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With Fourier convention $c_n=N^{-1/2}{\sum }_k{e}^{-ikn}{c}_k$, the Transverse-field Ising model has $H=J{\sum }_k{\Psi }_k^{†}[(g-\cos k){\sigma }_z-\sin k{\sigma }_x]{\Psi }_k$. Expanding the spinor retains the constant $-JgN$ and pairs $k$ with $-k$. The physical quasiparticle energy is twice the positive matrix eigenvalue times $J$, because both momentum partners occur in the Nambu sum.

Take $2M$ base Ising spins on a ring and add one apex Ising spin over every other base bond. Give all base and apex bonds the antiferromagnetic energy $Js{s}^{\prime }$ with $J>0$. There are $3M$ sites, and adjacent triangles are connected by a single undecorated bridge rather than sharing a corner. With $K=J/(k_{B}T)$, summing the apex spins yields alternating spin-chain transfer matricesTheir common symmetric eigenvector gives the leading cell eigenvalue $3+e^{-4K}+e^{-2K}+3e^{2K}$. A minimum-energy triangle has one frustrated bond and energy $-J$; its connecting bridge can have energy $-J$. Requiring the bridge to be antiparallel leaves the counting transfer matrix $\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$ on successive first base spins. Its eigenvalues $3,1$ give ground-state degeneracy $g_M=3^M+1$ and ground-state energy $-2JM$. The residual entropy per site is $k_B\ln 3/3$.

The Ising model is a spin model with an Ising spin ${\sigma }_v\in \{-1,1\}$ at every graph vertex $v$. For a finite graph with vertex set $V$ and edge set $E$, a uniform coupling $J$ and field parameter $h$ give the HamiltonianEach edge is counted once. Here $T>0$ is temperature, $k_B$ is the Boltzmann constant, and $\beta$ is inverse temperature; $J$ and $h$ have units of energy. For magnetic applications, $h$ includes the magnetic moment multiplying the applied magnetic field. The canonical ensemble assigns the Boltzmann distributionThe partition function $Z$ makes the probabilities sum to one. Positive $J$ rewards equal neighboring Ising spins, modeling ferromagnetism; negative $J$ rewards opposite neighbors, giving an antiferromagnetic Ising model. Positive $h$ rewards $+1$, while negative $h$ rewards $-1$.^[ref]

A single local rule can organize a whole graph: rewarding agreement on each edge makes all-equal Ising spin patterns favorable even though no edge directly couples distant vertices. The four-spin Ising ring makes this visible with only sixteen microstates. It also separates three different questions: whether neighbors agree, whether the mean spin magnetization is nonzero, and how large its thermal fluctuations are.

A motivating experiment is the University of Colorado Boulder's Curie-temperature demonstration. An iron rod suspended from a twisting wire on a stand is initially held near a permanent magnet by ferromagnetism. A gas burner supplied from a propane cylinder heats the rod; the readout is its motion away from the magnet, followed by return as it cools. Repeated cooling supplies a reversible comparison on the same sample. The source explicitly calls for heating the rod while keeping the magnet cool, since heating the magnet can reduce its strength. The inference is that warming the sample weakens its magnetic attraction across the Curie temperature. This is a qualitative force readout, with no calibrated temperature or magnetization trace supplied: it motivates competition between alignment and thermal fluctuations, but does not identify $J$, measure individual Ising spins, or realize a four-spin Ising ring.^[ref]

Ernst Ising's 1925 paper examined a linear chain of two-orientation magnetic elements with neighboring interactions. In its calculation for a large chain, the zero-field mean magnetization vanishes and the model has no spontaneous magnetization. The inspected English translation states this immediately after equation (8) in §2. This historical result concerns the chain; it does not establish the behavior of a two-dimensional Ising model.^[ref]

The Ising ring bond correlation is the neighboring two-point correlation function $c_1=⟨{\sigma }_i{\sigma }_{i+1}⟩$ in the four-spin Ising ring. All four edges are equivalent. At zero field each mean Ising spin is zero by spin inversion symmetry, so $c_1$ also equals the connected correlation function. Agreement gives bond product $+1$ and disagreement $-1$; thus $P({\sigma }_i={\sigma }_{i+1})=(1+c_1)/2$.

Differentiating the finite partition function with respect to $K=\beta J$ inserts the sum of four bond products. ThereforeHere $U$ is internal energy. At $K=(\ln 2)/4$, the class-weight calculation gives $c_1=(4-1)/17=3/17$ and $U=-12J/17$. The agreement probability for a selected bond is $10/17$, despite the exactly zero mean spin magnetization.

A second derivative gives the energy variance, and hence the heat capacity $C_H=dU/dT$ at fixed $J$ and $h=0$:To see the variance identity directly, a derivative of an expected value under weights $e^{KS}$ inserts $S-⟨S⟩$, where $S={\sum }_i{\sigma }_i{\sigma }_{i+1}=-H/J$. The factor $K^2$ follows from $dK/dT=-K/T$. At the same exact example, $\mathrm{Var}(H)/J^2=1216/289$. The heat capacity vanishes both as $K\to 0$ and as $K\to \infty$: at high temperature its dimensionless factor $K^2$ vanishes, while at low temperature excited microstates become exponentially rare.

Define the instantaneous spin magnetization $m=\frac{1}{4}{\sum }_i{\sigma }_i$. Global reversal sends $m$ to $-m$ without changing the zero-field energy. Pairing every microstate with its reversed partner proves $⟨m⟩=0$ at every positive temperature, even when almost all probability lies on the two fully aligned patterns. A vanishing mean does not imply weak alignment: the limiting probability distribution is half at $m=+1$ and half at $m=-1$.

The Ising ring magnetization fluctuations measure this distinction through variance. In the $d=0$ class, both patterns have $m^2=1$. Among the $d=2$ patterns, eight have $m^2=1/4$ and four have $m^2=0$; the two $d=4$ patterns also have $m^2=0$. ConsequentlyIt equals $6/17$ at $K=(\ln 2)/4$, approaches $1/4$ for four independent fair Ising spins at $K\to 0$, and approaches $1$ as $K\to \infty$. Unlike the energy variance, the spin magnetization variance remains large in the low-temperature symmetric mixture.

The same thermal fluctuation controls the field response. Allow $h$ to vary before setting it to zero; differentiating the Boltzmann distribution inserts $\beta {\sum }_i{\sigma }_i=4\beta m$, givingThis is a spin-normalized magnetic susceptibility with respect to the energy-valued parameter $h$; conversion to physical magnetization and magnetic field requires the magnetic moment and sample volume factors.

For finite $N$, $Z$ is a finite positive sum of exponential functions. Its free energy $-k_{B}T\log Z$ is a real analytic function for $T>0$, so the smooth heat capacity peak here is not a critical phase transition. Spontaneous magnetization concerns the order of limits: first a thermodynamic limit, then removal of a symmetry-breaking field. The four-spin Ising ring alone establishes neither that limit nor any two-dimensional Ising model result.^[ref]

Nuclear quantum states in different nuclear isobars belonging to the same isospin multiplet, with equal total isospin and approximately the same spatial and spin structure. An isospin raising or lowering operator connects the states in the limit of exact isospin symmetry. Nuclear Coulomb energy and other isospin-breaking effects shift their energies and modify this ideal relationship.

With fixed pressure $p$, let $A=\mu p/\mathcal{R}$, so $\rho =A/T$. A material mass coordinate has ${\partial }_x=\rho {\partial }_m$ and the material time derivative becomes ${\partial }_t{|}_m$. The energy and continuity equations then give the displayed equation with $\tau =Ct$ and $C=(\gamma -1)(\mu /\mathcal{R}{)}^{2}p/\gamma$. A mass integral from a fixed origin is material only when the mass flux there vanishes; otherwise subtract its time-integrated flux to label parcels. Uniform pressure is a low-Mach approximation when sound crossing is faster than thermal evolution.

An isolated attosecond pulse has one dominant short burst within the observation window, with any satellite emission quantified. A broad optical spectrum alone cannot distinguish one burst from several. Selecting a generation interval and an appropriate spectral band can suppress neighboring bursts, but the resulting temporal contrast must still be tested.

For a simple calculation independent of any apparatus, let $\mathcal{E}(t)=f(t)+r{e}^{i\theta }f(t-\Delta )$, with two nonoverlapping copies of a short optical pulse. Its optical spectrum has intensityA satellite produces spectral fringes of angular spacing $2\pi /\Delta$ when the two bursts are coherent and resolved. For chosen field-amplitude ratio $r=0.2$, the satellite/main energy ratio is $r^2=0.04$, while fringe visibility is $2r/(1+r^2)\simeq 0.3846$. Weak satellite energy can thus leave substantial fringes. Averaging an uncontrolled $\theta$ removes the interference term without removing the satellite: absence of fringes alone does not prove isolation.

An isolated system exchanges neither matter nor energy with its surroundings. Its total energy is fixed, while the Second law of thermodynamics requires its total entropy to be nondecreasing. Its internal energy alone is fixed when bulk kinetic energy and external potential energy also remain unchanged. For example, an initially spinning stirrer inside an insulated rigid vessel can lose kinetic energy while the contents gain internal energy, with no external input.

An isolated system need not yet be in thermodynamic equilibrium. Two enclosed bodies at different temperatures can exchange heat internally until they reach thermal equilibrium; their combined energy stays constant. Their internal transfer is absent from the outer boundary ledger. A vacuum-insulated container is only an approximation to isolation: residual thermal conduction and thermal radiation can still carry energy across its boundary.

An isothermal expansion increases the volume of a thermodynamic system at fixed temperature. For a reversible ideal gas path, the gas delivers positive work and absorbs positive heat, with $Q=W_{\mathrm{by},\mathrm{rev}}=nRT\ln (V_f/V_i)$ and $\Delta U=0$. The heat bath supplies the expansion energy; without sufficient heat transfer, doing work instead lowers the temperature. This transfer is the hot step of a Carnot cycle, and it is why an expanding gas need not cool when its surroundings continuously replenish its internal energy.

An insulated free expansion into an evacuated compartment is different. In the idealization with no heat exchange, no external work and no container energy change, $Q=W_{\mathrm{by}}=0$. Since internal energy of an ideal gas depends only on temperature, its final equilibrium temperature equals its initial value. The transient need not have a uniform temperature or pressure, so equal endpoint temperatures do not make it a reversible isotherm. Do not apply the logarithmic isothermal ideal-gas work formula to this irreversible process. Its entropy still increases by $\Delta S=nR\ln (V_f/V_i)$; entropy change in free expansion evaluates this state function difference using a hypothetical reversible path.^[ref]

For a real gas, even equal-temperature equilibrium endpoints need not have equal internal energy. An insulated free expansion can therefore change the temperature, while a controlled isothermal expansion must provide the heat needed for both its internal energy change and its delivered work. The internal energy at constant temperature article gives the Van der Waals equation example.^[ref]

For the one-mole Van der Waals equation with constant parameters, integrate $dF=-pdV$ along an isotherm to obtain the displayed difference. The attractive contribution has a positive change on expansion, whereas the translational term lowers free energy. Thermodynamic consistency independently gives $\Delta U=a(1/V_1-1/V_2)$ and $\Delta S=R\ln [(V_2-b)/(V_1-b)]$. These homogeneous-branch expressions are not a substitute for equilibrium convexification through a subcritical coexistence region.

For a reversible thermodynamic process that changes the volume of a fixed amount of ideal gas at absolute temperature $T$, the work done by the gas isHere $n$ is the amount of substance, $R$ the molar gas constant, $p$ the pressure, and $\ln$ the natural logarithm. The ideal gas law supplies $p=nRT/V$; the integral is the signed area under the pressure-volume curve. Isothermal expansion gives positive work; isothermal compression gives negative work. A convention counting work done on the gas instead has $W_{\mathrm{on}}=-W_{\mathrm{by}}$.^[ref]

For a calculated example, let $n=1\mathrm{mol}$, $T=300\mathrm{K}$ and $V_i=10\mathrm{L}$, $V_f=20\mathrm{L}$. The pressure falls from approximately $249.4$ to $124.7\mathrm{kPa}$, andTraversing the same isothermal process backward changes both transfer signs: $W_{\mathrm{by},\mathrm{rev}}=Q\approx -1.729\mathrm{kJ}$. Neither direction changes internal energy of an ideal gas. These stipulated values illustrate the energy balance; they are not experimental measurements.

The shaded geometric area is positive in both panels; the direction of the integral supplies the sign of work. Knowing only the two equilibrium endpoints does not justify shading this curve for an irreversible process. For an idealized piston moving against a specified external resisting pressure, boundary work is $W_{\mathrm{by}}=\int p_{\mathrm{ext}}dV$. Substitution of the equilibrium gas pressure requires the appropriate reversible limit. In free expansion into vacuum, $p_{\mathrm{ext}}=0$ and this boundary work is zero.

An equilibrium statistical ensemble with fixed particle number, pressure and temperature, allowing volume and energy to fluctuate. With a temperature-independent volume measure, its joint weight is proportional to $e^{-\beta (E+PV)}$. Its normalization and any temperature-dependent prefactor must be specified when taking thermodynamic derivatives.

An ideal compact volume source has $M_{ij}(t)=M(t){\delta }_{ij}$. In an unbounded homogeneous isotropic solid its far-field body radiation is P only. Boundaries and heterogeneity can create converted SV and surface-wave energy; a real explosion can have nonisotropic components.

For a small displacement field $\mathbf{u}$, define $G_{ij}={\partial }_i{u}_j$ and the infinitesimal strain tensor $\epsilon =(G+G^T)/2$. The isotropic linear elasticity energy density isHere $\lambda ,\mu$ are Lamé parameters and the colon is the Frobenius inner product. The energy depends on symmetric strain, so an infinitesimal rigid rotation costs no elastic energy. A term involving only $G:G^T$ does not have this property. ^[ref]

Differentiation gives the Cauchy stress tensorVarying $\int WdV$ with fixed boundary displacement and no body force yields $\nabla \cdot \sigma =0$, orThese are the static Navier-Cauchy equations.

In two dimensions, separating trace and deviatoric strain givesThus $\mu >0$ and $\lambda +\mu >0$ make every nonzero two-dimensional strain cost positive energy. Rigid translations and infinitesimal rigid rotations remain zero-strain motions.

The polarized ground state of an isotropic Heisenberg model with ferromagnetic exchange is an exact state. Its normal-ordered harmonic Hamiltonian has no pair-creation terms, no vacuum magnon occupation and no additional zero-point shift. An antiferromagnetic Néel state instead generates anomalous terms, a squeezed Bogoliubov transformation vacuum, a negative energy correction and reduced staggered magnetization. Neither statement licenses omitting normal-ordering constants from an arbitrary spin-wave model with anisotropy or a different reference state.

Ivar Giaever used electron tunnelling through a thin insulating barrier to measure the superconducting energy gap. The available electronic states on either side determine which tunnelling processes contribute at a given voltage. A gap suppresses low-energy current, so the current-voltage curve becomes a spectroscopic probe of the superconducting state. Following the feature with temperature connects an electrical measurement with a microscopic excitation energy, rather than merely demonstrating low resistance.^[ref]

A physicist.^[ref][ref]

A Jablonski diagram arranges molecular energy eigenstates vertically and draws arrows for changes between them. In the usual closed-shell molecular example, $S_0$ is the electronic ground state, $S_1$ and $S_2$ are electronic excited states with singlet state spin, and $T_1$ has triplet state spin. Each electronic state also has levels of molecular vibration. The diagram records possible processes, not molecular paths in space.^[ref]

Photon absorption can populate an upper level of molecular vibration in $S_1$. Relaxation removes excess energy without photon emission; fluorescence commonly starts from relaxed $S_1$.^[ref] The drawing also includes relaxation from a higher electronic excited state and a competing spin-changing route to a triplet state.

The Jacobi constant is a conserved quantity for the circular restricted three-body problem. In barycentric coordinates rotating at constant angular frequency $n$, defineThe equation of motion in a rotating frame is ${\dot{\mathbf{v}}}_{\mathrm{rot}}+2\mathbf{\Omega }\times {\mathbf{v}}_{\mathrm{rot}}=\nabla U$. Taking its dot product with ${\mathbf{v}}_{\mathrm{rot}}$ proves $d(|{\mathbf{v}}_{\mathrm{rot}}{|}^2/2-U)/dt=0$: the Coriolis force does no work.

Only positions satisfying $2U-C\ge 0$ are accessible. The boundary gives a zero-velocity surface; openings near Lagrange points allow transit between regions but do not force it. Ordinary inertial-frame energy can change even while $C$ remains constant.^[ref]

For a mechanical system whose imposed motion is uniform rotation with angular velocity $\omega$ about a fixed axis, a conserved rotating-frame energy can exist even when the inertial mechanical energy $E$ changes. With angular momentum $L_z$ about that axis, it is $J=E-\omega L_z$. For a bead constrained to rotate with $\dot{\theta }=\omega$, this becomesFor $z=r^2/(2a)$ in uniform gravity, division by mass yieldsThis is the first integral that controls the bead's radial motion. Different normalizations and signs are used for the related Jacobi constant in celestial mechanics.