Microcanonical Ensemble

The microcanonical ensemble represents an isolated quantum system whose exact conserved quantities are fixed and whose energy is restricted to a specified shell. For a fixed-particle-number Hamiltonian $H_N$, choose the interval

$$I_{E,\Delta E} = \left[ E-\frac{\Delta E}{2}, E+\frac{\Delta E}{2} \right].$$

Let

$$\Pi_{E,\Delta E} = \mathbf 1_{I_{E,\Delta E}}(H_N)$$

be the spectral projector onto that interval. If

$$W(E,\Delta E,N,V) = \operatorname{Tr}_{\mathcal H_N} \Pi_{E,\Delta E}$$

is finite and nonzero, the microcanonical density operator is

$$\rho_{\mathrm{mc}} = \frac{ \Pi_{E,\Delta E} }{ W(E,\Delta E,N,V) }.$$

The uniform weights are a statistical assignment from the stated constraints. Isolation alone does not imply them: it preserves any initially unequal energy populations as well.

Helpful background. The ensemble comparison explains the physical constraints; spectral decomposition supplies the projectors used here.

The Energy-Shell Projector

Suppose the Hamiltonian has spectral resolution

$$H_N = \sum_a E_aP_a,$$

where $P_a$ projects onto the eigenspace of energy $E_a$. Then

$$\Pi_{E,\Delta E} = \sum_{E_a\in I_{E,\Delta E}} P_a.$$

If

$$g_a = \operatorname{Tr}P_a$$

is the degeneracy of $E_a$, the shell dimension is

$$W = \sum_{E_a\in I_{E,\Delta E}} g_a.$$

The microcanonical state therefore assigns equal weight per orthonormal state inside the shell:

$$\rho_{\mathrm{mc}} = \frac{1}{W} \sum_{E_a\in I_{E,\Delta E}} P_a.$$

A finite quantum spectrum is restricted to a window of width $\Delta E$. The spectral projector retains every eigenspace in the window, $W=\operatorname{Tr}\Pi_{E,\Delta E}$ counts its dimension, and $\rho_{\mathrm{mc}}=\Pi_{E,\Delta E}/W$ is uniform on that subspace.

Uniform probabilities within the shell

The eigenvalues of $\rho_{\mathrm{mc}}$ are $1/W$ on the shell and zero outside. Consequently it is positive, has unit trace, and has purity $\operatorname{Tr}\rho_{\mathrm{mc}}^2=1/W$. It is pure exactly when $W=1$. Because the shell projector is a function of $H_N$, it commutes with $H_N$ and the state is stationary.

Uniform weights maximize the entropy under the shell-support constraint; the exercise below gives the relative-entropy proof. This explains the assignment without claiming that time evolution randomizes the weights. See maximum entropy for comparison with constraints on mean energy.

Microcanonical Expectation Values

For an observable $A$,

$$\langle A\rangle_{\mathrm{mc}} = \operatorname{Tr} \left( \rho_{\mathrm{mc}}A \right) = \frac{1}{W} \operatorname{Tr} \left( \Pi_{E,\Delta E}A \right).$$

In the energy basis,

$$\langle A\rangle_{\mathrm{mc}} = \frac{1}{W} \sum_{E_a\in I_{E,\Delta E}} \operatorname{Tr} \left( P_aA \right).$$

Off-diagonal matrix elements between distinct energy eigenspaces do not contribute to this equilibrium average. Degenerate eigenspaces are included through their full projectors, so the formula is basis independent inside each degeneracy.

Energy Support and Fluctuations

Every energy in the shell lies between

$$E_- = E-\frac{\Delta E}{2}$$

and

$$E_+ = E+\frac{\Delta E}{2}.$$

Consequently,

$$E_- \leq \langle H_N\rangle_{\mathrm{mc}} \leq E_+.$$

The energy is not necessarily one exact eigenvalue. Its variance satisfies the support bound

$$\operatorname{Var}_{\mathrm{mc}}(H_N) \leq \frac{(\Delta E)^2}{4}.$$

Thus “fixed energy” means fixed to the declared shell. If the shell contains several eigenvalues, an energy measurement still has a distribution over those values.

Exact-Energy Special Case

If the energy is known exactly and $E_a$ is an eigenvalue, use

$$\Pi_{E_a} = P_a.$$

The exact-energy microcanonical state is

$$\rho_{E_a} = \frac{P_a}{g_a}.$$

For a nondegenerate eigenvalue,

$$g_a=1,$$

so

$$\rho_{E_a} = |E_a\rangle\langle E_a|$$

is pure. For a degenerate eigenvalue, the exact-energy ensemble is maximally mixed within the degenerate eigenspace unless further conserved quantities or preparation information resolve it.

This distinction matters in small systems. A single nondegenerate eigenstate has no large state count from which smooth thermodynamics can be inferred.

How Wide Should the Shell Be?

The shell width is part of the ensemble definition. A useful many-body window should be:

1. broad compared with the local level spacing;
2. narrow compared with the macroscopic energy scale;
3. stable under modest changes of its center and width;
4. restricted to the correct symmetry and conserved-charge sector.

Schematically, one wants

$$\delta E_{\mathrm{level}} \ll \Delta E \ll E_{\mathrm{macro}}.$$

Along a thermodynamic sequence, a common requirement is

$$W(E,\Delta E) \longrightarrow \infty$$

while

$$\frac{\Delta E}{L} \longrightarrow 0.$$

Here $L$ counts sites or particles along a fixed-density sequence, and the microscopic energy unit is fixed. Using width per constituent avoids dependence on the arbitrary energy zero. The two limits are compatible because the many-body level spacing often becomes extremely small as system size grows.

There is no universal shell width that works for every Hamiltonian, energy, and observable. Near spectral edges, gaps, mobility edges, first-order coexistence, or sparse symmetry sectors, the required checks can be especially delicate.

State Counting and Density of States

Define the cumulative state count

$$\Gamma_N(E) = \operatorname{Tr}_{\mathcal H_N} \Theta(E-H_N).$$

For a discrete spectrum, $\Gamma_N(E)$ is a staircase. Its distributional derivative is the density of states

$$\omega_N(E) = \operatorname{Tr}_{\mathcal H_N} \delta(E-H_N).$$

For the closed interval used here, the exact count is

$$W(E,\Delta E)=\Gamma_N(E_+)-\Gamma_N(E_-^-), \qquad E_\pm=E\pm\Delta E/2.$$

Here $\Gamma_N(E)=\#\{E_a\leq E\}$ includes degeneracies, and $\Gamma_N(E_-^-)$ counts states strictly below the lower endpoint. Subtracting $\Gamma_N(E_-)$ would incorrectly exclude a level lying exactly at that endpoint.

If a coarse-grained density of states varies slowly across the window,

$$W(E,\Delta E) \simeq \omega_N(E)\Delta E.$$

For a finite exact spectrum, $\omega_N(E)$ is a sum of delta functions. Differentiating or taking logarithms therefore requires a shell, smoothing prescription, or thermodynamic limit. See density of states for the spectral methods behind this counting.

Microcanonical Entropy

The shell entropy is

$$S_{\mathrm{sh}}(E,\Delta E,N,V) = k_{\mathrm B}\ln W(E,\Delta E,N,V).$$

Because $W$ is dimensionless, the logarithm is well defined.

The von Neumann entropy of the microcanonical density operator gives the same result:

$$\begin{aligned} S_{\mathrm{vN}} \left( \rho_{\mathrm{mc}} \right) &= -k_{\mathrm B} \operatorname{Tr} \left( \rho_{\mathrm{mc}} \ln\rho_{\mathrm{mc}} \right) \\ &= k_{\mathrm B}\ln W. \end{aligned}$$

This equality relies on the uniform nonzero eigenvalues $1/W$.

Entropy Conventions

Several related quantities are called microcanonical entropy.

Shell-count entropy

The finite-window definition is

$$S_{\mathrm{sh}} = k_{\mathrm B}\ln W(E,\Delta E).$$

It makes the energy resolution explicit and is the primary convention on this page.

Density-of-states entropy

For a smoothed density of states, one may define

$$S_{\omega}(E) = k_{\mathrm B} \ln \left[ \omega(E)\varepsilon_0 \right],$$

where the reference energy $\varepsilon_0$ makes the logarithm dimensionless.

When $\omega(E)$ is nearly constant across the shell,

$$S_{\mathrm{sh}} \simeq S_\omega +k_{\mathrm B} \ln \left( \frac{\Delta E}{\varepsilon_0} \right).$$

Cumulative-count entropy

Another convention is

$$S_{\Gamma}(E) = k_{\mathrm B}\ln\Gamma(E).$$

For regular macroscopic systems, these definitions can share the same leading extensive entropy while differing by subextensive or finite-size terms. For small systems, bounded spectra, and questions about negative temperature, their derivatives can differ qualitatively.

Always state the convention before quoting a microcanonical temperature or heat capacity.

Thermodynamic Temperature

Temperature is derived in the microcanonical ensemble. Let

$$\sigma(E,V,N) \equiv \frac{S(E,V,N)}{k_{\mathrm B}}$$

be a smooth dimensionless thermodynamic entropy. The inverse temperature is

$$\beta_{\mathrm{mc}} = \left( \frac{\partial\sigma}{\partial E} \right)_{V,N} = \frac{1}{k_{\mathrm B}} \left( \frac{\partial S}{\partial E} \right)_{V,N}.$$

Equivalently,

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

This derivative presupposes a smooth entropy. For a finite staircase count, one must specify finite differences, interpolation, smoothing, or a limiting sequence.

Pressure and Chemical Potential

The microcanonical entropy representation has differential

$$dS = \frac{1}{T}dE +\frac{P}{T}dV -\frac{\mu}{T}dN.$$

Thus

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

and

$$-\frac{\mu}{T} = \left( \frac{\partial S}{\partial N} \right)_{E,V}.$$

Because $N$ is discrete in a finite quantum system, particle-addition derivatives are often implemented as finite differences. The thermodynamic chemical potential emerges after a suitable large-system limit.

Heat Capacity and Entropy Curvature

At fixed $V$ and $N$,

$$\frac{1}{T} = \frac{\partial S}{\partial E}.$$

Differentiating gives

$$-\frac{1}{T^2} \frac{\partial T}{\partial E} = \frac{\partial^2S}{\partial E^2}.$$

Therefore the microcanonical heat capacity is

$$C_{V,N} = \left( \frac{\partial E}{\partial T} \right)_{V,N} = -\frac{1}{ T^2 \left( \partial^2S/\partial E^2 \right)_{V,N} }.$$

In terms of the dimensionless entropy,

$$\frac{C_{V,N}}{k_{\mathrm B}} = -\frac{ \beta_{\mathrm{mc}}^2 }{ \left( \partial^2\sigma/\partial E^2 \right)_{V,N} }.$$

A concave entropy has

$$\frac{\partial^2S}{\partial E^2} < 0,$$

and therefore positive heat capacity. Convex regions can produce negative microcanonical heat capacity. Such behavior requires careful interpretation and may occur in finite systems with interface effects or in nonadditive long-range systems, where canonical and microcanonical descriptions can be nonequivalent.

Why Equal Temperatures Describe Equilibrium

Consider two weakly coupled systems $A$ and $B$ with fixed total energy

$$E_{\mathrm{tot}} = E_A+E_B.$$

The number of compatible states near a division $E_A$ is proportional to

$$W_A(E_A) W_B(E_{\mathrm{tot}}-E_A).$$

Its logarithm is

$$\frac{S_{\mathrm{tot}}(E_A)}{k_{\mathrm B}} = \frac{S_A(E_A)}{k_{\mathrm B}} +\frac{ S_B(E_{\mathrm{tot}}-E_A) }{ k_{\mathrm B} }.$$

At an interior maximum of the total entropy,

$$\frac{dS_{\mathrm{tot}}}{dE_A} = 0.$$

Therefore

$$\left( \frac{\partial S_A}{\partial E_A} \right) = \left( \frac{\partial S_B}{\partial E_B} \right),$$

which gives

$$T_A = T_B.$$

Equality of entropy slopes is a stationary-point condition. A stable interior equilibrium also requires $S_A^{\prime\prime}+S_B^{\prime\prime}<0$; otherwise that stationary point need not be the most probable division.

Connection to the Canonical Ensemble

The canonical partition function can be written as a transform of the density of states:

$$Z_N(\beta) = \int dE\, \omega_N(E) e^{-\beta E}.$$

Writing

$$\omega_N(E) \sim \exp \left[ \sigma_N(E) \right]$$

gives

$$Z_N(\beta) \sim \int dE\, \exp \left[ \sigma_N(E)-\beta E \right].$$

For a large regular system, a saddle point $E_*$ satisfies

$$\left. \frac{\partial\sigma_N}{\partial E} \right|_{E_*} = \beta.$$

This is exactly the matching condition

$$\beta_{\mathrm{mc}}(E_*) = \beta_{\mathrm c}.$$

The canonical ensemble concentrates near the microcanonical energy whose entropy slope matches the bath temperature. Concavity, additivity, finite-size corrections, and phase coexistence determine how accurate this saddle description is.

Isolation Preserves the Energy Distribution

For any state evolving under a time-independent Hamiltonian,

$$\rho(t) = e^{-iHt/\hbar} \rho(0) e^{iHt/\hbar},$$

the probability of each energy eigenspace is conserved:

$$\operatorname{Tr} \left[ P_a\rho(t) \right] = \operatorname{Tr} \left[ P_a\rho(0) \right].$$

If the initial state lies entirely inside the shell,

$$\rho(0) = \Pi_{E,\Delta E} \rho(0) \Pi_{E,\Delta E},$$

then it remains inside the shell:

$$\rho(t) = \Pi_{E,\Delta E} \rho(t) \Pi_{E,\Delta E}.$$

Isolation therefore justifies energy-shell support. It does not force equal weights inside that support.

Pure States Do Not Become the Mixed Shell State

If

$$|\psi(0)\rangle = \sum_a P_a|\psi(0)\rangle$$

is pure, unitary evolution preserves purity:

$$\operatorname{Tr}\rho(t)^2 = 1.$$

When $W>1$,

$$\operatorname{Tr}\rho_{\mathrm{mc}}^2 = \frac{1}{W} < 1.$$

Hence the exact global state cannot converge in density operator to $\rho_{\mathrm{mc}}$ under closed unitary dynamics.

What can happen is subtler:

• expectation values of selected observables can approach microcanonical values;
• time fluctuations can become small;
• small subsystems can have approximately thermal reduced states;
• coarse-grained descriptions can lose access to phase information.

These are statements about observables, reductions, or approximations, not literal global mixing.

Time Average and the Diagonal Ensemble

For a time-independent Hamiltonian, the infinite-time average is

$$\overline\rho = \lim_{\tau\to\infty} \frac{1}{\tau} \int_0^\tau dt\, \rho(t).$$

Using the spectral projectors of $H$,

$$\overline\rho = \sum_a P_a\rho(0)P_a.$$

This dephases coherences between distinct energies while preserving the initial weights in each eigenspace.

In general,

$$\overline\rho \neq \rho_{\mathrm{mc}}.$$

The diagonal ensemble becomes microcanonically predictive for an observable only when its preserved weights, the observable’s eigenstate matrix elements, and the shell width make the difference irrelevant.

Typical Pure States in a Shell

Let $|\psi\rangle$ be Haar-random in a shell of dimension $W$. Define

$$A_{\mathcal S} = \Pi_{E,\Delta E} A \Pi_{E,\Delta E}.$$

The average pure-state expectation is exactly microcanonical:

$$\mathbb E_\psi \langle\psi|A|\psi\rangle = \frac{ \operatorname{Tr}A_{\mathcal S} }{ W } = \langle A\rangle_{\mathrm{mc}}.$$

For Hermitian $A$, the Haar variance is

$$\operatorname{Var}_\psi \left( \langle\psi|A|\psi\rangle \right) ={} \frac{ \operatorname{Tr} \left( A_{\mathcal S}^2 \right) }{ W(W+1) } - \frac{ \left( \operatorname{Tr}A_{\mathcal S} \right)^2 }{ W^2(W+1) }.$$

In particular,

$$\operatorname{Var}_\psi \left( \langle A\rangle \right) \leq \frac{ \|A\|^2 }{ W+1 }.$$

When $W$ is large, most shell vectors give nearly the same expectation value for a fixed bounded observable.

This is typicality, not a statement that every physically prepared state is Haar-random or that dynamics explores the shell uniformly.

Canonical Typicality for a Subsystem

Let the isolated system split into a small subsystem $S$ and a much larger environment $B$. The reduced microcanonical state is

$$\rho_S^{\mathrm{mc}} = \operatorname{Tr}_B \rho_{\mathrm{mc}}.$$

For most pure vectors in a sufficiently large constrained shell, the reduced state

$$\rho_S(\psi) = \operatorname{Tr}_B |\psi\rangle\langle\psi|$$

is close to $\rho_S^{\mathrm{mc}}$ under suitable dimension conditions. If the environment density of states has the required smooth thermodynamic form and the coupling is weak, $\rho_S^{\mathrm{mc}}$ can in turn be approximately canonical.

This route explains how a globally pure, fixed-energy state can look thermal locally. It does not imply that the global pure state equals the mixed microcanonical state.

When eigenstates give thermal predictions

The eigenstate thermalization hypothesis relates nearby eigenstates’ local matrix elements to smooth thermodynamic values. Together with dephasing and a suitably narrow initial energy distribution, this can explain agreement with microcanonical averages. It is a condition on the Hamiltonian and observables, not a consequence of isolation alone. Integrability and other dynamical constraints may prevent that agreement; see relaxation and thermalization.

Conserved Charges and Symmetry Sectors

Energy may not be the only exact constraint. Suppose

$$[H,Q_j]=0.$$

If the preparation has definite values of the charges $Q_j$, the microcanonical projector should be restricted accordingly:

$$\Pi = \Pi_{E,\Delta E} \Pi_{q_1} \Pi_{q_2} \cdots$$

for mutually commuting projectors.

Examples include:

• particle number;
• total magnetization;
• crystal momentum;
• parity;
• total angular momentum;
• superselection charges.

Mixing dynamically disconnected sectors can inflate $W$ and give wrong averages. When there are extensively many relevant conserved quantities, an ordinary energy-only ensemble may be insufficient.

Example: Independent Two-Level Sites

Consider $L$ sites with

$$H = \varepsilon \sum_{i=1}^{L} n_i,$$

where

$$n_i \in \{0,1\}.$$

The energy with exactly $m$ excited sites is

$$E_m = m\varepsilon.$$

Its degeneracy is

$$W_m = \binom{L}{m}.$$

The exact-energy microcanonical state is

$$\rho_m = \frac{P_m}{ \binom{L}{m} },$$

where $P_m$ projects onto the $m$-excitation subspace.

Entropy

The shell entropy is

$$S_m = k_{\mathrm B} \ln \binom{L}{m}.$$

For

$$x = \frac{m}{L}$$

and large $L$, Stirling’s approximation gives

$$\frac{S_m}{k_{\mathrm B}} \simeq L \left[ -x\ln x -(1-x)\ln(1-x) \right].$$

Temperature

Treating $x$ as continuous,

$$\beta_{\mathrm{mc}} = \frac{\partial(S/k_{\mathrm B})}{\partial E} = \frac{1}{\varepsilon} \ln \left( \frac{1-x}{x} \right).$$

Thus:

$$\begin{array}{c|c} 0<x<1/2 & \beta_{\mathrm{mc}}>0 \\ x=1/2 & \beta_{\mathrm{mc}}=0 \\ 1/2<x<1 & \beta_{\mathrm{mc}}<0 \end{array}$$

The negative-$\beta$ branch is possible because the spectrum is bounded above by $L\varepsilon$. It corresponds to population inversion under the shell-entropy convention.

One-site reduced state

Permutation symmetry gives

$$\langle n_i\rangle_m = \frac{m}{L} = x.$$

The one-site state is

$$\rho_i = (1-x)|0\rangle\langle0| +x|1\rangle\langle1|.$$

For every $L$ and $0<m<L$, this one-site density operator agrees exactly with a canonical two-level state whose inverse temperature satisfies

$$x = \frac{1}{ 1+e^{\beta\varepsilon} }.$$

This fitted one-site temperature agrees with the leading thermodynamic entropy slope, but need not equal a chosen finite-difference temperature at finite $L$. Two-site correlations still distinguish the ensembles.

Constraint-induced correlations

For distinct sites,

$$\langle n_i n_j\rangle_m = \frac{ m(m-1) }{ L(L-1) }.$$

The connected correlation is

$$\begin{aligned} \langle n_i n_j\rangle_{m,c} &= \langle n_i n_j\rangle_m -\langle n_i\rangle_m \langle n_j\rangle_m \\ &= -\frac{ x(1-x) }{ L-1 }. \end{aligned}$$

Exact fixed total excitation number creates a weak anticorrelation. It vanishes for fixed $x$ as $L\to\infty$, illustrating how local canonical behavior can coexist with an exact global microcanonical constraint.

Negative temperatures

The finite spin example has fewer shell states as energy approaches its upper edge, so the shell-entropy derivative becomes negative. Negative $\beta$ gives larger weights to higher energies. In canonical language the necessary condition is $\operatorname{Tr}e^{-\beta H}<\infty$: a finite-dimensional spin Hamiltonian satisfies it for either sign of $\beta$, whereas an ordinary kinetic-energy spectrum unbounded above does not.

An upper bound alone is not sufficient in an infinite-dimensional space: if infinitely many levels remain in a bounded energy interval, their positive weights still give an infinite trace. A physical negative-temperature description also requires equilibration within the bounded degrees of freedom and sufficiently slow leakage to unbounded ones. The nuclear-spin experiment of Purcell and Pound (1951) is an early realization. Population inversion by itself does not demonstrate equilibration.

Keep the entropy convention explicit. The monotone cumulative count cannot have the negative slope of the shell count, so the two definitions cannot be substituted into a finite-system temperature formula interchangeably.

Exercises

Bound the shell energy variance

Let the spectral support of a state lie in $[E_-,E_+]$. Prove

$$\operatorname{Var}(H) \leq \frac{(E_+-E_-)^2}{4}.$$

Apply the result to a shell of width $\Delta E$.

Solution

On the supported subspace,

$$(H-E_-)(E_+-H) \geq 0.$$

Taking the expectation gives

$$\langle H^2\rangle \leq (E_-+E_+)\langle H\rangle -E_-E_+.$$

Let $m=\langle H\rangle$. Then

$$\begin{aligned} \operatorname{Var}(H) &= \langle H^2\rangle-m^2 \\ &\leq (E_+-m)(m-E_-). \end{aligned}$$

The product on the right is maximal at

$$m = \frac{E_-+E_+}{2},$$

where it equals

$$\frac{(E_+-E_-)^2}{4}.$$

For

$$E_+-E_-=\Delta E,$$

one obtains

$$\operatorname{Var}(H) \leq \frac{(\Delta E)^2}{4}.$$

Maximum entropy in the shell

Let $\sigma$ be any density operator supported in a $W$-dimensional shell. Use relative entropy to prove

$$S(\sigma) \leq k_{\mathrm B}\ln W.$$

When is equality attained?

Solution

The uniform shell state is

$$\rho_{\mathrm{mc}} = \frac{\Pi}{W}.$$

For $\sigma=\Pi\sigma\Pi$,

$$\ln\rho_{\mathrm{mc}} = -\ln W$$

on the support. Positivity of relative entropy gives

$$\begin{aligned} 0 &\leq D \left( \sigma \middle\| \rho_{\mathrm{mc}} \right) \\ &= \operatorname{Tr}(\sigma\ln\sigma) +\ln W \\ &= -\frac{S(\sigma)}{k_{\mathrm B}} +\ln W. \end{aligned}$$

Hence

$$S(\sigma) \leq k_{\mathrm B}\ln W.$$

Equality in quantum relative entropy holds exactly when

$$\sigma = \rho_{\mathrm{mc}}.$$

Fixed-excitation temperature

For $L$ two-level sites with $E_m=m\varepsilon$, show using a finite difference that

$$\beta_{m+1/2} \equiv \frac{ S_{m+1}-S_m }{ k_{\mathrm B}\varepsilon } = \frac{1}{\varepsilon} \ln \left( \frac{L-m}{m+1} \right).$$

Interpret its sign.

Solution

Since

$$S_m = k_{\mathrm B} \ln \binom{L}{m},$$

the entropy difference is

$$\begin{aligned} S_{m+1}-S_m &= k_{\mathrm B} \ln \left[ \frac{ \binom{L}{m+1} }{ \binom{L}{m} } \right] \\ &= k_{\mathrm B} \ln \left( \frac{L-m}{m+1} \right). \end{aligned}$$

Dividing by $\varepsilon k_{\mathrm B}$ gives the result.

The inverse temperature is positive when

$$L-m>m+1,$$

approximately $m<L/2$ for large $L$. It is negative above half filling because the degeneracy decreases toward the upper spectral edge. Near the midpoint it passes through zero, corresponding to infinite temperature in the large-$L$ interpolation.

Constraint-induced correlation

In the same $m$-excitation ensemble, derive

$$\langle n_i n_j\rangle_c = -\frac{x(1-x)}{L-1}, \qquad x=\frac{m}{L},$$

for $i\neq j$.

Solution

Every configuration with $m$ excited sites has equal weight. The probability that site $i$ is excited is

$$\langle n_i\rangle = \frac{m}{L} = x.$$

The probability that two specified distinct sites are both excited is

$$\langle n_i n_j\rangle = \frac{ \binom{L-2}{m-2} }{ \binom{L}{m} } = \frac{ m(m-1) }{ L(L-1) }.$$

Therefore

$$\begin{aligned} \langle n_i n_j\rangle_c &= \frac{ m(m-1) }{ L(L-1) } -\frac{m^2}{L^2} \\ &= -\frac{ m(L-m) }{ L^2(L-1) } \\ &= -\frac{x(1-x)}{L-1}. \end{aligned}$$

The anticorrelation enforces the exact global excitation number and vanishes locally as $L\to\infty$ at fixed $x$.

Diagonal ensemble versus microcanonical ensemble

Let a nondegenerate three-level Hamiltonian have eigenstates $|1\rangle$, $|2\rangle$, and $|3\rangle$, all inside one chosen shell. The initial state is

$$|\psi(0)\rangle = \sqrt{\frac{1}{2}}|1\rangle +\sqrt{\frac{1}{3}}|2\rangle +\sqrt{\frac{1}{6}}|3\rangle.$$

Compute the time-averaged state and compare it with the microcanonical state.

Solution

Time averaging removes coherences between distinct energies:

$$\overline\rho = \frac{1}{2}|1\rangle\langle1| +\frac{1}{3}|2\rangle\langle2| +\frac{1}{6}|3\rangle\langle3|.$$

The shell dimension is $W=3$, so the microcanonical state is

$$\rho_{\mathrm{mc}} = \frac{1}{3} \sum_{n=1}^{3} |n\rangle\langle n|.$$

Thus

$$\overline\rho \neq \rho_{\mathrm{mc}}.$$

Isolation and dephasing preserve the unequal initial populations. The two states can still give similar values for a restricted observable if its diagonal matrix elements are nearly constant across the shell, which is the kind of additional condition supplied by ETH or narrow-shell smoothness.

References

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