Ground-state energy

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 $E_0$.

The quantum variational principle estimates the lowest energy by choosing a trial wavefunction and computing its expectation value, without solving the Schrödinger equation exactly. Minimizing over a manageable family of trial wavefunctions gives an upper bound; enlarging that family can improve the estimate.

Let $H$ be a self-adjoint operator bounded below, and put $E_{\ast }=\inf \sigma (H)$. For $\Vert \psi \Vert =1$ in its operator domain $D(H)$,Trial states may also lie in the form domain $Q(H)=D((H-bI{)}^{1/2})$, where $b\le E_{\ast }$. Then use the quadratic form of a positive quantum Hamiltonian shifted by $b$:This extends the expectation value beyond $D(H)$ without requiring $H\psi$ to exist. Equality for a normalized state holds exactly in the energy eigenspace at $E_{\ast }$, if that eigenvalue exists. ^[ref].

A short proof makes the direction of the bound memorable. If $H$ has a complete orthonormal basis of energy eigenstates ${\varphi }_n$, writeThe expectation value is a weighted average of allowed energies:Equality requires every occupied energy eigenstate to have energy $E_{\ast }$. ^[ref]. With continuous spectrum, the spectral theorem replaces the sum by an integral against the scalar spectral measure; the same nonnegative-difference argument works.

For an unnormalized trial wavefunction, divide by its squared norm: $R_H[\psi ]=q_H[\psi ]/\Vert \psi {\Vert }^2$ is the form version of the Rayleigh quotient. The quantum variational principle supplies an upper bound on $E_{\ast }$; it supplies neither a lower bound on $E_{\ast }$ nor, by itself, a guarantee that a bound state exists.

To see the method at work, choose the real Gaussian wave packetThe Gaussian integral gives wavefunction normalization and the two useful position averages:Thus its position probability density is a normal distribution with standard deviation $1/\sqrt{2a}$: large $a$ means narrow confinement. In units $\hslash =m=1$, integration by parts and ${\psi }_a^{\prime }=-ax{\psi }_a$ give the kinetic energyThese smooth, rapidly decaying trial wavefunctions belong to the operator domains of both polynomial Quantum Hamiltonians below. Narrowing the Gaussian wave packet increases its kinetic energy; widening it increases the potential energy in either confining potential.

For the quantum harmonic oscillator, take $\hslash =m=\omega =1$:The trial values at $a=1/2,1,2$ are respectively $0.625,0.5,0.625$. The derivative $E_{\mathrm{ho}}^{\prime }(a)=(1-a^{-2})/4$ changes sign at $a=1$, where kinetic energy and potential energy are both $1/4$. Here the trial family contains an exact ground state:The zero-point energy remains positive even at the best width. For comparison, removing the potential leaves $E(a)=a/4\to 0$ as $a\to 0$: the free Quantum Hamiltonian on the line has spectral bottom zero but no normalizable zero-energy energy eigenstate. A sequence of ever broader trial wavefunctions need not approach an actual bound state.

For a pure quartic oscillator, keep $\hslash =m=1$ and set the quartic coefficient to one:The trial values at $a=1,2,3$ are respectively $1,0.6875,0.833333\dots$. Setting the derivative to zero givesThe sign of this derivative changes from negative to positive at the optimum, giving the smallest energy within this Gaussian wave packet family:At this width, $⟨T⟩\approx 0.454280$ and $⟨V⟩\approx 0.227140$, so $2⟨T⟩=4⟨V⟩$, as expected from the power-law quantum virial theorem for a scale-optimized trial wavefunction. The number $0.681420$ is an upper estimate of the ground-state energy, not an asserted exact energy eigenvalue. Indeed,depends on $x$ for every $a>0$, so no member of this trial family is an energy eigenstate. Nevertheless, a single width parameter already turns a difficult Schrödinger equation into a useful, explicit energy estimate.

For $V=-A{e}^{-\mu r}/r$, the normalized exponential with scale $\alpha$ gives $E(\alpha )={\hslash }^2{\alpha }^2/(2m)-4A{\alpha }^3/(\mu +2\alpha {)}^2$. Choosing $\alpha =\mu /2$ proves binding if $\mu <Am/{\hslash }^2$. Failure to obtain a negative trial expectation outside this range does not prove absence of a bound state. At equality a small positive admixture of the scale-$\mu$ exponential has a negative cross matrix element and proves binding too.

For a bound stationary state in a homogeneous potential $V(\lambda x)={\lambda }^{n}V(x)$, scaling a normalized wavefunction gives stationarity of ${\lambda }^2⟨T⟩+{\lambda }^{-n}⟨V⟩$ at $\lambda =1$. Hence $2⟨T⟩=n⟨V⟩$.