Pauli matrices and the two-qubit exchange operator

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.