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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.

Products of noncommuting linear operators depend on their order. For position and momentum, $xp-px=i\hslash$; substituting classical commuting expressions into a product can therefore lose an additive quantum term. For example, expansion of the oscillator product $a^{†}a$ includes a commutator contribution $-1/2$, leading to the correct zero-point energy in $H=\hslash \omega (a^{†}a+1/2)$.

For one quantum harmonic oscillator of frequency $\Omega$, $Z=e^{-x/2}/(1-e^{-x})$ with $x=\hslash \Omega /(k_{B}T)$. Differentiating $\ln Z$ gives $U=\hslash \Omega [1/2+1/(e^x-1)]$ and $C_V=k_B{x}^2{e}^x/(e^x-1{)}^2$. Zero-point energy leaves $U\to \hslash \Omega /2$ as $T\to 0$ but contributes no heat capacity. Excitation freeze-out makes $C_V$ exponentially small, while high temperature gives $U\sim k_{B}T$ and $C_V\to k_B$.

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.

A researcher whose joint trapped-ion experiment prepared motion near the ground state in three dimensions. The measured sideband strengths revealed the remaining excitation after resolved sideband cooling. Such state diagnostics distinguish low thermal occupation from the unavoidable zero-point energy of a confined quantum particle.^[ref]

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.

The three-dimensional isotropic harmonic oscillator has the same oscillator angular frequency $\omega$ in all three directions:The three independent quantum harmonic oscillators give energy eigenvalues $E_N=\hslash \omega (N+3/2)$, where $N=n_x+n_y+n_z$ and the three occupation numbers are nonnegative integers. The zero-point energy is the sum of the three one-dimensional contributions.

For fixed $N$, choosing all triples gives energy-level degeneracy $(N+1)(N+2)/2$. For example, $N=1$ has the three excitations $(1,0,0)$, $(0,1,0)$ and $(0,0,1)$. Anisotropic oscillator frequencies generally split this degeneracy. The model is useful near a rotationally symmetric minimum of a smooth three-dimensional potential energy.

The normalized eigenfunction ${\psi }_0(x)=\mathrm{sech}x/\sqrt{2}$ has zero eigenvalue under the Sine-Gordon kink fluctuation operator. It is proportional to ${\partial }_x{\varphi }_K$ and comes from shifting the collective coordinate of the kink. Its frequency is zero, so it contributes no oscillator zero-point energy; it must be handled separately from a Gaussian functional determinant.

An internal qubit uses two distinguishable electronic or hyperfine structure quantum states, denoted $|g⟩,|e⟩$. The ion's centre-of-mass motion has a different Hilbert space. In one approximately harmonic direction, a number state $|n⟩$ has energy $E_n=\hslash \nu (n+1/2)$, where $\nu$ denotes a motional angular frequency, not ordinary frequency. The tensor product of quantum systems gives $|g,n⟩$. Cooling towards $n=0$ prepares motion; it does not automatically prepare the internal qubit. Nor does the ground state mean a stationary point particle: zero-point energy and position spread remain.

For several ions, Coulomb forces couple their displacements into collective normal modes. Each selected normal mode has its own quantum harmonic oscillator number states, while each ion retains internal qubit states. A pulse addressing one ion can change a shared mode. This supplies a route to correlations between distant internal quantum states; it also makes residual motional excitation and unwanted excitation of other normal modes relevant. A single-ion pulse calculation alone establishes no experimental multi-ion quantum gate performance.

Vacuum energy is the energy assigned to a theory's vacuum state. In a free bosonic field description it includes the sum of oscillator zero-point energies, formally $\frac{1}{2}{\sum }_{\mathbf{k}}\hslash {\omega }_{\mathbf{k}}$. The continuum limit of that expression is generally divergent and requires a prescription, rather than direct identification with a finite measured energy density.

Adding a constant to a nongravitational Hamiltonian changes an overall evolution phase but not transition probabilities. Energy differences between boundary configurations can nevertheless have physical effects, as in the Casimir effect. When gravity is included, a vacuum energy density contributes to the gravitational source and is related to the cosmological constant. Normal ordering removes a reference-theory constant by convention; it does not solve every interacting or gravitational vacuum-energy problem.

A vacuum fluctuation is a nonzero quantum variance in a vacuum state. For a single electromagnetic normal mode, dimensionless optical quadratures $a+a^{†}$ and $-i(a-a^{†})$ have zero means and unit variances in the vacuum state, despite zero photon number. The mode's zero-point energy is $\hslash \omega /2$.

The harmonic-oscillator entropy contribution for one vibrational coordinate, with $x={\theta }_{\mathrm{vib}}/T$. Contributions from independent normal modes add, with one term for each component of a degenerate mode. A temperature-independent zero-point energy does not change this entropy.

Vibrational excitation raises the energy associated with internal nuclear motion in a molecule. It can occur within one electronic state; changing vibrational energy does not necessarily promote an electron. For a diatomic molecule, the simplest coordinate is a change in the separation of its two atomic nuclei.

Near a stable equilibrium bond length, the quantum harmonic oscillator model haswith adjacent spacing $\hslash \omega$. The $v=0$ state already has zero-point energy; vibrational excitation refers to raising the level above that ground state.^[ref] Anharmonicity changes the spacing and becomes important away from the potential minimum.

For a chosen vibrational spacing $\tilde{\nu }=1000\mathrm{c}{\mathrm{m}}^{-1}$, $\Delta E=hc\tilde{\nu }\approx 0.124\mathrm{eV}$, corresponding to wavelength $10\mu \mathrm{m}$. In thermal equilibrium, the population ratio of adjacent nondegenerate harmonic levels isAt $300\mathrm{K}$ it is about $0.0083$. This stipulated example illustrates why a high-frequency mode can be mostly in its vibrational ground state even though the sample is not at zero temperature.

A motivating apparatus is the University of Florida HCl and DCl FTIR experiment: an infrared transmission spectrum is measured using Fourier transform spectroscopy and a matched background. The observed rotational-vibrational bands constrain stretching energies and isotope dependence.^[ref] Whether a vibrational transition absorbs infrared light also depends on vibrational infrared activity, not just its energy spacing.

A short excitation pulse can create a nonequilibrium vibrational population. Subsequent vibrational relaxation redistributes that energy through other modes or the surroundings. Measuring a population change, a transient spectrum and energy deposited as heat are related but distinct experimental observations.

A vibrational partition function sums over molecular vibrational energy levels. A single quantum harmonic oscillator, with energies measured relative to its ground level and vibrational temperature ${\theta }_v=h\nu /k$, has $q_{\mathrm{vib}}=(1-e^{-{\theta }_v/T}{)}^{-1}$. Retaining its absolute zero-point energy multiplies this by $e^{-{\theta }_v/(2T)}$. That factor shifts the internal energy but contributes no entropy.