--- title: Two-State Systems, Paramagnets, and Negative Temperature module: The Microcanonical Ensemble moduleNumber: 3 lessonNumber: 4 order: 304 summary: > The ideal two-state paramagnet has a multiplicity counted by the binomial coefficient, an entropy that is an inverted dome in the energy, and a temperature read from the slope $1/T=\partial S/\partial E$. Because the energy is bounded above, the slope changes sign past the entropy maximum: a population-inverted spin system has a negative absolute temperature, which is hotter than any positive temperature. Nuclear-spin experiments and lasers realize the inverted state. topics: [The Microcanonical Ensemble] sources: - book: Schroeder ref: "Ch. 2 — The Second Law; §2.1–2.3 The Two-State Paramagnet, and Ch. 3 — Interactions and Implications; §3.3" - book: Reif ref: "Ch. 3 — Statistical Thermodynamics; §3.10" - book: Pathria & Beale ref: "Ch. 3 — The Canonical Ensemble; §3.9–3.10" draft: false --- The ideal gas is a continuous system whose count required a high-dimensional volume. The opposite extreme is a system with a discrete, finite set of microstates, and the cleanest example is a collection of spins that each point either along or against a magnetic field. Its multiplicity is a binomial coefficient, its entropy a closed-form dome, and its temperature the slope of that dome. The finiteness has one striking consequence absent from the gas: the energy is bounded above, so the entropy can decrease with energy, and the temperature can be negative. ## The two-state paramagnet Consider $N$ localized, non-interacting magnetic moments, each of magnitude $\mu$, in a uniform magnetic field of strength $B$. Every moment has two orientations: it points along the field, with energy $-\mu B$, or against it, with energy $+\mu B$. Write $N_\uparrow$ for the number aligned with the field and $N_\downarrow$ for the number against it, so $N=N_\uparrow+N_\downarrow$. The total energy is $$ E = -\mu B\,(N_\uparrow - N_\downarrow) = -\mu B\,(N - 2N_\downarrow), $$ which ranges from $-N\mu B$ when every moment is aligned to $+N\mu B$ when every moment is anti-aligned. The energy is bounded both below and above — the feature that separates this system from the gas, whose kinetic energy has no upper limit. The moments are localized and therefore distinguishable by their lattice sites, so no $N!$ appears. A microstate is a list of orientations, and the number of microstates with a given energy is the number of ways to choose which $N_\uparrow$ of the $N$ sites are aligned, $$ \Omega(N_\uparrow) = \binom{N}{N_\uparrow} = \frac{N!}{N_\uparrow!\,N_\downarrow!}. $$ This binomial multiplicity is largest when the two populations are equal and smallest when one orientation is fully occupied, mirroring the fact that there is exactly one way to align every moment but astronomically many ways to align half. $$ % caption: The binomial multiplicity $\Omega(N_\uparrow)=\binom{N}{N_\uparrow}$ is sharply peaked at $N_\uparrow=N/2$, where the two orientations are equally populated; the aligned and anti-aligned extremes are unique microstates. \begin{tikzpicture}[>=Latex,font=\footnotesize,scale=1.0] \definecolor{acc}{HTML}{4A6FA5} \draw[->,black] (0,0)--(6.4,0) node[right,black]{up-spins}; \draw[->,black] (0,0)--(0,4.0) node[above,black]{count}; \draw[acc,very thick] plot[domain=0.3:5.7,samples=70] (\x,{3.4*exp(-1.4*(\x-3.0)^2)}); \draw[black,dashed] (3.0,0)--(3.0,3.4); \node[black,below] at (3.0,0) {$\frac{N}{2}$}; \node[black,below] at (0.4,0) {$0$}; \node[black,below] at (5.7,0) {$N$}; \node[acc] at (4.9,2.4) {binomial peak}; \end{tikzpicture} $$ ## Entropy as a dome in the energy The entropy is $S=k\ln\Omega$. With Stirling's approximation $\ln N!\approx N\ln N-N$, and writing $x=N_\uparrow/N$ for the aligned fraction, $$ \frac{S}{Nk} = -x\ln x - (1-x)\ln(1-x). $$ This is the binary entropy function. It vanishes at $x=0$ and $x=1$, where the system is fully ordered and has a single microstate, and it reaches its maximum $S/Nk=\ln2$ at $x=\tfrac12$, where the two orientations are equally likely and the multiplicity is largest. Every moment then carries one bit of orientational uncertainty, and $N$ moments carry $Nk\ln2$ of entropy. The energy fixes the aligned fraction through $E=-N\mu B(2x-1)$, so $x=\tfrac12(1-E/N\mu B)$. Substituting turns the entropy into a function of energy alone, an inverted dome: $S=0$ at both energy extremes $E=\pm N\mu B$ and a maximum at $E=0$, where the populations are equal. On the low-energy side the entropy rises with energy; on the high-energy side it falls. That descending branch has no counterpart in the gas and is the origin of everything that follows. $$ % caption: The paramagnet entropy $S(E)$ is a dome: zero at the fully ordered extremes $E=\pm N\mu B$ and maximal at $E=0$; the rising branch has positive slope (positive $T$) and the falling branch negative slope (negative $T$). \begin{tikzpicture}[>=Latex,font=\footnotesize,scale=1.0] \definecolor{acc}{HTML}{4A6FA5} \draw[->,black] (0,0)--(8.2,0) node[right,black]{$E$}; \draw[->,black] (2.4,0)--(2.4,3.9) node[above,black]{$S$}; \draw[acc,very thick] plot[domain=0.55:4.25,samples=60] (\x,{3.3*sqrt(1-((\x-4.25)/3.7)^2)}); \draw[black,very thick] plot[domain=4.25:7.95,samples=60] (\x,{3.3*sqrt(1-((\x-4.25)/3.7)^2)}); \filldraw[black] (4.25,3.3) circle (1.8pt); \node[black] at (4.25,3.7) {max}; \node[acc] at (1.7,2.6) {$T>0$}; \node[black] at (6.8,2.6) {$T<0$}; \node[black,below] at (0.55,0) {aligned}; \node[black,below] at (7.95,0) {anti-aligned}; \draw[black,dashed] (4.25,0)--(4.25,3.3); \node[black,below] at (4.25,0) {$E=0$}; \end{tikzpicture} $$ ## Temperature from the slope The microcanonical temperature is $1/T=(\partial S/\partial E)_{N}$. Differentiating the entropy with respect to $x$ and dividing by $\d E/\d x=-2N\mu B$, $$ \frac{1}{T} = \frac{\partial S/\partial x}{\partial E/\partial x} = \frac{Nk\,\ln\!\frac{1-x}{x}}{-2N\mu B} = \frac{k}{2\mu B}\ln\!\frac{x}{1-x} = \frac{k}{2\mu B}\ln\!\frac{N_\uparrow}{N_\downarrow}. $$ Inverting gives the population ratio, $$ \frac{N_\uparrow}{N_\downarrow} = e^{2\mu B/kT} = e^{-\Delta\varepsilon/kT}, \qquad \Delta\varepsilon = \varepsilon_\downarrow-\varepsilon_\uparrow = 2\mu B, $$ the Boltzmann ratio for two levels split by $2\mu B$: the lower-energy aligned state is more populated at any positive temperature, and the ratio approaches unity as $T\to\infty$. The microcanonical derivation reproduces the canonical Boltzmann factor without ever invoking a reservoir, because the two-level population ratio is fixed by counting alone. > **Worked example.** Electron spins in a field $B=1\ \mathrm{T}$ at > $T=300\ \mathrm{K}$. Each spin has $\mu\approx\mu_B=9.274\times10^{-24}\ > \mathrm{J\,T^{-1}}$, so the level splitting is > $2\mu B=1.85\times10^{-23}\ \mathrm{J}$, against a thermal energy > $kT=4.14\times10^{-21}\ \mathrm{J}$. The population ratio is > $$ > \frac{N_\uparrow}{N_\downarrow} = e^{2\mu B/kT} > = e^{4.5\times10^{-3}} \approx 1.0045. > $$ > The aligned excess is only about $0.45\%$: at room temperature the thermal energy > dwarfs the magnetic splitting, so a paramagnet is very weakly polarized and its > magnetization is nearly linear in $B$, the microscopic origin of Curie's law. > Reaching strong polarization requires $2\mu B\gtrsim kT$, that is, high fields and > low temperatures. The sign of $1/T$ tracks the sign of $\ln(N_\uparrow/N_\downarrow)$, which is the slope of the entropy dome. - **Positive temperature, $E<0$.** More moments are aligned than anti-aligned, $N_\uparrow>N_\downarrow$, the entropy rises with energy, and $1/T>0$. Adding energy flips aligned moments and opens up more microstates, the ordinary situation. - **Infinite temperature, $E=0$.** The populations are equal, the entropy is at its peak, the slope vanishes, and $1/T=0$. This is $T=\pm\infty$: the two signs meet. - **Negative temperature, $E>0$.** More moments are anti-aligned than aligned, $N_\uparrow=Latex,font=\footnotesize,scale=1.0] \definecolor{acc}{HTML}{4A6FA5} \draw[->,black] (0,-1.9)--(8.2,-1.9) node[right,black]{$E$}; \draw[->,black] (4.1,-2.6)--(4.1,2.4) node[above,black]{$\frac{1}{T}$}; \draw[acc,very thick] plot[domain=0.75:4.1,samples=60] (\x,{0.9*ln((4.1-0.55)/(\x-0.55))-0.0}); \draw[black,very thick] plot[domain=4.1:7.45,samples=60] (\x,{0.9*ln((4.1-0.55)/(\x-0.55))-0.0}); \filldraw[black] (4.1,0) circle (1.8pt); \node[black] at (5.35,0.35) {slope zero}; \node[acc] at (2.0,1.65) {$\frac{1}{T}>0$}; \node[black] at (6.5,-1.35) {$\frac{1}{T}<0$}; \node[black,below] at (4.1,-2.05) {$E=0$}; \end{tikzpicture} $$ ## Negative temperature and population inversion Negative absolute temperature is not colder than zero; it is hotter than any positive temperature, including $+\infty$. The correct ordering of hotness is by the parameter $\beta=1/kT$, which decreases monotonically from $+\infty$ (at $T=0^{+}$) through $0$ (at $T=\pm\infty$) to $-\infty$ (at $T=0^{-}$). A system at negative $T$ has $\beta<0$, below the $\beta=0$ of an infinitely hot positive-$T$ system, so energy flows from the negative-$T$ system into any positive-$T$ system brought into contact with it — the direction that increases total entropy. Ranked by increasing hotness, the sequence runs $$ T=0^{+}\ \to\ T=+\infty\ \to\ T=-\infty\ \to\ T=0^{-}. $$ A negative temperature requires two conditions. The energy spectrum must be **bounded above**, so that the entropy can turn over and decrease; a gas, with unbounded kinetic energy, can never reach it. And the system must be driven into **population inversion**, with more moments in the upper state than the lower, which does not happen by ordinary heating — heating only pushes the populations toward equality at $T=\infty$. Inversion is produced by an external operation, such as suddenly reversing the field so the previously aligned moments become anti-aligned. $$ % caption: Ordinary populations (left) have more moments in the lower level, giving $T>0$; a population-inverted system (right) has more in the upper level, giving $T<0$, and is hotter than any positive temperature. \begin{tikzpicture}[>=Latex,font=\footnotesize,scale=1.0] \definecolor{acc}{HTML}{4A6FA5} \draw[black,very thick] (0,0.6)--(2.6,0.6); \draw[black,very thick] (0,2.4)--(2.6,2.4); \node[black,left] at (0,0.6) {lower}; \node[black,left] at (0,2.4) {upper}; \filldraw[black] (0.5,0.6) circle (3pt); \filldraw[black] (1.0,0.6) circle (3pt); \filldraw[black] (1.5,0.6) circle (3pt); \filldraw[black] (2.0,0.6) circle (3pt); \filldraw[black] (0.7,2.4) circle (3pt); \node[black] at (1.3,3.0) {$T>0$}; \draw[black,very thick] (6.0,0.6)--(8.6,0.6); \draw[black,very thick] (6.0,2.4)--(8.6,2.4); \node[black,left] at (6.0,0.6) {lower}; \node[black,left] at (6.0,2.4) {upper}; \filldraw[acc] (6.7,0.6) circle (3pt); \filldraw[acc] (6.5,2.4) circle (3pt); \filldraw[acc] (7.0,2.4) circle (3pt); \filldraw[acc] (7.5,2.4) circle (3pt); \filldraw[acc] (8.0,2.4) circle (3pt); \node[acc] at (7.3,3.0) {$T<0$}; \end{tikzpicture} $$ Negative-temperature states are realized wherever a nearly isolated set of levels can be inverted faster than it equilibrates with its surroundings. Nuclear-spin systems in solids are the classic case: the spin–spin equilibration time can be far shorter than the spin–lattice relaxation time, so a suddenly reversed field leaves the spins internally equilibrated at a negative temperature for a measurable interval before they leak energy to the lattice.[^purcell] The same inverted population is the working condition of a laser, where more atoms occupy the upper lasing level than the lower, and stimulated emission amplifies rather than absorbs the light. In each case the negative temperature is a property of the inverted subsystem alone, meaningful only because that subsystem has a bounded spectrum and equilibrates within itself. ## Summary - The ideal two-state paramagnet of $N$ moments in a field has multiplicity $\Omega=\binom{N}{N_\uparrow}$ and entropy $S/Nk=-x\ln x-(1-x)\ln(1-x)$ with $x=N_\uparrow/N$, peaking at $S=Nk\ln2$ where the populations are equal. - The energy $E=-N\mu B(2x-1)$ is bounded between $\pm N\mu B$, so $S(E)$ is a dome; the slope gives $1/T=(k/2\mu B)\ln(N_\uparrow/N_\downarrow)$ and the population ratio $N_\uparrow/N_\downarrow=e^{2\mu B/kT}$. - Positive $T$ corresponds to $E<0$ and a normal population; $E=0$ is $T=\pm\infty$; $E>0$ inverts the populations and gives negative $T$, which is hotter than any positive temperature since energy flows out of it. - Negative temperature requires an energy spectrum bounded above and an externally produced population inversion, realized in nuclear-spin systems and lasers. [^schroeder-para]: Schroeder, _An Introduction to Thermal Physics_ (Addison-Wesley, 2000), Ch. 2 §2.1–2.3 counts the two-state paramagnet and Ch. 3 §3.3 develops its temperature and negative-temperature behavior; companion page . [^purcell]: The negative-temperature nuclear-spin experiment is Purcell & Pound, "A Nuclear Spin System at Negative Temperature," _Phys. Rev._ **81**, 279 (1951). The counting and temperature analysis follow Reif, _Fundamentals of Statistical and Thermal Physics_ (Waveland reprint, 2009), §3.10, and Pathria & Beale, _Statistical Mechanics_ (4th ed., Elsevier, 2021), §3.9–3.10.