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For a nondegenerate parity-symmetric ground state in an external electric field, write the perturbation as $V=-E\hat{d}$. The first-order state correction givesThe positive coefficient $\alpha$ is the electric polarizability along the applied direction. Meanwhile second-order nondegenerate perturbation theory gives $\delta E_0^{(2)}=-\alpha E^2/2$. Thus the induced electric dipole moment satisfies $d=-2\delta E_0^{(2)}/E$, not $-\delta E_0^{(2)}/E$. The Hellmann–Feynman theorem independently gives $d=-\partial E_0/\partial E$.

For a Transverse-field Ising model whose field term is $-2Jg{\sum }_n{S}_n^z$, the Hellmann–Feynman theorem gives the displayed relation. Differentiating $E_0=-J{\sum }_k\sqrt{1+g^2-2g\cos k}$ gives ${\sum }_n⟨S_n^z⟩={\sum }_k(g-\cos k)/(2\sqrt{1+g^2-2g\cos k})$. This is polarization along the imposed field, not spontaneous order along the exchange axis. In the thermodynamic limit it is zero per site at g=0, tends to1/2 at large positive g and equals $1/\pi$ at g=1.

The vacuum of the diagonal Bogoliubov quasiparticles has the displayed energy. In the thermodynamic bulk, $E_0/N=-J(2\pi {)}^{-1}{\int }_{-\pi }^{\pi }\sqrt{1+g^2-2g\cos k}dk$. The Hellmann–Feynman theorem then gives the field magnetization ${\sum }_n⟨S_n^z⟩={\sum }_{k}J(g-\cos k)/{ϵ}_k$. Sector-dependent finite-chain boundary terms must be restored when exact finite-size levels are needed.

Introduce a bookkeeping parameter $\lambda$ in $H=H_0+\lambda H_1$ and expand $|{\varphi }_n⟩=|{\psi }_n⟩+\lambda |{\eta }_n⟩+O({\lambda }^2)$ and ${\mathcal{E}}_n=E_n+\lambda \Delta E_n+O({\lambda }^2)$. Fix phase and normalization at this order by $⟨{\psi }_n|{\eta }_n⟩=0$. The first-order Schrödinger equation isangular momentum projection onto $⟨{\psi }_n|$ gives $\Delta E_n=⟨{\psi }_n|H_1|{\psi }_n⟩$. angular momentum projection onto a different eigenstate givesCompleteness therefore yields the first-order nondegenerate perturbation theory resultNonzero energy denominators are essential. The displayed state has unit norm through first order; exact normalization would change terms of second order. A continuum, if present, contributes its corresponding integral as well as any discrete sum.

For the oscillator in a static electric field, electrostatic potential energy is $H_1=-qϵx$ and $\omega =\sqrt{k/m}$. parity gives $⟨0|x|0⟩=0$, and the supplied Gaussian wavefunctions giveSince $E_0-E_1=-\hslash \omega$, the two-state first-order ground state isThus $⟨x⟩=2\eta x_{\mathrm{zp}}+O({ϵ}^2)$, and the induced electric dipole isIt is aligned with the field for either sign of $q$. The harmonic-oscillator electric polarizability is $q^2/k$. The position operator connects $|0⟩$ only to $|1⟩$, so the requested two-state calculation contains the complete first-order dipole response, rather than dropping other first-order terms.

Complete the square to find the exact electric-field displacement of a quantum harmonic oscillator:The exact normalized ground wavefunction is ${\psi }_0(x-x_0)$, with energy $\hslash \omega /2-q^2{ϵ}^2/(2k)$. Its mean displacement is $x_0$, soThe first-order dipole equals the exact dipole, because the restoring force is exactly linear and translating a harmonic well does not change its width or frequency. This agreement of one expectation value does not make the truncated wavefunction exact. The Hellmann–Feynman theorem provides an independent check: $d=-\partial E_0/\partial ϵ$.

For perturbation theory to approximate the state, $|\eta |\ll 1$, or $|x_0|\ll 2x_{\mathrm{zp}}$. The exact translated ground state has the original-oscillator expansionIt is a coherent state with mean occupation ${\eta }^2$. Large fields require many oscillator levels, whereas the first-order state retains only two. The normalized two-state truncation of a displaced oscillator instead giveswhich incorrectly tends towards zero for large $|\eta |$ instead of following the displaced well. In position space the truncated polynomial times Gaussian also develops a node absent from the exact ground state. Taking a linearized formula outside its small-state-correction regime is therefore invalid even though its dipole coefficient happens to be exact.

The electron's electric dipole moment is $\mathbf{d}=-e⟨x⟩\hat{\mathbf{x}}$. It vanishes in the unperturbed ground energy eigenstate. The first correction to the wavefunction is linear in the electric field and has the opposite parity; its interference with the unperturbed state therefore gives a nonzero position expectation value of first order.

In detail, the normalized ground quantum state, to the order needed, isThe normalization correction starts at second order and does not affect the linear term below. The two cross terms giveThe absence of an $E^2$ term follows from the parity operator: $⟨x⟩$ changes sign under field reversal. Comparing with the second-order nondegenerate perturbation theory energy sum yieldsThus the induced electric dipole moment points along the electric field, with positive electric polarizability. The induced dipole moment and quadratic energy shift obey $\delta {ϵ}_0^{(2)}=-{\alpha }_{\mathrm{pol}}{E}^2/2$. This factor of one half accounts for the energy needed to induce the dipole. As a check, the Hellmann–Feynman theorem gives $d_x=-\partial {ϵ}_0/\partial E$, with the same result.

The Hellmann–Feynman theorem gives ${\partial }_g{E}_0=⟨{\partial }_{g}H⟩=-2J{\sum }_n⟨S_n^z⟩$. Differentiating the Ising-chain ground-state energy yields ${\partial }_g{E}_0=-J{\sum }_k(g-\cos k)/r_k$. HenceThis is also obtained from the occupations of the Bogoliubov quasiparticle vacuum: $⟨c_k^{†}{c}_k⟩=[1-(g-\cos k)/r_k]/2$ and $S_n^z=1/2-n_n$. At an exact zero mode the finite-sector limiting ground state specifies the occupation. In the bulk limits, the magnetization per site is zero at $g=0$, tends to $1/2$ as $g\to \infty$, and equals $1/\pi$ at $g=1$: there the integrand $(1-\cos k)/r_k=|\sin (k/2)|$.

Differentiate the eigenvalue equation along the assumed smooth normalized eigenbranch:Since $A$ is real symmetric, $x^{T}A=\lambda x^T$. Multiplying the differentiated equation by $x^T$ and using $x^{T}x=1$ cancels the terms containing $x^{\prime }$:Thus the finite-dimensional Hellmann–Feynman theorem givesEquivalently, differentiating the normalization gives $2x^T{x}^{\prime }=0$ and eliminates those terms from the derivative of $\lambda =x^{T}Ax$. The proof needs the smooth eigenbranch assumed in the question; the smallest eigenvalue need not be differentiable at an arbitrary crossing without that assumption.

The Hellmann–Feynman theorem applied to the source field normalization givesDifferentiating the Ising-chain ground-state energy, using ${\partial }_g{ϵ}_k=4J^2(g-\cos k)/{ϵ}_k$, therefore provesThis is the Ising-chain field derivative and transverse polarization, not its spontaneous x-order parameter. It tends to zero at g=0 and to N/2 at large positive field. The critical transverse polarization of an Ising chain is finite: in the thermodynamic critical limit this gives $⟨S^z⟩=1/\pi$. A zero-energy isolated momentum is treated by the appropriate limiting state, rather than by assigning a value to a literal 0/0 summand.

Differentiating $E_k\simeq \sqrt{{\hslash }^{3}q\alpha k/2}/m$ gives the displayed group velocity when $q\alpha k>0$. Its sign is the sign of $q\alpha$, equivalently of k on that branch. Writing only a positive square root of $q\alpha /k$ would lose this sign. Exactly, the Hellmann–Feynman theorem gives $v_y=⟨\hslash k-q\alpha x^2/2⟩/m$. Evaluating this small drift from a localized wavefunction requires enough anharmonic accuracy: a leading centred Gaussian alone need not reproduce it.

An electric field shifts and can split the energy eigenvalues of an atom. This is the Stark effect; its hydrogen atom ground-state case is quadratic in a weak field.

The observable in Stark effect spectroscopy is a change in a spectral line, rather than a direct reading of an individual energy eigenvalue. If an upper energy eigenstate $u$ emits a photon into a lower energy eigenstate $l$, the Planck constant $h$ relates its frequency to the difference:Here $\mathcal{E}$ denotes the applied electric field component along a chosen axis. Both levels can move, and electric-dipole selection rules for hydrogen decide which pairs produce observable spectral lines. This makes Stark effect spectroscopy a probe of energy eigenstates and their electric dipole moments: the slope $-\partial E/\partial \mathcal{E}$ is the corresponding dipole component by the Hellmann–Feynman theorem. The comparison with Zeeman splitting is useful: there an applied magnetic field supplies the perturbation.