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The Araki–Lieb inequality bounds the difference of the two subsystem Von Neumann entropies by their joint Von Neumann entropy. In finite dimensions, all terms are finite:

For a bipartite density operator ${\rho }_{AB}$,

To prove one direction, take a purification of a density operator ${\rho }_{AB}$ using a system $C$. Equality of complementary entropies gives $S(A)=S(BC)$ and $S(C)=S(AB)$. Then Subadditivity of Von Neumann entropy gives $S(A)\le S(B)+S(AB)$; exchanging $A$ and $B$ proves the absolute-value bound. A Bell state has $S(AB)=0$ despite nonzero equal marginal entropies. ^[ref]

For the transposition map $\Theta :M_d\to M_d$, $\Theta (X)=X^T$, the induced trace norm is one, while the diamond norm isThe first equality follows because transposition preserves every singular value and hence the trace norm.

For the lower bound, apply $\Theta \otimes {\mathrm{id}}_d$ to the normalized maximally entangled state projector $\rho =|{\Phi }_d⟩⟨{\Phi }_d|$. Its partial transpose is $F/d$, with $F$ the swap operator. All $d^2$ singular values of $F$ are one, so $\Vert F/d{\Vert }_1=d$ while $\Vert \rho {\Vert }_1=1$.

For the upper bound, partial transpose preserves the Frobenius norm $\Vert \cdot {\Vert }_2$. On matrices of size $d^2$,A $d$-dimensional ancilla suffices in the definition of the diamond norm, proving equality. ^[ref]

For $d>1$, transposition is not completely positive: $F/d$ has a negative eigenvalue on the antisymmetric subspace. Thus this is a mathematical example of norm amplification by an entangled ancilla; transposition itself is not a physically implementable quantum channel.

For an exact dimensionless Hamiltonian $H$ and trial Hamiltonian $H_0$, the free energies obeywhere $F=-\ln Z$, $F_0=-\ln Z_0$, and the expectation value uses the normalized trial equilibrium state. A physical Hamiltonian is multiplied by $\beta =1/(k_{B}T)$ in these formulas; multiplying the bound by $k_{B}T$ restores physical free energy units.

For classical configurations, the identity $e^{-F}=e^{-F_0}⟨e^{-(H-H_0)}{⟩}_0$ and the Jensen inequality for the convex exponential function give $\ln ⟨e^{-(H-H_0)}{⟩}_0\ge -⟨H-H_0{⟩}_0$. Taking minus the logarithm proves the bound.

The same bound holds for finite-dimensional, noncommuting quantum Hamiltonians, although the preceding exponential-factorization identity cannot generally be used. Define the density matrices $\rho =e^{-H}/Z$ and ${\rho }_0=e^{-H_0}/Z_0$. With natural logarithms, nonnegativity of quantum relative entropy givesRearrangement proves the quantum result without requiring $[H,H_0]=0$. Equality holds when ${\rho }_0=\rho$; in this finite-dimensional setting that means $H-H_0$ is a scalar multiple of the identity operator.^[ref]

A tractable trial family therefore supplies upper bounds that can be minimized over its parameters. This variational minimum need not reproduce exact correlations or phase transition locations.

Suppose a preparation chooses normalized pure states $|{\alpha }_0⟩,|{\alpha }_1⟩$ with equal prior probabilities. Write ${\rho }_j=|{\alpha }_j⟩⟨{\alpha }_j|$. The optimal minimum-error quantum measurement assigns the positive and negative eigenspaces of ${\rho }_0-{\rho }_1$ to guesses 0 and 1, respectively. Its success probability isZero-eigenvalue directions may be assigned arbitrarily. Orthogonal quantum states give certainty, while identical quantum states allow only probability $1/2$. This is minimum-error discrimination; it does not introduce an inconclusive outcome.

For unequal prior probabilities $p_0,p_1$ with $p_0+p_1=1$, use the positive and negative eigenspaces of $p_0{\rho }_0-p_1{\rho }_1$ instead. For two pure states,Thus identical states reduce to guessing the more likely preparation. Omitting the prior weights would generally select the wrong measurement. ^[ref]

Suppose one equally likely action returns a known normalized pure state $|s⟩$ unchanged, while another returns the maximally mixed state $I/d$. A basic measurement in any orthonormal basis containing $|s⟩$ distinguishes them with success $1-1/(2d)$: guess the pure-state action on its basis label and the reset action on the remaining $d-1$ labels. The first action succeeds with probability one and the second with probability $(d-1)/d$.

The Holevo–Helstrom theorem also proves optimality for these returned states. The Hermitian difference $|s⟩⟨s|-I/d$ has eigenvalues $1-1/d$ and $-1/d$ with multiplicity $d-1$, hence trace norm $2(1-1/d)$. Its optimal equal-prior discrimination success is $1/2+\Vert {\rho }_0-{\rho }_1{\Vert }_1/4$. Random computational-basis permutations fix the uniform superposition, so they supply an example of the invariant action. This compares the returned states in the stated unassisted protocol; it makes no assertion about a more general channel-discrimination problem with an entangled ancillary probe.^[ref]

For two $d$-dimensional systems, $d\ge 2$, an isotropic quantum state has the formThe eigenvalue on the maximally entangled state is $[1+(d^2-1)p]/d^2$; all orthogonal directions have eigenvalue $(1-p)/d^2$. Thus positive semidefiniteness requiresNegative $p$ can therefore describe a valid density matrix; the interval $0\le p\le 1$ is only its convex-mixture subfamily.

The defining symmetry is invariance under $U\otimes \overline{U}$ for every unitary operator $U$. It differs from the $U\otimes U$ invariance of a Werner state. The overlap $F=⟨{\Phi }^{+}|{\rho }_p|{\Phi }^{+}⟩$ equals $[1+(d^2-1)p]/d^2$.

Within the physical range, the state is a separable quantum state exactly when $p\le 1/(d+1)$, equivalently $F\le 1/d$. Indeed, partial transpose sends $\Phi$ to the swap operator divided by $d$; on its antisymmetric subspace the resulting eigenvalue is $[1-(d+1)p]/d^2$. Above the threshold it is negative, proving entanglement. Below it, an isotropic twirling channel averages product states using correlated local unitary operators and Haar measure. Product-state overlaps cover $0\le F\le 1/d$, so these averages give separable quantum states throughout that interval.^[ref]

For $d=2$, the valid range is $-1/3\le p\le 1$, and entanglement begins above $p=1/3$. The point $p=1$ is a Bell state; $p=0$ is maximally mixed.

Prepare the uniform pure stateEvery permutation matrix fixes this vector, so RandomPermute returns $|s⟩$ with certainty. Choose an orthonormal basis with first vector $|{\psi }_0⟩=|s⟩$, for example the quantum Fourier transform basis. Under RandomReset, the returned state is $I/d$, which gives each of the $d$ measurement labels probability $1/d$.

Guess RandomPermute on label 0 and RandomReset on every other label. ThenThis is invariant-pure-state discrimination against a maximally mixed reset. It also agrees with the Holevo–Helstrom theorem: the eigenvalues of $|s⟩⟨s|-I/d$ are $1-1/d$ once and $-1/d$ with multiplicity $d-1$, giving optimal equal-prior success $1/2+\Vert {\rho }_P-{\rho }_R{\Vert }_1/4=1-1/(2d)$. This confirms optimal discrimination of these returned states, in addition to constructing the requested strategy.^[ref] For $d=1$, the actions coincide and the formula correctly gives random-guessing success $1/2$.

The Von Neumann entropy of a finite-dimensional density operator measures uncertainty in its eigenvalue distribution:where $p_i$ are the eigenvalues and $0{\log }_{2}0=0$. Base-two logarithms give bits. A pure state has Von Neumann entropy zero, while the maximally mixed state $I/d$ has ${\log }_{2}d$ bits. Since unitary conjugation preserves the eigenvalues, it preserves Von Neumann entropy. ^[ref]

For a bipartite pure state, the Von Neumann entropy of either reduced density operator quantifies its entanglement. That interpretation requires purity of the joint state. A Bell state has one bit of marginal Von Neumann entropy, but so does the separable state $(|00⟩⟨00|+|11⟩⟨11|)/2$. The latter mixture has classical correlations and no entanglement. Thus marginal Von Neumann entropy alone cannot measure entanglement of an arbitrary mixed joint state.