Finite two-qubit density matrices and separability

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.