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A bond-dissociation enthalpy is the molar enthalpy change for the specified homolysis at temperature $T$, with reactant and products in specified standard states. Its usual units are $\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. The superscript $\circ$ specifies standard states; it does not fix $T$ to room temperature.^[ref]

Write $\delta h_i(T)=H_{m,i}^{\circ }(T)-H_{m,i}^{\circ }(0)$ for each species' molar thermal enthalpy increment. With $D_0$ an energy per parent molecule,Here $N_{\mathrm{A}}$ is the Avogadro constant. The thermal increments include translational kinetic energy, molecular rotation, thermally excited molecular vibration and electronic states when populated, and the $pV$ contribution to enthalpy. They are additional to the zero-point energies already included in $D_0$. If values of standard enthalpy of formation at the same $T$ are available, Hess's law gives the same quantity as ${\Delta }_f{H}^{\circ }(A)+{\Delta }_f{H}^{\circ }(B)-{\Delta }_f{H}^{\circ }(AB)$.

Units alone do not identify the convention. Multiplying either $D_e$ or $D_0$ by the Avogadro constant gives molar energy, but does not add thermal enthalpy. One electronvolt per dissociation corresponds to $96.485332\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. A spectroscopic wavenumber $\tilde{\nu }$ represents energy $hc\tilde{\nu }$, using the Planck constant $h$ and speed of light $c$; $1\mathrm{c}{\mathrm{m}}^{-1}$ corresponds to $0.011962657\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$.^[ref]

The Einstein model of a solid represents the vibrations of $N$ chemical atoms by $3N$ independent one-dimensional quantum harmonic oscillators, all with the same positive angular frequency $\omega$. Equivalently, each chemical atom has three independent oscillator directions. The common angular frequency is a simplifying model assumption, not a claim about every vibration of a real solid. Hold the volume, $N$ and $\omega$ fixed. Define the characteristic temperature ${\theta }_E=\hslash \omega /k_B$ and the dimensionless ratio $x={\theta }_E/T$. The vibrational internal energy isThe first term is the zero-point energy $3N\hslash \omega /2$. It is constant at fixed $\omega$, so it contributes no heat capacity at constant volume. The temperature-dependent term givesThis is the vibrational heat capacity at constant volume of the model, with no expansion or other contributions included. ^[ref]

For a worked comparison, choose the model parameter ${\theta }_E=300\mathrm{K}$. The normalized heat capacity at constant volume depends only on $T/{\theta }_E$:

At $T={\theta }_E$, one mole of chemical atoms therefore has vibrational heat capacity at constant volume $3R(0.92067)\simeq 22.965\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$, using the unrounded model value. Even at twice the characteristic temperature, the model remains about $2.06\%$ below the Dulong-Petit law limit. These are calculations for the ideal Einstein solid, not material measurements.

A short derivation uses the thermal partition function of a quantum harmonic oscillator. The energy eigenvalues $E_n=\hslash \omega (n+1/2)$ and their Boltzmann factors form a geometric series, giving the one-oscillator canonical partition function $z=e^{-x/2}/(1-e^{-x})$. The mean occupation is $1/(e^x-1)$, hence the displayed $U$ after summing over $3N$ independent quantum harmonic oscillators. Taking its derivative with $dx/dT=-x/T$ gives the displayed heat capacity at constant volume. ^[ref]

The two limits explain the shape. At high temperature, a Taylor series in small $x$ givesAt low temperature, $1-e^{-x}\to 1$, soThe zero-point energy remains even when the thermal response is tiny: freeze-out means that changing temperature scarcely changes the occupations, not that the chemical atoms have zero energy. This is the useful lesson of the Einstein model of a solid: the classical counting rule reappears when many levels are accessible, while discrete excitation spacings explain its failure upon cooling.

The exponential low-temperature decay is also the model's limitation. Real three-dimensional crystalline solids have acoustic phonon branches with arbitrarily small angular frequencies in the bulk limit. The Debye model includes these collective normal modes and predicts a low-temperature vibrational contribution $C_V\propto T^3$, which the single-frequency Einstein solid cannot reproduce. This qualification concerns the vibrational contribution; additional excitations can change a material's total heat capacity. ^[ref]

Einstein introduced the quantum treatment of solid vibrations in his 1907 paper Die Plancksche Theorie der Strahlung und die Theorie der spezifischen Wärme. Its historical importance is that quantum excitation energies explained why the classical Dulong-Petit law could fail; the modern quantum harmonic oscillator formulation above includes the zero-point energy explicitly. ^[ref]

The equipartition theorem says that each independent quadratic term in a Hamiltonian described by classical mechanics has mean energy $k_{B}T/2$ in a canonical ensemble. Here $k_B$ is the Boltzmann constant and $T>0$ the absolute temperature. Count terms in the energy, rather than vaguely counting everything that can move: one harmonic oscillator has one position degree of freedom but TWO quadratic terms, one in momentum and one in displacement.

A precise sufficient setting is $H(y,\xi )=a{y}^2/2+H_0(\xi )$ with $a>0$, where $y$ is one of the canonical variables, $\xi$ denotes all remaining variables, and the phase space integration measure factors as $dyd\xi$. The range of $y$ is the whole real line, independently of $\xi$; there is no constraint coupling its allowed values to other variables. The partition function must be finite, and the relevant expected values must exist. The Boltzmann factor then givesThe probability density function ${\rho }_y$ is a normal distribution with variance $k_{B}T/a$. A larger coefficient narrows the fluctuations enough to cancel its greater energy cost. This is why the mean contribution is independent of mass or spring constant.^[ref]

For a proof outline, the one-variable Gaussian integral is $Z_y={\int }_{-\infty }^{\infty }{e}^{-\beta a{y}^2/2}dy=\sqrt{2\pi /(\beta a)}$, so $-{\partial }_{\beta }\ln Z_y=1/(2\beta )$. More generally, integration by parts in a canonical variable gives $⟨y{\partial }_{y}H⟩=k_{B}T$ when its boundary term $[y{e}^{-\beta H}]$ vanishes and the integrals converge. For $a{y}^2/2$ this is twice its mean energy. At a finite hard boundary the boundary term need not vanish; hard-wall correction to harmonic equipartition illustrates the resulting departure. A constraint, a nonquadratic potential energy, or a noncanonical integration measure therefore requires its own calculation. The theorem is an ensemble average, not equality of instantaneous energies, and applying it to a recorded trajectory requires equilibration and sufficient sampling.

For a classical one-dimensional harmonic oscillator, with positive mass $m$ and spring constant $k$,The origin is the equilibrium position. Its heat capacity is $k_B$ at fixed $m,k$, not $k_B/2$. Although the two mean energies agree, their measured amplitudes do not: $⟨p^2⟩=m{k}_{B}T$ and $⟨x^2⟩=k_{B}T/k$. In a harmonic optical trap, the position measurement observes the potential energy contribution; it does not measure both contributions by observing $x$ alone.^[ref]

For $N$ particles in a three-dimensional classical monatomic gas, neglect interactions and thermally accessible internal excitations. Its three translational momentum terms give internal energy $U=3N{k}_{B}T/2$, heat capacity at constant volume $C_V=3N{k}_B/2$, and heat capacity at constant pressure $C_p=C_V+N{k}_B=5N{k}_B/2$. Per mole, replace $N{k}_B$ by the molar gas constant $R$. For example, using stipulated $R=8.314\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$, the molar heat capacity at fixed volume is $12.471\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$. These results assume a dilute, nonrelativistic ideal gas with classical translational statistics, rather than a quantum degeneracy regime.

For an ideal gas of diatomic molecules, modeled as two chemical atoms with a bond, there are three translational quadratic terms, two accessible rotational kinetic energy terms for changes of bond orientation, and one bond-stretching harmonic oscillator with two quadratic terms. Rotation about the bond axis is not an additional active mode in this linear-molecule model. The three familiar limits of heat capacity at constant volume areCorresponding molar heat capacities are $3R/2$, $5R/2$ and $7R/2$. Counting the stretching harmonic oscillator only once as $k_{B}T/2$ would give the wrong last value.^[ref]

The frozen regimes require quantum mechanics. In the simple molecular rotation model, $E_j=k_B{\theta }_{r}j(j+1)$ with state degeneracy $2j+1$, ${\theta }_r={\hslash }^2/(2I{k}_B)$, and moment of inertia $I$. For the bond-stretching quantum harmonic oscillator, ${\theta }_v=\hslash \omega /k_B$ with angular frequency $\omega$ and $\hslash =h/(2\pi )$ from the Planck constant. When ${\theta }_r\ll {\theta }_v$, the three limits above correspond to $T\ll {\theta }_r$, ${\theta }_r\ll T\ll {\theta }_v$, and $T\gg {\theta }_v$, provided translational motion remains classical. They are smooth crossovers, not abrupt switches. Real molecules can have different allowed rotational populations; at high temperature, thermal dissociation, internal excitations and nonharmonic bond motion can invalidate the fixed-molecule model before its last limit is reached.^[ref]

For a quantum harmonic oscillator with fixed $\omega$, put $z={\theta }_v/T$. Its exact mean energy and heat capacity areAt $T\gg {\theta }_v$, the mean energy approaches $k_{B}T$ and $C_{\mathrm{osc}}$ approaches $k_B$. At $T\ll {\theta }_v$, thermal excitation is exponentially suppressed and $C_{\mathrm{osc}}$ tends to zero, although the zero-point energy $\hslash \omega /2$ survives. Frozen thus means little thermal response, not zero motion or zero energy.^[ref]

A motivating measurement is thermal calibration of an optical trap. A focused laser through a microscope objective confines a bead in water; imaging or a quadrant photodetector records a position time series. Gieseler and colleagues describe a polystyrene bead of radius $1.03\mu \mathrm{m}$ at $293.15\mathrm{K}$, a $1064\mathrm{nm}$ trapping laser, and photodetector acquisition at $10^5\mathrm{Hz}$. Their recorded bead continues undergoing Brownian motion; higher trapping power produces a narrower equilibrium position distribution. Infer the centered variance, not a squared distance from an arbitrary camera origin:The normal distribution position histogram and agreement with dynamical calibration methods support the local harmonic optical trap interpretation. For a stipulated example, $T=300\mathrm{K}$ and ${\sigma }_x=20\mathrm{nm}$ imply $k\simeq 1.04\times 10^{-5}N/m=0.0104pN/nm$. This calculated scale is not an experimental reading.^[ref]

Position calibration is essential: if detector output $v$ has conversion $x=s(v-v_0)$, its physical variance is $s^2\mathrm{Var}(v)$. An unknown scale $s$ cannot be determined together with $k$ from that one variance. A complementary calibration demonstrated by Tolić-Nørrelykke and colleagues combines the power spectrum of thermal motion with the bead response to known sinusoidal motion of a translation stage. The driven response supplies an additional length scale, and their method does not require bead size, liquid dynamic viscosity or wall distance as inputs. It still uses a thermal dynamical model; agreement between methods is a consistency check, not a theory-free proof of equipartition theorem.^[ref]

The readout itself can hide fluctuations. Wong and Halvorsen used a $1064\mathrm{nm}$ optical trap and adjustable-exposure CCD imaging of a bead in water. They varied exposure and trapping power, held the bead away from the nearest wall, adjusted illumination to maintain comparable images, and measured tracking noise using a stuck bead. The measured variance fell as exposure increased; corrected results agreed with their time-averaging model. For a stationary overdamped harmonic optical trap, with optical trap relaxation time $\tau$, an exposure of length $W$ measures ${\bar{x}}_W=W^{-1}{\int }_0^{W}x(t)dt$. Its ideal noise-free variance isFor $\alpha >0$, $S(\alpha )<1$: using the blurred variance in the uncorrected formula overestimates $k$ by $1/S(\alpha )$. At $W=\tau$, the calculated factor is $e/2\simeq 1.36$. Additive tracking noise instead raises the measured variance and biases the inferred $k$ downward if left uncorrected.^[ref]

Useful controls are to record over many optical trap relaxation times, compare blocks for drift, calibrate the position scale, measure fixed-bead noise, vary exposure, and check the position histogram for departures from a normal distribution. A sparse sequence of accurate instantaneous positions can estimate an equilibrium variance; a long exposure averages away motion even if many frames are collected. Flow, appreciable laser heating, escape from the harmonic region or nonconservative forces can break the assumed thermal equilibrium. Inferring $k$ by imposing equipartition theorem is a calibration, whereas testing the relation requires an independently established stiffness and position scale. The beautiful practical connection is that microscopic randomness can set a force scale: once calibrated, a small constant applied force displaces the mean by $f/k$. The same quadratic-term counting predicts macroscopic heat capacity and makes microscopic fluctuations useful measurements.

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 dissociation energy starts at the lowest vibrational quantum state, with the lowest allowed rotational quantum state and the fragment states explicitly specified. For a diatomic molecule whose fragments are ground-state chemical atoms, omitting any additional rotational correction,where $E_{\mathrm{ZP}}$ is the vibrational zero-point energy measured upward from the well minimum. The initial molecule already has this energy, so less additional energy is needed to reach the same final limit.^[ref]

For fragments that are themselves molecules, include their zero-point energies too. Define $D_e$ as the electronic energy difference between separately relaxed products and reactant; then, with consistent rotational conventions,Thus “subtract the parent's zero-point energy” is sufficient for atomic products but generally incomplete for molecular products. An excited final quantum state raises its channel threshold by its excitation energy; that threshold does not become the ground-product $D_0$ just because a detector selects it.

Quantizing the stretching coordinate gives a quantum harmonic oscillator with reduced mass $\mu$:Here $v$ is the vibrational quantum number and $\hslash =h/(2\pi )$ the reduced Planck constant. Consecutive energy eigenvalues have gap $\hslash \omega$; the ground state has zero-point energy $\hslash \omega /2$ above the well minimum. The offset $V_0$ cancels from every transition energy.

A short proof uses creation and annihilation operators: with $\ell =\sqrt{\hslash /(\mu \omega )}$, write $q=\ell (a+a^{†})/\sqrt{2}$. The Quantum Hamiltonian becomes $\hat{H}-V_0=\hslash \omega (a^{†}a+1/2)$. Nonnegativity of $a^{†}a$ gives the lowest level; repeated application of $a^{†}$ raises its energy by $\hslash \omega$ each time. This also gives $⟨q{⟩}_v=0$ andThe ground state therefore has a nonzero spread of bond lengths, even though its mean stretching displacement vanishes. A stationary energy eigenstate has a time-independent position probability density; the classical oscillation curve is not a trajectory of an atomic nucleus in that state.

The parabolic model formally extends $q$ to the whole real line, whereas an actual bond length satisfies $r>0$. It is useful for low levels concentrated near $r_e$, with typical stretching displacements much smaller than $r_e$. It fails near breakup of the chemical bond and when neglected anharmonicity becomes substantial.

Keep the same bond force constant $k$ and approximate the nuclear masses of ${}^1\mathrm{H}$, ${}^2\mathrm{H}$, and ${}^{35}\mathrm{Cl}$ by $1u$, $2u$, and $35u$, respectively. Here $u$ is the unified atomic mass unit, and $\mathrm{D}$ denotes deuterium. These are chosen mass number approximations, not measured atomic masses or natural-abundance averages. With the same chlorine isotope in both molecules,This is the isotope scaling of diatomic vibration prediction. Doubling the light partner's mass nearly doubles the reduced mass, lowering the frequency by about $28.3\%$. The factor differs from $1/\sqrt{2}$ because the chlorine partner also moves.

For the same small initial stretch $A$ and zero velocity, both models start with the same internal mechanical energy $k{A}^2/2$, but the DCl oscillation period is longer by $\sqrt{72/37}\approx 1.394972$. In the centre of mass frame, $u_{\mathrm{H}}=-35q/36$ and $u_{\mathrm{Cl}}=q/36$ for HCl; $u_{\mathrm{D}}=-35q/37$ and $u_{\mathrm{Cl}}=2q/37$ for DCl. These ratios explain which partner moves and are separate from the change in frequency.

The common-$k$ assumption belongs to the leading Born-Oppenheimer approximation: fixed nuclear positions and electric charges determine the electronic potential energy, while nuclear masses govern motion on that curve. Quantitative comparison requires actual atomic masses, departures from that approximation, and separation of anharmonicity and molecular rotation from the measured band. Even with common $k$, the quantum harmonic oscillator predicts a different zero-point energy for each isotope.^[ref]

For a fixed amount of substance and composition of classical ideal gas, internal energy depends only on temperature: $U=U(T,n)$, with no independent volume dependence. This does not require constant heat capacity at constant volume. Between equilibrium endpoints,Here $C_V$ is total heat capacity at constant volume, $C_{V,\mathrm{m}}$ is molar heat capacity, and $U/m$ for sample mass $m$ is specific internal energy, whose change is ${\int }_{T_1}^{T_2}{c}_V(T)dT$. These endpoint relations apply even when the intervening process is not at fixed volume; they do not say that transferred heat always equals $\Delta U$. With constant $C_V$, the safe statement is $U(T)=U(T_{\ast })+C_V(T-T_{\ast })$. Writing $U=C_{V}T$ additionally chooses the reference energy; it does not prove that a physical gas remains classical and ideal down to zero temperature. ^[ref]

A thermodynamic proof outline uses the first law of thermodynamics in equilibrium form, $dU=TdS-pdV$, and the Maxwell relation $(\partial S/\partial V{)}_T=(\partial p/\partial T{)}_V$:The last step uses the ideal gas law and the molar gas constant $R$, with fixed composition and amount of substance. Microscopically, separated ideal gas particles store no intermolecular interaction energy; their translational kinetic energy and any internal molecular energy are determined by temperature. Negligible intermolecular interactions still allow a molecule to have internal vibrational potential energy.

An original analytical model makes the distinction between an ideal gas and constant heat capacity concrete. Assume classical translation, two fully active rotational terms, and one independent quantum harmonic oscillator per diatomic molecule, with fixed angular frequency $\omega$. Define the vibrational temperature scale $\Theta =\hslash \omega /k_B$, using the reduced Planck constant $\hslash =h/(2\pi )$ and Boltzmann constant $k_B$. Summing its thermal occupations givesThe constant $U_{\ast }$ includes the chosen reference and zero-point energy, whose temperature derivative vanishes. When $T\ll \Theta$ the vibrational contribution to heat capacity is exponentially small; when $T\gg \Theta$ it tends to $R$ per mole. This model assumes rotations remain active throughout the interval and excludes dissociation, electronic excitation and real gas corrections. Its low-$T/\Theta$ edge is not a model of rotational freezing or of an actual substance at absolute zero.^[ref]

For this stipulated model, heating from $T_1=0.2\Theta$ to $T_2=\Theta$ gives $\Delta U/(nR\Theta )\simeq 2.57519$. Holding the frozen-vibration value $C_{V,\mathrm{m}}=5R/2$ throughout would give $2$, underestimating the required internal energy increase by about $22.3\%$ of the model's full value. The integral measures all thermal storage activated along the interval. At fixed volume with only heat input, this is also the supplied heat; at fixed pressure, expansion work must be added through Mayer's relation.

Joule's free-expansion experiment tests the independence from volume in a different way: an insulated rigid vessel allows gas into an evacuated compartment, so no external work is delivered and ideally no heat is exchanged. The first law of thermodynamics gives $\Delta U=0$; positive $C_V$ then predicts equal initial and final equilibrium temperatures for the ideal gas. The apparatus, thermal-capacity controls and limitations of this inference are described in Joule's free-expansion experiment and temperature sensitivity of a free-expansion experiment. This endpoint test differs from the reversible adiabatic expansion of a loaded piston, which delivers work and lowers internal energy.

A molecular nitrogen experiment used a $0.5\mathrm{mm}$ pulsed nozzle and two skimmers. A $5\mathrm{cm}$ tube raised the rotational temperature. Opposed vacuum-ultraviolet laser pulses excited selected molecular levels and, about $10\mathrm{ns}$ later, selectively ionized excited nitrogen fragments. A microchannel plate detector, phosphor screen and camera recorded radii.^[ref]

The measured squared radii were fitted against molecular excitation energies, referenced to the parent ground state:The intercept at zero radius gave the excited-product threshold; the fitted slope absorbed the detector scale. A wavemeter monitored the excitation source; scanning the probe covered the recoil Doppler shift. Subtracting the atomic excitation energies converted the intercept to ground-product $D_0$.^[ref]

An apertured grounded plate and detector timing rejected faster competing channels. Selecting fragment electronic states distinguished nearby thresholds. Agreement of $D_e$ across isotopes after zero-point energy correction tested consistency. The method still requires correctly assigned levels, adequate near-threshold signal and stable imaging; a signal drop alone can reflect another channel or a barrier.^[ref]

This measurement determines a state-resolved ground-state dissociation energy. Obtaining a finite-temperature bond-dissociation enthalpy requires the thermal increments above; obtaining a bond potential-well depth requires the zero-point energy correction. The three quantities describe different starting conditions for the same specified separation.

For a canonically quantized real scalar field, the quadratic Hamiltonian contains $\frac{1}{2}{\sum }_{\mathbf{p}}{E}_{\mathbf{p}}$ in a regulated box. Normal ordering subtracts this zero-point energy, leavingThe vacuum momentum is zero with an inversion-symmetric regulator. The canonical commutation relation then gives $[P^{\mu },a_{\mathbf{p}}^{†}]=p^{\mu }{a}_{\mathbf{p}}^{†}$, identifying the momentum and energy added by one particle. The un-subtracted energy differs by a constant and is not literally the same operator.

Use units $\hslash =c=1$ and the Minkowski metric $g=\mathrm{diag}(1,-1,-1,-1)$. The canonical momentum of the real scalar field is $\pi =\partial \mathcal{L}/\partial \dot{\varphi }=\dot{\varphi }$. Canonical quantization replaces the classical fields by operator-valued distributions and imposes the equal-time canonical commutation relationsThese relations are imposed on an initial slice, with smeared fields understood. In the Heisenberg picture, simultaneous unitary time evolution of both operators preserves their commutators. Since the momentum is $\dot{\varphi }$, the requested equal-time commutator is $[\varphi (t,\mathbf{x}),\dot{\varphi }(t,\mathbf{y})]=i{\delta }^{(3)}(\mathbf{x}-\mathbf{y})$.

The Legendre transform in mechanics, applied pointwise, gives the canonical Hamiltonian density of a real scalar field. ThusTo obtain its dynamics rather than assume them, use the Heisenberg equation of motion $\dot{O}=i[H,O]$. The momentum-square term gives $[H,\varphi ]=-i\pi$. The field-square term gives $[H,\pi ]=i{m}^2\varphi$, while differentiating the Dirac delta function in the gradient term and integrating by parts gives $[H,\pi ]=-i{\nabla }^2\varphi +i{m}^2\varphi$. ConsequentlyBoundary terms vanish for suitably decaying smeared fields. This is the Klein-Gordon equation as an operator identity.

A spatial Fourier transform reduces the Klein-Gordon equation to ${\ddot{\varphi }}_{\mathbf{p}}+E_p^2{\varphi }_{\mathbf{p}}=0$, where $E_p=\sqrt{{\mathbf{p}}^2+m^2}$. Its two frequency branches and the Hermitian adjoint condition giveThe invariant-normalized scalar mode extraction can be written particularly simply asSubstituting the mode expansion of a free field proves these inversion formulas: the opposite-frequency coefficient is proportional to $E_p-E_q$ and vanishes after spatial Fourier inversion. Alternatively they are the corresponding Klein-Gordon inner products. Taking their commutators using the equal-time fields gives $(E_p+E_q)(2\pi {)}^3{\delta }^{(3)}(\mathbf{p}-\mathbf{q})$ in the mixed case. In the two equal-type cases the delta function enforces $\mathbf{q}=-\mathbf{p}$, multiplying $E_p-E_q=0$. ThereforeConversely these mode commutators reproduce $[\varphi ,\dot{\varphi }]=i{\delta }^{(3)}$: the two frequency contributions each supply half the spatial Dirac delta function.

For the covariant oscillator Hamiltonian of a real scalar field, insert the mode expansion of a free field into $H$. Spatial integration sets the two momenta equal in the mixed terms and opposite in the two-creator or two-annihilator terms. The latter have coefficient $-E_p^2+{\mathbf{p}}^2+m^2=0$. The mixed terms giveThe constant $E_0$ is the divergent zero-point energy; a finite box and a momentum cutoff make its meaning explicit. Normal ordering sets the free vacuum energy to zero and leaves all commutators unchanged. Directly applying the mode commutators to the mixed expression yieldsChoose the Fock vacuum by $a(p)|0⟩=0$. Each creation operator adds a boson of energy $E_p$ and four-momentum $p$, while each annihilation operator removes one. Products of commuting creation operators give bosonic statistics from commuting creation operators and unrestricted occupation numbers. The real scalar field has one spin-zero species, identical to its antiparticle. Its one-particle states obey $⟨q|p⟩=(2\pi {)}^{3}2E_p{\delta }^{(3)}(\mathbf{p}-\mathbf{q})$, precisely the relativistic normalization of a one-particle state.

For the Feynman propagator, put $z=x-y$ and contract the two fields in the Fock vacuum. Only $a{a}^{†}$ survives, giving the Wightman functionThe stipulated time-ordered product convention is $i{\Delta }_F(z)=\theta (z^0)W(z)+\theta (-z^0)W(-z)$. In a contour integral over $p^0$, the Feynman i-epsilon prescription places the positive-energy pole just below the real axis and the negative-energy pole just above it. For $z^0>0$ close below, clockwise, obtaining $-i{e}^{-i{E}_p{z}^0}/(2E_p)$; for $z^0<0$ close above, obtaining $-i{e}^{i{E}_p{z}^0}/(2E_p)$. Changing $\mathbf{p}\mapsto -\mathbf{p}$ in the second contribution therefore provesActing with $\square +m^2$ multiplies the integrand by $-(p^2-m^2)$. The remaining factor tends to $-1$ as a distribution, so Fourier inversion givesThe sign also follows from the derivative jump of a free scalar time-ordered two-point function: the first time derivative of $i{\Delta }_F$ jumps by $⟨[\dot{\varphi },\varphi ]⟩=-i{\delta }^{(3)}$, so that of ${\Delta }_F$ jumps by $-{\delta }^{(3)}$.

The RNS string adds worldsheet Majorana fermions ${\psi }^{\mu }$ to the coordinate fields $X^{\mu }$. The new fields are spinors on the worldsheet but vectors under the target Lorentz group; this distinction matters when interpreting their oscillator states. In a flat target, a convenient Lorentzian convention for the free gauge-fixed action iswhere ${\rho }^a$ are two-dimensional gamma matrices. The bosonic and fermionic kinetic terms are related by worldsheet supersymmetry. Before fixing local worldsheet supersymmetry, one introduces a worldsheet gravitino along with the worldsheet metric. After superconformal gauge fixing, the residual matter constraints are the Virasoro constraints and the constraints generated by the superconformal current $G\propto \psi \cdot \partial X$.

Each real worldsheet Majorana fermion contributes $1/2$ to the chiral central charge, while each coordinate contributes one. The reparameterization bc system contributes $-26$, and fixing local worldsheet supersymmetry also produces commuting superconformal ghosts of weights $(3/2,-1/2)$, with central charge $11$. ThusIn light-cone gauge in string theory there are eight physical transverse coordinate fields and eight transverse worldsheet Majorana fermions.

On the doubled spatial interval of an open string, the two allowed fermionic periodicities areThey arise from the relative signs of the left-right fermion gluing conditions at the two ends. Thus the Ramond sector has integer-mode oscillators $d_n$, including $d_0$, and the Neveu–Schwarz sector has half-integer oscillators $b_r$, $r\in \mathbb{Z}+1/2$, without zero modes. In either sector,A closed string has independent left- and right-moving periodicities, giving four sector choices rather than two.

The Ramond zero-mode Clifford algebra isHence its ground states carry a Spinor representation of the Lorentz group. The zero-mode supercurrent constraint reduces on a ground state to the Dirac equation $p_{\mu }{\Gamma }^{\mu }u=0$. In the Neveu–Schwarz sector the oscillator vacuum instead carries a scalar representation. Consequently the open string NS sector describes spacetime bosons, while the R sector describes spacetime fermions. Exciting a target-vector oscillator does not change this statistics assignment.

The string intercepts follow from the transverse zero-point energy. Per boson the integer-mode contribution is $\frac{1}{2}\zeta (-1)=-1/24$. Per fermion it is the negative of the corresponding half-sum. Since $\zeta (-1,1/2)=1/24$, the eight transverse pairs giveThus $a_{\mathrm{NS}}=1/2$, $a_{\mathrm{R}}=0$, and the open string mass formula is ${\alpha }^{\prime }{M}^2=N-a$. This distinguishes the lowest states explicitly:

The supersymmetric GSO projection keeps odd oscillator fermion number in the NS sector. Defining $F_{\mathrm{osc}}$ so that the NS vacuum has $F_{\mathrm{osc}}=0$, its projector isIt removes the scalar tachyon and keeps the massless vector. In the R sector, fermion parity contains the zero-mode chirality operator, so one may writeHere $F_{\mathrm{osc}}$ counts nonzero-mode fermionic excitations. On the ground states, this keeps one Majorana-Weyl spinor chirality: 16 real covariant components, or eight on-shell polarizations after the Dirac equation. The projected massless open-string states are an eight-polarization gauge vector and an eight-polarization gaugino. The compatible projected tower exhibits spacetime supersymmetry; unprojected worldsheet supersymmetry by itself did not eliminate the tachyon or equalize these massless counts.

For a closed string, both chiral zero-mode constraints must hold:This shifted closed-string level matching is particularly important in the mixed sectors. The lowest states and their spacetime nature are as follows.

For type IIA superstring theory, the two R sector ground-state chiralities are opposite. The Ramond–Ramond potentials can be represented by $C_1,C_3$, with transverse counts $8+56=64$. The two mixed-sector gravitinos have opposite chiralities, and the ten-dimensional theory is nonchiral. For type IIB superstring theory, the two R sector chiralities agree; the Ramond–Ramond potentials are $C_0,C_2,C_4$, with self-dual five-form field strength, giving $1+28+35=64$. Its two gravitinos have the same chirality. Thus both projected type II theories have 128 massless bosonic and 128 massless fermionic polarizations and no tachyon, while their chiralities and form spectra differ.

If unoriented open strings are included in the complete theory, the supersymmetric example is type I string theory. Worldsheet orientation reversal projects the closed type IIB superstring theory spectrum and identifies the two mixed sectors; the surviving closed fields include the graviton, dilaton, a Ramond–Ramond potential $C_2$, and one gravitino and dilatino. The open states form vector and gaugino multiplets with Chan-Paton factors; tadpole and anomaly cancellation select $SO(32)$. This is an additional orientation projection, not part of the GSO projection itself. The heterotic string instead pairs one supersymmetric chiral side with a bosonic chiral side carrying an internal gauge sector; it does not have a Ramond choice on both sides.

The string sigma model in a spacetime metric is the Polyakov actionTake conformal gauge $h_{ab}=\mathrm{diag}(-1,1)$, and choose spatial period $2\pi$ for the closed string. Substitution of the plane-wave metric givesThe stipulated light-cone gauge in string theory $X^{+}=\tau$ sets ${\dot{X}}^{+}=1$ and $X^{\prime +}=0$. Its longitudinal term $-T\int {\dot{X}}^{-}$ is a total time derivative. The transverse string sigma model in an isotropic plane wave is thereforeThe resulting Euler-Lagrange equations are the massive wave equationsThese conventions fix the mass term directly from $X^{+}=\tau$. More generally $X^{+}=\kappa \tau$ would replace $m$ by $|m\kappa |$, with $p^{+}=2\pi T\kappa$ for this spatial period.

Use the real orthonormal Fourier basisFor the plane-wave string mode frequencies, let ${\omega }_0=|m|$ and ${\omega }_{n,c}={\omega }_{n,s}=\sqrt{n^2+m^2}$. The complete real mode expansion for nonzero frequencies isEvery term obeys the wave equation and is periodic. Orthonormality and completeness let its coefficients match arbitrary appropriate initial position and velocity data. For $m=0$, replace the zero-mode bracket by $q_0^I(0)+\tau p_0^I/T$; the zero mode becomes a free particle rather than an oscillator.

The canonical momentum density is ${\Pi }_I=T{\dot{X}}^I$. The transverse Hamiltonian isWrite $X^I={\sum }_a{q}_a^I{f}_a$ and $p_a^I=T{\dot{q}}_a^I$. Since $\int f_a{f}_b={\delta }_{ab}$ and $\int f_a^{\prime }{f}_b^{\prime }=n_a^2{\delta }_{ab}$, substitution gives the plane-wave string mode HamiltonianThis is a collection of independent harmonic oscillators. The last equality follows by substituting the sine-and-cosine time dependence; each oscillator energy is constant.

For canonical quantization of each nonzero-frequency mode, putThen $H_{\perp }={\sum }_{I,a}{\omega }_a(a_a^{I†}{a}_a^I+1/2)$. Opposite travelling-wave oscillator combinations give the equivalent formFor $m\ne 0$, the formal zero-point energy is $E_0=\frac{24}{2}[|m|+2{\sum }_{n\ge 1}\sqrt{n^2+m^2}]$. At $m=0$, replace the zero-mode occupation term by ${\sum }_I(p_0^I{)}^2/(2T)$. The zero-point expression is divergent and needs a regulator or a specified normal ordering convention; no mass-independent intercept has been assumed.

The remaining Virasoro constraints determine the longitudinal coordinate:Periodicity of $X^{-}$ imposes ${\int }_0^{2\pi }{\Pi }_I{X}^{\prime I}d\sigma =0$, the closed-string level matching condition. Labelling the travelling oscillators by their opposite worldsheet momenta gives ${\sum }_{n\ge 1,I}n(N_n^I-{\tilde{N}}_n^I)=0$. The light-cone energy $p^{-}=-p_{+}$ equals $T\int ({\dot{X}}^{-}+m^2{X}^I{X}^I)d\sigma =H_{\perp }$, consistent with the positive mass term in the Hamiltonian.