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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]

Use an orthonormal real Fourier series basis on a circle of length $\beta$, with ${\nu }_r=2\pi r/\beta$. Retain the constant coefficient and the sine/cosine pair for each $1\le r\le N$. The oscillator action becomes $\frac{1}{2}{\omega }^2{q}_0^2+\frac{1}{2}{\sum }_r({\omega }^2+{\nu }_r^2)(q_r^2+s_r^2)$. A product of real Gaussian integrals gives the displayed functional determinant, with $A_N$ independent of $\omega$. Ratios for two frequencies converge regardless of $A_N$. The absolute limit requires its normalization: $A_N={\beta }^{-1}{\prod }_r{\nu }_r^2$ gives the thermal partition function of a quantum harmonic oscillator in the limit. Multiplying this choice by an oscillating positive sequence preserves all frequency ratios but destroys the absolute limit.

For the quantum harmonic oscillator $H=\hslash \omega (a^{†}a+1/2)$, its normal symbol is $H_N=\hslash \omega (\bar{\psi }\psi +1/2)$. The Weyl ordering symbol of $a^{†}a$ is $\bar{\psi }\psi -1/2$, so $H_W=\hslash \omega \bar{\psi }\psi =p^2/(2m)+m{\omega }^2{q}^2/2$. Switching from adjacent-label coherent-state time slicing to a midpoint phase-space path integral must retain this ordering correction. Both correctly normalized constructions give the thermal partition function of a quantum harmonic oscillator $Z=[2\sinh (\beta \hslash \omega /2){]}^{-1}$; retaining the normal-symbol constant after switching prescriptions would count the zero-point shift twice.

For the quantum harmonic oscillator, set $\mu =0$ and $\Omega =\hslash \omega$. The normal symbol is $H_N(\bar{\psi },\psi )=\Omega (\bar{\psi }\psi +1/2)$, so the thermal coherent-state time slicing givesThe connection with a phase-space path integral uses the canonical real coordinatesTheir derivative term satisfies $\bar{\psi }\dot{\psi }=-ip\dot{q}/\hslash$ up to total derivatives that vanish for periodic paths. The normal and Weyl symbols of a harmonic oscillator must be distinguished: in midpoint phase-space time slicing, the Weyl ordering symbol of $a^{†}a$ is $\bar{\psi }\psi -1/2$. Thus the appropriate midpoint Hamiltonian is $H_W=p^2/(2m)+m{\omega }^2{q}^2/2$, with no additional constant. The change from the adjacent-label normal prescription to the midpoint prescription includes this ordering correction.

In physical imaginary time $u=\hslash \tau$, the resulting phase-space path integral isGaussian momentum integration in a phase-space path integral produces the usual oscillator configuration-space path integral. An exact check follows directly from the coherent kernel $⟨{\psi }^{\prime }|e^{-\beta \Omega a^{†}a}|\psi ⟩=\exp (e^{-\beta \Omega }{\bar{\psi }}^{\prime }\psi )$:This is the thermal partition function of a quantum harmonic oscillator. Keeping the $\Omega /2$ normal-ordering constant a second time after switching to the Weyl symbol would double count the zero-point energy.

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.

The frequency-independent measure factor cancels in the ratio:Upon analytic continuation in frequency, this has poles at ${\omega }_1=0,\pm 2\pi ir/\beta$ and zeros at ${\omega }_2=0,\pm 2\pi ir/\beta$, for $1\le r\le N$, apart from cancellations when numerator and denominator vanish together. They match those of $\sinh (\beta {\omega }_2/2)/\sinh (\beta {\omega }_1/2)$. To justify equality, rather than merely matching this divisor, use the hyperbolic-sine infinite productIt follows from pairing the Weierstrass product for the reciprocal gamma function at opposite imaginary arguments and using the Gamma reflection formula. Normal convergence on compact sets follows from $\sum r^{-2}<\infty$. Thus, for positive real frequencies,The cutoff ratio determines the frequency dependence, agreeing with the thermal partition function of a quantum harmonic oscillator.

The printed assertion about an absolute ${\lim }_N{\mathcal{Z}}_N$ needs a normalization qualification: independence of $A_N$ from $\omega$ alone does not ensure existence of a nonzero limit. Set $B_N=A_N{\prod }_{r=1}^N{\nu }_r^{-2}$. Then ${\mathcal{Z}}_N=B_N{\omega }^{-1}{\prod }_r(1+{\omega }^2/{\nu }_r^2{)}^{-1}$, so if $B_N\to B\ne 0$, its limit is $B\beta /[2\sinh (\beta \omega /2)]$. The choice $B_N=1/\beta$ gives the required limit. Conversely, $B_N=(1+(-1{)}^N/2)/\beta$ is a positive frequency-independent normalization with no absolute limit, though every ratio above still converges. Matching zeros and poles alone also permits nonvanishing entire functions; the convergent normalized product is what rules them out here.

For excitation energy $\Delta =\hslash \omega$, the geometric series gives the thermal partition function of a quantum harmonic oscillator with its zero-point energy subtracted:Thus $U\to 0$ at low temperature and approaches $k_{B}T$ at high temperature. Restoring the zero-point term multiplies $Z$ by $e^{-\beta \Delta /2}$ and adds $\Delta /2$ to $U$, without changing the thermal occupation probabilities.

To count electromagnetic modes, put a large cavity in a periodic box of volume $V$. One transverse polarization has wavevector-state density $V/(2\pi {)}^3$; there are two independent polarizations. In the shell $k\to k+dk$,With $\omega =ck$, the electromagnetic mode density in a box becomesThe continuum expression is the large-cavity leading term; using standing-wave boundary conditions gives the same bulk count. Both photon polarizations must be included. This is density per angular frequency, not per ordinary frequency.

Each mode is an oscillator with zero photon chemical potential in thermal equilibrium. Multiplying its mean thermal energy by the mode density per volume gives the angular-frequency Planck energy densityAt small $\hslash \omega /(k_{B}T)$ it reduces to the Rayleigh–Jeans result $k_{B}T{\omega }^2/({\pi }^2{c}^3)$; exponential suppression at high frequency prevents a divergent thermal energy. With $x=\hslash \omega /(k_{B}T)$,The integral is the stipulated ${\pi }^4/15$. This is a volume energy density, $a{T}^4$; for isotropic cavity radiation the outward black-surface flux is $cu/4={\sigma }_{\mathrm{SB}}{T}^4$, with ${\sigma }_{\mathrm{SB}}={\pi }^2{k}_B^4/(60{\hslash }^3{c}^2)$. Distinguishing flux, spectral density and total volume energy avoids extra factors of $c$, $4$ or $2\pi$.