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Expanding the Hamiltonian about a static classical solution removes the linear term by its Euler-Lagrange field equation. For a canonical real scalar field with potential $U$, the quadratic fluctuation operator is $\mathcal{H}=-{\partial }_X^2+U^{\prime \prime }({\phi }_{\mathrm{cl}})$. Positive-frequency normal modes become quantum harmonic oscillators under canonical quantization, yielding the Gaussian contribution to a one-loop soliton mass correction. A negative eigenvalue signals an instability, while a physical zero mode in field theory must be replaced by a collective coordinate.

For a weakly damped well, quantizing the conservative part of current-biased Josephson phase dynamics gives the local Hamiltonian operatorThe tilted potential energy is unbounded downhill. Its trapped energy levels are therefore long-lived resonances with escape lifetimes, rather than globally bound stationary states. Close to ${\varphi }_m$, set $x=\varphi -{\varphi }_m$ and expandKeeping only the quadratic term gives the quantum harmonic oscillator resultIt is useful for states localized near the minimum, with energy well below $\Delta U$. It does not count all the resonances accurately near escape. The omitted terms make successive spacings unequal. For the chosen example, the three plotted reference energies divided by $h$ are $1.44804$, $4.34412$ and $7.24020\mathrm{GHz}$ above the minimum.

A small microwave bias contributes $\delta U=-(\hslash /2e)I_{\mu }(t)\varphi$. A one-photon transition between energy levels is resonant when $hf=E_{n+1}-E_n$. Exciting a higher energy level can strongly increase its escape probability. Measuring microwave absorption indirectly through escape therefore connects an invisible quantum phase state to a visible electric potential difference. A classical oscillation can also show resonance, so the mere existence of one peak is insufficient: the bias dependence and resolved unequal spacings are the discriminating predictions.

A quantum harmonic chain has quadratic couplings between canonical coordinates and momenta. A normal mode transformation turns it into independent quantum harmonic oscillators. A continuum elastic action with time coefficient $\chi$ and spatial stiffness $\kappa$ has long-wave frequency ${\omega }^2=(\kappa /\chi )q^2$, the structure of a smooth condensate phase action.

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.

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.

The simplest off-resonant explanation is a vibrating polarizability. Let $Q(t)=Q_0\cos (2\pi {\nu }_{v}t)$ be a small normal mode coordinate of molecular vibration, and use one scalar component of polarizability, $\alpha (Q)\simeq {\alpha }_0+{\alpha }_0^{\prime }Q$. For incident electric field $E(t)=E_0\cos (2\pi {\nu }_{0}t)$, the induced electric dipole moment isThe product-to-sum formula produces the two displaced frequencies. A nonzero derivative ${\alpha }_0^{\prime }$ gives vibrational Raman scattering in this linear model. Real polarizability is generally a tensor, and the selected components depend on incident and observed wave polarization. This condition differs from vibrational infrared activity, which requires a changing molecular electric dipole moment without an applied optical electric field. A vibration can therefore appear in one measurement and be absent in the other.

This classical modulation argument explains line locations but supplies equal sideband wave amplitudes; it does not explain their thermal asymmetry or the nonzero Stokes Raman scattering from a vibrational ground state. Those require the quantum harmonic oscillator population and transition strengths. The far-off-resonance drawing uses dashed virtual intermediate levels: they represent terms in a scattering probability amplitude, not a long-lived occupied electronic energy eigenstate. Near an electronic resonance, a more detailed treatment of the intermediate response is needed.

The lowest even quantum harmonic oscillator quantum state has density $\rho (y)=e^{-y^2/{\tau }^2}/(\sqrt{\pi }\tau )$, and the lowest odd quantum state has density $2y^2\rho (y)/{\tau }^2$. Since ${\rho }^{\prime }=-2y\rho /{\tau }^2$, integration by parts gives ${\mathbb{E}}_{\mathrm{F}}[f]={\mathbb{E}}_{\mathrm{B}}[f+y{f}^{\prime }]$ whenever the boundary terms vanish. For $f(y)=y^n$, this proves the displayed identity. Odd moments vanish in both quantum states by parity; even moments quantify the larger spatial spread in the odd quantum state. In particular the relative-coordinate variance is three times as large, and the odd density vanishes at particle coincidence.

Displacing a one-dimensional quantum harmonic oscillator produces a linear force term and a scalar shift relative to the unshifted Hamiltonian. The scalar adds only a common phase. The linear term couples number states separated by one excitation. Arbitrary linear driving of the ground state creates a coherent state; a first-order expansion of that state contains only the vacuum and first excited state.^[ref]

A spin-boson model couples a two-state quantum system to a collection of quantum harmonic oscillators. The oscillator modes provide environmental degrees of freedom that can exchange energy with the spin and change its reduced density matrix. One transverse-coupling convention isThe choice of coupling axis matters: here the Pauli matrix ${\sigma }_x$ flips the energy-basis spin states. Coupling only through ${\sigma }_z$ has a different effect, since it commutes with the free spin Hamiltonian operator.

For an isotropic three-dimensional quantum harmonic oscillator, $V(r)=m{\omega }^2{r}^2/2$ makes $r^{-1}{V}^{\prime }(r)=m{\omega }^2$. The spin-orbit energy shift in a central potential is therefore independent of the radial excitation. For spin one-half, the $l=0,j=1/2$ shift is zero, while the two $l=1$ shifts are $-{\hslash }^2{\omega }^2/(2m{c}^2)$ for $j=1/2$ and $+{\hslash }^2{\omega }^2/(4m{c}^2)$ for $j=3/2$.

These shifts are quadratic in trap frequency. Their degeneracy-weighted sum in the $l=1$ manifold is zero, since the multiplicities are two and four. They are perturbative corrections rather than full oscillator energies; a nonrelativistic treatment requires them to be small compared with the orbital spacings.

In Stokes Raman scattering, matter gains internal energy and the photon loses the same amount. For a quantum harmonic oscillator representing one molecular vibration, $E_v=h{\nu }_v(v+1/2)$ with $v=0,1,\dots$. The transition $v\to v+1$ gives ${\nu }_s={\nu }_0-{\nu }_v$, requiring ${\nu }_0>{\nu }_v$ for an outgoing photon. Even a molecule initially in its vibrational ground state can undergo this process. The shared name does not turn it into the absorption/emission-band Stokes shift used for fluorescence.

For a single mode in thermal equilibrium, its quantum harmonic oscillator energy ladder gives a geometric distributionIts probability generating function is $F(z)=(1-q)/(1-qz)$. Differentiating twice gives $F^{\prime \prime }(1)=2q^2/(1-q{)}^2=2{\mu }^2$, hence $g^{(2)}(0)=2$. At the chosen mean $\mu =2$, $q=2/3$, $p_0=1/3$, $p_1=2/9$, $p_2=4/27$, and $\mathrm{Var}N=6$. Rare high occupations increase the pair count even though the mean matches the other preparations. The geometric weights also arise from a Gaussian mixture of coherent states.^[ref]

Single-mode selection is essential to this factor of two. If $M$ independent equally populated thermal modes are collected together, let $N={\sum }_{j=1}^M{N}_j$ with $⟨N_j⟩=\mu /M$. Adding their variances gives $\mathrm{Var}N=\mu +{\mu }^2/M$, hence $g^{(2)}(0)=1+1/M$. Averaging over many unresolved modes can make thermal pair counts approach the independent-count reference.

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.

At fixed positive bias ratio $s=I_b/I_c$ with $0<s<1$, the Josephson coupling energy and work of the ideal bias source combine intoThis is the effective potential energy for an unwrapped Gauge-invariant Josephson phase; it includes the bias source and decreases by $2\pi s{E}_J$ on advancing one period. A local minimum and the adjacent downhill barrier areTo verify the barrier, solve $U^{\prime }=E_J(\sin \varphi -s)=0$, select $U^{\prime \prime }({\varphi }_m)>0$ and $U^{\prime \prime }({\varphi }_b)<0$, and subtract their potential energies. The small-oscillation angular frequency is fixed by the curvature:Thus the barrier and oscillation frequency both vanish as $s\to 1^{-}$, but with different powers. For $ϵ=1-s\ll 1$, expanding $\arccos (1-ϵ)$ and $\sqrt{2ϵ-{ϵ}^2}$ givesThe first approximation is for a nearly critical bias. At $s=0$ the exact downhill barrier is $2E_J$; at $s\ge 1$ there is no confining local minimum in this model.

For those stipulated parameters, direct evaluation gives $E_J/h=496.684\mathrm{GHz}$, $\Delta U/h=29.7670\mathrm{GHz}$, and ${\omega }_p/(2\pi )=2.89608\mathrm{GHz}$. Hence $\Delta U/(\hslash {\omega }_p)=10.2784$. The nearly critical barrier approximation gives $29.6165\mathrm{GHz}$, about $0.51\%$ below the exact result. These are original circuit calculations, not parameters assigned to the historical samples.

The weighted strain density in Ornstein-Uhlenbeck multiplicative amplification evolves under ${\mathcal{L}}_{n}p=(\kappa /2)p^{\prime \prime }+{\tau }^{-1}(\sigma p{)}^{\prime }+n\sigma p$. Setting $p=e^{-{\sigma }^2/(2\kappa \tau )}\psi$ eliminates its first derivative. Shifting and scaling to $x=(\sigma -\kappa n{\tau }^2)/\sqrt{\kappa \tau }$ gives ${\psi }_{xx}+[1+\kappa n^2{\tau }^3-2\gamma \tau -x^2]\psi =0$. The quantum harmonic oscillator levels yield the displayed eigenvalues. Its ground state is the dominant positive mode; higher sign-changing modes may occur in a transient expansion.

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.

Trapped-ion quantum control prepares an ion's internal quantum state, couples it to quantized motion with laser pulses, and measures the result through state-dependent fluorescence. Its appeal is access to both a controllable two-level system and a quantum harmonic oscillator in one apparatus.

Confinement, preparation, control and readout solve different problems. The Paul trap's RF electric field confines the ion through its RF-averaged trapping potential; it still produces Paul trap micromotion. Doppler cooling and resolved sideband cooling reduce motional excitation. Optical pumping selects an internal quantum state. A coherent pulse then changes amplitudes, while electron shelving converts an internal population into a detectable stream of photons. A bright/dark count by itself does not reveal a quantum superposition's relative phase; an additional controlled rotation before detection can make that phase affect the measured population.

Consider an original ideal calculation with no heating, decoherence, detuning or spectator modes. In the ordered basis $(|g,n⟩,|e,n-1⟩)$ for $n\ge 1$, the red motional sideband has $H_r=\hslash \eta \Omega \sqrt{n}{\sigma }_x/2$. Here ${\sigma }_x$ is the Pauli matrix exchanging the two basis states. Since ${\sigma }_x^2=I$, its unitary time evolution isThus $U_n^{†}{U}_n=I$, and excitation from $|g,n⟩$ has probability $P_r(n,t)={\sin }^2(\theta \sqrt{n}/2)$. For $n=0$ it vanishes. The blue motional sideband similarly gives $P_b(n,t)={\sin }^2(\theta \sqrt{n+1}/2)$. A red Rabi pi pulse calibrated for $n=1$ has $\theta =\pi$; it is not a Rabi pi pulse for every occupied $n$.

Choose $\eta =0.08$, $\Omega /(2\pi )=100\mathrm{kHz}$ and $\nu /(2\pi )=2\mathrm{MHz}$. The $n=1$ red Rabi pi pulse lasts $t_{\pi }=\pi /(\eta \Omega )=62.5\mu \mathrm{s}$. Let motion have the Boltzmann distribution of a quantum harmonic oscillator, with chosen mean occupation $\bar{n}=0.25$:Here $\sum p_n=1$, $\sum n{p}_n=\bar{n}$ and $p_0=0.8$. Reindex the red sum using $n=m+1$ and $p_{m+1}=r{p}_m$: $P_r(t)=r{P}_b(t)$ for equal resonant pulse parameters, even outside the short-pulse limit. At times with $P_b>0$, the ratio $R=P_r/P_b$ therefore gives $\bar{n}=R/(1-R)$. A nonthermal distribution, unequal drive strengths, detuning or state-detection bias can invalidate that thermometer. Ground-state population cannot generally be deduced from the mean alone without specifying the distribution.

The chosen pulse gives $P_r\simeq 0.181393$ and $P_b\simeq 0.906967$. These are internal excitation probabilities, not quantum gate fidelities or cooling-cycle efficiencies. An independent matrix exponentiation of each two-state block agrees with the sine formula and preserves normalization. Summing $n=0,\dots ,79$ omits probability $r^{80}<1.21\times 10^{-56}$; because every sine squared lies between zero and one, this also bounds each omitted response. The thermal relation supplies a separate dimensional check:Here $k_B$ is the Boltzmann constant. Substitution recovers $\bar{n}=0.25$; the thermal root-mean-square expansion parameter is $\eta \sqrt{2\bar{n}+1}\simeq 0.098$. This checks the populated part of the Lamb-Dicke regime, while exponentially rare large $n$ still exceed any fixed small-parameter criterion. The pulse is a leading-order model; it does not bound corrections from the full coupling or experimental noise.

The spatial ground orbital of a three-dimensional quantum harmonic oscillator has $l=0$ and energy $3\hslash {\omega }_0/2$; its first excited shell has $l=1$ and energy $5\hslash {\omega }_0/2$. Two identical electrons both in the ground orbital must form the singlet state, giving one state at total energy $3\hslash {\omega }_0$.

At energy $4\hslash {\omega }_0$, one electron occupies each shell. There are two ground-shell spin orbitals and six first-excited-shell spin orbitals, hence twelve Slater determinants. Equivalently, $0\otimes 1$ gives $L=1$; symmetric spatial combinations with the singlet state give three states, while antisymmetric spatial combinations with the triplet states give nine. Distinct occupied orbitals allow both spatial exchange symmetries, so both spin sectors must be kept.