Negative Temperature

Cool a gas and its temperature falls toward absolute zero; heat it and the temperature climbs without bound. That picture is complete only for systems whose energy has no ceiling — an ordinary gas can always absorb more kinetic energy. But a system with a bounded energy spectrum, such as a collection of spins in a magnetic field, has a highest possible energy as well as a lowest. Pump enough energy in and you reach a regime where adding still more energy lowers the entropy, because the spins are being forced into the unique top state. The thermodynamic definition of temperature, $1/T=(\partial S/\partial E)$ , then turns negative. The system is at a negative absolute temperature. Far from being colder than absolute zero, a negative-temperature system is hotter than any positive temperature. The natural ordering of "hotness" is not $T$ but $1/T=\partial S/\partial E$ , which decreases smoothly from $+\infty$ (at $T=+0$ ) through zero (at $T=\pm\infty$ ) to $-\infty$ (at $T=-0$ ). On that scale a body at $T=-100\ \mathrm K$ sits above every positive temperature: put it in contact with anything at positive $T$ and energy flows out of it. This article derives the entropy dome $S(E)$ of the two-level system, shows exactly where $\beta=\partial S/\partial E$ changes sign, and works out the temperature ordering and the direction of heat flow — always with the second law intact. It owns the two-level $S(E)$ construction and the sign-flip of $\beta$ ; the laser-gain and optical-lattice applications are named and left to their own articles.

⚡ Key Takeaways

Negative absolute temperature exists only for a system with a bounded energy spectrum (an upper as well as a lower bound) that is thermally isolated and internally equilibrated — spins in a field are the canonical example.
Temperature is defined by $1/T=(\partial S/\partial E)_{V,N}$ . When adding energy reduces entropy, this slope is negative and $T<0$ .
For $N$ spin- $\tfrac12$ moments the entropy $S(E)$ is a dome: it rises from zero at the ground state, peaks at $S=Nk\ln 2$ when the energy is central, and falls back to zero at the unique top state.
$T<0$ means a population inversion: upper levels are more occupied than lower ones, the reverse of the ordinary Boltzmann ladder.
The hotness order is $+0<\dots<+\infty\equiv-\infty<\dots<-0$ : a negative- $T$ body is hotter than any positive- $T$ body, and $-0\ \mathrm K$ is the hottest state of all.
Key idea: a bounded spectrum lets entropy turn over, and once $\partial S/\partial E<0$ the same definition that gives ordinary temperatures gives a negative one — which is hot, not cold.

💡 The Core Insight

Temperature is the reciprocal slope of entropy versus energy, $1/T=\partial S/\partial E$ . A system whose energy is bounded above has an entropy that rises, peaks, and falls as energy increases; on the falling side $\partial S/\partial E<0$ , so $T<0$ — a population-inverted state that is thermodynamically hotter than any $T>0$ .

The physical picture

Imagine $N$ magnetic moments, each of which can point only "down" (energy $-\varepsilon$ , aligned with the field) or "up" (energy $+\varepsilon$ , anti-aligned). The total energy runs from $-N\varepsilon$ (all down, the unique ground state) to $+N\varepsilon$ (all up, the unique top state). At the bottom and top there is exactly one microstate, so the entropy is zero. In between, the energy can be shared among the spins in enormously many ways, and the entropy is large. The number of arrangements — and hence the entropy — is greatest when the energy sits in the middle, with roughly half the spins up and half down. Entropy versus energy is therefore not a monotonically rising curve but a dome. Now feed energy into the isolated spin system. Starting from the ground state, each bit of added energy flips more spins up, increasing the disorder: entropy rises, and the temperature $1/T=\partial S/\partial E$ is positive, exactly as for an ordinary body. But once you pass the halfway point, adding energy means flipping spins from the crowded middle toward the sparse, ordered top — the arrangements become fewer, entropy decreases, and the slope $\partial S/\partial E$ goes negative. The temperature has passed through infinity and come back on the negative side.

Figure

Entropy dome S(E) for a bounded spin system, with the slope 1/T changing sign

There is one decisive consequence. Because $1/T$ runs continuously from $+\infty$ down to $-\infty$ as the energy climbs, the temperature scale does not wrap around through zero. It wraps around through infinity. The hottest a positive-temperature body can get is $T\to+\infty$ ; the very next state, one notch more energetic, is $T=-\infty$ , and from there the temperature rises toward $T=-0$ . So the states just below the top of the dome, at small negative $T$ , are the most energetic and the "hottest" of all.

Figure

Level populations: ordinary Boltzmann, infinite temperature, and inverted

The core equations

Temperature is defined not as "average kinetic energy" — that picture is a special case — but through the entropy, which is why it can be negative. The starting point is the entropy–energy relation and its slope.

\frac{1}{T} = \left(\frac{\partial S}{\partial E}\right)_{V,N} \qquad (1)

where: $S$ is the entropy, $E$ the internal energy at fixed volume $V$ and particle number $N$ , and $T$ the absolute temperature. This is the operational definition of temperature; ordinary systems have entropy increasing with energy, so $\partial S/\partial E>0$ and $T>0$ . Nothing in the definition forbids a negative slope. The owner of this relation is the article on temperature, pressure, and chemical potential as entropy derivatives. For $N$ independent spin- $\tfrac12$ moments with single-spin energies $\pm\varepsilon$ , write the fraction pointing up as $n_+$ and down as $n_-=1-n_+$ . The energy per spin is $u\equiv E/(N\varepsilon)$ , and $n_\pm=(1\pm u)/2$ .

\frac{S}{Nk} = -\Big[\tfrac{1+u}{2}\ln\tfrac{1+u}{2} + \tfrac{1-u}{2}\ln\tfrac{1-u}{2}\Big], \qquad u=\frac{E}{N\varepsilon}\in[-1,+1] \qquad (2)

where: $S$ is the total entropy, $k$ Boltzmann's constant, and $u$ the dimensionless energy per spin. This is the mixing entropy of $Nn_+$ up-spins among $N$ sites; it vanishes at $u=\pm1$ (one arrangement) and is largest at $u=0$ . It is the standard $S=k\ln\Omega$ multiplicity, worked out in the Boltzmann-entropy article. Differentiating (2) and inserting it into (1) gives the temperature directly in terms of the energy.

\frac{1}{T} = \frac{k}{2\varepsilon}\,\ln\!\frac{1-u}{1+u} \qquad (3)

where: the logarithm is positive for $u<0$ (low energy, $T>0$ ), zero for $u=0$ ( $T=\pm\infty$ ), and negative for $u>0$ (high energy, $T<0$ ). Equation (3) is the whole story of negative temperature compressed to one line: the sign of $T$ is inherited from the sign of $u$ , i.e. from whether the system sits below or above the midpoint of its bounded energy band. Equivalently, in canonical equilibrium the level occupations follow the Boltzmann factor, whose reading at $T<0$ is a population inversion.

\frac{n_+}{n_-} = e^{-2\varepsilon/kT} = \frac{1+u}{1-u} \qquad (4)

where: $n_+/n_-$ is the ratio of upper- to lower-level occupation. For $T>0$ it is less than one (lower level fuller); for $T<0$ it exceeds one (upper level fuller). The crossover $n_+/n_-=1$ is $T=\pm\infty$ .

The physical picture in numbers

Fix the spin gap by choosing $\varepsilon/k=100\ \mathrm K$ (a full level splitting $2\varepsilon/k=200\ \mathrm K$ ). Then equation (3) reads $1/T=(1/200\ \mathrm K)\ln[(1-u)/(1+u)]$ , so $T$ is fixed once you know how much energy the isolated spins hold.

Derivation AThe entropy dome and the sign of the temperature

The goal is to establish, with no hand-waving, that the entropy of a bounded two-level system is a concave dome in the energy and that the temperature defined by $1/T=\partial S/\partial E$ is positive on the low-energy half, infinite at the centre, and negative on the high-energy half. The strategy is to write $S$ as the exact multiplicity entropy of $N_+$ up-spins among $N$ , convert to the intensive variable $u=E/N\varepsilon$ , differentiate once to get $1/T$ , and differentiate again to confirm concavity so that each energy corresponds to a single temperature. The only subtlety is bookkeeping of the two logarithmic terms, whose $\ln$ arguments must stay between 0 and 1.

Step 1. Count microstates. With $N_+=N(1+u)/2$ spins up, the multiplicity is $\Omega=\binom{N}{N_+}$ . By the Stirling approximation $\ln\Omega = -N[n_+\ln n_+ + n_-\ln n_-]$ with $n_\pm=(1\pm u)/2$ , giving equation (2) via $S=k\ln\Omega$ . Both $n_+$ and $n_-$ lie in $[0,1]$ , so each $\ln$ is $\le 0$ and $S\ge0$ , vanishing only at $u=\pm1$ .

Step 2. Differentiate once. Using $dS/dE=(dS/du)(du/dE)$ with $E=N\varepsilon u$ , so $du/dE=1/(N\varepsilon)$ , and $\frac{d}{du}\frac{S}{Nk}=-\tfrac12\ln\frac{1+u}{1-u}$ , one obtains $\dfrac{1}{T}=\dfrac{\partial S}{\partial E}=\dfrac{k}{2\varepsilon}\ln\dfrac{1-u}{1+u}$ , equation (3).

Step 3. Read the signs. For $-1<u<0$ the ratio $(1-u)/(1+u)>1$ , its logarithm is positive, and $T>0$ . At $u=0$ the logarithm is zero, so $1/T=0$ , i.e. $T=\pm\infty$ — the peak of the dome. For $0<u<1$ the ratio is less than one, the logarithm is negative, and $T<0$ . The temperature therefore climbs from $+0$ through $+\infty$ , jumps to $-\infty$ , and rises to $-0$ as the energy sweeps from $-N\varepsilon$ to $+N\varepsilon$ .

Step 4. Confirm single-valuedness. The second derivative $\frac{d^2}{du^2}\frac{S}{Nk}=-\frac{1}{1-u^2}<0$ for all $|u|<1$ : the entropy is strictly concave, so $1/T$ is strictly decreasing in $u$ and each energy has exactly one temperature. There is no ambiguity — every point on the dome has a well-defined (possibly negative) $T$ . Verified: the companion verifier reproduces $S(u)$ and $dS/du=\tfrac12\ln[(1-u)/(1+u)]$ symbolically, confirms the sign flip at $u=0$ (values $+0.5493$ at $u=-0.5$ , $0$ at $u=0$ , $-0.5493$ at $u=+0.5$ ), and checks strict concavity $S''<0$ .

Derivation BWhy a negative-temperature body is hotter than any positive temperature

The aim here is to prove the counterintuitive ordering: a system at $T<0$ gives up energy to a system at any $T>0$ , so it is thermodynamically hotter, and to show this follows from the second law rather than contradicting it. The strategy is to bring two bounded subsystems into thermal contact, demand that the total entropy not decrease when a small energy $\delta E$ flows between them, and read off the direction of flow in terms of $\beta=1/T$ . The one thing to keep straight is that the natural "hotness" variable is $\beta=\partial S/\partial E$ , not $T$ : $\beta$ is monotonic across the whole range while $T$ is not.

Step 1. Set up contact. Two isolated subsystems $A$ and $B$ can exchange energy at fixed total $E_A+E_B$ . The total entropy is $S=S_A(E_A)+S_B(E_B)$ . Move $\delta E$ from $A$ to $B$ : $\delta S = (\partial S_B/\partial E_B - \partial S_A/\partial E_A)\,\delta E = (\beta_B-\beta_A)\,\delta E$ , where $\beta\equiv1/T=\partial S/\partial E$ .

Step 2. Impose the second law. Equilibrium (maximum entropy) requires $\delta S\ge0$ for the spontaneous direction. Energy flows from $A$ to $B$ ( $\delta E>0$ ) precisely when $\beta_B>\beta_A$ , i.e. from the subsystem with the smaller $\beta$ to the one with the larger $\beta$ . "Hotter" means smaller $\beta=1/T$ , and equilibrium is $\beta_A=\beta_B$ .

Step 3. Order the whole scale. As energy increases, $\beta$ decreases monotonically from $+\infty$ (at $T=+0$ ) through $0$ (at $T=\pm\infty$ ) to $-\infty$ (at $T=-0$ ). Since hotness is decreasing $\beta$ , the hotness order is $T=+0$ (coldest) $<\dots< +\infty \equiv -\infty <\dots< T=-0$ (hottest). Every negative temperature has a smaller $\beta$ than every positive temperature, so it is hotter.

Step 4. Check a concrete pair. Take $A$ at $T_A=-100\ \mathrm K$ ( $\beta_A=-0.01/k\ \mathrm K^{-1}$ ) and $B$ at $T_B=+300\ \mathrm K$ ( $\beta_B=+0.00333/k\ \mathrm K^{-1}$ ). Since $\beta_B>\beta_A$ , energy flows $A\to B$ ; the negative-temperature body cools by giving up energy (its entropy rises as it slides down the dome) and the positive-temperature body warms (its entropy also rises). Total entropy increases — the second law is obeyed, and $A$ was the hotter of the two. Verified: the verifier evaluates $\beta_A,\beta_B$ for the $-100\,\mathrm K$ / $+300\,\mathrm K$ pair, confirms the flow direction $A\to B$ , and checks that both $\delta S_A$ and $\delta S_B$ are positive so $\delta S_{\rm tot}>0$ .

Limiting cases

Each limit of equation (3) is a physical check on the construction.

Case	Result	Meaning
Ground state $u\to-1$	$S\to0$ , $T\to+0$	All spins down; a unique microstate, the coldest positive temperature.
Central energy $u=0$	$S=Nk\ln2$ , $T=\pm\infty$	Half up, half down; maximum entropy; infinite temperature (equal populations).
Top state $u\to+1$	$S\to0$ , $T\to-0$	All spins up; a unique microstate, the hottest state of all.
Small excitation $u\to-1^+$	$T>0$ , $S$ rising	Ordinary low-temperature behaviour; entropy grows as energy is added.

⚠️ Thinking negative temperature is "below absolute zero" or "colder than cold"

It is the opposite. Absolute zero is approached as $\beta=1/T\to+\infty$ (energy at its floor); a negative temperature has $\beta<0$ , which is at the other end of the hotness scale, past $T=+\infty$ . You never pass through $T=0$ to reach negative temperatures — you pass through $T=\pm\infty$ by adding energy to a population-inverted, bounded system. A $-1\ \mathrm K$ system is hotter than a $+10^{9}\ \mathrm K$ system. The word "negative" refers only to the sign of the number $T$ ; the physically monotonic hotness variable is $\beta=\partial S/\partial E$ , which runs smoothly from $+\infty$ to $-\infty$ with no discontinuity at $T=\pm\infty$ . Negative temperatures also require a bounded spectrum and thermal isolation from any unbounded degrees of freedom (like translational motion): a real spin system reaches $T<0$ only because the spins equilibrate among themselves far faster than they leak energy to the lattice.

Worked examples

Example 1 — A population-inverted spin system. Take $N$ spin- $\tfrac12$ moments with $\varepsilon/k=100\ \mathrm K$ , holding energy $u=+0.6$ per spin (above the midpoint). Find the entropy, the temperature, and the occupation ratio.

Step 1. Occupations: $n_+=(1+0.6)/2=0.8$ , $n_-=(1-0.6)/2=0.2$ — the upper level is four times as full as the lower one, a genuine inversion.

Step 2. Entropy from (2): $S/Nk = -[0.8\ln0.8 + 0.2\ln0.2] = 0.5004$ , i.e. about $0.72$ of the maximum $\ln2=0.693$ .

Step 3. Temperature from (3): $1/T = (1/200\ \mathrm K)\ln[(1-0.6)/(1+0.6)] = (1/200)\ln(0.25) = -0.00693\ \mathrm K^{-1}$ , so $T = \mathbf{-144.3\ K}$ . The occupation ratio $n_+/n_-=0.8/0.2=\mathbf{4}$ matches $e^{-2\varepsilon/kT}=e^{200/144.3}=4.0$ . The takeaway: with more than half the spins in the upper level, the temperature is negative and the system is hotter than any positive-temperature body, even though it holds only a modest fraction of its maximum energy.

Example 2 — Heat flow between a negative- and a positive-temperature body. Bring subsystem $A$ at $T_A=-100\ \mathrm K$ into thermal contact with subsystem $B$ at $T_B=+300\ \mathrm K$ . Which way does energy flow, and does the total entropy increase?

Step 1. Compute the hotness variables: $\beta_A=1/T_A=-0.0100\ k^{-1}\mathrm K^{-1}$ and $\beta_B=1/T_B=+0.00333\ k^{-1}\mathrm K^{-1}$ . Since $\beta_B>\beta_A$ , a small energy $\delta E$ flows from $A$ to $B$ .

Step 2. Entropy change for $\delta E=1\ \mathrm{J}$ transferred $A\to B$ : $\delta S_A = -\delta E/T_A = -1/(-100) = +0.0100\ \mathrm{J/K}$ and $\delta S_B = +\delta E/T_B = +1/300 = +0.00333\ \mathrm{J/K}$ .

Step 3. Total: $\delta S_{\rm tot} = +0.0100+0.00333 = +0.0133\ \mathrm{J/K} > 0$ . Both subsystems gain entropy; the process is spontaneous and irreversible. The takeaways: energy flows out of the $-100\ \mathrm K$ body into the $+300\ \mathrm K$ body — the negative-temperature system is the hotter one — and the total entropy rises, so the second law holds throughout.

norman ramsey

1915–2011 · American physicist who put negative temperature on a rigorous thermodynamic footing

Norman Foster Ramsey Jr. spent most of his career at Harvard, where in 1949 he invented the method of separated oscillatory fields, the technique underlying the caesium atomic clock and the hydrogen maser, for which he shared the 1989 Nobel Prize in Physics. In a 1956 paper (Physical Review 103, 20) he analyzed the thermodynamics of systems capable of negative absolute temperature, clarifying that such states are legitimate equilibria of bounded-spectrum systems, that they are hotter than any positive temperature, and that the second and third laws remain intact once "hotness" is read as $-1/T$ . His treatment settled a debate opened by the 1951 Purcell–Pound nuclear-spin experiment and remains the standard reference. Ramsey also helped found Brookhaven and Fermilab and chaired the committee that measured the neutron electric dipole moment.

⏳ History

1926

— Fermi and Dirac formulate the quantum statistics of identical particles

1949

— Onsager notes negative-temperature configurations of quantized point vortices in two dimensions

1951

— Purcell and Pound produce and measure a negative spin temperature in a lithium fluoride crystal, published in Phys. Rev. 81, 279

1956

— Ramsey gives the thermodynamic theory of negative temperature in Phys. Rev. 103, 20

1957

— Abragam and Proctor develop spin-temperature theory for coupled nuclear-spin systems

2013

— Braun and co-workers realize a negative absolute temperature for motional degrees of freedom of ultracold atoms in an optical lattice, Science 339, 52

Where this lives today

Negative temperature is not a curiosity confined to textbooks. Every laser depends on a population inversion — more atoms or molecules in an excited state than the ground state — which is exactly a negative spin (or level) temperature for the lasing transition; the optical gain medium is, for that transition, a negative-temperature system, and pumping is the act of driving it past $T=\pm\infty$ . Nuclear-spin systems in solids routinely reach negative temperatures because the spins equilibrate among themselves in milliseconds while leaking energy to the lattice over seconds, giving a genuine spin temperature that can be flipped negative by reversing the field faster than the spins can follow. In 2013 an ultracold-atom experiment engineered a bounded kinetic-energy band in an optical lattice and prepared atoms at negative temperature, extending the concept from spins to motional degrees of freedom and stirring fresh debate about which entropy definition (Boltzmann or Gibbs) best captures the effect. The deeper measure-theoretic questions and the laser-gain dynamics are taken up in their own articles; here the point is that the sign of $\partial S/\partial E$ is a physical, measurable quantity.

Going further

References. Ramsey, "Thermodynamics and Statistical Mechanics at Negative Absolute Temperatures," Phys. Rev. 103, 20 (1956) · Purcell & Pound, "A Nuclear Spin System at Negative Temperature," Phys. Rev. 81, 279 (1951) · Kittel & Kroemer, Thermal Physics, 2nd ed., Ch. 1 and the negative-temperature appendix · Reif, Fundamentals of Statistical and Thermal Physics, §9.3 · Braun et al., "Negative Absolute Temperature for Motional Degrees of Freedom," Science 339, 52 (2013).

Provenance. A verifier should reproduce: (1) the entropy $S/Nk$ from equation (2) and its peak $\ln2$ at $u=0$ ; (2) the derivative $dS/du=\tfrac12\ln[(1-u)/(1+u)]$ with the sign flip ( $+0.5493$ , $0$ , $-0.5493$ at $u=-0.5,0,+0.5$ ); (3) strict concavity $S''=-1/(1-u^2)<0$ ; (4) Example 1 at $u=0.6$ : $S/Nk=0.5004$ , $T=-144.3\ \mathrm K$ (for $\varepsilon/k=100\ \mathrm K$ ), $n_+/n_-=4$ ; (5) Example 2 flow direction $A\to B$ and $\delta S_{\rm tot}=+0.0133\ \mathrm{J/K}>0$ .

Builds on: Temperature, Pressure, and Chemical Potential as Entropy Derivatives · Boltzmann Entropy · The Two-Level Paramagnet and the Brillouin Function · The Schottky Anomaly. Leads to: Second Law of Thermodynamics and Entropy · Population Inversion and Laser Gain · Negative Temperature in Optical Lattices.

Is a negative temperature really hotter than an infinite temperature?

Yes, and it is worth being careful about what "hotter" means. Two bodies in contact exchange energy so as to maximize total entropy, and energy flows from the one with the smaller $\beta=\partial S/\partial E=1/T$ to the one with the larger $\beta$ . A negative-temperature body has $\beta<0$ , which is smaller than the $\beta>0$ of any positive-temperature body, so it gives up energy to any positive-temperature partner — including one at $T=+10^{9}\ \mathrm K$ . Since giving up energy while being hotter is exactly what "hotter" means, a negative-temperature body outranks every positive one. The apparent paradox dissolves once you plot the monotonic variable $\beta$ rather than $T$ : $\beta$ decreases smoothly through zero (at $T=\pm\infty$ ) with nothing special happening, while $T$ merely has a coordinate singularity there.

Why can't an ordinary gas have a negative temperature?

Because its energy is unbounded above. A gas can always absorb more kinetic energy — there is no highest-energy state — so its entropy keeps rising with energy and $\partial S/\partial E$ stays positive forever. Negative temperature needs the entropy to turn over, which requires a ceiling on the energy. Spins in a field have exactly two (or a few) levels per particle and hence a maximum total energy, so their entropy peaks and falls; that is what makes them the standard realization. If you couple the spins to unbounded translational motion, energy leaks away into the motion and the negative-temperature state relaxes back through $T=\pm\infty$ to an ordinary positive temperature — which is why the spins must be thermally isolated on the timescale of the experiment.

Does negative temperature violate the second or third laws?

No. The second law is that total entropy does not decrease; the worked example shows it strictly increases when a negative- and a positive-temperature body equilibrate. Negative-temperature equilibria are perfectly good entropy maxima at fixed energy for a bounded system. The third law concerns the approach to $T=+0$ (the ground state, $\beta\to+\infty$ ), which is at the cold end of the scale and is untouched by anything happening at the hot, negative end. If anything, the existence of negative temperatures reinforces the third-law picture that $\beta$ , not $T$ , is the physically fundamental quantity: absolute zero is $\beta=+\infty$ , unreachable from either side.

What exactly is a population inversion, and how is it made?

A population inversion is a state in which a higher-energy level is more occupied than a lower one, the reverse of the ordinary Boltzmann ladder where lower levels are always fuller. Equation (4) shows this is precisely the condition $T<0$ . In nuclear magnetic resonance it is produced by suddenly reversing the applied field: the spins, which had relaxed to alignment (lower level fuller), find themselves anti-aligned (upper level fuller) faster than they can re-equilibrate, so they momentarily sit at a negative spin temperature. In a laser it is produced by optical or electrical pumping that lifts atoms into an excited state faster than they decay. Either way, "inverted" and "negative temperature" describe the same physics from two angles.

If T = ±∞ at the midpoint, is the temperature undefined there?

The temperature $T$ has a coordinate singularity at the midpoint, but nothing physical is singular. The entropy, the energy, the populations, and the hotness variable $\beta=1/T$ are all perfectly smooth as the system passes through $u=0$ : $\beta$ simply crosses zero. Saying " $T=+\infty$ and $T=-\infty$ are the same state" is just the statement that $\beta=0$ is a single point. This is a good reason to regard $\beta$ as the natural thermodynamic coordinate; the familiar $T$ is a convenient but non-monotonic reparametrization that misbehaves only at this one harmless point.

Can you extract work using a negative-temperature reservoir?

In principle yes, and quite effectively, because a negative-temperature reservoir is so hot. A Carnot engine running between a negative-temperature hot reservoir and a positive-temperature cold one has an efficiency expression that can formally exceed one, reflecting that the "hot" reservoir can dump energy while gaining entropy. In practice the negative-temperature state is metastable and finite — the spins hold a bounded amount of energy and relax — so it behaves like a charged battery rather than an inexhaustible furnace. The useful lesson is conceptual: because entropy and energy are the primary quantities, engines and heat flow are governed by $\beta=\partial S/\partial E$ , and the ordinary intuition "heat flows from high $T$ to low $T$ " must be rephrased as "from low $\beta$ to high $\beta$ " once bounded systems are in play.

Is negative temperature the same idea as the Schottky anomaly?

They are two faces of the same bounded two-level system. The Schottky anomaly is what the heat capacity of that system looks like as it is warmed from $T=0$ : a bump that peaks and falls because the finite level spacing lets the populations saturate. Negative temperature is what happens if you keep pushing energy in past $T=+\infty$ , driving the populations into inversion. Both phenomena exist only because the spectrum is bounded; an unbounded system has neither a Schottky bump nor a negative temperature. Reading the two articles together gives the complete thermal portrait of a finite-level system — its heat capacity on the way up and its negative-temperature regime beyond the top.