Finite Pauli dimer spectrum and exact separability threshold

Use the usual complex field $\mathbb{C}$; its scalar metric is $d_{\mathbb{C}}(z,w)=\mathrm{\mid }z-w\mathrm{\mid }$ from The Euclidean metric, convergence, Cauchy sequences, and continuity on the complex plane. On ${\mathbb{C}}^2$, use the standard basis $e_0=(1,0)$, $e_1=(0,1)$ and inner product $⟨x,y⟩=x_0\overline{y_0}+x_1\overline{y_1}$, linear in its first variable. On ${\mathbb{R}}^3$, define $a\cdot b:=a_x{b}_x+a_y{b}_y+a_z{b}_z$, $\mathrm{\mid }a\mathrm{\mid }:=\sqrt{a\cdot a}$ using the unique nonnegative square root from Existence and uniqueness of $n$-th roots: a unique $a^{1\mathrm{/}n}\ge 0$ with $(a^{1\mathrm{/}n}{)}^n=a$, and the standard unit vectors ${\mathbf{e}}_x=(1,0,0)$, ${\mathbf{e}}_y=(0,1,0)$, and ${\mathbf{e}}_z=(0,0,1)$.

For a finite complex matrix $A=(a_{jk})$, define its adjoint by $A^{\ast }:=(\overline{a_{kj}}{)}_{j,k}={\bar{A}}^T$, where the bar is entrywise complex conjugation from Real and imaginary parts, complex conjugation, and modulus and $T$ is transpose. The matrix is Hermitian exactly when $A=A^{\ast }$. For vectors $u,v$ in a finite-dimensional complex inner-product space $V$ with the stated linear-first convention, define the rank-one operator $\mathrm{\mid }u⟩⟨v\mathrm{\mid }:V\to V$ by

$\mathrm{\mid }u⟩⟨v\mathrm{\mid }(x):=⟨x,v⟩u\mathrm{.}$

This action is linear in $x$ under the linear-first convention.

Use the ordered tensor-product basis $(e_0\otimes e_0,e_0\otimes e_1,e_1\otimes e_0,e_1\otimes e_1)$ of ${\mathbb{C}}^2\otimes {\mathbb{C}}^2$. Its inner product is the linear-first product inner product, specified on elementary tensors by $⟨u\otimes v,u^{\mathrm{\prime }}\otimes v^{\mathrm{\prime }}⟩=⟨u,u^{\mathrm{\prime }}⟩⟨v,v^{\mathrm{\prime }}⟩$ and extended sesquilinearly. For two $2\times 2$ matrices $A,B$, their tensor product acts by $(A\otimes B)(u\otimes v)=Au\otimes Bv$ and is extended linearly.

On elementary tensors this definition gives $(A_1\otimes B_1)(A_2\otimes B_2)=(A_1{A}_2)\otimes (B_1{B}_2)$. The product inner product also gives $(A\otimes B{)}^{\ast }=A^{\ast }\otimes B^{\ast }$: both sides have the adjoint pairing on elementary tensors, and those tensors span the product space.

Write $I_2:=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$ for the identity on ${\mathbb{C}}^2$, use $I:=I_2$ when its domain is clear, and set $I_4:=I_2\otimes I_2$. Define the everywhere-defined matrices

${\sigma }_x=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right),{\sigma }_y=\left(\begin{array}{cc}0& -i\\ i& 0\end{array}\right),{\sigma }_z=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)\mathrm{.}$

Define the dimensionless exchange matrix

$C:={\sigma }_x\otimes {\sigma }_x+{\sigma }_y\otimes {\sigma }_y+{\sigma }_z\otimes {\sigma }_z\mathrm{.}$

Direct conjugate-transpose and multiplication of the displayed matrices give ${\sigma }_x^{\ast }={\sigma }_x$, ${\sigma }_y^{\ast }={\sigma }_y$, ${\sigma }_z^{\ast }={\sigma }_z$, and ${\sigma }_x^2={\sigma }_y^2={\sigma }_z^2=I_2$. In the ordered tensor-product basis,

$C=\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& -1& 2& 0\\ 0& 2& -1& 0\\ 0& 0& 0& 1\end{array}\right),$

so $C^{\ast }=C$ as well. These matrices have their full finite-dimensional domains. For a finite matrix $A$, $\mathrm{tr}(A)$ denotes the sum of its diagonal entries in the displayed basis.

Use the single-qubit space and ordered two-qubit basis from Pauli matrices and the two-qubit exchange operator. A one-qubit density matrix is a Hermitian $2\times 2$ matrix $\rho$ such that $⟨\rho x,x⟩\ge 0$ for every $x\in {\mathbb{C}}^2$ and $\mathrm{tr}(\rho )=1$. A two-qubit density matrix is defined in the same way on ${\mathbb{C}}^2\otimes {\mathbb{C}}^2$, using the four-dimensional trace.

The tensor product of two one-qubit density matrices is again a two-qubit density matrix. To verify positivity, write any positive semidefinite Hermitian matrix as $R=\left(\begin{array}{cc}a& z\\ \bar{z}& b\end{array}\right)$; testing the quadratic form on the two coordinate vectors gives $a,b\ge 0$. If $a>0$, completing the square gives

$⟨R(x,y),(x,y)⟩=a{\mid x+\frac{z}{a}y\mid }^2+(b-\frac{\mathrm{\mid }z{\mathrm{\mid }}^2}{a})\mathrm{\mid }y{\mathrm{\mid }}^2,$

so positivity implies $b-\mathrm{\mid }z{\mathrm{\mid }}^2\mathrm{/}a\ge 0$ and

$R=S^{\ast }S,S=\left(\begin{array}{cc}\sqrt{a}& z\mathrm{/}\sqrt{a}\\ 0& \sqrt{b-\mathrm{\mid }z{\mathrm{\mid }}^2\mathrm{/}a}\end{array}\right)\mathrm{.}$

If $a=0$, positivity forces $z=0$: if $z\mathrm{\ne }0$, choose $y=1$ and $x=-tz$ with real $t>b\mathrm{/}(2\mathrm{\mid }z{\mathrm{\mid }}^2)$, giving quadratic form $b-2t\mathrm{\mid }z{\mathrm{\mid }}^2<0$. Then $R=S^{\ast }S$ for $S=\mathrm{diag}(0,\sqrt{b})$. The nonnegative square roots exist by Existence and uniqueness of $n$-th roots: a unique $a^{1\mathrm{/}n}\ge 0$ with $(a^{1\mathrm{/}n}{)}^n=a$ through the preceding matrix conventions. Thus for one-qubit densities ${\rho }_A=S_A^{\ast }{S}_A$ and ${\rho }_B=S_B^{\ast }{S}_B$, the tensor-product adjoint identity from Pauli matrices and the two-qubit exchange operator gives

${\rho }_A\otimes {\rho }_B=(S_A\otimes S_B{)}^{\ast }(S_A\otimes S_B),$

which is Hermitian and positive semidefinite. In the displayed product basis, $\mathrm{tr}(X\otimes Y)={\sum }_{i,j=0}^1{X}_{ii}{Y}_{jj}=\mathrm{tr}(X)\mathrm{tr}(Y)$ for $2\times 2$ matrices $X,Y$. Therefore $\mathrm{tr}({\rho }_A\otimes {\rho }_B)=1$.

A product density matrix is ${\rho }_A\otimes {\rho }_B$ for one-qubit density matrices ${\rho }_A,{\rho }_B$. A two-qubit density matrix is separable if it is a finite convex combination ${\sum }_{j=1}^m{t}_j{\rho }_{A,j}\otimes {\rho }_{B,j}$, where $t_j\ge 0$ and ${\sum }_{j=1}^m{t}_j=1$; these convex combinations are again density matrices by linearity of the trace and quadratic form. A two-qubit density matrix that is not separable is entangled. This finite-convex-combination convention is the one used by the local threshold proof.

For $C$ from Pauli matrices and the two-qubit exchange operator, “$v$ is an eigenvector with eigenvalue $\lambda$” has the exact meaning from Eigenvalues, eigenvectors, eigenspaces $E_{\lambda }(T)=\ker (T-\lambda I)$, and the spectrum ${\sigma }_F(T)$ of an endomorphism: $v\mathrm{\ne }0$ and $Cv=\lambda v$. The singlet vector $s=(e_0\otimes e_1-e_1\otimes e_0)\mathrm{/}\sqrt{2}$ has eigenvalue $-3$, and the three orthonormal triplet vectors $e_0\otimes e_0$, $(e_0\otimes e_1+e_1\otimes e_0)\mathrm{/}\sqrt{2}$, and $e_1\otimes e_1$ have eigenvalue $1$. Here $\sqrt{2}$ is the unique positive square root supplied by Existence and uniqueness of $n$-th roots: a unique $a^{1\mathrm{/}n}\ge 0$ with $(a^{1\mathrm{/}n}{)}^n=a$. These four vectors are an orthonormal basis of ${\mathbb{C}}^2\otimes {\mathbb{C}}^2$.

Every Hermitian trace-one $2\times 2$ matrix $\rho$ has a unique vector $a=(a_x,a_y,a_z)\in {\mathbb{R}}^3$ such that

$\rho =\frac{1}{2}(I+a_x{\sigma }_x+a_y{\sigma }_y+a_z{\sigma }_z)\mathrm{.}$

Conversely, for every $a\in {\mathbb{R}}^3$ this formula defines a Hermitian trace-one matrix, and that matrix is positive semidefinite exactly when $\mathrm{\mid }a\mathrm{\mid }\le 1$, where $\mathrm{\mid }a\mathrm{\mid }=\sqrt{a\cdot a}$ is the Euclidean norm defined in Pauli matrices and the two-qubit exchange operator. Thus the one-qubit density matrices of Finite two-qubit density matrices and separability are exactly the displayed matrices with $\mathrm{\mid }a\mathrm{\mid }\le 1$. Their coordinates are $a_j=\mathrm{tr}(\rho {\sigma }_j)$.

For every separable two-qubit density matrix $\rho$ and exchange operator $C$ from Pauli matrices and the two-qubit exchange operator, $\mathrm{tr}(\rho C)\ge -1.$

For $n\in \{\pm {\mathbf{e}}_x,\pm {\mathbf{e}}_y,\pm {\mathbf{e}}_z\}$, with the standard vectors from Pauli matrices and the two-qubit exchange operator, put ${\rho }_n=(I+n_x{\sigma }_x+n_y{\sigma }_y+n_z{\sigma }_z)\mathrm{/}2$. The matrix ${\rho }_{\ast }:=\frac{1}{6}{\sum }_{n\in \{\pm e_x,\pm e_y,\pm e_z\}}{\rho }_n\otimes {\rho }_{-n}=\frac{1}{4}(I_4-\frac{1}{3}C)$ is a separable two-qubit density matrix.

Use $C$ from Pauli matrices and the two-qubit exchange operator, with singlet projection $P_s=\mathrm{\mid }s⟩⟨s\mathrm{\mid }$ and triplet projection $P_t=I_4-P_s$, where $s=(e_0\otimes e_1-e_1\otimes e_0)\mathrm{/}\sqrt{2}$. For each dimensionless $x\ge 0$, define the finite two-qubit density matrix ${\rho }_x:=\frac{e^{3x}{P}_s+e^{-x}{P}_t}{e^{3x}+3e^{-x}}\mathrm{.}$ Then ${\rho }_x$ is separable exactly when $x\le (\log 3)\mathrm{/}4$, and it is entangled exactly when $x>(\log 3)\mathrm{/}4$. In particular, the equality state at $x=(\log 3)\mathrm{/}4$ is separable.

Let $e_0,e_1$ and ${\sigma }_x,{\sigma }_y,{\sigma }_z,C$ be as in Pauli matrices and the two-qubit exchange operator, with ${\sigma }_z{e}_0=e_0$ and ${\sigma }_z{e}_1=-e_1$. By Spectrum of the two-qubit Pauli exchange matrix, the normalized singlet $s=(e_0\otimes e_1-e_1\otimes e_0)\mathrm{/}\sqrt{2}$ has $C$-eigenvalue $-3$, and the three orthonormal triplets have eigenvalue $1$. Adopt the two-spin Hamiltonian as a finite matrix on ${\mathbb{C}}^2\otimes {\mathbb{C}}^2$, with full domain $D(H_{J,h})={\mathbb{C}}^2\otimes {\mathbb{C}}^2$: $H_{J,h}:=JC-h({\sigma }_z\otimes I+I\otimes {\sigma }_z),$ where $J,h\in \mathbb{R}$ are energies in joules. For inverse temperature $\beta >0$ in J${}^{-1}$, use the canonical Gibbs preparation only after normalizing its finite trace. Its singlet energy is $-3J$; its triplet energies are $J-2h$, $J$, and $J+2h$. Here $\cosh y:=(e^y+e^{-y})\mathrm{/}2$. $Z(\beta ,J,h)=e^{3\beta J}+e^{-\beta J}(1+2\cosh (2\beta h))>0.$ At $h=0$ and $J>0$, the Gibbs density matrix is entangled exactly when $\beta J>(\log 3)\mathrm{/}4$, and is separable at and below equality. This exact threshold is a finite-state quantum-model conclusion. For every fixed $J,h$, $Z$ and the free energy $-{\beta }^{-1}\log Z$ are real analytic for $\beta >0$; the entanglement threshold is not a finite-temperature thermodynamic singularity. At zero field the entropy tends to $0$ as $\beta \to \mathrm{\infty }$ for $J>0$, to $k_B\log 3$ for $J<0$, and to $k_B\log 4$ for $J=0$. For $J>0$, a nonzero field makes the singlet and the field-aligned triplet cross at $\mathrm{\mid }h\mathrm{\mid }=2J$ at zero temperature; this ground-space crossing does not make $Z$ singular at finite positive temperature.

Periodic commuting Ising model. For $N\ge 2$, let ${\mathcal{H}}_N=({\mathbb{C}}^2{)}^{\otimes N}$ with its product inner product, and let $Z_i$ act as ${\sigma }_z$ on tensor factor $i$ and as identity on the others; each $Z_i$ acts on all of ${\mathcal{H}}_N$. With periodic indexing $Z_{N+1}=Z_1$, count exactly the $N$ indexed bonds, so $N=2$ has two bonds. For real energies $J,h$ and $\beta >0$ in J${}^{-1}$, adopt the finite matrix $H_N^{\mathrm{q}}=-J{\sum }_{i=1}^N{Z}_i{Z}_{i+1}-h{\sum }_{i=1}^N{Z}_i,D(H_N^{\mathrm{q}})={\mathcal{H}}_N\mathrm{.}$ For each site set $\mathrm{\mid }+1⟩:=e_0$ and $\mathrm{\mid }-1⟩:=e_1$, so ${\sigma }_z\mathrm{\mid }{s}_i⟩=s_i\mathrm{\mid }{s}_i⟩$; define $\mathrm{\mid }{s}_1,\dots ,s_N⟩:=\mathrm{\mid }{s}_1⟩\otimes \cdots \otimes \mathrm{\mid }{s}_N⟩$. Its energy equals $H_N^{\mathrm{c}\mathrm{l}}(s)=-J{\sum }_i{s}_i{s}_{i+1}-h{\sum }_i{s}_i$. Thus the quantum Gibbs density is diagonal in this basis, and its partition function and diagonal probabilities agree exactly with the specified classical finite ring. In particular, when $J\ge 0$ and $h=0$, its ${\sigma }_z$ correlations are those of Finite periodic Ising-ring correlations and exact decay. Use $\sinh x=(e^x-e^{-x})\mathrm{/}2$, $\cosh x=(e^x+e^{-x})\mathrm{/}2$, and $\tanh x=\sinh x\mathrm{/}\cosh x$ from The six hyperbolic functions and their natural domains, where $\cosh x>0$ for every real $x$, and put ${\lambda }_{\pm }=e^{\beta J}\cosh (\beta h)\pm \sqrt{e^{2\beta J}{\sinh }^2(\beta h)+e^{-2\beta J}}\mathrm{.}$ Then $N^{-1}\log Z_N^{\mathrm{q}}\to \log {\lambda }_{+}$, with every fixed finite order of derivatives converging locally uniformly on compact subsets of $(0,\mathrm{\infty })\times {\mathbb{R}}^2$. This is the exact commuting one-dimensional model, not a theorem for noncommuting interactions.

Global three-sector countermodel. For $N\ge 1$ and $ϵ>0$ joules, let ${\mathcal{K}}_N=\mathbb{C}\oplus \mathbb{C}\oplus ({\mathbb{C}}^2{)}^{\otimes 2N}$ with its orthogonal direct-sum inner product, and write $I_{4^N}$ for the identity on the final $4^N$-dimensional summand. Let $\mathrm{\mid }A⟩,\mathrm{\mid }B⟩$ be the standard unit vectors in the first and second summands, and let $\mathrm{\mid }s⟩$ range over the standard product basis in the final summand. Define the finite matrix $H_N^{\mathrm{g}}=0\oplus Nϵ\oplus 2NϵI_{4^N}$ on the full domain ${\mathcal{K}}_N$, so the basis energies are $E(A)=0$, $E(B)=Nϵ$, and $E(s)=2Nϵ$. For $0\le \mathrm{\Delta }<Nϵ$, define the spectral-shell basis subset $S_{N,\mathrm{\Delta }}:=\{v:\mathrm{\mid }E(v)-Nϵ\mathrm{\mid }\le \mathrm{\Delta }\}$ and its uniform state ${\rho }_{N,\mathrm{\Delta }}^{\mathrm{s}\mathrm{h}\mathrm{e}\mathrm{l}\mathrm{l}}:=\mathrm{\mid }{S}_{N,\mathrm{\Delta }}{\mathrm{\mid }}^{-1}{\sum }_{v\in S_{N,\mathrm{\Delta }}}\mathrm{\mid }v⟩⟨v\mathrm{\mid }$. For $\beta >0$ this finite quantum model has $Z_N^{\mathrm{g}}=1+e^{-\beta Nϵ}+4^N{e}^{-2\beta Nϵ}\mathrm{.}$ For every real $\beta$ as an algebraic finite-weight extension, the normalized exponential-tilt weight of the one-dimensional middle sector is at most $(2^{N+1}+1{)}^{-1}$; the shell state above assigns that sector probability $1$. The term “probability” here is the trace of the state against the middle-sector projection $\mathrm{\mid }B⟩⟨B\mathrm{\mid }$. Also $\frac{1}{N}\log Z_N^{\mathrm{g}}⟶\max \{0,\log 4-2\beta ϵ\},0\le \frac{1}{N}\log Z_N^{\mathrm{g}}-\max \{0,\log 4-2\beta ϵ\}\le \frac{\log 3}{N}\mathrm{.}$ The limiting function has a corner at ${\beta }_c=(\log 2)\mathrm{/}ϵ>0$ J${}^{-1}$. For $\beta >0$, define the canonical mean energy per member of the global model by

${\bar{E}}_N(\beta ):=\frac{Nϵe^{-\beta Nϵ}+2Nϵ4^N{e}^{-2\beta Nϵ}}{N{Z}_N^{\mathrm{g}}(\beta )}\mathrm{.}$

The limit $u(\beta ):={\lim }_{N\to \mathrm{\infty }}{\bar{E}}_N(\beta )$ exists for all $\beta >0$ and equals $2ϵ$ for $0<\beta <{\beta }_c$, $ϵ$ at $\beta ={\beta }_c$, and $0$ for $\beta >{\beta }_c$. Thus the one-sided limits of the thermodynamic-limit mean energy are $2ϵ$ as $\beta \uparrow {\beta }_c$ through positive values and $0$ as $\beta \downarrow {\beta }_c$ from above. The finite partition expressions are real analytic as functions of $\beta \in \mathbb{R}$ for fixed remaining model parameters; this is a regularity statement about the finite formulas, while the canonical preparation and mean-energy claims use only $\beta >0$. This global, size-dependent family is not a product of one fixed finite one-site alphabet and makes no claim about local or short-range matter.