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In atomic units, use the radial trial wavefunction $\psi (r)=\max (1-r/\beta ,0)$ with $\beta >0$. Its radial norm integral is $\int r^2{\psi }^{2}dr={\beta }^3/30$, its gradient contribution is $\int r^2{\psi }^{\prime 2}dr=\beta /3$, and its Coulomb contribution is $-2\int r{\psi }^{2}dr=-{\beta }^2/6$. In the hydrogen radial Rayleigh quotient, this gives the displayed energy, minimized at $\beta =4$. It is above the exact hydrogen ground-state energy $-1/2$, as required by the quantum variational principle. The derivative jumps at $r=\beta$, but the function has a square-integrable weak first derivative and is an admissible trial for the energy quadratic form. One must use that form rather than assume that a classical second derivative exists at the join.

For $\kappa >0$ and the radial exponential trial state $e^{-\alpha r}$, $\alpha >0$, put $a={\hslash }^2/(m\kappa )$. Direct radial integration gives the Rayleigh quotientThe kinetic expectation is ${\hslash }^2{\alpha }^2/(2m)$; the potential expectation follows by splitting its radial integral at $R$. The common prefactor cancels from the stationary condition, giving $a\alpha =(1+2\alpha R)e^{-2\alpha R}$. Its positive solution is unique, since the left side increases and the right side decreases. For $\epsilon =R/a\to 0$, it satisfies $a\alpha =1-2{\epsilon }^2+O({\epsilon }^3)$.

The quantum variational principle makes $E(\alpha )$ an upper bound for the ground energy. The explicit choice $\alpha =1/a$ gives $E_0(R)\le (\kappa /a)[1/2-\{1-(1+\epsilon )e^{-2\epsilon }\}/\epsilon ]$, whose first correction to $-\kappa /(2a)$ is $2\kappa R^2/(3a^3)$. The true energy is no lower than the point-Coulomb value because capping makes the potential less attractive.

The trial $x{e}^{-bx}$, $b>0$, satisfies the wall condition in the half-line square-root potential well. Its norm is $1/(4b^3)$, kinetic energy is ${\hslash }^2{b}^2/(2m)$ and potential energy is $(15a/16)\sqrt{\pi /2}{b}^{-1/2}$. The optimal $b$ satisfies $b^{5/2}=(ma/{\hslash }^2)(15/32)\sqrt{\pi /2}$. At the minimum, $⟨V⟩=4⟨T⟩$ and $E_{\mathrm{var}}=5{\hslash }^2{b}^2/(2m)=1.876493E_{\ast }$. The quantum variational principle makes this an upper bound on the ground-state energy.

A ground state is a normalized energy eigenstate that attains the lowest possible energy of a quantum system. Precisely, let the Hamiltonian operator $H$ be a self-adjoint operator bounded below on a Hilbert space, and write $E_{\ast }=\inf \sigma (H)$ for the infimum of its spectrum. A ground state is a unit vector ${\psi }_0$ in the operator domain $D(H)$ satisfyingThus a ground state exists exactly when the spectral bottom $E_{\ast }$ is an energy eigenvalue. The lowest energy eigenvalue among some listed bound states is insufficient if other parts of the spectrum lie below it. The ground-state subspace collects all minimizing energy eigenstates; it can have ground-state degeneracy. The quantum variational principle characterizes $E_{\ast }$ as the infimum of normalized energy expectation values. ^[ref].

For the one-dimensional quantum harmonic oscillator, the ground state has energy $\hslash \omega /2$ above the minimum of the potential energy. This zero-point energy is compatible with being the lowest allowed energy: a localized wavefunction cannot simultaneously have zero kinetic energy. For a free particle on the whole line, the spectral bottom is zero but is unattained by any normalizable wavefunction; see ground-state energy. A lower bound therefore need not describe a physically preparable energy eigenstate.

The ground state supplies a reference for excitation costs and for the quantum variational principle. A ground state is also a stationary state under a time-independent Hamiltonian operator: its quantum phase changes, while its position probability density stays fixed. A nonzero ground-state energy does not imply motion along a classical trajectory, and ground-state degeneracy means there need not be a unique lowest quantum state.

The ground-state energy $E_0$ is the energy eigenvalue attained by a ground state. For a self-adjoint operator $H$ bounded below, distinguish this attained value from the always-defined spectral bottom $E_{\ast }=\inf \sigma (H)$. When a ground state exists, $E_0=E_{\ast }$. Some usage calls $E_{\ast }$ the ground-state energy even when it is unattained; here an unattained bottom is explicitly identified as an infimum. The quantum variational principle statesApproaching this infimum with trial wavefunctions does not ensure that a minimizing wavefunction exists. ^[ref].

For the whole-line free particle with mass $m>0$,Its spectrum is continuous and contains no energy eigenvalues represented by normalizable wavefunctions. The Fourier transform makes $H$ multiplication by ${\hslash }^2{k}^2/(2m)$; a fixed energy restricts $k$ to a set of zero length, on which a nonzero square-integrable function cannot be supported. ^[ref]. An explicit family of normalized Gaussian wave packets demonstrates how the bottom is approached:The formula follows from ${\psi }_s^{\prime }=-x{\psi }_s/(2s^2)$, integration by parts, and $⟨x^2{⟩}_s=s^2$. Every finite width gives positive kinetic energy. At zero energy, $H\psi =0$ would require an affine function on the whole line, which is never a nonzero normalizable wavefunction. Spreading a Gaussian wave packet indefinitely produces an unattained infimum, rather than a zero-energy ground state.

An additive choice of energy origin changes the numerical ground-state energy without changing the ground-state subspace. For $H^{\prime }=H+CI$, every energy eigenvalue becomes $E^{\prime }=E+C$, while all differences $E_j-E_i$ stay fixed. Time evolution gains only the global phase $e^{-iCt/\hslash }$. This makes excitation costs and spectroscopic differences more intrinsic than the isolated number $E_0$.

A product of two normalized $1s$ orbitals proportional to $e^{-\zeta r}$, combined with a singlet state, gives the fixed-nucleus helium atom trial energy $E(\zeta )={\zeta }^2-4\zeta +5\zeta /8$ in atomic units. Minimization gives $\zeta =27/16$ and $E=-729/256$ Hartree. Nuclear charge remains two in the Hamiltonian operator; varying $\zeta$ changes only the trial orbitals. The reduced orbital charge describes screening, while explicit dependence on electron separation can further lower the trial energy under the quantum variational principle.

A free scalar Klein-Gordon equation with rest mass $m$ isHere $c$ is the speed of light and $\hslash$ the reduced Planck constant. In the static limit, comparison with the static Yukawa Green function identifiesThis $\lambda$ is the reduced Compton wavelength, rather than $h/(mc)$; the two differ by $2\pi$. It is an exponential decay length of this exchange field, not the geometric size of a nucleon or an exact radius of a bound state. This article uses $\lambda$ for a length; the 1935 paper used that letter for its inverse.

For a deliberately rounded worked example, stipulate $\hslash c=197.3\mathrm{MeV}\mathrm{fm}$ and $m{c}^2=140\mathrm{MeV}$. ThenA femtometre is $10^{-15}\mathrm{m}$. The chosen rest energy is a convenient pion-scale input; neither it nor the resulting Yukawa potential is asserted to be an exact measured nuclear interaction. Holding $\hslash c$ fixed, chosen rest energies $70$, $140$ and $280\mathrm{MeV}$ give lengths $2.819$, $1.409$ and $0.7046\mathrm{fm}$. Doubling the exchange rest mass halves the decay length.

Range alone does not guarantee binding: the potential energy must overcome the kinetic energy cost of localizing a wavefunction. The existing exponential variational bound for the Yukawa potential makes this precise. For the same toy $V$, with reduced mass $M$, choosing a normalized exponential wavefunction proportional to $e^{-\alpha r}$ with $\alpha =\kappa /2$ yieldsFor additional stipulated inputs $M{c}^2=470\mathrm{MeV}$ and $A=100\mathrm{MeV}\mathrm{fm}$, this gives $⟨H⟩\approx -3.66\mathrm{MeV}$. A negative value proves a bound state in this scalar model by the quantum variational principle; it is an upper bound on its ground-state energy, not a prediction for the deuteron. A nonnegative trial value would not prove the absence of binding.

For an infinite square well of width $a$, the normalized trial $\tilde{\psi }=\sqrt{30/a^5}x(a-x)$ obeys both Dirichlet boundary conditions and has no interior nodes of a wavefunction. Its kinetic energy isThe ratio to the exact ground-state energy is $10/{\pi }^2=1.01321\dots$. Its expansion in the normalized sine eigenfunctions has coefficients $4\sqrt{60}/({\pi }^3{n}^3)$ for odd $n$ and zero for even $n$. This supplies a concrete check of the quantum variational principle and shows that a small energy error can coexist with excited-state admixtures.

The quantum variational principle turns the ground-state problem into minimization of a Rayleigh quotient:The Hamiltonian must be a self-adjoint operator bounded below, and the nonzero trial state must have finite form energy. Expanding a normalized trial state in exact energy eigenstates gives $E[\psi ]={\sum }_n|c_n{|}^2{E}_n$, a weighted average no smaller than $E_0$. Equality requires support only in the ground-state eigenspace. The spectral version also accommodates continuous spectrum; a lowest normalizable eigenstate is not required merely to bound the spectral infimum.

The variational method chooses a trial family with the correct boundary conditions, fermionic exchange symmetry and other required quantum numbers, evaluates its energy and minimizes over its parameters. It gives an upper bound on ground-state energy. Enlarging a nested trial family can only improve or preserve its optimal bound. A low energy does not by itself establish a highly accurate wavefunction, especially near degeneracy; energy errors are second order in small admixtures orthogonal to a nondegenerate exact ground state.

For example, a normalized Gaussian trial for a one-dimensional quantum harmonic oscillator is ${\psi }_{\beta }=(\beta /\pi {)}^{1/4}{e}^{-\beta x^2/2}$, with $\beta >0$. Its energy isThe two terms balance kinetic localization cost against the potential cost of spreading out. This Gaussian variational energy for the harmonic oscillator happens to attain the exact answer because the family contains the exact ground state.

A trial expansion $\psi ={\sum }_j{c}_j{\chi }_j$ in a finite, possibly nonorthogonal basis gives the generalized eigenvalue problem $H\mathbf{c}=ES\mathbf{c}$, with $H_{ij}=⟨{\chi }_i|H|{\chi }_j⟩$ and $S_{ij}=⟨{\chi }_i|{\chi }_j⟩$. Its lowest Ritz value minimizes the quotient in that subspace. For excited states, the excited-state variational constraint imposes exact orthogonality to the lower eigenstates to obtain a corresponding upper bound; the Courant–Fischer min-max principle yields upper bounds for ordered Ritz eigenvalues from a common subspace. orthogonality only to an independently approximate ground state does not automatically provide an excited-state bound: residual true-ground-state contamination can lower the trial energy below the desired excited energy. Symmetry can impose exact orthogonality when the relevant states have different symmetry sectors.

For a self-adjoint Hamiltonian operator bounded below with ground-state energy $E_0$, every admissible nonzero trial wavefunction has a Rayleigh quotient at least $E_0$. To see this explicitly in a complete orthonormal eigenbasis, write $|\psi ⟩={\sum }_n{c}_n|n⟩$ with $H|n⟩=E_n|n⟩$. ThensoFor continuous spectral components, the spectral theorem replaces the sum by an integral with the same nonnegative integrand. Equality requires the trial quantum state to lie entirely in the ground-state subspace. The variational method chooses a tractable family, evaluates the quotient and minimizes over its parameters. This gives an upper bound; enlarging the family can only improve that bound. The trial must belong to the operator domain for the operator expression above, or have finite energy in the corresponding quadratic-form version of the quantum variational principle.

For the oscillator, use the standard kinetic operator $T=-{\hslash }^2{d}^2/(2md{x}^2)$: the original PDF omits the superscript $2$ on the derivative in its numerator. Take $a>0$, and normalize the scaled exact ground state asThe factor $\sqrt{\lambda }$ is optional in a quotient but makes the calculation transparent. Changing variable to $u=\lambda x$ and differentiating the scaled wavefunction givesAt $\lambda =1$ this trial is the exact minimizing state, so its energy $\mathcal{E}(\lambda )={\lambda }^2{T}_0+{\lambda }^{-2}{V}_0$ has zero derivative there. Hence $2T_0-2V_0=0$. Since $E_0=T_0+V_0$,This is the quantum virial identity from scaling for a quadratic potential, and hence the oscillator form of the virial theorem.

For the inverse-power attraction, take $a>0$ and suppose, for contradiction, that an exact ground state exists with separately finite kinetic energy $T_0>0$ and potential mean $U_0<0$, admitting the same dilation variations. Homogeneity now givesStationarity at $\lambda =1$ would require $2T_0+p{U}_0=0$. Its second derivative would then beFor $p>2$ this is negative, contradicting a ground-state minimum. More strongly, one does not need a presumed exact state: choose any normalized smooth test function supported away from the singular origin. Its $T$ and $U$ are finite, and its scaled quotient satisfiesThe unregularized attraction therefore exhibits fall to the centre for a supercritical inverse-power potential and has no stable ground state.

There is a necessary qualification to the PDF's broader wording about “no bound state”. Under the usual finite-energy virial theorem assumptions, a putative negative-energy bound state would satisfywhich contradicts being below the continuum threshold at zero. Thus no negative-energy bound state satisfying those regularity assumptions is possible. But this singular differential expression needs a specified domain at $x=0$; separately divergent kinetic and potential means or boundary conditions not preserved by dilation invalidate that ordinary calculation. The domain assumptions in singular-potential virial arguments must therefore be stated: the scaling proof establishes absence of a semibounded ground-state problem, rather than a domain-independent theorem about every formal square-integrable solution. A short-distance regularization would be a different potential.

Expand the trial in a complete orthonormal eigenbasis of the Quantum Hamiltonian, $\tilde{\Psi }={\sum }_n{c}_n{\varphi }_n$, with $\hat{H}{\varphi }_n=E_n{\varphi }_n$. Orthonormality givesEvery summand in the numerator is nonnegative because $E_n\ge E_0$. This proves the quantum variational principle for the stated discrete expansion; the trial energy is a weighted mean of exact energy eigenvalues.

The quantum variational principle says that every admissible nonzero trial wavefunction obeyswhere $E_0$ is the ground-state energy. Thus the Rayleigh quotient is an upper bound. Equality holds when the trial lies entirely in the ground-state subspace.

The trial is even under reflection about the midpoint. The normalized box eigenfunctions satisfy ${\varphi }_n(a-x)=(-1{)}^{n+1}{\varphi }_n(x)$, so orthogonality between even and odd functions gives $c_n=0$ for even $n$. If the higher odd contributions $n\ge 5$ are neglected, the weighted mean from the quantum variational principle becomesRearranging yields the two-level variational contamination estimateThe required assumption concerns the neglected contributions to the energy as well as their small probabilities. Numerically the two-level estimate is $0.0016542$. It is not an exact identity: direct expansion of this parabolic trial gives $c_n\propto n^{-3}$ for odd $n$, hence the exact ratio is $1/729=0.0013717$. The omitted higher odd states account for the difference.

Comparing the variational method estimate with the exact ground-state energy givesThe estimate is about $1.32\%$ too high, consistently with the quantum variational principle. Its closeness does not mean that the trial wavefunction is an exact energy eigenfunction.

Let $\{{\varphi }_j\}$ be a complete orthonormal eigenbasis of the Hamiltonian, with $H{\varphi }_j=E_j{\varphi }_j$ and $E_j\ge E_0$. Expand a nonzero admissible trial wavefunction as $\tilde{\Psi }={\sum }_j{c}_j{\varphi }_j$. ThenConsequentlyEvery numerator term is nonnegative and the denominator is positive. Equality requires the trial to lie wholly in the ground-state subspace; it need not be an individually chosen eigenvector when the ground energy is degenerate. This is the quantum variational principle, and wavefunction normalization before calculation is optional because the quotient removes the overall scale.

The quantum variational principle states that the energy expectation of an admissible trial wavefunction is at least the ground state energy:For the discrete, complete orthonormal basis assumed here, expand in Hamiltonian operator eigenfunctions, $f={\sum }_n{c}_n{\varphi }_n$, with $\hat{H}{\varphi }_n=E_n{\varphi }_n$ and $E_n\ge E_0$. ThenEvery summand is nonnegative. Equality holds precisely when all occupied components lie in the ground eigenspace. Thus minimizing trial energies supplies upper bounds on $E_0$, with exact equality if the trial family contains a ground state.

For a self-adjoint Hamiltonian bounded below, every nonzero admissible trial wavefunction satisfies the quantum variational principleHere $E_0$ is the lowest exact energy, and the trial function must obey the physical boundary conditions and have finite relevant integrals. Equality holds for a wavefunction in the ground-state eigenspace. The quotient is unchanged by normalization or any nonzero overall scalar.

Expand the trial wavefunction in a complete orthonormal set of exact eigenfunctions, $\psi ={\sum }_n{c}_n{\varphi }_n$, with $\hat{H}{\varphi }_n=E_n{\varphi }_n$ and $E_n\ge E_0$. The coefficients may depend on any chosen variational parameters. Orthogonality removes the cross terms in both integrals, givingConsequentlyThis proves the quantum variational principle. The denominator is positive because the trial function is nonzero. Equality requires every occupied exact state to have $E_n=E_0$, allowing for ground-state degeneracy. With a continuous part of the spectrum, the corresponding nonnegative spectral integral gives the same conclusion.

Exact solutions of the Schrödinger equation are unavailable for most interacting or realistic molecular systems. The quantum variational principle turns the search for a ground-state wavefunction into optimization of a computable energy quotient in a tractable trial family. The result is an upper bound: enlarging a nested trial space can only lower or preserve the optimized energy, providing a systematic convergence check. Trial functions can incorporate boundary conditions and known symmetry before calculation. Near an exact normalized ground state, a small orthogonal wavefunction error changes the energy only to second order, which explains why useful energies can result from imperfect trial functions. This does not ensure that every other observable is equally accurate, nor that a restricted trial family captures correlation adequately. Excited states require orthogonality or appropriate symmetry restrictions; unrestricted minimization seeks the ground state, as the preceding calculation demonstrates.

Direct evaluation of the Rayleigh quotient exposes a problem with the requested minimization. Using the supplied Gaussian integrals,ThusThe eigenfunction value $a=1$ is a maximum, not a minimum. Indeed $\mathcal{E}(1)=5/2$, whereas $\mathcal{E}(0)=11/6$ is already lower. There is no finite minimizing $a$ in this family: its infimum is $1/2$, approached as $|a|\to \infty$, when the normalized trial wavefunction approaches the ground state. This is consistent with the quantum variational principle. The original printed instruction in this subpart is therefore inconsistent as an unconstrained minimization; a stationary Rayleigh quotient need not be a minimum.

The intended excited state can instead be obtained by imposing an excited-state variational constraint. Orthogonality to the ground state requireswhich fixes $a=1$. Every member of the family is even, so it is already orthogonal to the odd $n=1$ state. With the ground state excluded, the lowest allowed oscillator energy is therefore $E_2=5/2$. In the exact oscillator basis, $\psi$ is proportional to $\sqrt{2}{\varphi }_2+(1-a){\varphi }_0$, making both the maximum and the orthogonality correction transparent.