and using normalization gives

$$\begin{aligned} \frac{d\mathcal E}{d\lambda} ={}& \langle\phi\rvert H'\lvert\phi\rangle \\ &+ 2\operatorname{Re} \langle\phi'\rvert \bigl(H-\mathcal E\bigr) \lvert\phi\rangle. \end{aligned}$$

Define the residual

$$\lvert r\rangle = \bigl(H-\mathcal E\bigr) \lvert\phi\rangle.$$

Then the difference between the true energy derivative and the naive Hellmann–Feynman expectation is controlled by the response overlap

$$2\operatorname{Re} \langle\phi'\vert r\rangle.$$

For an exact eigenstate, $\lvert r\rangle=0$. For an approximate state, a small energy error alone does not guarantee a small force error: the residual can couple strongly to the parameter derivative of the state.

This identity is a practical diagnostic. Report both the eigenvalue residual and the stability of the derivative under improvement of the variational space or basis.

Variational Stationarity

Internal ansatz gradients, tangent metrics, Hessians, and moving-basis overlap derivatives are developed in Variational Parameters. This section isolates the derivative with respect to an external physical parameter.

Suppose a normalized trial state depends on internal parameters $\boldsymbol\alpha$ and on an external parameter $\lambda$. Let

$$\mathcal E = \mathcal E(\boldsymbol\alpha,\lambda).$$

Along optimized parameters $\boldsymbol\alpha_*(\lambda)$,

$$\frac{d\mathcal E_*}{d\lambda} = \frac{\partial\mathcal E}{\partial\lambda} + \sum_i \frac{\partial\mathcal E}{\partial\alpha_i} \frac{d\alpha_{*,i}}{d\lambda}.$$

If the optimization is fully stationary,

$$\left. \frac{\partial\mathcal E} {\partial\alpha_i} \right|_{\boldsymbol\alpha_*} =0,$$

so response of the optimized coefficients does not contribute explicitly to the first derivative:

$$\frac{d\mathcal E_*}{d\lambda} = \left. \frac{\partial\mathcal E}{\partial\lambda} \right|_{\boldsymbol\alpha_*}.$$

This is an envelope-theorem form of the same cancellation. It does not imply that every approximate calculation obeys the bare Hellmann–Feynman formula. The variational manifold itself may depend on $\lambda$, the optimization may be incomplete, or some orbital and basis degrees of freedom may not be stationary.

Molecular Forces

Within the Born–Oppenheimer electronic problem, let $\mathbf R_A$ be a nuclear coordinate and write the potential-energy surface as

$$U(\mathbf R) = E_{\mathrm e}(\mathbf R) + V_{\mathrm{NN}}(\mathbf R).$$

For an exact normalized electronic eigenstate on a fixed representation,

$$\begin{aligned} \mathbf F_A ={}& -\nabla_{\mathbf R_A}U \\ ={}& -\left\langle\Psi_{\mathrm e}\left\vert \nabla_{\mathbf R_A}H_{\mathrm e} \right\vert\Psi_{\mathrm e}\right\rangle \\ &- \nabla_{\mathbf R_A}V_{\mathrm{NN}}. \end{aligned}$$

This is the molecular-force interpretation emphasized by Feynman. It says that an exact electronic energy gradient can be evaluated from the derivative of the electronic Hamiltonian, without explicitly differentiating the exact wavefunction.

The formula does not make molecular force calculations trivial. Electronic states are approximate, atom-centered basis functions move with the nuclei, electron correlation is truncated, and near-degenerate surfaces can be non-smooth or require a multistate description.

Parameter-Dependent Bases and Pulay Terms

Consider a finite nonorthogonal basis $\{\lvert\chi_\mu(\lambda)\rangle\}$ and the generalized eigenproblem

$$H(\lambda)c(\lambda) = E(\lambda)S(\lambda)c(\lambda),$$

with

$$c^\dagger S c=1.$$

Here

$$H_{\mu\nu} = \langle\chi_\mu\vert H \lvert\chi_\nu\rangle, \qquad S_{\mu\nu} = \langle\chi_\mu\vert\chi_\nu\rangle.$$

Differentiating the generalized eigenvalue equation gives

$$E' = c^\dagger \bigl(H'-ES'\bigr) c.$$

$H'$ is the total matrix derivative: it includes the operator derivative and derivatives of the basis functions. The $-ES'$ term accounts for the changing overlap metric. In atom-centered electronic-structure calculations, the basis-response contributions are commonly called Pulay terms.

Three formulas must not be conflated:

1. The exact Hilbert-space theorem uses $\langle\Psi\vert\partial_\lambda H\vert\Psi\rangle$ for an exact eigenstate in a fixed representation.
2. The finite generalized eigenproblem uses $c^\dagger(H'-ES')c$ with total matrix derivatives.
3. A bare expectation of only the operator derivative can miss basis motion, incomplete variational response, or both.

Pulay’s force analysis explains why apparently well-converged energies can still yield inaccurate gradients when the basis response is neglected.

Changing Domains and Boundary Conditions

The theorem’s derivative is not merely the derivative of a differential expression. The operator includes its domain and boundary conditions.

Consider a particle in an infinite well on $0\lt x\lt L$:

$$H_L = -\frac{\hbar^2}{2m} \frac{d^2}{dx^2},$$

with Dirichlet conditions at $x=0$ and $x=L$. Its energies are

$$E_n(L) = \frac{ \pi^2\hbar^2n^2 }{ 2mL^2 }.$$

The differential expression contains no visible $L$, but its domain does. Setting $\partial H_L/\partial L=0$ would incorrectly predict $dE_n/dL=0$.

Map the interval to the fixed coordinate $y=x/L$, with $0\lt y\lt1$. The unitarily transformed Hamiltonian is

$$\widetilde H(L) = -\frac{\hbar^2}{2mL^2} \frac{d^2}{dy^2}.$$

Now

$$\frac{\partial\widetilde H}{\partial L} = -\frac{2}{L}\widetilde H,$$

and the theorem gives

$$\frac{dE_n}{dL} = -\frac{2E_n}{L},$$

in agreement with direct differentiation. The fixed-domain transformation has made the hidden boundary dependence explicit.

Worked Example: Oscillator Width from an Energy Derivative

Let

$$H(\omega) = \frac{p^2}{2m} + \frac{1}{2}m\omega^2x^2, \qquad \omega\gt0.$$

The exact energies are

$$E_n(\omega) = \hbar\omega \left(n+\frac{1}{2}\right).$$

Differentiate the Hamiltonian:

$$\frac{\partial H}{\partial\omega} = m\omega x^2.$$

The Hellmann–Feynman theorem yields

$$\hbar \left(n+\frac{1}{2}\right) = m\omega \langle n\vert x^2\vert n\rangle.$$

Therefore

$$\langle n\vert x^2\vert n\rangle = \frac{\hbar}{m\omega} \left(n+\frac{1}{2}\right).$$

Multiplying by $m\omega^2/2$ gives

$$\left\langle \frac{1}{2}m\omega^2x^2 \right\rangle_n = \frac{E_n}{2}.$$

The kinetic and potential expectations are equal, consistent with the Virial Theorem. The example illustrates a common use: differentiate a known spectrum to recover an expectation value without integrating the wavefunction explicitly.

When the Theorem Does Not Apply Directly

The elementary Hermitian statement must be modified or replaced for:

• non-normalizable continuum states without an appropriate box, wave-packet, or spectral regularization;
• resonances described by non-Hermitian effective Hamiltonians, which require left and right eigenvectors;
• parameter-dependent inner products or metrics not included in the derivative;
• moving domains or boundary conditions not mapped to a fixed space;
• a sorted eigenvalue at a non-differentiable crossing;
• approximate states whose residual-response or basis-response terms are nonzero;
• time-dependent forces away from an instantaneous stationary eigenstate.

These are changes of hypothesis, not paradoxes. State which spectral object and derivative are being used before applying the formula.

Practical Checklist

1. Define the parameter. State what is held fixed when $\lambda$ changes.
2. Track a branch. Use symmetry or overlap, not only sorted energy order.
3. Check degeneracy. Diagonalize $PH'P$ in a degenerate subspace.
4. Fix the representation. Expose moving measures, bases, boundaries, and domains.
5. Separate exact and approximate formulas. Compute residual and response terms when the state is not exact.
6. Differentiate every explicit dependence. Include higher-order Hamiltonian terms, overlap matrices, and external potentials.
7. Validate the gradient. Compare with symmetric finite differences over a converged step-size window.
8. Report numerical layers. Distinguish eigensolver, basis, correlation, and finite-difference errors.

Common Mistakes

• Writing a partial derivative of the eigenvalue where a total branch derivative is intended.
• Differentiating the wavefunction explicitly even though the exact eigenvalue equation already cancels its first-order response.
• Applying the theorem to an arbitrary vector in a degenerate eigenspace.
• Treating the sorted ground-state energy as differentiable through a level crossing.
• Using an approximate state while omitting the residual-response term.
• Calling all gradient corrections “Pulay terms” without identifying basis, overlap, orbital, or correlation response.
• Ignoring parameter dependence in boundary conditions or the operator domain.
• Assuming the theorem gives derivatives of arbitrary observables; those usually require the state derivative.
• Forgetting the sign convention between an energy gradient and its conjugate force.

Exercises

1. Prove the theorem without choosing a phase

Derive the Hellmann–Feynman formula directly from the differentiated eigenvalue equation and explain why no condition on $\langle n\vert n'\rangle$ is needed.

Solution

Differentiate:

$$H'\lvert n\rangle + H\lvert n'\rangle = E_n'\lvert n\rangle + E_n\lvert n'\rangle.$$

Left-multiplication by $\langle n\rvert$ gives

$$\begin{aligned} \langle n\vert H'\vert n\rangle &+ E_n\langle n\vert n'\rangle \\ &= E_n' + E_n\langle n\vert n'\rangle. \end{aligned}$$

The state-derivative terms cancel directly, leaving

$$E_n' = \langle n\vert H'\vert n\rangle.$$

A parameter-dependent phase changes $\langle n\vert n'\rangle$, but only inside the two terms that cancel. No parallel-transport phase convention is required.

2. Obtain an oscillator expectation value

Use the theorem with $\lambda=m$ rather than $\omega$ for

$$H(m) = \frac{p^2}{2m} + \frac{1}{2}m\omega^2x^2$$

at fixed $\omega$. Recover a relation between kinetic and potential energy.

Solution

The exact energy $E_n=\hbar\omega(n+1/2)$ is independent of $m$, so

$$\frac{dE_n}{dm}=0.$$

The Hamiltonian derivative is

$$\frac{\partial H}{\partial m} = -\frac{p^2}{2m^2} + \frac{1}{2}\omega^2x^2.$$

Therefore

$$-\left\langle\frac{p^2}{2m^2}\right\rangle + \left\langle\frac{1}{2}\omega^2x^2\right\rangle =0.$$

Multiplying by $m$ gives

$$\left\langle\frac{p^2}{2m}\right\rangle = \left\langle\frac{1}{2}m\omega^2x^2\right\rangle.$$

3. Derive the curvature formula

For a simple discrete eigenvalue, derive

$$\begin{aligned} E_n'' ={}& \langle n\vert H''\vert n\rangle \\ &+ 2\sum_{m\ne n} \frac{ \lvert\langle m\vert H'\vert n\rangle\rvert^2 }{ E_n-E_m }. \end{aligned}$$

Solution

Differentiate the first-derivative theorem:

$$\begin{aligned} E_n'' ={}& \langle n\vert H''\vert n\rangle \\ &+ 2\operatorname{Re} \langle n'\vert H'\vert n\rangle. \end{aligned}$$

Insert completeness into the response term. The real part of the parallel component vanishes because normalization implies

$$\operatorname{Re} \langle n\vert n'\rangle=0.$$

For $m\ne n$,

$$\langle m\vert n'\rangle = \frac{ \langle m\vert H'\vert n\rangle }{ E_n-E_m }.$$

Substitution gives the stated sum. A continuous spectrum adds its spectral integral.

4. Resolve a degenerate derivative

At $\lambda=0$, let

$$H(\lambda) = E_0I+ \lambda \bigl(a\sigma_z+b\sigma_x\bigr).$$

Find the two branch derivatives and explain why $\langle\uparrow_z\vert H'\vert\uparrow_z\rangle=a$ is not, in general, one of them.

Solution

Inside the degenerate subspace,

$$H' = \begin{pmatrix} a&b\\ b&-a \end{pmatrix}.$$

Its eigenvalues are

$$w_\pm = \pm\sqrt{a^2+b^2}.$$

Thus

$$\left. \frac{dE_\pm}{d\lambda} \right|_{\lambda=0} = \pm\sqrt{a^2+b^2}.$$

The vector $\lvert\uparrow_z\rangle$ is not an eigenvector of $H'$ when $b\ne0$. Its expectation $a$ depends on a basis choice within the degenerate eigenspace and does not label a differentiable eigenbranch.

5. Quantify failure for an approximate state

Let $\lvert\phi(\lambda)\rangle$ be normalized with residual $\lvert r\rangle=(H-\mathcal E)\lvert\phi\rangle$. Show that

$$\left| \frac{d\mathcal E}{d\lambda} - \langle\phi\vert H'\vert\phi\rangle \right| \le 2\lVert\phi'\rVert\lVert r\rVert.$$

Solution

The approximate-state identity gives

$$\frac{d\mathcal E}{d\lambda} - \langle\phi\vert H'\vert\phi\rangle = 2\operatorname{Re} \langle\phi'\vert r\rangle.$$

Use

$$\lvert\operatorname{Re}z\rvert \le \lvert z\rvert$$

and the Cauchy–Schwarz inequality:

$$\lvert\langle\phi'\vert r\rangle\rvert \le \lVert\phi'\rVert\lVert r\rVert.$$

Combining them gives the bound. A small residual helps only when the state response is also controlled.

6. Repair the moving-wall paradox

For the infinite well of width $L$, explain why differentiating the differential expression at fixed $x$ gives the wrong answer and verify the result after mapping to $y=x/L$.

Solution

At fixed $x$, the expression

$$-\frac{\hbar^2}{2m} \frac{d^2}{dx^2}$$

has no explicit $L$ dependence, but the domain and the boundary condition at $x=L$ do. The elementary fixed-domain hypothesis is therefore violated.

After mapping to $0\lt y\lt1$,

$$\widetilde H(L) = -\frac{\hbar^2}{2mL^2} \frac{d^2}{dy^2},$$

so

$$\frac{\partial\widetilde H}{\partial L} = -\frac{2}{L}\widetilde H.$$

The theorem now gives

$$\frac{dE_n}{dL} = -\frac{2E_n}{L} = -\frac{\pi^2\hbar^2n^2}{mL^3},$$

which agrees with direct differentiation of $E_n(L)$.

Cross-Links

• Time-Independent Perturbation Theory for the chapter-level framework.
• First-Order Energy Corrections for diagonal shifts, symmetry, and examples.
• Second-Order Energy Corrections for curvature and response sums.
• Degenerate Perturbation Theory for branch derivatives inside a multiplet.
• Variational Principle for stationarity and energy bounds.
• Berry Connection for the phase-sensitive parallel part of state derivatives.
• Molecular Physics for rotational, vibrational, and electronic applications.
• Hellmann–Feynman Theorem Reference Card for compact lookup.

References

• H. Hellmann, Einführung in die Quantenchemie, Franz Deuticke, Leipzig and Vienna, 1937.
• R. P. Feynman, “Forces in Molecules,” Physical Review 56, 340–343 (1939), doi:10.1103/PhysRev.56.340.
• T. Kato, Perturbation Theory for Linear Operators, 2nd ed., Springer, 1976; reprint 1995, doi:10.1007/978-3-642-66282-9.
• P. Pulay, “Ab Initio Calculation of Force Constants and Equilibrium Geometries in Polyatomic Molecules. I. Theory,” Molecular Physics 17, 197–204 (1969), doi:10.1080/00268976900100941.
• F. M. Fernández, “Generalization of the Hellmann–Feynman Theorem,” Physics Letters A 374, 819–822 (2010), doi:10.1016/j.physleta.2009.12.005.
• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Vol. II, Wiley, 1977.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• T. Helgaker, P. Jørgensen, and J. Olsen, Molecular Electronic-Structure Theory, Wiley, 2000, Chapters 10 and 14.