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Use natural units and the Minkowski metric ${\eta }_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$. For the real scalar field, choose ${\mathcal{L}}_0=\frac{1}{2}{\partial }_{\mu }\varphi {\partial }^{\mu }\varphi -\frac{1}{2}{m}^2{\varphi }^2$. The conjugate momentum is $\pi =\dot{\varphi }$. Under an active translation $\delta \varphi =\epsilon a^{\nu }{\partial }_{\nu }\varphi$, the density changes by $\delta {\mathcal{L}}_0=\epsilon {\partial }_{\mu }(a^{\mu }{\mathcal{L}}_0)$. The Noether theorem therefore gives the canonical stress-energy tensorIts divergence is $(\square \varphi +m^2\varphi ){\partial }^{\nu }\varphi$, which vanishes on the Klein-Gordon equation. With vanishing flux at spatial infinity, the charges are conserved. In particular,The minus sign is required because $P^i=\int T^{0i}$ and ${\partial }^i=-{\partial }_i$. These are the four-momentum of a free real scalar field.

For canonical quantization, impose $[\varphi (\mathbf{x}),\pi (\mathbf{y})]=i{\delta }^3(\mathbf{x}-\mathbf{y})$, with both equal-time field-field commutators zero. Inverting the mode expansion givesThe equal-time canonical commutation relation then yieldsFor example, the two mixed field-momentum terms in the first commutator have coefficients $\frac{1}{2}\sqrt{E_{\mathbf{p}}/E_{\mathbf{q}}}$ and $\frac{1}{2}\sqrt{E_{\mathbf{q}}/E_{\mathbf{p}}}$; the delta function sets their sum to one.

Insert the mode expansion into the quadratic energy. Spatial integration supplies $(2\pi {)}^3{\delta }^3(\mathbf{p}\pm \mathbf{q})$. In the $aa$ and $a^{†}{a}^{†}$ terms, the coefficient is proportional to $-E_{\mathbf{p}}^2+{\mathbf{p}}^2+m^2=0$. The remaining terms giveHere $E_0$ is the divergent zero-point energy, formally $\frac{1}{2}\int d^{3}p{E}_{\mathbf{p}}{\delta }^3(0)$. Thus the vacuum-free energy formula requires normal ordering, or equivalently subtraction of this constant. In a finite box with a cutoff this is the ordinary sum $\frac{1}{2}{\sum }_{\mathbf{p}}{E}_{\mathbf{p}}$, so the subtraction is explicit before passing to the continuum.

Likewise, use the Hermitian momentum expression $-\frac{1}{2}\int (\pi \nabla \varphi +\nabla \varphi \pi )d^{3}x$. Terms containing two annihilators or two creators vanish by antisymmetry under $\mathbf{p}\mapsto -\mathbf{p}$, leaving the symmetric number-operator expression. Its vacuum term is zero with an inversion-symmetric regulator. Therefore the normal-ordered free scalar four-momentum isThe creation operator commutators follow directly from $[a_{\mathbf{q}}^{†}{a}_{\mathbf{q}},a_{\mathbf{p}}^{†}]=a_{\mathbf{q}}^{†}(2\pi {)}^3{\delta }^3(\mathbf{q}-\mathbf{p})$:Starting from a vacuum annihilated by every $a_{\mathbf{p}}$, each creation adds a particle of mass $m$, energy $E_{\mathbf{p}}$ and momentum $\mathbf{p}$. The real scalar field has one spin-zero species: its antiparticle is the same species. Products of creation operators commute, so multiparticle states are invariant under exchange of their labels. In a normalized discrete mode, $(a^{†}{)}^r|0⟩/\sqrt{r!}$ exists for every $r\ge 0$; there is no exclusion restriction. These are bosonic statistics from commuting creation operators and prove Bose-Einstein statistics. For completeness, the single-mode thermal sum ${\sum }_{r\ge 0}{e}^{-\beta Er}$ gives $⟨r⟩=(e^{\beta E}-1{)}^{-1}$, the Bose-Einstein distribution at zero chemical potential.

To keep the paper's coupling convention explicit, write $g=\sqrt{\beta }$ and use physical coordinates $X=x/m$, $T=t/m$. Assume $\beta >0$ and weak coupling $g\ll 1$. The canonical real scalar field is $\phi =\varphi /g$, and the physical potential isThe symbol $g$ is the coupling usually appearing in canonical Sine-Gordon theory; the paper's $\beta$ is its square. The vacua have $g\phi =2\pi n$. A static Sine-Gordon kink has $g{\phi }_K=4\arctan e^{m(X-X_0)}$ and satisfies $\frac{1}{2}{\phi }_K^{\prime 2}=U({\phi }_K)$. Since $1-\cos (g{\phi }_K)=2{\mathrm{sech}}^2[m(X-X_0)]$, its classical mass isThe topological sector is important: this energy is measured relative to a vacuum, and the kink joins distinct vacua at the two spatial ends. Expanding around a spatially constant vacuum cannot construct this state by any finite-order perturbation in $g$.

For the general one-loop soliton mass correction, start with a canonical scalar field potential $U$ and a stable static soliton ${\phi }_K$. Write $\phi ={\phi }_K+\eta$. The term linear in $\eta$ vanishes by the classical Euler-Lagrange field equation. The quadratic fluctuation Hamiltonian isChoose a common large box and a common finite-mode regularization in quantum field theory. Expand the nonzero normal modes as $\eta ={\sum }_n{q}_n{f}_n$, with ${\mathcal{H}}_K{f}_n={\omega }_n^2{f}_n$ and normalized eigenfunctions. Each pair $(q_n,p_n)$ is a quantum harmonic oscillator contributing ground-state energy $\hslash {\omega }_n/2$. In the vacuum, replace ${\mathcal{H}}_K$ by ${\mathcal{H}}_0=-{\partial }_X^2+U^{\prime \prime }({\phi }_{\mathrm{vac}})$. Subtract the two ground-state energies and add the local counterterms evaluated on the soliton relative to the vacuum. This derives the general formulaThe prime excludes exact zero modes in field theory from oscillator quantization; their zero frequencies contribute no zero-point energy, but their role in mode counting must not be forgotten. The two sums mean a paired finite regulator, not separate subtractions of divergent answers. Equivalently the nonzero-mode term is the regulated difference of the square-root traces of the two fluctuation operators. Discrete bound states and continuum modes both contribute. This is a vacuum-subtracted soliton mass and the first term in the semiclassical soliton mass expansion.

Translation gives a zero mode in field theory because differentiating the static equation yields ${\mathcal{H}}_K{\phi }_K^{\prime }=0$. A Gaussian oscillator or an unprimed functional determinant is inappropriate along this flat direction. Replace its amplitude by the position collective coordinate $X_0(T)$ and require the residual fluctuation to obey $\int \eta {\phi }_K^{\prime }dX=0$, preventing double counting. The associated change-of-variables Jacobian supplies the zero-mode normalization. At low speed the collective-coordinate effective Lagrangian for a soliton is $-M_{\mathrm{cl}}+\frac{1}{2}{M}_{\mathrm{cl}}{\dot{X}}_0^2+\cdots$; quantizing the position gives the soliton momentum and its translational states, not an extra oscillator rest energy. More generally, every physical continuous modulus needs a collective coordinate; gauge directions require gauge fixing rather than additional physical states.

The Sine-Gordon kink fluctuation operator is particularly simple:while $D{D}^{†}=-{\partial }_X^2+m^2={\mathcal{H}}_0$. This supersymmetric factorization of the one-soliton potential shows stability and generates all nonzero eigenfunctions from vacuum plane waves. The sole normalizable bound state is the translational zero mode of a sine-Gordon kink, proportional to $\mathrm{sech}m(X-X_0)$; there is no positive-frequency internal bound oscillator. The continuum has $\omega (k)=\sqrt{k^2+m^2}$ and no reflection. Applying $D^{†}$ to $e^{ikX}$ gives a transmission amplitudeThis scattering phase shift changes the density of continuum modes. The high-frequency vacuum subtraction cancels the extensive vacuum contribution but still leaves a logarithmic ultraviolet divergence.

The finite part also requires consistent mode-number regularization of soliton masses. The periodic-box phase-shift quantization condition is $k_{n}L+\delta (k_n)=2\pi n$. Match $2N+1$ modes: the kink has its translation mode plus the two continuum modes at each $n=1,\dots ,N$, while the vacuum has the $k=0$ oscillator of frequency $m$ and the corresponding continuum pairs. With $\hslash =1$ and $\Lambda =2\pi N/L$, expanding $k_n-k_n^{(0)}=-\delta (k_n^{(0)})/L+o(L^{-1})$ givesIntegration by parts uses $\delta (0)=\pi$ and yieldsThe cutoff surface term for a Sine-Gordon kink tends to $-m/\pi$. It cannot be dropped merely because $\delta (\Lambda )\to 0$: $\omega (\Lambda )$ grows at the same time.

A renormalization condition must specify which mass and coupling are held fixed. For vacuum normal ordering, or cancellation of the vacuum tadpole diagram with the elementary mass fixed at $m$, the quartic interaction gives the mass countertermEvaluating this Sine-Gordon vacuum tadpole counterterm on the kink givesThe logarithmic ultraviolet divergence cancels, leaving the renormalized one-loop Sine-Gordon kink mass in the stated vacuum scheme:This example illustrates why a zero mode in field theory must be treated as a collective coordinate, why vacuum subtraction alone need not remove ultraviolet divergences, and why the finite relation between the two regulators matters. Different finite counterterms amount to different definitions of the renormalized parameters; an unregulated frequency difference without a renormalization condition is not a physical mass prediction.

Choose boundary states with nonzero overlap with the lowest-energy state in the vacuum sector $Q=0$ and in the one-kink topological sector $Q=1$. In a finite spatial box, a fixed scalar field configuration eigenstate may be used formally, with temporal endpoints equal to the chosen configuration. More generally, smear the endpoints with wavefunctionals ${\Psi }_Q$. Their unitary time evolution kernels areThe inner scalar field path integral remains in the chosen topological sector, with $\varphi (+\infty )-\varphi (-\infty )=2\pi Q$. A zero-total-momentum projection can be included to select the rest state; alternatively, the translational prefactor does not change the large-time exponential. After a Wick rotation to physical Euclidean time $\tau$, the energy eigenstate expansion gives $K_Q^E(\tau )\sim C_Q{e}^{-E_Q\tau }$. Thus the exact vacuum-subtracted soliton mass isEquivalently it is $i{T}^{-1}\log (K_1/K_0)$ at large time with a damping prescription. The ratio subtracts the vacuum energy; the Hamiltonian operator and action here are the fully regulated and renormalized ones, not merely their classical approximations.

With the dimensionless coordinates of this paper, the correctly normalized classical action isFor the static Sine-Gordon kink, ${\varphi }_K^{\prime }=2\mathrm{sech}x$ and $1-\cos {\varphi }_K=2{\mathrm{sech}}^{2}x$, so $M_{\mathrm{cl}}=8m/{\beta }^2$. Write $\eta =\delta \varphi$. Expanding and integrating by parts givesThe first variation is ${\beta }^{-2}\int ({\varphi }_K^{\prime \prime }-\sin {\varphi }_K)\eta$, plus boundary terms. It vanishes because the kink satisfies the Euler-Lagrange field equation and the fluctuations have the prescribed temporal endpoints and admissible spatial boundary behavior. This is the principle of stationary action, not a symmetry assumption about $\eta$.

The printed expansion omits ${\beta }^{-2}$ despite the stated Lagrangian density. Its displayed form is the expansion of ${\beta }^{2}S$; it is not the physical $S$ at arbitrary coupling. Alternatively, writing $\eta =\beta \chi$ puts the quadratic term in canonical normalization, while leaving the classical term $-8mT/{\beta }^2$ unchanged. This normalization repair does not change ${\Delta }_x$ or the physical fluctuation frequencies.

The Sine-Gordon kink fluctuation operator has the useful factorizationIt is nonnegative, and $A{\psi }_0=0$ gives the normalized translational zero mode of a sine-Gordon kink:A displacement $a$ changes ${\varphi }_K(x-a)$ by $-a{\varphi }_K^{\prime }(x)$. Thus the zero mode in field theory is the position collective coordinate of the kink, reflecting translation invariance. It has no restoring force and no oscillator zero-point energy. Integrate that collective coordinate separately rather than inserting a zero factor into the Gaussian functional determinant.

In the Gaussian fluctuation approximation, the formal oscillator contribution to the one-loop soliton mass correction, before adding any counterterm, isUse a common regulator for the two sums, include all discrete modes, and treat the translation mode as above. The vacuum sum is essential: subtracting only classical vacuum energy would leave an extensive oscillator energy. A periodic fluctuation and its derivative are matched at the two ends of the large box; the one-kink background lies in the twisted topological sector, and its infinite-line profile is accurate up to exponentially small boundary corrections.

For a continuum scattering wavefunction, equality of its two asymptotic values gives the periodic-box phase-shift quantizationAway from the threshold, expand at a matched mode number:Since consecutive free wave numbers are separated by $2\pi /L$, replacing the continuum-mode sum by an integral proves the displayed continuum contribution:The ultraviolet cutoff is retained until the counterterm is added.

There is a finite threshold issue if this expression is identified with the complete oscillator correction. It can be settled directly using the factorization: a continuum eigenfunction is $f_k\propto (\tanh x-ik)e^{ikx}$. Its transmission phase obeyswith the odd phase branch that tends to zero at large $|k|$. For $k>0$, the continuum roots have labels $n=1,2,\dots$; there is no periodic continuum root at $k=0$, since the limiting eigenfunction $\tanh x$ has opposite signs at the two ends. The bound state at ${\omega }_0=0$ replaces the free $k=0$ oscillator with ${\omega }_0^{(0)}=1$. Consequently, in mode-number regularization of soliton masses,The PDF's continuum-only formula misses this finite $-m/2$ under this standard phase and mode-counting convention. It has the correct logarithmic ultraviolet divergence, but the missing term is not suppressed by large $L$. Changing the phase branch requires changing the mode labels and endpoint terms consistently; it cannot erase a physical mode from the formal spectrum sum.

At high momentum, $\delta (k)=2/k+O(k^{-3})$, so both expressions have divergent part $-(m/\pi )\log \Lambda$. The canonical field $\phi =\varphi /\beta$ has a quartic interaction with coupling $-m^2{\beta }^2$. Its vacuum tadpole diagram shifts the squared mass by $-m^2{\beta }^{2}J(\Lambda )/4$, whereThe Sine-Gordon vacuum tadpole counterterm has $\delta m^2=+m^2{\beta }^{2}J/4$ and adds potential energy density $\delta m^2(1-\cos \varphi )/{\beta }^2$. Since $\int (1-\cos {\varphi }_K)dx=4$, its vacuum-subtracted kink energy iswhich cancels the logarithmic ultraviolet divergence. Finite parts require a specified renormalization condition. As a consistency check, with this tadpole subtraction and matched mode-number cutoff, integration by parts yieldsThe complete semiclassical soliton mass is then $8m/{\beta }^2-m/\pi +O(m{\beta }^2)$ in that convention. This last finite result uses the bound-mode term and is additional to the requested ultraviolet cancellation.

For $\omega >0$, the quantum harmonic oscillator has energies $E_r=\omega (r+1/2)$, $r=0,1,\dots$, in units $\hslash =1$. For positive inverse temperature $\beta$, taking the trace in the energy basis and summing a geometric series givesThis is the thermal partition function of a quantum harmonic oscillator, including its zero-point energy. The assumptions $\beta ,\omega >0$ ensure convergence; at $\omega =0$ the free particle on the noncompact line instead has an infinite spatial-volume factor.

In the displayed version of light-cone gauge in string theory, the oscillators have $D-2$ transverse components andThe center-of-mass Poisson brackets remain $\{x^m,p_n\}={\delta }_n^m$, and the transverse oscillator Poisson brackets are $\{{\alpha }_j^a,{\alpha }_k^b\}=-ij{\delta }^{ab}{\delta }_{j+k,0}$. Other independent brackets vanish. With $\hslash =1$, canonical quantization gives the canonical commutation relationsThe oscillator Fock vacuum at momentum $p$ is defined by ${\alpha }_k^a|0;p⟩=0$ for $k>0$. It is the ground state of one string, rather than the empty spacetime vacuum. Set $a_k^a={\alpha }_k^a/\sqrt{k}$ for $k>0$. The normal ordering prescription gives the string level operatorA Fock state basis is obtained by applying ${\prod }_{k,a}((a_k^a{)}^{†}{)}^{n_{ka}}/\sqrt{n_{ka}!}$ to $|0;p⟩$, with string level operator eigenvalue $N={\sum }_{k,a}k{n}_{ka}$. In particular, its level-one states areThey transform as the transverse vector of the little group rotation subgroup $SO(D-2)$, precisely the vector-particle polarizations of a massless vector. A massive vector would instead require $D-1$ vector-particle polarizations. This conclusion uses a quantization compatible with the Lorentz group of the bosonic string theory.

The classical mass constraint alone has no quantum zero-point energy shift. Its quantum version includes the normal-ordering constant of a string $a$:Masslessness at level one fixes $a=1$, soThus the ground state is a tachyon. Without the quantum ordering shift, the displayed classical constraint would give $M_1^2=2\pi T$ and would not support the stated massless interpretation. In the usual transverse vacuum regularization, $a=(D-2)/24$; consistency with $a=1$ also gives the critical dimension of string theory $D=26$.

In a box of volume $V$, the electromagnetic mode density in a box, including the two transverse polarizations, isEach photon has energy $\hslash \omega =\hslash ck$, and the given mean occupation number is $(e^{\hslash ck/T}-1{)}^{-1}$. Omitting the temperature-independent zero-point energy, the total energy is thereforeWith the change of variables $x=\hslash ck/T$,

The term ${\hat{p}}^{,2}/2$ is the target-space zero-mode kinetic energy. The constant $-1/12$ is the zero-point energy of the left- and right-moving oscillators: each chiral boson contributes $-1/24$. Finally,count the energies of the two independent sets of string oscillators; one excitation of mode $n$ has energy $n$.

Planck's law gives the spectral radiance of an ideal blackbody in thermal equilibrium at absolute temperature $T>0$. It predicts how much power is carried in each small frequency or wavelength interval, per projected area and per solid angle. The whole curve is determined by $T$: no material-specific parameter enters this ideal Planck function.

Writing $\nu$ for frequency in hertz and $\lambda$ for wavelength in metres, the two forms of the Planck function areHere $h$ is the Planck constant, $k_B$ the Boltzmann constant, and $c$ the speed of light in vacuum. Their exact values in the International System of Units are $h=6.62607015\times 10^{-34}\mathrm{J}\mathrm{s}$, $k_B=1.380649\times 10^{-23}\mathrm{J}{\mathrm{K}}^{-1}$, and $c=299792458\mathrm{m}{\mathrm{s}}^{-1}$. ^[ref]

The units matter: $B_{\nu }$ is measured in $\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}\mathrm{H}{\mathrm{z}}^{-1}$, whereas $B_{\lambda }$ is measured in $\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}{\mathrm{m}}^{-1}$. Both are spectral radiance, rather than radiant exitance or energy density. To report $B_{\lambda }$ per micrometre, multiply its per-metre value by $10^{-6}$; a micrometre-wide interval is $10^{-6}$ metres wide.

For a computed example, at $T=6000\mathrm{K}$ and $\lambda =0.500\mu \mathrm{m}$, the exponent is $hc/(\lambda k_{B}T)\approx 4.79592$. Substitution into Planck's law gives $B_{\lambda }\approx 3.17569\times 10^{13}\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}{\mathrm{m}}^{-1}$, or $3.17569\times 10^7\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}$ per micrometre. At the same wavelength, the illustrative temperatures $3000\mathrm{K}$ and $4500\mathrm{K}$ give $2.60268\times 10^5$ and $6.37793\times 10^6$ in the latter units. These are ideal predictions for stipulated temperatures.

The key change of variables is $\nu =c/\lambda$. Corresponding positive intervals carry the same power, so $B_{\lambda }d\lambda =B_{\nu }|d\nu |$ andEqual-width wavelength intervals have unequal widths in frequency; the factor $c/{\lambda }^2$ compensates for this. Merely replacing the horizontal-axis labels of $B_{\nu }$ by $c/\nu$ does not produce $B_{\lambda }$. This explains the frequency and wavelength peaks of Planck's law.

A short derivation uses the Boltzmann distribution for a single electromagnetic mode with thermal energies $nh\nu$, $n=0,1,\dots$. Put $q=e^{-h\nu /(k_{B}T)}$. The geometric series gives $Z={\sum }_{n\ge 0}{q}^n=(1-q{)}^{-1}$ and mean thermal energy $h\nu q/(1-q)$. The electromagnetic mode density in a box, including both independent polarization of an electromagnetic wave states of a photon, is $8\pi {\nu }^2/c^3$ per volume per unit frequency. Multiplication gives spectral energy density $u_{\nu }=8\pi h{\nu }^3/[c^3(e^{h\nu /(k_{B}T)}-1)]$; isotropy gives $B_{\nu }=c{u}_{\nu }/(4\pi )$. This counts thermal excitation energy, excluding zero-point energy. ^[ref]

Planck's law connects quantum mechanics to a measurable radiation curve. Its small-$h\nu /(k_{B}T)$ limit is the Rayleigh-Jeans law, $B_{\nu }\approx 2k_{B}T{\nu }^2/c^2$; its large-$h\nu /(k_{B}T)$ limit is the Wien approximation, $B_{\nu }\approx (2h{\nu }^3/c^2)e^{-h\nu /(k_{B}T)}$. The suppression at high frequency makes the total energy density finite. Integrating the curve gives the Stefan–Boltzmann law, while locating its peak gives Wien's displacement law. Thus one formula predicts both the spectral shape and the total emission.

Expand every transverse closed string coordinate in a real orthonormal Fourier basis of the spatial circle. The canonical momenta are $p_{Ia}=T{\dot{q}}_{Ia}$, and spatial orthogonality reduces the field Hamiltonian to the displayed independent harmonic oscillators. Canonical quantization gives ${\omega }_a(a_{Ia}^{†}{a}_{Ia}+1/2)$ for each nonzero-frequency mode; a zero-frequency mode retains the free particle term $p_{Ia}^2/(2T)$. The formal zero-point energy requires a regulator; it is not automatically the massless flat-background intercept. Opposite travelling-wave oscillator combinations diagonalize the translation generator and satisfy closed-string level matching.

The quantum harmonic oscillator describes a particle of mass $m>0$ in a quadratic potential energy with angular frequency $\omega >0$. In the position representation its Quantum Hamiltonian isThe model is useful far beyond a literal spring: a Taylor series about a stable equilibrium with positive potential curvature has this quadratic term as its leading nonconstant approximation. The pure quartic oscillator has zero curvature at its minimum, so that particular approximation does not apply. ^[ref].

In units with $\hslash =1$, creation and annihilation operators giveHere $N$ is the number operator. Restoring the reduced Planck constant, the energy eigenvalues are $E_n=\hslash \omega (n+1/2)$ for $n=0,1,2,\dots$: neighboring levels are separated by $\hslash \omega$, and the ground-state energy is the zero-point energy $\hslash \omega /2$.

With length scale $\ell =\sqrt{\hslash /(m\omega )}$ and dimensionless coordinate $u=x/\ell$, the ground state is the normalized Gaussian wave packetFor $\hslash =m=\omega =1$, this is precisely the $a=1$ trial wavefunction in the quantum variational principle. The balance of kinetic energy and potential energy therefore recovers the exact ground-state energy using only Gaussian integrals.

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.

Take an invented diatomic molecule with a Morse potential, setting the separated-atom energy to zero:Here $a$ is an inverse length, $r_e$ the equilibrium bond length, and ${\omega }_e=a\sqrt{2D_e/\mu }$, with reduced mass $\mu$. These stipulated parameters describe a simple model, not a measured chemical substance. For the one-dimensional Morse oscillator, energy eigenvalues above the minimum areConsequently,The harmonic estimate $A/2=0.1000\mathrm{eV}$ is slightly larger than this model's exact zero-point energy. The physical bound-level restriction matters: the quadratic expression is not an infinite ladder of vibrational quantum states.

For a separate thermal illustration using this same stipulated $D_0$, assume an ideal gas, fully active molecular rotation of the parent, negligible thermal molecular vibration, and negligible electronic excitation for all species. Each monatomic product contributes $\delta h=5RT/2$; the diatomic molecule contributes $7RT/2$. Thus the net increment is $3RT/2$, with molar gas constant $R$. At $T=300\mathrm{K}$,This last result follows from the stated thermal model, not a universal correction for every chemical bond. Frozen rotations, accessible excited states, different fragment composition, or a condensed phase change the thermal increments.

For a quantum harmonic oscillator with $\hslash =1$, $\omega >0$ and inverse temperature $\beta >0$, the canonical partition function is ${\sum }_{r\ge 0}{e}^{-\beta \omega (r+1/2)}$. The geometric series gives the displayed formula. Keeping the zero-point energy is essential to its prefactor.