Maximum Entropy Principle

The maximum entropy principle assigns the density operator with the largest von Neumann entropy among all states consistent with specified information. For Hermitian constraint operators $A_a$ and target expectation values $a_a$, the finite-dimensional problem is

$$\rho_\star = \underset{\rho}{\operatorname{arg\,max}} \left\{ \mathsf S(\rho) \,\middle|\, \begin{array}{l} \rho\geq0,\quad \operatorname{Tr}\rho=1,\\ \operatorname{Tr}(\rho A_a)=a_a \end{array} \right\},$$

where

$$\mathsf S(\rho) = - \operatorname{Tr} \left( \rho\ln\rho \right)$$

is dimensionless entropy. Physical entropy is $S=k_{\mathrm B}\mathsf S$.

If the feasible set contains a full-rank state, the answer belongs to an exponential family:

$$\rho_\star = \frac{ \exp\left(-\sum_a\lambda_a A_a\right) }{ Z(\boldsymbol\lambda) }, \qquad Z(\boldsymbol\lambda) = \operatorname{Tr} \exp\left(-\sum_a\lambda_a A_a\right).$$

The multipliers $\lambda_a$ are determined by the target values, not chosen freely. A mean-energy constraint gives the canonical state. Mean energy and mean particle number give the grand-canonical state after the number multiplier is identified with chemical potential.

This is an inference statement conditional on a Hilbert space, a trace domain, and a constraint set. It does not by itself show that a physical system thermalizes, that the retained constraints are complete, or that a reservoir realizes the inferred state.

Helpful background. Density operators supply positivity and normalization; entropy and relative entropy supply the optimization tools.

Constrained quantum states

The constraints specify what is known about a preparation. For example, $\langle\sigma_z\rangle=1/2$ specifies a qubit’s average polarization; it does not specify each measurement outcome or its transverse components. Maximum entropy makes the least additional commitment in the precise sense of maximizing $\mathsf S$ on the compatible set. Jaynes’s inference formulation and Wichmann’s incomplete-measurement analysis develop this distinction.

In finite dimension, the density operators form a compact convex set. Affine expectation-value constraints select a closed convex subset. If that subset is nonempty, continuity of entropy guarantees a maximum, and strict concavity makes its state unique. The state can lie on the boundary and have zero eigenvalues; finite exponential multipliers describe the full-rank case.

For a qubit with $r_z=m=1/2$, entropy is largest at $r_x=r_y=0$. This slice shows $r_y=0$; the full feasible set is a disk in the Bloch ball. Circular entropy contours are a property of the qubit Bloch representation, not a general distance rule for density matrices.

General Finite-Dimensional Problem

Let $A_1,\ldots,A_m$ be Hermitian operators on a $d$-dimensional Hilbert space. Define the feasible set

$$\mathcal C(\mathbf a) = \left\{ \rho\geq0: \operatorname{Tr}\rho=1,\ \operatorname{Tr}(\rho A_a)=a_a \right\}.$$

The target vector

$$\mathbf a = (a_1,\ldots,a_m)$$

must lie in the convex set of attainable expectation values. If $\mathcal C(\mathbf a)$ is empty, no inference method can produce a state satisfying all constraints exactly.

Introduce the generalized potential

$$K(\boldsymbol\lambda) = \sum_{a=1}^{m} \lambda_aA_a,$$

the partition function

$$Z(\boldsymbol\lambda) = \operatorname{Tr} e^{-K(\boldsymbol\lambda)},$$

and the exponential state

$$\sigma_{\boldsymbol\lambda} = \frac{ e^{-K(\boldsymbol\lambda)} }{ Z(\boldsymbol\lambda) }.$$

The central task is to find multipliers satisfying

$$\operatorname{Tr} \left( \sigma_{\boldsymbol\lambda}A_a \right) = a_a$$

for every independent constraint.

Relative-Entropy Proof of Optimality

The cleanest proof avoids delicate variations at zero eigenvalues. For any feasible $\rho$, consider the quantum relative entropy

$$D \left( \rho\Vert\sigma_{\boldsymbol\lambda} \right) = \operatorname{Tr} \left[ \rho \left( \ln\rho - \ln\sigma_{\boldsymbol\lambda} \right) \right].$$

Because

$$\ln\sigma_{\boldsymbol\lambda} = - \sum_a\lambda_aA_a - \ln Z,$$

one obtains

$$D \left( \rho\Vert\sigma_{\boldsymbol\lambda} \right) ={} - \mathsf S(\rho) + \sum_a \lambda_a a_a + \ln Z.$$

Nonnegativity of relative entropy gives the upper bound

$$\mathsf S(\rho) \leq \ln Z(\boldsymbol\lambda) + \sum_a\lambda_a a_a.$$

If $\sigma_{\boldsymbol\lambda}$ itself satisfies the constraints, then

$$\mathsf S \left( \sigma_{\boldsymbol\lambda} \right) = \ln Z(\boldsymbol\lambda) + \sum_a\lambda_a a_a,$$

so every feasible state obeys

$$\mathsf S(\rho) \leq \mathsf S \left( \sigma_{\boldsymbol\lambda} \right).$$

Equality holds only when

$$D \left( \rho\Vert\sigma_{\boldsymbol\lambda} \right) = 0,$$

which implies $\rho=\sigma_{\boldsymbol\lambda}$. This proves both optimality and uniqueness.

The same calculation gives the exact entropy-gap identity

$$\mathsf S \left( \rho_\star \right) - \mathsf S(\rho) = D \left( \rho\Vert\rho_\star \right)$$

for every state $\rho$ sharing the retained expectation values with a full-rank exponential maximizer $\rho_\star$.

Lagrange-Multiplier Derivation

For an interior solution, vary the functional

$$\mathcal L[\rho] ={} - \operatorname{Tr}(\rho\ln\rho) - \alpha \left( \operatorname{Tr}\rho-1 \right) - \sum_a\lambda_a \left[ \operatorname{Tr}(\rho A_a)-a_a \right].$$

At a full-rank $\rho$,

$$\delta\mathsf S = - \operatorname{Tr} \left[ \delta\rho \left( \ln\rho+\mathbb I \right) \right].$$

Stationarity under arbitrary Hermitian variations gives

$$\ln\rho = - (1+\alpha)\mathbb I - \sum_a\lambda_aA_a.$$

Exponentiating,

$$\rho = e^{-(1+\alpha)} \exp \left( - \sum_a\lambda_aA_a \right).$$

Normalization fixes

$$e^{1+\alpha} = Z(\boldsymbol\lambda),$$

and hence

$$\rho = \frac{ \exp\left(-\sum_a\lambda_aA_a\right) }{ Z(\boldsymbol\lambda) }.$$

This stationarity derivation is useful, but it assumes the solution lies in the interior of state space. The relative-entropy proof also makes clear why the stationary state is the global maximum.

Determining the Multipliers

Define

$$\psi(\boldsymbol\lambda) = \ln Z(\boldsymbol\lambda).$$

Then

$$\frac{\partial\psi}{\partial\lambda_a} = - \operatorname{Tr} \left( \rho_{\boldsymbol\lambda}A_a \right) = - \langle A_a\rangle_{\boldsymbol\lambda}.$$

This derivative remains valid when the constraint operators do not commute. One uses the operator identity

$$\frac{\partial e^{-K}}{\partial\lambda_a} = - \int_0^1 e^{-(1-s)K} A_a e^{-sK} \,ds$$

and cyclicity of the trace.

The multiplier equations are therefore

$$\nabla_{\boldsymbol\lambda} \psi(\boldsymbol\lambda) = - \mathbf a.$$

For centered operators

$$\Delta A_a = A_a-\langle A_a\rangle\mathbb I,$$

the Hessian is the Kubo–Mori covariance matrix

$$\frac{\partial^2\psi} {\partial\lambda_a\partial\lambda_b} = \int_0^1 \operatorname{Tr} \left( \rho^s \Delta A_a \rho^{1-s} \Delta A_b \right) ds.$$

It is positive semidefinite, so $\psi$ is convex. If all constrained operators commute with $\rho$, this reduces to the ordinary covariance

$$\frac{\partial^2\psi} {\partial\lambda_a\partial\lambda_b} = \langle \Delta A_a\Delta A_b \rangle.$$

Zero modes of the covariance matrix signal redundant constraints or directions that do not change the state.

Entropy as a Dual Function

For any multiplier vector, define

$$g(\boldsymbol\lambda;\mathbf a) = \psi(\boldsymbol\lambda) + \boldsymbol\lambda\mathbin{\cdot}\mathbf a.$$

The relative-entropy inequality shows that $g$ is an upper bound on the entropy of every feasible state. For feasible finite-dimensional moment constraints,

$$\mathsf S_{\max}(\mathbf a) = \inf_{\boldsymbol\lambda} \left[ \psi(\boldsymbol\lambda) + \boldsymbol\lambda\mathbin{\cdot}\mathbf a \right].$$

When a full-rank feasible state exists, the infimum is attained at finite multipliers (up to redundant directions). At this optimum,

$$\frac{\partial g}{\partial\lambda_a} = 0$$

is exactly the moment-matching equation. The envelope relation is

$$\frac{\partial\mathsf S_{\max}} {\partial a_a} = \lambda_a.$$

Thus the Lagrange multipliers are variables conjugate to the retained expectation values. If the Kubo–Mori covariance matrix $G$ is invertible,

$$\frac{\partial\lambda_a}{\partial a_b} = - \left( G^{-1} \right)_{ab},$$

which expresses concavity of the optimized entropy.

Thermodynamic Legendre transforms are concrete realizations of this dual structure. Thermodynamic Potentials develops their natural variables, differentials, and Maxwell relations.

Energy and particle-number constraints

With only mean energy $U=\operatorname{Tr}(\rho H)$ specified, the exponential state is $\rho_\beta=e^{-\beta H}/Z$. The optimized entropy satisfies

$$\frac{\partial S_{\max}}{\partial U}=k_{\mathrm B}\beta=\frac1T.$$

This identifies the energy multiplier as inverse temperature when equilibrium thermodynamics applies. It has units of inverse energy. For a bounded finite-dimensional spectrum, the target mean energy can require positive, zero, or negative $\beta$. The entropy maximum at fixed energy is valid for all these cases. The associated free-energy functional is minimized at positive temperature and maximized at negative temperature; see Thermal Density Operators.

When mean particle number is also constrained, use independent multipliers:

$$\rho_{\beta,\gamma}=\frac{e^{-\beta H-\gamma N}}{\mathcal Z(\beta,\gamma)},\qquad U=-\left(\partial_\beta\ln\mathcal Z\right)_\gamma, \quad \overline N=-\left(\partial_\gamma\ln\mathcal Z\right)_\beta.$$

Here $\gamma$ is dimensionless. For $\beta\ne0$, writing $\gamma=-\beta\mu$ produces $e^{-\beta(H-\mu N)}$ and chemical potential $\mu$ has energy units. At $\beta=0$ with nonzero $\gamma$, no finite $\mu$ gives this parametrization; the independent multipliers remain well defined. At fixed $\mu$, differentiating with respect to $\beta$ gives $U-\mu\overline N$, since both terms in the exponent change.

The grand-canonical ensemble uses a full Fock-space trace and a conserved particle number. Within one exact-$N$ sector the number multiplier cancels out of the normalized state. Constraining a nonconserved observable is mathematically allowed, but it does not generally produce a stationary equilibrium state or a chemical potential.

Exact Constraints Are Not Mean Constraints

Suppose $P$ projects onto an allowed subspace of dimension

$$r = \operatorname{Tr}P.$$

Imposing the exact support condition

$$P\rho P = \rho$$

does not introduce a finite expectation-value multiplier. The maximum-entropy state on the subspace is

$$\rho_P = \frac{P}{r}.$$

Indeed, for any state supported in $P\mathcal H$,

$$D \left( \rho\middle\Vert\frac{P}{r} \right) = \ln r-\mathsf S(\rho) \geq0.$$

For an energy-shell projector $P_{E,\Delta}$, this gives

$$\rho_{\mathrm{mc}} = \frac{ P_{E,\Delta} }{ \operatorname{Tr}P_{E,\Delta} }.$$

By contrast, fixing only $\langle H\rangle=U$ allows support on many energies and produces $e^{-\beta H}/Z$.

Constraint	What is exact?	Typical maximizing state
$P\rho P=\rho$	allowed subspace	uniform on that subspace
$\langle H\rangle=U$	mean energy	canonical exponential state
$N\rho=N_0\rho$	number sector	state restricted to $\mathcal H_{N_0}$
$\langle N\rangle=\overline N$	mean number	state spanning number sectors

The Microcanonical Ensemble explains energy-window conventions and shell thermodynamics.

Interior and Boundary Solutions

For finite multipliers in finite dimension,

$$\rho_{\boldsymbol\lambda} \propto e^{-K(\boldsymbol\lambda)}$$

is full rank. Therefore a rank-deficient maximizer cannot generally be represented with finite multipliers.

Consider a Hamiltonian with lowest energy $E_0$ and ground-space projector $P_0$. The boundary target

$$\langle H\rangle = E_0$$

forces the state into the ground space. Maximum entropy gives

$$\rho_\star = \frac{ P_0 }{ \operatorname{Tr}P_0 }.$$

It is reached as the limit

$$\rho_\star = \lim_{\beta\to+\infty} \frac{e^{-\beta H}}{Z(\beta)}.$$

For a nondegenerate ground state this limit is pure; for a degenerate ground space it is the uniform mixture unless further constraints distinguish ground states.

Boundary behavior matters numerically. A solver may send one or more multipliers toward infinity rather than converge to a finite point. That can be the correct signal of a rank-deficient optimum rather than a failure of the entropy principle.

Feasibility and Redundant Constraints

Before solving for multipliers, check whether the target moments are possible. For one Hermitian operator $A$,

$$\lambda_{\min}(A) \leq \langle A\rangle \leq \lambda_{\max}(A).$$

With several operators, separate interval checks are necessary but may not be sufficient. Their joint expectation-value region can impose additional inequalities.

Constraints are redundant when one retained operator is an affine combination of the others. If

$$A_2 = cA_1+d\mathbb I,$$

then every normalized state satisfies

$$\langle A_2\rangle = c\langle A_1\rangle+d.$$

If the targets violate this relation, the feasible set is empty. If they obey it, the maximizing state is unique but the pair $(\lambda_1,\lambda_2)$ is not: only the combination multiplying $A_1$ affects the normalized state.

In computations, remove affine redundancies or use a gauge convention before inverting a susceptibility matrix.

Classical Reduction

If all constrained operators commute, choose a common eigenbasis:

$$A_a|i\rangle = a_{a,i}|i\rangle.$$

The exponential state is diagonal,

$$\rho_\star = \sum_i p_i|i\rangle\langle i|,$$

with

$$p_i = \frac{ e^{-\sum_a\lambda_a a_{a,i}} }{ \sum_j e^{-\sum_a\lambda_a a_{a,j}} }.$$

The quantum problem then reduces to the classical maximum-Shannon-entropy distribution over the joint eigenvalues. Within an unresolved degenerate joint eigenspace, the maximizing state is proportional to the identity.

Off-diagonal coherence cannot improve the entropy while preserving only commuting moments. Dephasing in the common eigenbasis preserves those moments and cannot decrease the entropy.

Noncommuting expectation values

For a full-rank maximizer, the same exponential construction applies even when the constrained operators do not commute.

Expectation values of noncommuting observables can be constrained simultaneously. They need not possess simultaneous sharp values. The finite-dimensional variational answer remains

$$\rho_\star = \frac{ e^{-\sum_a\lambda_aA_a} }{ Z },$$

where the exponential is taken after forming the operator sum.

In general,

$$e^{-\lambda A-\eta B} \neq e^{-\lambda A}e^{-\eta B}$$

when $[A,B]\neq0$. Product factorization would change the state and can even destroy Hermiticity if used carelessly.

The inferred state need not commute with each constrained observable:

$$[\rho_\star,A_a] \neq 0$$

can occur even though all target expectation values are reproduced. Ordinary covariances must then be replaced by the Kubo–Mori matrix in multiplier-response formulas.

There is also a physical distinction between inference and equilibrium. If a constrained charge fails to commute with $H$, the exponential state may fail to be stationary:

$$[H,\rho_\star] \neq 0.$$

Noncommuting conserved charges require a more careful thermodynamic framework, including what conservation means for the composite system and which exchanges are allowed. The formal maximum-entropy calculation is valid more broadly than the equilibrium interpretation attached to it.

Example: One Qubit Constraint

Write a qubit state as

$$\rho = \frac{1}{2} \left( \mathbb I + \mathbf r\mathbin{\cdot}\boldsymbol\sigma \right), \qquad |\mathbf r|\leq1.$$

Suppose the only measured moment is

$$\langle\sigma_z\rangle = m, \qquad -1\leq m\leq1.$$

The constraint fixes $r_z=m$ but leaves $r_x$ and $r_y$ unknown. The eigenvalues are

$$p_\pm = \frac{1\pm|\mathbf r|}{2},$$

and the entropy decreases as $|\mathbf r|$ increases away from zero. For fixed $r_z=m$, the smallest possible Bloch-vector length is obtained at

$$r_x=r_y=0.$$

Therefore

$$\rho_\star = \frac{1}{2} \left( \mathbb I+m\sigma_z \right).$$

The same state has exponential form

$$\rho_\star = \frac{ e^{-\lambda\sigma_z} }{ 2\cosh\lambda },$$

with

$$m = - \tanh\lambda, \qquad \lambda = - \operatorname{artanh}m.$$

At $m=0$, no polarization is known and $\rho_\star=\mathbb I/2$. At $m=\pm1$, the target lies on the boundary and the maximizing state is pure; the multiplier diverges.

Example: Two Noncommuting Qubit Constraints

Now retain

$$\langle\sigma_x\rangle=x, \qquad \langle\sigma_z\rangle=z.$$

Feasibility requires

$$x^2+z^2 \leq 1.$$

The unmeasured component $r_y$ only increases $|\mathbf r|$ and lowers the entropy. Hence

$$\rho_\star = \frac{1}{2} \left( \mathbb I+x\sigma_x+z\sigma_z \right).$$

Let

$$r = \sqrt{x^2+z^2}, \qquad \widehat{\mathbf r} = \frac{(x,0,z)}{r}$$

for $0<r<1$. Then

$$\rho_\star = \frac{ \exp \left[ \operatorname{artanh}(r) \, \widehat{\mathbf r} \mathbin{\cdot} \boldsymbol\sigma \right] }{ 2\cosh \left[ \operatorname{artanh}(r) \right] }.$$

This is of the required form with a multiplier vector antiparallel to the retained Bloch vector. At $r=0$, the answer is $\mathbb I/2$; at $r=1$, it is a pure boundary state obtained by a divergent-multiplier limit. The noncommutation

$$[\sigma_x,\sigma_z] = - 2i\sigma_y$$

does not obstruct expectation-value inference. It does obstruct interpreting $x$ and $z$ as simultaneously sharp eigenvalues.

Maximum Entropy Relative to a Prior State

Plain maximum entropy uses the maximally mixed state as its implicit finite-dimensional reference because

$$D \left( \rho\middle\Vert\frac{\mathbb I}{d} \right) = \ln d-\mathsf S(\rho).$$

If a full-rank reference state $\sigma$ represents information that should be retained before imposing new moments, a natural generalization minimizes

$$D(\rho\Vert\sigma)$$

over the feasible set. The interior solution is

$$\rho_\star = \frac{ \exp \left( \ln\sigma-\sum_a\lambda_aA_a \right) }{ \operatorname{Tr} \exp \left( \ln\sigma-\sum_a\lambda_aA_a \right) }.$$

When $\sigma$ commutes with all $A_a$, this resembles a classical reweighting. In the noncommuting case,

$$\exp \left( \ln\sigma-\lambda A \right)$$

must not be replaced by $\sigma e^{-\lambda A}$.

This relative-entropy projection makes “least biased” explicitly conditional on a reference. It is also indispensable in infinite-dimensional settings, where no normalized state proportional to the identity may exist.

Constraint-Based Coarse Graining

Suppose an exact state $\rho$ is replaced by the maximum-entropy state that preserves selected moments:

$$\operatorname{Tr}(\rho_\star A_a) = \operatorname{Tr}(\rho A_a).$$

Then

$$\mathsf S(\rho_\star) \geq \mathsf S(\rho).$$

The entropy increase measures distinctions discarded by retaining only those moments. For a full-rank exponential maximizer,

$$\mathsf S(\rho_\star) - \mathsf S(\rho) = D(\rho\Vert\rho_\star).$$

This replacement is generally nonlinear in $\rho$ because the multipliers depend nonlinearly on the retained expectation values. It should not automatically be treated as a physical quantum channel or a dynamical law. Entropy in Quantum Statistical Mechanics distinguishes thermal, entanglement, diagonal, and coarse-grained entropies.

Inference and physical equilibration

Maximum entropy does not select the physically relevant constraints or prove that a system thermalizes. These require a preparation, a model of conserved quantities and exchange, or a dynamical argument. For example, the qubit estimate with only $\langle\sigma_z\rangle$ retained sets transverse components to zero. Other compatible preparations can have nonzero transverse components: zero is a prediction of this assignment, not a fact implied by the measurement alone.

Noncommuting conserved quantities also admit exponential-state descriptions, but physically realizing their joint approximate constraints takes additional reasoning; see Yunger Halpern et al. (2016). Relaxation and Thermalization addresses the dynamical question.

In infinite dimension, first check that the exponential is trace class, the moments and entropy are meaningful, and a maximizing state actually exists. Entropy can have an unattained supremum or be unbounded above. A uniform prior on an infinite-dimensional Hilbert space is not a normalized density operator. Confinement or a finite-volume regularization may make a Gibbs trace finite; dividing a divergent trace by volume does not itself define a normalized global state.

Exercises

The maximally mixed state

Show that $\mathbb I/d$ uniquely maximizes von Neumann entropy on a $d$-dimensional Hilbert space when normalization is the only constraint.

Solution

For any density operator $\rho$,

$$\begin{aligned} D \left( \rho\middle\Vert\frac{\mathbb I}{d} \right) &= \operatorname{Tr} \left[ \rho \left( \ln\rho+\ln d \right) \right] \\ &= \ln d-\mathsf S(\rho). \end{aligned}$$

Relative entropy is nonnegative, so

$$\mathsf S(\rho) \leq \ln d.$$

Equality holds only if

$$\rho = \frac{\mathbb I}{d}.$$

Qubit polarization

Maximize the entropy of a qubit subject to

$$\langle\sigma_z\rangle = \frac{1}{2}.$$

Find the density matrix, its entropy, and the multiplier $\lambda$ in

$$\rho_\star \propto e^{-\lambda\sigma_z}.$$

Solution

The maximum-entropy state has no unconstrained transverse Bloch components:

$$\rho_\star = \frac{1}{2} \left( \mathbb I+\frac{1}{2}\sigma_z \right).$$

In the $\sigma_z$ basis,

$$\rho_\star = \begin{pmatrix} 3/4&0\\ 0&1/4 \end{pmatrix}.$$

Its dimensionless entropy is

$$\mathsf S(\rho_\star) = - \frac{3}{4}\ln\frac{3}{4} - \frac{1}{4}\ln\frac{1}{4}.$$

Because $m=-\tanh\lambda$,

$$\lambda = - \operatorname{artanh} \left( \frac{1}{2} \right) = - \frac{1}{2}\ln3.$$

Exact ground energy

A finite-dimensional Hamiltonian has ground energy $E_0$ with degeneracy $g_0$ and at least one higher energy level. Maximize entropy subject to the exact mean-energy target $\langle H\rangle=E_0$. Explain why the answer is not a finite-temperature Gibbs state.

Solution

Since $H-E_0\mathbb I\geq0$,

$$\operatorname{Tr} \left[ \rho \left( H-E_0\mathbb I \right) \right] = 0$$

forces $\rho$ to have support only in the kernel of $H-E_0\mathbb I$, namely the ground space. Entropy is maximized uniformly on that $g_0$-dimensional subspace:

$$\rho_\star = \frac{P_0}{g_0}, \qquad \mathsf S(\rho_\star) = \ln g_0.$$

Every finite $\beta$ gives a full-rank Gibbs state and assigns nonzero weight to excited levels. The solution is the boundary limit

$$\rho_\star = \lim_{\beta\to+\infty} \frac{e^{-\beta H}}{Z(\beta)}.$$

Two noncommuting qubit moments

Suppose

$$\langle\sigma_x\rangle = \frac{1}{3}, \qquad \langle\sigma_z\rangle = \frac{2}{3}.$$

Find the maximum-entropy state and verify that it is positive.

Solution

Maximum entropy sets the unconstrained component $\langle\sigma_y\rangle$ to zero:

$$\rho_\star = \frac{1}{2} \left( \mathbb I + \frac{1}{3}\sigma_x + \frac{2}{3}\sigma_z \right).$$

In the $\sigma_z$ basis,

$$\rho_\star = \begin{pmatrix} 5/6&1/6\\ 1/6&1/6 \end{pmatrix}.$$

The Bloch-vector length is

$$r = \frac{\sqrt5}{3} < 1,$$

so the eigenvalues

$$p_\pm = \frac{1}{2} \left( 1\pm\frac{\sqrt5}{3} \right)$$

are both positive. The state is therefore a valid full-rank density operator even though $\sigma_x$ and $\sigma_z$ do not commute.

Redundant constraints

Let $A_2=2A_1+3\mathbb I$. Determine the feasibility condition on targets $(a_1,a_2)$ and show why the multipliers in

$$\rho \propto e^{-\lambda_1A_1-\lambda_2A_2}$$

are not unique.

Solution

Every normalized state satisfies

$$\langle A_2\rangle = 2\langle A_1\rangle+3,$$

so feasibility requires

$$a_2 = 2a_1+3.$$

One also needs $a_1\in[\lambda_{\min}(A_1),\lambda_{\max}(A_1)]$. Together these conditions are sufficient, because every value in that interval is an expectation of $A_1$. The exponent can be rearranged as

$$- \lambda_1A_1 - \lambda_2A_2 ={} - \left( \lambda_1+2\lambda_2 \right) A_1 - 3\lambda_2\mathbb I.$$

The identity term cancels from the normalized density operator. Therefore the state depends only on

$$\lambda_{\mathrm{eff}} = \lambda_1+2\lambda_2.$$

Infinitely many multiplier pairs give the same unique maximum-entropy state.

Individually allowed but jointly impossible

Can a qubit satisfy $\langle\sigma_x\rangle=\langle\sigma_z\rangle=0.9$? Each target lies in the spectrum’s convex hull $[-1,1]$.

Solution

No. Positivity requires the Bloch-vector length to be at most one, whereas $r_x^2+r_z^2=1.62>1$ even before adding $r_y^2$. Separate bounds on each moment do not ensure joint feasibility. Inserting these targets into $\frac12(I+0.9\sigma_x+0.9\sigma_z)$ would give a negative eigenvalue $(1-\sqrt{1.62})/2$.

References

• R. Balian, From Microphysics to Macrophysics: Methods and Applications of Statistical Physics, volumes I and II, Springer (1991–1992), chapters on statistical inference and equilibrium ensembles.
• E. T. Jaynes, “Information Theory and Statistical Mechanics. II”, Physical Review 108, 171–190 (1957).
• E. T. Jaynes, “Information Theory and Statistical Mechanics”, Physical Review 106, 620–630 (1957).
• R. K. Pathria and P. D. Beale, Statistical Mechanics, 4th ed., Elsevier (2021), chapters 2–5.
• D. Petz, Quantum Information Theory and Quantum Statistics, Springer (2008), chapters on entropy, relative entropy, and exponential families.
• A. Wehrl, “General Properties of Entropy”, Reviews of Modern Physics 50, 221–260 (1978).
• E. H. Wichmann, “Density Matrices Arising from Incomplete Measurements”, Journal of Mathematical Physics 4, 884–896 (1963).
• N. Yunger Halpern, P. Faist, J. Oppenheim, and A. Winter, “Microcanonical and Resource-Theoretic Derivations of the Thermal State of a Quantum System with Noncommuting Charges”, Nature Communications 7, 12051 (2016).