Normal density states and spectral moments

By [F1] and [F2], write $\rho ={\sum }_{j\in J}{p}_j\mathrm{\mid }{u}_j⟩⟨u_j\mathrm{\mid }$, where $J$ is finite or countable, $p_j>0$, and the $u_j$ are orthonormal. The positive-operator trace formula [F3] gives ${\sum }_j{p}_j=\mathrm{Tr}\rho =1$ and $\mathrm{\parallel }\rho {\mathrm{\parallel }}_1=1$. For every finite subset $F\subseteq J$, the remainder $\rho -{\sum }_{j\in F}{p}_j\mathrm{\mid }{u}_j⟩⟨u_j\mathrm{\mid }$ is positive trace class with trace norm $1-{\sum }_{j\in F}{p}_j$. Thus these finite partial sums converge to $\rho$ in trace norm as the finite-subset net exhausts $J$; no basis of $\ker \rho$ is used.

By [F5], $\rho A$ is trace class. The trace-norm expansion from step 1.1 is the nuclear representation $\rho ={\sum }_{j\in J}{T}_{p_j{u}_j,u_j}$, since ${\sum }_j\mathrm{\parallel }{p}_j{u}_j\mathrm{\parallel }\mathrm{\parallel }{u}_j\mathrm{\parallel }={\sum }_j{p}_j=1$. The basis-independent trace [F4] and bounded cyclicity and its nuclear-series formula [F6] give $\mathrm{Tr}(\rho A)={\sum }_{j\in J}{p}_j⟨A{u}_j,u_j⟩\mathrm{.}$ The series is absolutely convergent because ${\sum }_j{p}_j\mathrm{\mid }⟨A{u}_j,u_j⟩\mathrm{\mid }\le \mathrm{\parallel }A\mathrm{\parallel }{\sum }_j{p}_j=\mathrm{\parallel }A\mathrm{\parallel }$. Thus ${\omega }_{\rho }$ is a well-defined bounded linear functional with $\mathrm{\mid }{\omega }_{\rho }(A)\mathrm{\mid }\le \mathrm{\parallel }A\mathrm{\parallel }$.

For $A=I$, step 2.1 gives ${\omega }_{\rho }(I)={\sum }_j{p}_j=1$. If $A\ge 0$, every $⟨A{u}_j,u_j⟩\ge 0$, so step 2.1 gives ${\omega }_{\rho }(A)\ge 0$. Therefore the trace pairing is a positive normalized bounded functional, as asserted.

Choose the ordered orthonormal basis of [F3]. In its coordinates $q(u)=⟨Mu,u⟩$ is a real continuous quadratic polynomial, bounded on the unit sphere by the finite sum of the absolute matrix entries. By [F4], enumerate all rational-coordinate vectors $y_k\in (\mathbb{Q}+i\mathbb{Q}{)}^D$, and put $v_k=y_k\mathrm{/}\mathrm{\parallel }{y}_k\mathrm{\parallel }$ when $y_k\mathrm{\ne }0$ and $v_k=e_1$ otherwise. For any unit $x$ and $\epsilon >0$, coordinatewise rational approximation gives $\mathrm{\parallel }y-x\mathrm{\parallel }<\min (\epsilon \mathrm{/}2,1\mathrm{/}2)$ for such a $y$. Then $y\mathrm{\ne }0$ and the reverse triangle inequality gives $\mathrm{\parallel }y\mathrm{/}\mathrm{\parallel }y\mathrm{\parallel }-x\mathrm{\parallel }\le \mathrm{\mid }1-\mathrm{\parallel }y\mathrm{\parallel }\mathrm{\mid }+\mathrm{\parallel }y-x\mathrm{\parallel }\le 2\mathrm{\parallel }y-x\mathrm{\parallel }<\epsilon$. Thus the $v_k$ are dense on the sphere; they are normalizations of rational-coordinate vectors, not necessarily rational-coordinate vectors themselves. The set $S:=\{q(u):\mathrm{\parallel }u\mathrm{\parallel }=1\}$ is nonempty because $e_1$ is a unit vector and is bounded above as just shown; [F5] gives $m:=\sup S\in \mathbb{R}$. For each $n\ge 1$ take the least index $k(n)$ with $q(v_{k(n)})>m-1\mathrm{/}n$, which exists by continuity and density. This deterministic rule uses no countable choice. By [F1] the bounded sequence $v_{k(n)}$ has a convergent subsequence; its limit has norm one and attains $m$.

Let $u$ be a maximizing unit vector and $w\perp u$. Differentiating $q((u+tw)\mathrm{/}\mathrm{\parallel }u+tw\mathrm{\parallel })$ at $t=0$ for real $t$ gives $2\mathrm{Re}⟨Mw,u⟩=0$. Replacing $w$ by $iw$ shows that the imaginary part also vanishes. Hermiticity gives $⟨Mu,w⟩=0$ for every $w\perp u$, so $Mu=q(u)u$ with real eigenvalue. Moreover $u^{\perp }$ is invariant, since $⟨Mw,u⟩=⟨w,Mu⟩=0$. Induction on $D$, including the one-dimensional case, gives an orthonormal eigenbasis with real eigenvalues.

For $M={\sum }_{\lambda }\lambda P_{\lambda }$, define $f(M)={\sum }_{\lambda }f(\lambda )P_{\lambda }$. The full eigenspaces and their orthogonal projections are intrinsic, so this definition is independent of the basis within each eigenspace. It supplies positive square roots for nonnegative matrices, exponentials for every Hermitian matrix, and the stated logarithm on the positive spectral subspace $S_M$ for every Hermitian $M$: the restriction has exactly the strictly positive eigenvalues. If $S_M=\{0\}$, the empty spectral sum is the unique endomorphism of that zero space. Finite coordinate sums give $\mathrm{Tr}(AB)={\sum }_{i,j}{A}_{ij}{B}_{ji}=\mathrm{Tr}(BA)$, hence invariance under unitary changes of orthonormal basis. Diagonalizing $\rho$ gives $\mathrm{Tr}(\rho A)={\sum }_j{p}_j⟨A{u}_j,u_j⟩$ for every complex matrix $A$, and in particular $\mathrm{Tr}(\rho B^{\ast }B)={\sum }_j{p}_j\mathrm{\parallel }B{u}_j{\mathrm{\parallel }}^2\ge 0$.

Put $a=\mathrm{Tr}(\rho X^{\ast }Y)$, $b=\mathrm{Tr}(\rho Y^{\ast }Y)$ and $c=\mathrm{Tr}(\rho X^{\ast }X)$. Step 3.1 gives $c+za+\bar{z}\bar{a}+\mathrm{\mid }z{\mathrm{\mid }}^{2}b\ge 0$ for all complex $z$. If $b>0$, take $z=-\bar{a}\mathrm{/}b$ to get $\mathrm{\mid }a{\mathrm{\mid }}^2\le bc$. If $b=0$, varying $z$ with arbitrarily large magnitude forces $a=0$, giving the same inequality. This proves the stated weighted Cauchy–Schwarz formula (whose form is conjugate-linear in $X$).

The density eigenvalues satisfy $p_j\ge 0$ and ${\sum }_j{p}_j=1$, so $\mathrm{Tr}({\rho }^2)={\sum }_j{p}_j^2\le 1$, with equality exactly when one eigenvalue is one and all others vanish. If a rank-one density $\rho =\mathrm{\mid }u⟩⟨u\mathrm{\mid }$ is $t\sigma +(1-t)\tau$ with $0<t<1$, then every $v\perp u$ has $0=t⟨\sigma v,v⟩+(1-t)⟨\tau v,v⟩$. Positivity and the square roots from step 3.1 give $\sigma v=\tau v=0$. Self-adjointness puts their ranges in $\mathbb{C}u$, and trace one gives $\sigma =\tau =\rho$. Conversely, two positive eigenvalues allow a sufficiently small opposite transfer of weight, yielding distinct densities ${\rho }_{+},{\rho }_{-}$ with midpoint $\rho$. Thus precisely the rank-one densities are extreme.

Choose an ordered orthonormal basis $e_1,\dots ,e_D$. Its matrix units form an ordered basis of the complex endomorphism space, since $A={\sum }_{i,j}{A}_{ij}{E}_{ij}$ uniquely. The operator norm is a norm: its triangle inequality and homogeneity follow by applying the vector norm to $Av$ for unit $v$, and norm zero implies $A{e}_j=0$ for all $j$. Matrix composition is associative and complex bilinear, and conjugate transpose is conjugate-linear, involutive and reverses products. By [F2] the norm is submultiplicative, complete, and satisfies $\mathrm{\parallel }{A}^{\ast }A\mathrm{\parallel }=\mathrm{\parallel }A{\mathrm{\parallel }}^2$. Thus all axioms of [F4] hold. The identity is nonzero and has norm one because $D\ge 1$. The C*-identity and submultiplicativity give $\mathrm{\parallel }A{\mathrm{\parallel }}^2\le \mathrm{\parallel }{A}^{\ast }\mathrm{\parallel }\mathrm{\parallel }A\mathrm{\parallel }$, and applying this also to $A^{\ast }$ gives $\mathrm{\parallel }{A}^{\ast }\mathrm{\parallel }=\mathrm{\parallel }A\mathrm{\parallel }$, including $A=0$.

For every vector $v$, $⟨(\mathrm{\parallel }A{\mathrm{\parallel }}^{2}I-A^{\ast }A)v,v⟩=\mathrm{\parallel }A{\mathrm{\parallel }}^2\mathrm{\parallel }v{\mathrm{\parallel }}^2-\mathrm{\parallel }Av{\mathrm{\parallel }}^2\ge 0$. This Hermitian matrix is positive semidefinite, so [F1] gives its positive square root. The same spectral calculus shows that matrix positivity is equivalent to being $B^{\ast }B$, agreeing with the algebraic positivity used in [F4].

Let $\omega$ be complex linear and positive, and define ${\rho }_{ij}:=\omega (E_{ji})$. Finite expansion gives $\omega (A)={\sum }_{i,j}{A}_{ij}\omega (E_{ij})=\mathrm{Tr}(\rho A)$. For a column vector $z$, take $B=e_1{z}^{\ast }$; then $B^{\ast }B=z{z}^{\ast }$, so $z^{\ast }\rho z=\mathrm{Tr}(\rho z{z}^{\ast })=\omega (B^{\ast }B)\ge 0$ is real. Substitution of $z=e_i$, $e_i+e_j$, and $e_i+i{e}_j$ shows that the diagonal is real and ${\rho }_{ji}=\overline{{\rho }_{ij}}$; hence $\rho$ is Hermitian and its nonnegative quadratic forms make it positive semidefinite. Taking $A=I$ gives $\mathrm{Tr}\rho =\omega (I)$. The formula on the matrix units forces every entry of any representing matrix, proving uniqueness.

Conversely let $\rho \ge 0$ and diagonalize it by [F1] as $\rho ={\sum }_j{p}_j\mathrm{\mid }{u}_j⟩⟨u_j\mathrm{\mid }$, $p_j\ge 0$. Then ${\omega }_{\rho }(A)={\sum }_j{p}_j⟨A{u}_j,u_j⟩$ is complex linear and ${\omega }_{\rho }(B^{\ast }B)={\sum }_j{p}_j\mathrm{\parallel }B{u}_j{\mathrm{\parallel }}^2\ge 0$. For every complex $A$, vector Cauchy–Schwarz gives $\mathrm{\mid }{\omega }_{\rho }(A)\mathrm{\mid }\le {\sum }_j{p}_j\mathrm{\parallel }A{u}_j\mathrm{\parallel }\le ({\sum }_j{p}_j)\mathrm{\parallel }A\mathrm{\parallel }$. Thus its functional norm is at most $\mathrm{Tr}\rho$, and evaluating at the norm-one identity gives the reverse inequality. This proves the norm formula, including $\rho =0$. The preceding reconstruction and this converse give the full bijection; imposing norm one is exactly imposing trace one by [F4], with no faithfulness requirement.

By [F1], choose an orthonormal eigenbasis of $K$ with real eigenvalues ${\lambda }_1,\dots ,{\lambda }_d$. Then $Z_{\beta }={\sum }_j{e}^{-\beta {\lambda }_j}$ is a finite sum of strictly positive real numbers, so $Z_{\beta }>0$. The operator ${\rho }_{\beta }$ is diagonal in this basis with positive entries $e^{-\beta {\lambda }_j}\mathrm{/}{Z}_{\beta }$, whose sum is one; [F2] therefore makes it a density operator.

Keep the arbitrary ordered orthonormal basis $(e_j)$ and its given reference copy $(e_j^{\mathrm{\prime }})$ from the Statement. Put $T=e^{-\beta K\mathrm{/}2}$, so the defining vector is ${\mathrm{\Psi }}_{\beta }=Z_{\beta }^{-1\mathrm{/}2}{\sum }_{j}T{e}_j\otimes e_j^{\mathrm{\prime }}$. Orthogonality of the reference vectors gives $\mathrm{\parallel }{\mathrm{\Psi }}_{\beta }{\mathrm{\parallel }}^2=Z_{\beta }^{-1}{\sum }_j\mathrm{\parallel }T{e}_j{\mathrm{\parallel }}^2=Z_{\beta }^{-1}\mathrm{Tr}(T^{\ast }T)=1$, because [F1] gives $T^{\ast }=T$ and $T^{\ast }T=e^{-\beta K}$, and step 1.1 gives its trace $Z_{\beta }$. This uses the given basis, without requiring it to diagonalize $K$.

In those same given bases, ${\mathrm{\Psi }}_{\beta }$ has coefficient $T_{j\mathrm{\ell }}\mathrm{/}\sqrt{Z_{\beta }}$ at $e_j\otimes e_{\mathrm{\ell }}^{\mathrm{\prime }}$. Thus [F3] gives the reduced matrix entry $Z_{\beta }^{-1}{\sum }_{\mathrm{\ell }}{T}_{j\mathrm{\ell }}\overline{T_{k\mathrm{\ell }}}=(T{T}^{\ast }{)}_{jk}\mathrm{/}{Z}_{\beta }=({\rho }_{\beta }{)}_{jk}$, since $T{T}^{\ast }=e^{-\beta K}$ by [F1]. Equivalently, the reduction is $Z_{\beta }^{-1}{\sum }_{\mathrm{\ell }}\mathrm{\mid }T{e}_{\mathrm{\ell }}⟩⟨T{e}_{\mathrm{\ell }}\mathrm{\mid }=T{T}^{\ast }\mathrm{/}{Z}_{\beta }$. The intrinsic characterization [F4] identifies this coordinate result with the finite partial trace, so the system marginal equals ${\rho }_{\beta }$ for every allowed original basis and reference copy. No basis independence of the purification vector itself is asserted.

On a compact interval of $s$, the matrix norms of $M(s)$ and $M^{\mathrm{\prime }}(s)$ are bounded. The exponential power series and its differentiated series converge uniformly there, since the derivative of $M^n$ is ${\sum }_{j=0}^{n-1}{M}^j{M}^{\mathrm{\prime }}{M}^{n-1-j}$, whose norm is at most $n\mathrm{\parallel }M{\mathrm{\parallel }}^{n-1}\mathrm{\parallel }{M}^{\mathrm{\prime }}\mathrm{\parallel }$. Expanding both exponentials in the integral, integrating term by term, and using ${\int }_0^1(1-u{)}^j{u}^{k}du=j!k!\mathrm{/}(j+k+1)!$ (obtained by integration by parts) gives the same differentiated series. This proves the Duhamel formula.

For finite matrices $C,D$, $\mathrm{Tr}(CD)={\sum }_{i,j}{C}_{ij}{D}_{ji}=\mathrm{Tr}(DC)$ by interchanging finite indices. Take the trace in step 1.1. This cyclicity makes every integrand trace equal to $\mathrm{Tr}(e^M{M}^{\mathrm{\prime }})$. Since $Z=\mathrm{Tr}{e}^M>0$, the scalar chain rule gives $(\log Z{)}^{\mathrm{\prime }}=\mathrm{Tr}(e^M{M}^{\mathrm{\prime }})\mathrm{/}Z=\mathrm{Tr}(\rho M^{\mathrm{\prime }})$. Setting $M^{\mathrm{\prime }}=X_a$ proves the first control derivative.

Differentiate $\mathrm{Tr}(\rho X_b)$ in the $X_a$ direction using step 1.1 and the quotient rule. The result is ${\int }_0^1\mathrm{Tr}({\rho }^{1-u}{X}_a{\rho }^u{X}_b)du-\mathrm{Tr}(\rho X_a)\mathrm{Tr}(\rho X_b)\mathrm{.}$ Because ${\rho }^{1-u}{\rho }^u=\rho$, cyclicity shows that replacing each $X$ by its centered version $\delta X$ changes the integral by exactly the subtracted product, giving the stated Hessian.

Diagonalize the positive definite $\rho$ with eigenvalues $p_i>0$. For Hermitian $X$, the quadratic form equals ${\sum }_{i,j}\mathrm{\mid }\delta X_{ij}{\mathrm{\mid }}^{2}L(p_i,p_j)$, where $L(a,b)={\int }_0^1{a}^{1-u}{b}^{u}du>0$. Thus it is real and nonnegative. For $a\mathrm{\ne }b$, direct integration gives $L(a,b)=(a-b)\mathrm{/}(\log a-\log b)$, and $L(a,a)=a$. Concavity of $\log$ gives $(1-u)\log a+u\log b\le \log ((1-u)a+ub)$, strictly when $a\mathrm{\ne }b$ and $0<u<1$. Exponentiating and integrating yields $L(a,b)\le (a+b)\mathrm{/}2$, strictly when $a\mathrm{\ne }b$. Substituting the matrix entries yields the upper bound by $\mathrm{Tr}(\rho (\delta X{)}^2)={\mathrm{Var}}_{\rho }(X)$. If $X$ commutes with $\rho$, its entries between distinct eigenspaces vanish, so equality holds. Conversely, if $X$ does not commute with $\rho$, some $\delta X_{ij}\mathrm{\ne }0$ has $p_i\mathrm{\ne }{p}_j$; its term is strictly below the arithmetic-mean bound while every other term is at most that bound, so the total inequality is strict. Thus equality holds exactly in the commuting case.

For $H(h)=H_0-hB$, put $M=-\beta H(h)$. Then $M^{\mathrm{\prime }}=\beta B$, so step 3.1 gives ${\mathrm{\partial }}_h\mathrm{Tr}({\rho }_{h}B)=\beta {\mathrm{Cov}}_{KM,{\rho }_h}(B,B)\ge 0$. Since $E\mapsto e^{-\beta E}\mathrm{/}Z(h)$ is injective on the real eigenvalues of $H(h)$, $H(h)$ and ${\rho }_h$ have the same eigenspaces; hence $[H(h),B]=0$ if and only if $[{\rho }_h,B]=0$. Step 4.1 therefore gives equality with $\beta {\mathrm{Var}}_{{\rho }_h}(B)$ exactly in the commuting case, and strict inequality otherwise. No ordinary-variance identity is inferred without this condition.

Since $\rho$ is the trace-one positive density of [F1], [F2] gives $\rho ={\sum }_j{p}_j\mathrm{\mid }{u}_j⟩⟨u_j\mathrm{\mid }$, $p_j>0$, with ${\sum }_j{p}_j=\mathrm{Tr}\rho =1$, and trace-norm convergence. For bounded $B$, [F3] makes $\rho B$ trace class. Applying the basis-independent trace to finite partial sums and passing to the trace-norm limit gives $\mathrm{Tr}(\rho B)={\sum }_j{p}_j⟨B{u}_j,u_j⟩$; its absolute sum is at most $\mathrm{\parallel }B\mathrm{\parallel }{\sum }_j{p}_j=\mathrm{\parallel }B\mathrm{\parallel }$. The bounded cyclicity statement in [F6] gives $\mathrm{Tr}(B\rho )=\mathrm{Tr}(\rho B)$.

The scalar mean is defined on the first absolute-moment domain stated above. If the second moment is finite, then $\mathrm{\mid }\lambda \mathrm{\mid }\le (1+{\lambda }^2)\mathrm{/}2$ and ${\mu }_{\rho }$ is a probability measure, so the first absolute moment is finite as well. Expanding $(\lambda -m_{\rho }(A){)}^2$ and using $\int 1d{\mu }_{\rho }=1$ gives the displayed variance identity. For a vector $u$, the PVM domain formula [F4] is the operator domain, while the first absolute moment defines the larger form domain $D(\mathrm{\mid }A{\mathrm{\mid }}^{1\mathrm{/}2})=\{u:\int \mathrm{\mid }\lambda \mathrm{\mid }d{E}_u(\lambda )<\mathrm{\infty }\}$. These definitions use scalar integrals and make no unbounded cyclic permutation.

Since $0<p_j\le 1$, every term $-p_j\log p_j$ is nonnegative, so the entropy sum is well defined in $[0,\mathrm{\infty }]$; multiplying by the given $k_B>0$ preserves that range. It can diverge: let $c^{-1}={\sum }_{n\ge 2}(n(\log n{)}^2{)}^{-1}$ and $p_n=c\mathrm{/}(n(\log n{)}^2)$. The normalizing series converges because the block $2^k\le n<2^{k+1}$ contributes at most $1\mathrm{/}(k\log 2{)}^2$, and ${\sum }_k{k}^{-2}$ converges by grouping dyadic blocks. For sufficiently large $n$, $-p_n\log p_n\ge c\mathrm{/}(2n\log n)$. Each dyadic block contributes at least $1\mathrm{/}(4(k+1)\log 2)$ to ${\sum }_{n}1\mathrm{/}(n\log n)$, so that entropy series diverges.

For each Borel set $C\subseteq \mathbb{R}$, [F5] and step 1.1 give ${\mu }_{\rho }(C)={\sum }_j{p}_j\mathrm{\parallel }{E}_A(C)u_j{\mathrm{\parallel }}^2\ge 0$ and ${\mu }_{\rho }(\mathbb{R})={\sum }_j{p}_j=1$. For pairwise disjoint $C_k$, the PVM property gives $\mathrm{\parallel }{E}_A({\bigcup }_k{C}_k)u_j{\mathrm{\parallel }}^2={\sum }_k\mathrm{\parallel }{E}_A(C_k)u_j{\mathrm{\parallel }}^2$, since the projected vectors are orthogonal. The nonnegative double series may be summed in either order; therefore ${\mu }_{\rho }({\bigcup }_k{C}_k)={\sum }_k{\mu }_{\rho }(C_k)$. This proves that ${\mu }_{\rho }$ is a probability measure and agrees with the canonical scalar-law interface [F7].

For densities ${\rho }_k$ and weights $t_k\ge 0$, ${\sum }_k{t}_k=1$, the partial sums ${\sum }_{k\le N}{t}_k{\rho }_k$ are Cauchy in trace norm because each $\mathrm{\parallel }{\rho }_k{\mathrm{\parallel }}_1=\mathrm{Tr}{\rho }_k=1$ and the tail norm is at most ${\sum }_{k>N}{t}_k$. Completeness in [F3] gives a trace-class limit. Its trace is one by [F2] and trace continuity; its quadratic forms are nonnegative as limits of nonnegative quadratic forms. Thus the countable mixture is again a normal density state and the convergence is in trace norm.