Lecture 9: Equilibrium Transition Weights

Many-body operators and correlations become predictive only after the preparation of the system is represented. When that preparation is incomplete or thermal, probabilities over quantum alternatives become part of the physical model. This lecture derives the required statistical mechanics locally from normalized weights and directed transitions. The same preparation logic later supports causal response, free-energy stability, superconducting self-consistency, reservoirs, and quantum-critical crossovers. Preparation uncertainty enters the mathematical state and changes observable averages and transition intensities.

A spin-dimer magnet provides a concrete two-level problem. In a TlCuCl3-type material, two coupled spins can form a singlet-like ground state separated from magnetic triplet excitations by an energy gap. At one selected momentum, field, and polarization, neutron scattering or ESR can record a transition line. Its position reflects an energy difference, but its area also depends on how often the initial level was occupied.

The Hamiltonian and transition operator do not determine that occupation. Consider two devices feeding the same two-level Hamiltonian: one emits an incoherent mixture of energy eigenstates, while the other emits a coherent superposition with the same energy populations. Can a measurement distinguish these preparations? The energy spectrum alone does not specify the state presented to the measurement.

We will represent incomplete preparation by a density operator and use the two-level anchor to derive normalized equilibrium weights and directed detailed-balance ratios from probability comparisons.

This reduction is intentionally bounded. A real dimer magnet has a dispersing triplet, degeneracies that a magnetic field can split, interactions between excitations, and finite linewidths. The two-level model is trusted for the logic of preparation weights and directed transition ratios, not for the full material spectrum.

The one-mode probability ratio first worked in Lecture 5 will be reused, but the calculation changes here: probabilities become entries of a density operator and later become weights of directed transitions.

The governing question is

How do we represent preparation by a density operator, and, after specifying thermal equilibrium, how do its probabilities weight the two directions of a measured transition?

Before calculating, predict whether the two preparation devices just described must give the same ${\hat\sigma_x}$ statistics merely because they share the same energy populations. Keep that prediction for the mixture-versus-superposition calculation. Later we will examine a different readout, the ordered two-time correlation, and determine which parts of the state it uses.

A correlation function answers the present question. A retarded response function answers the Lecture 10 question. They use the same energy levels, matrix elements, and equilibrium probabilities, but arrange them differently because an observation and an intervention are different protocols.

This lecture assumes quantum mechanics, including traces and Heisenberg-picture evolution, but does not assume a statistical-mechanics course. Temperature, equilibrium weights, the partition sum, and chemical potential are introduced only when a physical need for each appears. Lecture 10 will change the protocol by applying a weak source and asking how a later measurement changes.

The distinction is experimentally consequential. An equilibrium spectrum can contain both absorption and emission lines even when no external source is applied; a susceptibility instead answers how a specified source changes a specified readout. Temperature alters the two directed spectral weights, while selection rules can remove both. Later response calculations fail if these three ingredients—energy differences, matrix elements, and preparation weights—are collapsed into the single word “peak.”

For a two-level Hamiltonian, write its two energy eigenstates, evolve ${\hat\sigma_x}$ in the Heisenberg picture, and evaluate a trace in a chosen basis. Then explain in one sentence why the equal incoherent mixture of the two energy eigenstates is not the coherent superposition ${(|\mathrm g\rangle+|\mathrm e\rangle)/\sqrt2}$ . These are quantum-mechanical prerequisites. The equilibrium probabilities themselves are derived below; no prior statistical-mechanics ensemble is assumed.

Preparations as density operators

The density operator is the state assigned to a preparation, whereas one convex ensemble is only a representation of that operator. Equilibrium weights are probabilities and not quantum amplitudes. The finite-system identities below are exact once the equilibrium constraints are accepted. The reservoir-counting argument below motivates the canonical weights under stated equilibrium and large-reservoir assumptions; it does not explain microscopic relaxation to equilibrium. Extra conserved quantities or persistent driving can change the admissible state and therefore the transition ratios.

Mixtures as positive normalized operators

Historical lineage

[Lan27, pp. 430–441] — Early wave-mechanical discussion of statistical mixtures.
[Neu27a, pp. 245–272] — Statistical-operator formulation of quantum probabilities and ensembles.
[Neu27b, pp. 273–291] — Quantum-statistical ensembles, entropy, and canonical weighting.

Let ${\mathcal H_2=\operatorname{span}\{|\mathrm g\rangle,|\mathrm e\rangle\} \cong\mathbb C^2}$ , and declare the ordered orthonormal energy basis ${\mathcal B_{\mathrm E}=(|\mathrm g\rangle,|\mathrm e\rangle)}$ . Define the Pauli operators ${\hat\sigma_a:\mathcal H_2\to\mathcal H_2}$ by their coordinate matrices

${[\hat\sigma_x]}_{\mathcal B_{\mathrm E}}= \begin{pmatrix}0&1\\1&0\end{pmatrix},\qquad {[\hat\sigma_y]}_{\mathcal B_{\mathrm E}}= \begin{pmatrix}0&-\mathrm i\\ \mathrm i&0\end{pmatrix},\qquad {[\hat\sigma_z]}_{\mathcal B_{\mathrm E}}= \begin{pmatrix}1&0\\0&-1\end{pmatrix},$

and set ${\hat H_0:\mathcal H_2\to\mathcal H_2}$ to

$\hat H_0=-\frac{\hbar\Omega}{2}\hat\sigma_z, \qquad \Omega>0.$

The ground and excited states satisfy

$\hat H_0|\mathrm g\rangle =-\frac{\hbar\Omega}{2}|\mathrm g\rangle, \qquad \hat H_0|\mathrm e\rangle =+\frac{\hbar\Omega}{2}|\mathrm e\rangle.$

The upright subscripts are word abbreviations, not variables. The standard Pauli matrices, the ordered basis, and the minus sign in ${\hat H_0}$ are one convention: changing any one of them requires changing the others explicitly.

A density operator is the state object that assigns expectations. Its decomposition into ensembles of pure states is generally nonunique, although all such decompositions of the same operator give the same expectation functional.

Suppose a preparation device produces ${|\mathrm g\rangle}$ with probability ${p_{\mathrm g}}$ and ${|\mathrm e\rangle}$ with probability ${p_{\mathrm e}}$ , where ${p_{\mathrm g}+p_{\mathrm e}=1}$ . Repeating the experiment gives

$\langle \hat A\rangle =p_{\mathrm g}\langle\mathrm g|\hat A|\mathrm g\rangle +p_{\mathrm e}\langle\mathrm e|\hat A|\mathrm e\rangle.$

Let ${\mathcal H}$ be a complex Hilbert space. A density operator on ${\mathcal H}$ is a positive trace-class operator ${\hat\rho:\mathcal H\to\mathcal H}$ satisfying

$\hat\rho\geq0, \qquad \operatorname{Tr}\hat\rho=1.$

For every bounded observable ${\hat A}$ , its expectation is ${{\langle\hat A\rangle}_{\hat\rho} =\operatorname{Tr}(\hat\rho\hat A)}$ . An ensemble ${{\{(p_j,|\psi_j\rangle)\}}_j}$ of normalized vectors ${|\psi_j\rangle}$ , with ${p_j\geq0}$ and ${\sum_jp_j=1}$ , realizes

$\hat\rho=\sum_jp_j|\psi_j\rangle\langle\psi_j|,$

but this decomposition is generally not unique. The density operator is the state; a particular list of preparation stories is not extra system-only information. A rank-one density operator represents a pure state. A higher-rank density operator represents a mixed state; it cannot be written as one pure-state projector.

For the two-level preparation, the definition gives

$\hat\rho =p_{\mathrm g}|\mathrm g\rangle\langle\mathrm g| +p_{\mathrm e}|\mathrm e\rangle\langle\mathrm e|, \qquad \langle\hat A\rangle=\operatorname{Tr}(\hat\rho\hat A).$

Positivity and normalization are not decorative axioms. Hermiticity follows from positivity and makes expectations of Hermitian observables real; positivity prevents negative measurement probabilities; normalization makes total probability one.

Compare the equal mixture

$\hat\rho_{\mathrm{mix}} =\frac12|\mathrm g\rangle\langle\mathrm g| +\frac12|\mathrm e\rangle\langle\mathrm e| =\frac12\hat I$

with the coherent state ${|+\rangle=(|\mathrm g\rangle+|\mathrm e\rangle)/\sqrt2}$ . The latter has

${[\hat\rho_+]}_{\mathcal B_{\mathrm E}} =\frac12 \begin{pmatrix}1&1\\1&1\end{pmatrix},$

where ${\hat\rho_+=|+\rangle\langle+|}$ ; the mixture’s matrix in the same basis has zero off-diagonal entries. Accordingly, ${\operatorname{Tr}(\hat\rho_{\mathrm{mix}}\hat\sigma_x)=0}$ but ${\langle+|\hat\sigma_x|+\rangle=1}$ . A fixed but unknown relative phase still specifies a pure state. Randomizing that phase independently and uniformly between preparations gives the equal mixture after averaging. The density operator records the statistics of the specified preparation procedure.

Before reading Figure 9.1, cover the off-diagonal entries and decide whether an energy-basis measurement can distinguish the two preparations. Then uncover them and choose a noncommuting observable.

Basis-independent expectations and pure-state limits

Figure 9.1. An equal incoherent mixture and a coherent superposition share their energy populations but differ in off-diagonal coherence and therefore in a sigma-x measurement.

Energy populations alone cannot distinguish the preparations, but the ${\hat\sigma_x}$ readout can. The figure is therefore a measurement-decision diagram that separates information retained by the diagonal entries from information carried by coherence.

For any two-level state it is often useful to write

$\hat\rho=\frac12\left( \hat I+\mathbf r\mathbin{\cdot}\hat{\boldsymbol\sigma}\right), \qquad |\mathbf r|\leq1.$

Here ${\hat{\boldsymbol\sigma}:=(\hat\sigma_x,\hat\sigma_y,\hat\sigma_z)}$ is an operator-valued tuple on ${\mathcal H_2}$ . The components of the ordinary vector ${\mathbf r\in\mathbb R^3}$ are the three Pauli expectations. A pure state has ${|\mathbf r|=1}$ , whereas incomplete mixtures lie inside the Bloch sphere. The purity

$\operatorname{Tr}\hat\rho^2 =\frac{1+{|\mathbf r|}^2}{2}$

equals one only on the surface and equals ${1/2}$ for the equal mixture. Purity measures state mixedness and gives an immediate check that a proposed density matrix represents a valid mixed or pure preparation.

More generally, if an apparatus measures a projector ${\hat P_a}$ , the probability of outcome ${a}$ is

$\Pr(a)=\operatorname{Tr}(\hat\rho\hat P_a).$

Two laboratory ensembles that produce the same ${\hat\rho}$ give the same probabilities for every system-only measurement. They may have different preparation stories, but no calculation on the isolated system can distinguish those stories. This operational fact is why the density operator, not an ensemble narrative, is the state used below.

Equilibrium weights from minimal statistical mechanics

Put the two-level system in weak contact with a much larger environment and assume they reach thermal equilibrium, then isolate it before measuring. Stationarity alone would not establish this equilibrium assumption. The temperature ${T}$ describes the environment’s equilibrium energy scale. Define

$\beta=\frac{1}{k_{\mathrm B}T},$

so ${\beta E}$ is dimensionless. Equilibrium assigns relative weights

$\frac{p_m}{p_n} =\mathrm e^{-\beta(E_m-E_n)}.$

This relation says that a preparation beginning in a higher-energy state is exponentially less common. It is an equilibrium rule, not a statement about a transition rate and not a replacement for unitary time evolution.

If the upper state is less probable, has equilibrium changed the Hamiltonian or slowed the upward transition matrix element? No. The energies and matrix elements are unchanged; only the probability of starting in each state has been assigned. A response that calls ${\mathrm e^{-\beta(E_m-E_n)}}$ a transition rate must return to the preparation protocol before proceeding.

To motivate the ratio, assume weak coupling and equal probabilities for joint microstates in a narrow window of total energy ${E_{\mathrm{tot}}}$ . Let ${W_{\mathrm R}(E_{\mathrm R})}$ count reservoir states in the corresponding energy window. For a fixed system eigenstate ${|n\rangle}$ , energy conservation gives ${E_{\mathrm R}=E_{\mathrm{tot}}-E_n}$ , hence ${p_n\propto W_{\mathrm R}(E_{\mathrm{tot}}-E_n)}$ . Define reservoir entropy ${S_{\mathrm R}=k_{\mathrm B}\ln W_{\mathrm R}}$ , with temperature defined by ${\partial S_{\mathrm R}/\partial E_{\mathrm R}=1/T}$ . For an energy transfer ${\Delta E=E_m-E_n}$ , first-order expansion gives ${\ln(p_m/p_n)=\ln W_{\mathrm R}(E_{\mathrm R}-\Delta E)-\ln W_{\mathrm R}(E_{\mathrm R})\simeq-\beta\Delta E}$ . The large-reservoir approximation keeps its temperature nearly constant over the exchanged-energy range, neglecting curvature of the entropy. Exponentiation recovers the Boltzmann ratio; this counting argument does not describe relaxation.

Relative weights must be normalized. The normalization and its existence boundary belong to the definition, not to an afterthought.

For a discrete energy basis, fix any reference level ${r}$ . The ratio rule implies

$p_n\mathrm e^{+\beta E_n} =p_r\mathrm e^{+\beta E_r}=:C, \qquad p_n=C\mathrm e^{-\beta E_n}.$

Normalization determines the previously unknown constant rather than adding a new physical assumption:

$1=\sum_n p_n=C\sum_n\mathrm e^{-\beta E_n}, \qquad Z(\beta):=\sum_n\mathrm e^{-\beta E_n}, \qquad C=Z^{-1}.$

The spectral-calculus expression ${\mathrm e^{-\beta\hat H}=\sum_n \mathrm e^{-\beta E_n}|n\rangle\langle n|}$ packages these normalized probabilities into an operator and extends the construction beyond a chosen coordinate matrix.

Let ${\hat H}$ be a self-adjoint Hamiltonian on a complex Hilbert space ${\mathcal H}$ selected by the preparation’s fixed constraints, such as a fixed particle number. Energy may be exchanged with the reservoir during preparation; no fixed-energy constraint is imposed. For ${\beta={(k_{\mathrm B}T)}^{-1}>0}$ , suppose ${\mathrm e^{-\beta\hat H}}$ is trace class, so that

$0<Z(\beta):=\operatorname{Tr}\mathrm e^{-\beta\hat H}<\infty.$

The canonical equilibrium state determined by ${(\hat H,\beta)}$ is the density operator

$\hat\rho_\beta:=\frac{\mathrm e^{-\beta\hat H}}{Z(\beta)}.$

If ${\hat H|n\rangle=E_n|n\rangle}$ has a discrete eigenbasis, then ${Z=\sum_n\mathrm e^{-\beta E_n}}$ and ${p_n=\mathrm e^{-\beta E_n}/Z}$ . The state is stationary because ${[\hat\rho_\beta,\hat H]=0}$ . Replacing ${\hat H}$ by ${\hat H+C\hat I}$ multiplies numerator and denominator by the same scalar ${\mathrm e^{-\beta C}}$ , so the density operator is independent of the chosen energy zero. If the trace is infinite, this formula does not define a canonical density operator on ${\mathcal H}$ .

The defining properties can now be checked rather than inferred from the name. In the discrete case,

$\hat\rho_\beta =\sum_n\frac{\mathrm e^{-\beta E_n}}{Z}|n\rangle\langle n| \geq0, \qquad \operatorname{Tr}\hat\rho_\beta =\frac1Z\sum_n\mathrm e^{-\beta E_n}=1,$

and every spectral projector of ${\hat H}$ commutes with ${\hat H}$ , so ${[\hat\rho_\beta,\hat H]=0}$ . Positivity, unit trace, and stationarity are therefore consequences of the normalized spectral construction. The converse does not hold: a stationary state need not have Gibbs probabilities. The trace-class hypothesis is what makes the same statements meaningful when the Hilbert space is infinite dimensional.

Take three orthonormal states with energies ${(0,\Delta,\Delta)}$ , ${\Delta>0}$ , at fixed particle number and ${T>0}$ . Find the normalized equilibrium operator and extract its mean energy from the normalizer. Use ${\beta=1/(k_{\mathrm B}T)}$ ; check the second derivative of ${\ln Z}$ , not ${Z}$ , against the energy variance.

Put ${q=\mathrm e^{-\beta\Delta}}$ . Summing over states, including both members of the upper doublet, gives $Z=1+2q,\qquad {[\hat\rho_\beta]}_{\mathcal B} =\frac{1}{1+2q}\operatorname{diag}(1,q,q), \quad \mathcal B=(|0\rangle,|1\rangle,|2\rangle).$ The doublet’s total probability is ${2q/(1+2q)}$ , not the probability of one upper state. Direct expectation gives ${\langle\hat H\rangle=2\Delta q/(1+2q)}$ . Because this sum is finite and the Hamiltonian is independent of ${\beta}$ , it can be differentiated term by term: $-\partial_\beta\ln Z=\frac{2\Delta q}{1+2q}, \qquad \partial_\beta^2\ln Z=\frac{2\Delta^2q}{{(1+2q)}^2} =\langle\hat H^2\rangle-{\langle\hat H\rangle}^2.$ The variance identity is checked directly with ${\langle\hat H^2\rangle=2\Delta^2q/(1+2q)}$ . The density matrix is positive, has trace one and commutes with its diagonal Hamiltonian. As ${T\to0}$ it selects the lower state; as ${T\to\infty}$ each of the three states has probability ${1/3}$ , so the upper level has probability ${2/3}$ . An energy-zero shift multiplies all weights and ${Z}$ equally. These checks distinguish counting states from counting distinct energies. For an infinite spectrum one must separately justify trace convergence and differentiation. The next calculation returns to the nondegenerate two-level anchor, where populations also control a transverse measurement.

The ordinary scalar ${Z}$ is a normalizer called the canonical partition sum. No thermodynamic interpretation beyond normalized preparation weights is required here. Apply the definition with ${\hat H=\hat H_0}$ .

For ${\hat H_0=-(\hbar\Omega/2)\hat\sigma_z}$ , ${\Omega>0}$ , calculate the probabilities, polarization ${\langle\hat\sigma_z\rangle}$ , mean energy and energy variance at ${T>0}$ . The preparation exchanges energy but keeps the particle constraint fixed, with ${\beta=1/(k_{\mathrm B}T)}$ .

Use the canonical state for this preparation. Summing its two spectral weights gives

$Z=2\cosh\!\left(\frac{\beta\hbar\Omega}{2}\right),$

$p_{\mathrm g} =\frac{\mathrm e^{+\beta\hbar\Omega/2}} {2\cosh(\beta\hbar\Omega/2)}, \qquad p_{\mathrm e} =\frac{\mathrm e^{-\beta\hbar\Omega/2}} {2\cosh(\beta\hbar\Omega/2)}.$

Three checks give these symbols physical meaning. First, ${p_{\mathrm g}+p_{\mathrm e}=1}$ . Second, as ${T\to0}$ and ${\beta\to\infty}$ , the ground-state probability tends to one. Third, as ${T\to\infty}$ and ${\beta\to0}$ , the two probabilities tend to ${1/2}$ . Their difference is

${\langle\hat\sigma_z\rangle}_\beta =p_{\mathrm g}-p_{\mathrm e} =\tanh\!\left(\frac{\beta\hbar\Omega}{2}\right).$

This population difference is the dimensionless polarization of the anchor; a physical magnetic moment is a model-dependent constant times this expectation. It will later control the net response. It also gives two independent checks on the preparation. Direct evaluation yields

$\begin{aligned} {\langle\hat H_0\rangle}_\beta &=-\frac{\hbar\Omega}{2} \tanh\!\left(\frac{\beta\hbar\Omega}{2}\right),\\ {\langle\hat H_0^2\rangle}_\beta &={\left(\frac{\hbar\Omega}{2}\right)}^2,\\ \operatorname{Var}_\beta(\hat H_0) &={\left(\frac{\hbar\Omega}{2}\right)}^2 \operatorname{sech}^2\!\left(\frac{\beta\hbar\Omega}{2}\right). \end{aligned}$

At zero temperature both the variance and the excited-state probability vanish. At high temperature the mean energy tends to zero while the variance tends to ${{(\hbar\Omega/2)}^2}$ , as expected for equal sampling of the two energies. These checks would expose either a reversed Boltzmann exponent or a missing normalization factor before the response calculation uses the population difference.

The mean has energy units and the variance has squared-energy units. The partition sum gives an independent check. Keep ${\hat H_0}$ fixed and write ${Z=\sum_{n=\mathrm g,\mathrm e}\mathrm e^{-\beta E_n}}$ . Primes denote derivatives with respect to ${\beta}$ ; the finite sum can be differentiated term by term: $\begin{aligned} Z'&=-\sum_n E_n\mathrm e^{-\beta E_n},\qquad Z''=\sum_n E_n^2\mathrm e^{-\beta E_n},\\ -\partial_\beta\ln Z&=-\frac{Z'}{Z}=\langle\hat H_0\rangle,\\ \partial_\beta^2\ln Z&=\frac{Z''}{Z}-\left(\frac{Z'}{Z}\right)^2 =\langle\hat H_0^2\rangle-{\langle\hat H_0\rangle}^2 =\operatorname{Var}(\hat H_0). \end{aligned}$ For the two-level partition sum, these derivatives are $\begin{aligned} -\partial_\beta\ln\!\left[2\cosh\!\left(\frac{\beta\hbar\Omega}{2}\right)\right] &=-\frac{\hbar\Omega}{2}\tanh\!\left(\frac{\beta\hbar\Omega}{2}\right), \\ \partial_\beta^2\ln Z &=\left(\frac{\hbar\Omega}{2}\right)^2 \operatorname{sech}^2\!\left(\frac{\beta\hbar\Omega}{2}\right). \end{aligned}$ They reproduce the directly calculated mean and variance. The same method also applies to the degenerate preparation in Example 9.1. The Gibbs state here is diagonal in energy. Energy probabilities alone do not establish that a preparation has no coherence. Nor does every transverse statistic detect it: the particular ordered product calculated next is diagonal and depends only on populations.

Directed transitions in equilibrium correlations

Lehmann weights from operator order

Lecture 8 introduced correlations as expectations of ordered products at two times. Here the preparation is the canonical density operator: average the product ${\hat B(t)\hat B(0)}$ with that same state. Time dependence remains in the Heisenberg operators. Fourier transformation resolves the oscillations into transition frequencies. We will derive how each spectral weight combines an initial-state probability with a readout matrix element.

The mathematical object used here is an ordered correlation spectrum, not yet a response function. The eponym records Lehmann’s 1954 vacuum-field spectral representation, built by inserting complete momentum eigenstates into propagator matrix elements [Leh54, printed pp. 342–347, Eqs. (1)–(10)]. The finite-temperature, finite-system weights below are derived directly from the ordered thermal trace; they are a later many-body specialization and are not attributed to Lehmann’s paper in that broader form.

Let ${\hat H}$ be a self-adjoint Hamiltonian with a discrete orthonormal eigenbasis ${{\{|n\rangle\}}_n}$ , let ${\hat\rho_\beta}$ be its canonical equilibrium state, and let ${\hat B}$ be a bounded Hermitian readout. Define

$\hat B(t):=\mathrm e^{+\mathrm i\hat Ht/\hbar} \hat B\mathrm e^{-\mathrm i\hat Ht/\hbar}, \qquad C_{BB}(t):=\operatorname{Tr}\!\left[ \hat\rho_\beta\hat B(t)\hat B(0)\right].$

With the Fourier sign fixed as

$S_{BB}(\omega):= \int_{-\infty}^{\infty}\mathrm dt\, \mathrm e^{+\mathrm i\omega t}C_{BB}(t),$

the equilibrium transition spectrum is the positive distribution

$S_{BB}(\omega)=2\pi\sum_{n,m}p_n{|B_{nm}|}^2 \delta\!\left(\omega-\frac{E_m-E_n}{\hbar}\right), \quad B_{nm}:=\langle n|\hat B|m\rangle.$

For a finite-dimensional system this is a finite positive Borel measure on ${\mathbb R}$ . In an infinite discrete system the displayed series is admitted only when it defines a tempered positive measure. The operator order, the ${+\mathrm i\omega t}$ Fourier sign, and the initial-state weight ${p_n}$ are part of the definition. This unsymmetrized equilibrium spectrum is not a retarded response function [Kub57; MS59].

The spectral representation in the definition follows from the ordered trace; it is not an independent postulate. Insert the energy basis once between the two readout operators:

$\begin{aligned} C_{BB}(t) &=\sum_n p_n\langle n|\hat B(t)\hat B|n\rangle\\ &=\sum_{n,m}p_n \mathrm e^{-\mathrm i(E_m-E_n)t/\hbar} \langle n|\hat B|m\rangle \langle m|\hat B|n\rangle\\ &=\sum_{n,m}p_n{|B_{nm}|}^2 \mathrm e^{-\mathrm i(E_m-E_n)t/\hbar}. \end{aligned}$

Hermiticity supplies ${B_{mn}=B_{nm}^*}$ , while the operator order leaves ${p_n}$ attached to the initial ket ${|n\rangle}$ . With the declared Fourier sign,

$\int_{-\infty}^{\infty}\mathrm dt\, \mathrm e^{+\mathrm i\omega t} \mathrm e^{-\mathrm i(E_m-E_n)t/\hbar} =2\pi\delta\!\left( \omega-\frac{E_m-E_n}{\hbar} \right),$

which gives the displayed positive spectral measure term by term.

Use the Hamiltonian and equilibrium preparation of Example 9.2. Calculate the ordered correlation ${C_{xx}(t)=\operatorname{Tr}[\hat\rho_\beta\hat\sigma_x(t)\hat\sigma_x]}$ . For the anchor, take ${\hat B=\hat\sigma_x =|\mathrm g\rangle\langle\mathrm e| +|\mathrm e\rangle\langle\mathrm g|}$ .

Keep the operator order fixed and evolve each dyad before taking a trace; this exposes the sign of each transition frequency. Each dyad acquires the energy difference between its ket and bra. For example,

$\mathrm e^{+\mathrm i\hat H_0t/\hbar} |\mathrm g\rangle\langle\mathrm e| \mathrm e^{-\mathrm i\hat H_0t/\hbar} =\mathrm e^{+\mathrm i(E_{\mathrm g}-E_{\mathrm e})t/\hbar} |\mathrm g\rangle\langle\mathrm e| =\mathrm e^{-\mathrm i\Omega t} |\mathrm g\rangle\langle\mathrm e|.$

The reversed dyad carries the conjugate phase, so

$\hat\sigma_x(t) =\mathrm e^{-\mathrm i\Omega t} |\mathrm g\rangle\langle\mathrm e| +\mathrm e^{+\mathrm i\Omega t} |\mathrm e\rangle\langle\mathrm g|,$

and multiplication by the unevolved readout makes the initial-state projectors explicit:

$\hat\sigma_x(t)\hat\sigma_x =\mathrm e^{-\mathrm i\Omega t} |\mathrm g\rangle\langle\mathrm g| +\mathrm e^{+\mathrm i\Omega t} |\mathrm e\rangle\langle\mathrm e|.$

Since ${\hat\rho_\beta=p_{\mathrm g}|\mathrm g\rangle\langle\mathrm g| +p_{\mathrm e}|\mathrm e\rangle\langle\mathrm e|}$ , tracing each projector gives

$C_{xx}(t) =p_{\mathrm g}\mathrm e^{-\mathrm i\Omega t} +p_{\mathrm e}\mathrm e^{+\mathrm i\Omega t}.$

At ${t=0}$ , the result is one, agreeing with ${\hat\sigma_x^2=\hat I}$ and ${\operatorname{Tr}\hat\rho=1}$ . Also ${C_{xx}(-t)={C_{xx}(t)}^*}$ ; at zero temperature only the phase ${\mathrm e^{-\mathrm i\Omega t}}$ remains. The correlation is dimensionless but its Fourier transform has time units. These checks fix the relative signs before the following line-area calculation assigns physical directions. Symmetrizing at this stage would erase the directional population information. Because the displayed product is diagonal, any density operator with these same populations gives this same ${C_{xx}(t)}$ , even if it has off-diagonal coherence. Other observables, such as ${\langle\hat\sigma_x\rangle}$ , can distinguish such states. This correlation alone therefore certifies neither absence of coherence nor equilibrium.

Detailed balance

Fourier transform Example 9.3 with ${\mathrm e^{+\mathrm i\omega t}}$ , isolate each line and determine the reverse-to-forward integrated weight. Keep ${\Omega>0}$ , so the two windows can be separated.

Keep the ordered product so each weight retains its initial population. The ${+\mathrm i\omega t}$ Fourier sign maps ${\mathrm e^{-\mathrm i\Omega t}}$ to the positive-frequency line. For an ideal isolated system, ${S_{xx}}$ is a distribution, not an ordinary function:

(9.1)

$S_{xx}(\omega) =2\pi\!\left[ p_{\mathrm g}\delta(\omega-\Omega) +p_{\mathrm e}\delta(\omega+\Omega) \right].$

The Dirac delta distribution is defined by its action on a smooth test function ${f}$ :

$\int_{-\infty}^{\infty}\mathrm d\omega\, \delta(\omega-\Omega)f(\omega)=f(\Omega).$

It is therefore misleading to speak of the “height” of an ideal delta line. Its coefficient is an integrated spectral weight. A finite experimental or computed linewidth requires a separately declared broadening mechanism.

The positive-frequency line begins in ${|\mathrm g\rangle}$ and transfers energy ${+\hbar\Omega}$ to the system. The negative-frequency line begins in ${|\mathrm e\rangle}$ and describes the reverse energy transfer. Their matrix elements have equal magnitude. The coefficients in Equation (9.1) arise entirely from unequal initial-state probabilities. To compare the two lines without evaluating a distribution at a point, choose nonnegative test functions ${\chi_+,\chi_-\in C_{\mathrm c}^{\infty}(\mathbb R)}$ : smooth functions that vanish outside bounded frequency windows. Choose disjoint windows that isolate ${+\Omega}$ and ${-\Omega}$ , respectively, and functions that equal one at the isolated line they select. Define the line weights by distribution pairing:

$W_+:=\frac{1}{2\pi}\langle S_{xx},\chi_+\rangle=p_{\mathrm g}, \qquad W_-:=\frac{1}{2\pi}\langle S_{xx},\chi_-\rangle=p_{\mathrm e}, \qquad \frac{W_-}{W_+} =\mathrm e^{-\beta\hbar\Omega}.$

Here ${S_{xx}}$ is a distribution and each ${\chi_\pm}$ is an ordinary test function. The pairing ${\langle S_{xx},\chi\rangle}$ is the weighted integral ${\int S_{xx}(\omega)\chi(\omega)\,\mathrm d\omega}$ , interpreted through the delta distribution’s action, not an inner product of quantum states. In a resolved experiment the analogous quantities are integrated areas over two nonoverlapping frequency windows.

The sum ${W_++W_-=1}$ recovers the equal-time correlation, independently checking the ${2\pi}$ normalization. As ${T\to0}$ , the reverse area vanishes; as ${T\to\infty}$ , the two areas both approach one half. Equal transition-matrix magnitudes therefore coexist with unequal equilibrium line areas. The figure uses precisely this direction ledger; its arrow styling does not supply an additional intensity formula.

Before viewing Figure 9.2, assign the initial state to each frequency sign and recover which arrow loses weight as ${T\to0}$ . The solid/dashed distinction should remain sufficient if color is removed.

Figure 9.2. Equilibrium transitions and their detailed-balance ratio. Solid and dashed arrows distinguish the two directions without relying on color.

The figure labels the lower and upper states ${n}$ and ${m}$ ; for our two-level model these are ${\mathrm g}$ and ${\mathrm e}$ . The positive-frequency arrow starts from the lower state, and the reversed arrow requires the upper state to be occupied. In the two-level model, cooling removes the reversed weight. Their area ratio, rather than their drawn arrow length, encodes detailed balance.

Return to the dimer magnet. The positive-frequency neutron or ESR process can promote the singlet-like ground state into the selected magnetic excitation; the reversed process requires that excitation to be present initially. As the sample is cooled far below the gap, the reversed integrated weight is therefore exponentially suppressed. The ratio of the two directed areas can reveal the equilibrium factor even when an overall detector efficiency is unknown, provided the reversed matrix elements are related as assumed. An absolute area cannot be inferred from the Boltzmann factor alone because magnetic form factors, polarization selection, and instrumental acceptance remain [Hov54; ZS12].

A static susceptibility gives a complementary limit. Thermally populated magnetic levels can carry a field response, so a gapped singlet system has a strongly suppressed low-temperature susceptibility in the idealized regime. This connects the same equilibrium weights to a different observable. A complete material susceptibility additionally requires its dispersion, matrix elements, and interactions.

Cover the caption, write ${p_n}$ or ${p_m}$ beside the initial point of each arrow, and then write the integrated-area ratio ${W_-/W_+}$ without consulting the preceding display. If the ratio is inverted, diagnose whether the error entered in the initial-state label or in the Fourier-sign convention.

Seeing unequal sidebands without a circular test

What does the direction ledger look like in a measured record? Nuclear resonant inelastic X-ray scattering scans the incident photon energy around a nuclear resonance and counts delayed photons after nuclear excitation. It does not analyze the final photon energy as a neutron spectrometer analyzes the outgoing neutron. Locate the sample and readout in Figure 9.3 before interpreting the count asymmetry.

Figure 9.3. The iron sample and nuclear-resonant readout. The foil and bcc unit cell use separate schematic scales; foil thickness is exaggerated and all lattice sites represent iron. The signal is delayed-photon counts, not a momentum-resolved scattered-energy spectrum. [ZS12, fig. 1 and Sec. 3, pp. 604-606]

Above resonance, the incident photon has excess energy and can create phonons; below resonance, phonon annihilation must supply the missing energy. Which side should therefore be weaker at finite positive temperature? Compare the complete two-sided scan in Figure 9.4, first outside the elastic-dominated central band and then at comparable nonzero ${|E|}$ .

Figure 9.4. The complete published iron-foil scan at 298.0(2) kelvin is recovered on its stated 0.25-meV acquisition grid. Counts are totals in 45 seconds on a logarithmic scale. The central gray band is elastic-dominated and is not used for the sideband comparison; source fitted and resolution curves are excluded. [ZS12, fig. 2 and Sec. 3, pp. 604-606]

Adapted from Zhao and Sturhahn’s Figure 2 under CC BY 4.0; the axes and styling are redrawn. Every visible 0.25-meV acquisition bin from ${-39.5}$ to ${39.5}$ meV is represented by the recovered centre of the printed open-marker band. The sparse anchor path used by the recovery program only bounds its pixel search; it supplies no plotted ordinate. Pixel dispersion is not experimental uncertainty, and the source panel supplies no pointwise error bars.

The annihilation side is visibly weaker: near a 20-meV offset, the recovered trace is roughly 2000 versus 4400 counts. This is the direction of the thermal asymmetry, not an invitation to fit a temperature to the display. The original analysis subtracts background, handles the elastic contribution, and models the spectrum. Most importantly, it uses detailed balance to calibrate the energy scale at a separately measured temperature. Agreement after that calibration would not independently establish the relation that defined it. Here the record teaches the measured channels and that logical boundary, rather than serving as a precision test of the equilibrium theorem.

The same reasoning generalizes, but the assumptions and the kind of equality should be visible in the statement.

Spectral asymmetry as a preparation diagnostic

Under the hypotheses of Definition 9.3, the unsymmetrized spectrum of a Hermitian readout obeys

$S_{BB}(-\omega) =\mathrm e^{-\beta\hbar\omega}S_{BB}(\omega)$

as an equality of distributions on ${\mathbb R}$ . Thus each pair of isolated reversed lines has nonnegative integrated weights whose negative- to positive-frequency ratio is ${\mathrm e^{-\beta\hbar\omega}}$ for ${\omega>0}$ . A driven nonequilibrium state need not obey this relation [Kub57; MS59].

The index manipulation is worth doing once. Start from the negative-frequency spectrum:

$S_{BB}(-\omega) =2\pi\sum_{n,m}p_n{|B_{nm}|}^2 \delta\!\left( \omega+\frac{E_m-E_n}{\hbar}\right).$

Exchange the dummy indices ${n\leftrightarrow m}$ , use Hermiticity to set ${{|B_{mn}|}^2={|B_{nm}|}^2}$ , and write

$S_{BB}(-\omega) =2\pi\sum_{n,m}p_m{|B_{nm}|}^2 \delta\!\left( \omega-\frac{E_m-E_n}{\hbar}\right).$

On the delta support, ${p_m=p_n\mathrm e^{-\beta\hbar\omega}}$ , so the exponential can be factored out. The proof shows exactly where each assumption enters: equilibrium supplies the weight ratio, Hermiticity matches the reversed matrix element, and the delta distribution enforces the energy difference. The theorem above keeps the canonical preparation and its fixed particle constraint throughout.

Test whether stationarity forces detailed balance at one temperature. Supply three nondegenerate energies ${(0,\Delta,3\Delta)}$ , ${\Delta>0}$ , and a diagonal preparation with probabilities ${(1/2,1/3,1/6)}$ . A Hermitian readout has zero diagonal and couples only neighboring levels, with both matrix magnitudes equal to one. Determine stationarity and the reverse-to-forward weights of each transition pair; decide whether one inverse temperature fits both ratios.

Check stationarity separately from the proposed Gibbs ratios. The state is positive and normalized, and its diagonal form commutes with the Hamiltonian. It is therefore stationary.

Apply the ordered-trace construction, without inserting an equilibrium ratio. The line pair at ${\pm\Delta/\hbar}$ has weights ${(1/2,1/3)}$ ; the pair at ${\pm2\Delta/\hbar}$ has weights ${(1/3,1/6)}$ . Fitting each pair separately uses reverse/forward ratios ${2/3}$ and ${1/2}$ , respectively. A Boltzmann fit would require $\beta_{01}=\frac{\ln(3/2)}{\Delta},\qquad \beta_{12}=\frac{\ln2}{2\Delta}.$ They differ, since ${{(3/2)}^2\ne2}$ . No single inverse temperature describes both ratios, despite monotone populations and exact stationarity. Each line weight is nonnegative, and their sum is ${4/3}$ , equal to ${\langle\hat B^2\rangle}$ because the middle state has two outgoing channels. These normalization and operator checks exclude a spurious sign or missing transition as the cause. The failed step is the Gibbs weight ratio, not stationarity or Hermiticity. This example changes the preparation; the following reservoir return instead changes which quantities may exchange.

One can also test the common-temperature requirement graphically. Gibbs weights obey ${\ln p_n=-\beta E_n-\ln Z}$ , so their log probabilities lie on one straight line when plotted against energy. In Figure 9.5, the horizontal coordinate is ${E/\Delta}$ , and the slope would be ${-\beta\Delta}$ . The neighboring slopes are ${-\ln(3/2)}$ and ${-\ln2/2}$ . Their inequality is the same failed common-temperature test found above; stationarity alone places no such straight-line constraint on the probabilities.

Figure 9.5. A Gibbs preparation would put log probabilities on one straight line. The three stationary probabilities give different neighboring slopes. The faint extension shows the prediction from the first pair; it is a comparison line, not another prepared state.

Return now to the canonical equilibrium preparation and a Hermitian readout ${\hat B}$ . For an allowed transition ${|a\rangle\to|b\rangle}$ , with ${B_{ab}\ne0}$ and ${E_b-E_a=\hbar\omega_{ba}>0}$ , write only the two reversed spectral terms and their coefficient ratio. Identify the initial state in each term before comparing the weights.

After completing the attempt, check both contributions to the spectrum: ${S_{a\to b}(\omega)=2\pi p_a|B_{ab}|^2\delta(\omega-\omega_{ba})}$ and ${S_{b\to a}(\omega)=2\pi p_b|B_{ba}|^2\delta(\omega+\omega_{ba})}$ . Hermiticity gives ${|B_{ba}|^2=|B_{ab}|^2}$ , so the ratio of the nonzero coefficients is ${p_b/p_a=\mathrm e^{-\beta\hbar\omega_{ba}}}$ . A remaining matrix-element ratio signals an incorrect index exchange; a reversed exponential signals an incorrect initial-state assignment.

At low temperature, allowed positive-frequency absorption from the ground state survives while negative-frequency emission is suppressed unless an excited state has been prepared. At high temperature, frequencies small compared with ${k_{\mathrm B}T/\hbar}$ have nearly equal directed weights. These are predictions about spectral areas. Broadening the delta lines may change their shapes, but a normalized broadening model should preserve those areas.

Particle Exchange and Quantum-Mode Occupations

Some later examples exchange particles with a large reservoir. Then the reservoir controls temperature and a chemical potential ${\mu}$ , an energy assigned per exchanged particle. Here ${\hat H}$ is the isolated system Hamiltonian and conserves ${\hat N}$ . Particle exchange occurs through the reservoir coupling during preparation. The prepared ensemble can assign probabilities to several number sectors even though subsequent isolated evolution preserves each sector. The ensemble change must specify both the operator and its existence boundary. The reservoir argument motivates preparation weights without deriving equilibration dynamics.

Repeat the reservoir-counting argument, now allowing both energy and particles to be exchanged at fixed volume. With weak coupling and equal weights for joint microstates, a system state ${(E_n,N_n)}$ has probability proportional to ${W_{\mathrm R}(E_{\mathrm{tot}}-E_n,N_{\mathrm{tot}}-N_n)}$ . Define ${S_{\mathrm R}=k_{\mathrm B}\ln W_{\mathrm R}}$ , with ${{(\partial S_{\mathrm R}/\partial E_{\mathrm R})}_{N_{\mathrm R},V}=1/T}$ and ${{(\partial S_{\mathrm R}/\partial N_{\mathrm R})}_{E_{\mathrm R},V}=-\mu/T}$ . When the system gains ${(\Delta E,\Delta N)}$ , the reservoir loses those amounts. Its first-order entropy change gives

$\ln\frac{p_{m,N_m}}{p_{n,N_n}} =\frac{\Delta S_{\mathrm R}}{k_{\mathrm B}} \simeq-\beta\Delta E+\beta\mu\Delta N =-\beta(\Delta E-\mu\Delta N).$

Here ${\Delta E=E_m-E_n}$ and ${\Delta N=N_m-N_n}$ . The large reservoir keeps ${T}$ and ${\mu}$ nearly constant over the exchanged range. In this reservoir approximation, ${p_{n,N_n}=C\mathrm e^{-\beta(E_n-\mu N_n)}}$ . Define ${\Xi(\beta,\mu)=\sum_n\mathrm e^{-\beta(E_n-\mu N_n)}}$ , where the sum runs over the included system energy-number eigenstates, across all included number sectors. Normalization fixes ${C=\Xi^{-1}}$ .

Let ${\hat H}$ and ${\hat N}$ be self-adjoint operators on a Hilbert space ${\mathcal H}$ . Require them to commute strongly, meaning that their spectral projections commute. If ${\mathcal H}$ is finite-dimensional (for example, the Fock space of finitely many fermion modes), or more generally if both operators are bounded, strong commutation is equivalent to ${[\hat H,\hat N]=0}$ . For ${\beta>0}$ and ${\mu\in\mathbb R}$ , suppose ${\mathrm e^{-\beta(\hat H-\mu\hat N)}}$ is trace class and define

$0<\Xi(\beta,\mu):= \operatorname{Tr}\mathrm e^{-\beta(\hat H-\mu\hat N)}<\infty.$

The grand-canonical equilibrium state over all included particle-number sectors is

$\hat\rho_{\mathrm{gc}}:= \frac{\mathrm e^{-\beta(\hat H-\mu\hat N)}}{\Xi(\beta,\mu)}.$

In a joint eigenbasis its probability ratios obey

$\frac{p_{m,N_m}}{p_{n,N_n}} =\exp\!\left\{-\beta\left[ (E_m-E_n)-\mu(N_m-N_n)\right]\right\}.$

The chemical potential ${\mu}$ is a reservoir parameter that controls the mean particle number; it is not a probability amplitude or a one-particle wavefunction coefficient. If ${\Xi}$ diverges, the formula does not define a density operator for the declared ${(\beta,\mu)}$ and sector space.

For a Hermitian readout, consider an allowed resolved pair of transitions between joint energy-number eigenstates. If the readout conserves number, ${N_m=N_n}$ , and the chemical-potential terms cancel from the reversed-weight ratio. Otherwise retain ${\mu(N_m-N_n)}$ , as in the joint-state ratio above. This correction applies to each pair separately: contributions with different particle-number changes generally cannot share one factor after summation.

In a finite-dimensional system—or whenever differentiating the trace is justified—write ${\hat K:=\hat H-\mu\hat N}$ . Direct differentiation gives

$-\frac{\partial\ln\Xi}{\partial\beta} =\frac{\operatorname{Tr}(\hat K\mathrm e^{-\beta\hat K})}{\Xi} ={\langle\hat H-\mu\hat N\rangle}_{\mathrm{gc}}, \qquad \frac1\beta\frac{\partial\ln\Xi}{\partial\mu} =\frac{\operatorname{Tr}(\hat N\mathrm e^{-\beta\hat K})}{\Xi} ={\langle\hat N\rangle}_{\mathrm{gc}}.$

Strong commutation permits ${\hat N}$ to pass through the exponential in the second identity. These are generator identities for normalized preparation statistics; no additional thermodynamic interpretation is needed here.

Consider two independent fermionic modes with energies ${\varepsilon_1}$ and ${\varepsilon_2}$ , common reservoir ${(\beta,\mu)}$ , and ${\hat H=\varepsilon_1\hat n_1+\varepsilon_2\hat n_2}$ . Take ${\beta>0}$ . Determine the four occupation probabilities and verify the ${\mu}$ - and ${\beta}$ -derivative identities for the normalizer against the direct means of ${\hat N}$ and ${\hat H-\mu\hat N}$ .

Enumerate the four configurations first so that differentiating the normalizer can be checked against explicit probabilities. Define ${\xi_i=\varepsilon_i-\mu}$ and ${q_i=\mathrm e^{-\beta\xi_i}}$ . The configurations ${00,10,01,11}$ have weights ${1,q_1,q_2,q_1q_2}$ , so $\Xi=(1+q_1)(1+q_2),\qquad (p_{00},p_{10},p_{01},p_{11})=\frac{(1,q_1,q_2,q_1q_2)}{\Xi}.$ Counting occupied modes directly yields $\langle\hat N\rangle=\frac{q_1+q_2+2q_1q_2}{\Xi} =\frac{q_1}{1+q_1}+\frac{q_2}{1+q_2} =\frac1\beta\partial_\mu\ln\Xi.$ At fixed ${\mu}$ , differentiating with respect to ${\beta}$ gives $-\partial_\beta\ln\Xi =\xi_1\frac{q_1}{1+q_1}+\xi_2\frac{q_2}{1+q_2} =\langle\hat H-\mu\hat N\rangle.$ The same result follows by weighting the four constrained energies ${0,\xi_1,\xi_2,\xi_1+\xi_2}$ . This checks which energy the derivative returns. The four probabilities sum to one, ${0\leq\langle N\rangle\leq2}$ , and at high temperature all four probabilities tend to ${1/4}$ . At zero temperature each mode below ${\mu}$ is occupied and each above it is empty, away from the degeneracy ${\xi_i=0}$ . Factorization requires independent modes; adding an interaction ${U\hat n_1\hat n_2}$ changes the double-occupation weight and destroys the product normalizer. The single-mode return below isolates the state-counting distinction between fermions and bosons.

This is supporting material, not a second anchor. For one fermionic mode of energy ${\varepsilon}$ , put ${\xi:=\varepsilon-\mu}$ . The allowed occupations are ${n=0,1}$ , so the two normalized weights have denominator ${\Xi_{\mathrm F}=1+\mathrm e^{-\beta\xi}}$ , and

$\langle\hat n\rangle =\frac{0+\mathrm e^{-\beta(\varepsilon-\mu)}} {1+\mathrm e^{-\beta(\varepsilon-\mu)}} =\frac{1}{\mathrm e^{\beta(\varepsilon-\mu)}+1}.$

For one bosonic mode, ${n=0,1,2,\ldots}$ . Set ${q:=\mathrm e^{-\beta\xi}}$ . Normalization requires ${|q|<1}$ , or ${\xi>0}$ , and the geometric sum and its derivative give

$\Xi_{\mathrm B}=\sum_{n=0}^{\infty}q^n=\frac{1}{1-q}, \qquad \sum_{n=0}^{\infty}nq^n =q\frac{\mathrm d}{\mathrm dq}\frac{1}{1-q} =\frac{q}{{(1-q)}^2}.$

Therefore

$\langle\hat n\rangle =\frac{q/{(1-q)}^2}{1/(1-q)} =\frac{q}{1-q} =\frac{1}{\mathrm e^{\beta(\varepsilon-\mu)}-1}.$

For phonons, particle number is not conserved by the environment, so their equilibrium chemical potential is zero. These formulas interpret mode occupations encountered earlier in the course; the present lecture needs only their normalization and limiting behavior.

The mode formulas immediately predict unequal-time addition and removal weights. Let ${\hat H=\varepsilon\hat c^\dagger\hat c}$ for one fermionic mode. Its operator evolution is ${\hat c(t)=\mathrm e^{-\mathrm i\varepsilon t/\hbar}\hat c}$ . Write ${n_{\mathrm F}:=\langle\hat c^\dagger\hat c\rangle}$ for the mean fermion occupation just calculated. Using ${\hat c\hat c^\dagger=\hat I-\hat c^\dagger\hat c}$ ,

$\langle\hat c(t)\hat c^\dagger(0)\rangle =\bigl(1-n_{\mathrm F}\bigr) \mathrm e^{-\mathrm i\varepsilon t/\hbar}, \qquad \langle\hat c^\dagger(t)\hat c(0)\rangle =n_{\mathrm F} \mathrm e^{+\mathrm i\varepsilon t/\hbar}.$

The factor ${n_{\mathrm F}}$ requires the mode to be occupied before a particle can be removed; ${1-n_{\mathrm F}}$ requires it to be empty before one can be added. For a bosonic mode ${\hat b}$ , write ${n_{\mathrm B}:=\langle\hat b^\dagger\hat b\rangle}$ for its mean occupation. The commutator ${[\hat b,\hat b^\dagger]=\hat I}$ instead gives

$\langle\hat b(t)\hat b^\dagger(0)\rangle =\bigl(1+n_{\mathrm B}\bigr) \mathrm e^{-\mathrm i\varepsilon t/\hbar}, \qquad \langle\hat b^\dagger(t)\hat b(0)\rangle =n_{\mathrm B} \mathrm e^{+\mathrm i\varepsilon t/\hbar}.$

The extra one is spontaneous-plus-stimulated bosonic addition, whereas the fermionic factor expresses Pauli blocking. These are not new equilibrium postulates: they combine the normalized occupation probabilities with the operator algebra already established in Lectures 7–8. They will recur in later spectral functions.

From preparation weights to spectral asymmetry

Given an unfamiliar finite-level system and a Hermitian operator, reconstruct one positive- and one negative-frequency term of its ordered spectrum. Name the initial-state probability, the matrix element, and the delta distribution separately; then reverse the transition and derive the ratio of the two integrated weights. Finally state which step would fail for a driven nonequilibrium preparation. A line list without these directed labels is not yet a physical interpretation.

Without copying a displayed equation, state the chain from incomplete preparation to detailed balance in five moves. Your account must name the density operator, normalization, the three ingredients of one directed line, the reversed transition, and the equilibrium assumption. Then name the single changed constraint that produces the grand-canonical exponent. If any move can be omitted without changing your explanation, the causal chain has not yet been reconstructed.

Equilibrium correlations now tell us what the undriven system reveals. They do not tell us what happens after an external source is switched on. That causal question requires a new calculation.

The larger course role is the insertion of preparation between a Hamiltonian and an observation. Energy differences identify where transitions can occur; operator matrix elements identify which transitions a probe can see; and equilibrium probabilities identify how often the required initial states are available. Response theory, spectral functions, and transport will reuse this three-factor separation. A driven population that violates detailed balance is not a failure of quantum mechanics—it is evidence that the equilibrium preparation assumption no longer applies.

The calculation connects preparation to measurement: a preparation and its constraints select a positive trace-one density operator; that operator and a readout determine directed transition weights; detailed balance follows only with the equilibrium hypotheses. A different ensemble decomposition of the same density operator does not change expectations, whereas a merely stationary nonthermal state can violate the weight ratio. Thus stationarity, Gibbs form, and detailed balance must never be used as synonyms.

Problems

1. Mixture, coherence, and equilibrium

A two-level system is prepared in

${[\hat\rho(q,c)]}_{\mathcal B_{\mathrm E}}= \begin{pmatrix} q & c\\ c^* & 1-q \end{pmatrix}$

in the basis ${(|\mathrm g\rangle,|\mathrm e\rangle)}$ .

Find the restrictions on ${q}$ and ${c}$ imposed by positivity.
Compute the expectations of ${\hat\sigma_x,\hat\sigma_y,\hat\sigma_z}$ .
Determine the condition for stationarity under ${\hat H_0}$ .
Identify ${q}$ for canonical equilibrium and check both temperature limits. Explain why equilibrium fixes ${c=0}$ for nondegenerate levels.

2. Directed lines and detailed balance

Three nondegenerate states have energies ${E_0=0}$ , ${E_1=\Delta}$ , and ${E_2=3\Delta}$ , with ${\Delta>0}$ . A Hermitian operator ${\hat B}$ has only ${B_{01}=b}$ and ${B_{12}=c}$ , together with their conjugates, as nonzero off-diagonal matrix elements.

Write the canonical probabilities ${p_0,p_1,p_2}$ including their common normalizer.
Write ${S_{BB}(\omega)}$ as four delta-function lines. Label every line by its directed transition and equilibrium weight.
Verify ${S_{BB}(-\omega)=\mathrm e^{-\beta\hbar\omega}S_{BB}(\omega)}$ line by line for positive frequencies.
Take the limits ${T\to0}$ and ${T\to\infty}$ . Explain which change comes from level spacings, which from matrix elements, and which from equilibrium weights.

3. One quantum mode in a particle reservoir

Let ${\xi=\varepsilon-\mu}$ for one mode.

Sum the grand weights for a fermionic mode and derive its mean occupation.
Repeat for a bosonic mode, stating the condition on ${\xi}$ required for normalization.
Compare both occupations for ${\beta\xi\gg1}$ , and compare their behavior as ${\xi\to0^+}$ .
Explain why phonons may have ${\mu=0}$ without causing a divergence for a mode with ${\varepsilon=\hbar\omega>0}$ , and why the fermionic value at ${\xi=0}$ has a different interpretation.

References

[Kub57] Kubo, R. (1957). “Statistical-mechanical theory of irreversible processes. I. General theory and simple applications to magnetic and conduction problems.” Journal of the Physical Society of Japan 12 (6), 570–586. https://doi.org/10.1143/JPSJ.12.570.

[Lan27] Landau, L. D. (1927). “Das dämpfungsproblem in der wellenmechanik.” Zeitschrift für Physik 45 (5–6), 430–441. https://doi.org/10.1007/BF01343064.

[Leh54] Lehmann, H. (1954). “Über eigenschaften von ausbreitungsfunktionen und renormierungskonstanten quantisierter felder.” Il Nuovo Cimento 11 (4), 342–357. https://doi.org/10.1007/BF02783624.

[MS59] Martin, P. C., and Schwinger, J. (1959). “Theory of many-particle systems. i.” Physical Review 115 (6), 1342–1373. https://doi.org/10.1103/PhysRev.115.1342.

[Neu27b] Neumann, J. von. (1927). “Thermodynamik quantenmechanischer gesamtheiten.” Nachrichten von der Gesellschaft der Wissenschaften zu Göttingen, Mathematisch-Physikalische Klasse, 273–291. https://eudml.org/doc/59231.

[Neu27a] Neumann, J. von. (1927). “Wahrscheinlichkeitstheoretischer aufbau der quantenmechanik.” Nachrichten von der Gesellschaft der Wissenschaften zu Göttingen, Mathematisch-Physikalische Klasse, 245–272. https://eudml.org/doc/59230.

[Hov54] Van Hove, L. (1954). “Correlations in space and time and born approximation scattering in systems of interacting particles.” Physical Review 95 (1), 249–262. https://doi.org/10.1103/PhysRev.95.249.

[ZS12] Zhao, J. Y., and Sturhahn, W. (2012). “High-energy-resolution x-ray monochromator calibration using the detailed-balance principle.” Journal of Synchrotron Radiation 19 (4), 602–608. https://doi.org/10.1107/S0909049512022339.

Lecture 9: Equilibrium Transition Weights

9.1 Preparations as density operators

9.1.1 Mixtures as positive normalized operators

Historical lineage

9.1.2 Basis-independent expectations and pure-state limits

9.2 Equilibrium weights from minimal statistical mechanics

9.3 Directed transitions in equilibrium correlations

9.3.1 Lehmann weights from operator order

9.3.2 Detailed balance

9.3.3 Seeing unequal sidebands without a circular test

9.3.4 Spectral asymmetry as a preparation diagnostic

9.4 Particle Exchange and Quantum-Mode Occupations

9.5 From preparation weights to spectral asymmetry

9.6 Problems