Exact Two-Qubit Cartan Decomposition

By [F5] and [F7], $\det (U^{\ast })=\det ({\bar{U}}^{\mathsf{T}})=\det (\bar{U})=\overline{\det U}$. Then [F6] and $U^{\ast }U=I$ give $\mathrm{\mid }\det U{\mathrm{\mid }}^2=\det (U^{\ast }U)=1$. Choose $\delta$ with $\det U=e^{4i\delta }$ and put $W=e^{-i\delta }U$; applying the permutation formula to a scalar multiple of a $4\times 4$ matrix gives $\det W=e^{-4i\delta }\det U=1$, so $W\in SU(4)$. Use the magic basis $e_1=(\mathrm{\mid }00⟩+\mathrm{\mid }11⟩)\mathrm{/}\sqrt{2},e_2=i(\mathrm{\mid }00⟩-\mathrm{\mid }11⟩)\mathrm{/}\sqrt{2},e_3=i(\mathrm{\mid }01⟩+\mathrm{\mid }10⟩)\mathrm{/}\sqrt{2},e_4=(\mathrm{\mid }01⟩-\mathrm{\mid }10⟩)\mathrm{/}\sqrt{2},$ and let $Q$ have these columns. Direct inner products show each vector has norm one and distinct vectors are orthogonal, so $Q$ is unitary. Then $\widehat{W}=Q^{\ast }WQ$ is unitary with determinant one. The matrix $S={\widehat{W}}^T\widehat{W}$ is symmetric; since $S^{\ast }=\bar{S}$ and $\widehat{W}$ is unitary, $S^{\ast }S=I$, so $S$ is unitary.

Direct calculation from the definition of $S$ gives $S^T=S$ and $S^{\ast }S=I$. Write $S=P+iR$ with $P,R$ real symmetric. From $S\bar{S}=I$ we obtain $P^2+R^2=I$ and $PR=RP$. Apply [F2] first to $P$. If $Pv=\lambda v$, then $P(Rv)=R(Pv)=\lambda Rv$, so every eigenspace $E_{\lambda }$ of $P$ is invariant under $R$. The restriction $R{\mathrm{\mid }}_{E_{\lambda }}$ is self-adjoint for the inherited inner product, since $⟨Ru,v⟩=⟨u,Rv⟩$. Apply [F2] separately on each such finite-dimensional eigenspace; the union of these orthonormal eigenbases is an orthonormal common eigenbasis because distinct $P$ eigenspaces are orthogonal. Its columns form a real orthogonal $O$ simultaneously diagonalizing $P,R$, so $O^{T}SO=\mathrm{diag}({\zeta }_j)$ with $\mathrm{\mid }{\zeta }_j\mathrm{\mid }=1$. If $\det O=-1$, change the sign of one column; the diagonalization is unchanged and now $O\in SO(4)$. Choose unit square roots $d_j$ of each ${\zeta }_j$. By [F6]–[F7] and $\det \widehat{W}=1$, $\det S=\det ({\widehat{W}}^T)\det (\widehat{W})=1$, so ${\prod }_j{d}_j$ is either $1$ or $-1$; if it is $-1$, change the sign of one root. Then $D_0=\mathrm{diag}(d_j)$ and $K=O{D}_0{O}^T$ satisfy $K^T=K$, $K^2=S$, and $\det K=1$.

Put $R_0=\widehat{W}{K}^{-1}$. It is unitary and $R_0^T{R}_0=K^{-1}{\widehat{W}}^T\widehat{W}{K}^{-1}=K^{-1}S{K}^{-1}=I\mathrm{.}$ Thus $R_0^T=R_0^{\ast }$, which entrywise says $R_0$ is real. Also $\det R_0=1$, so $R_0\in SO(4)$, and $\widehat{W}=R_{0}O{D}_0{O}^T$.

Direct multiplication gives $X^2=Y^2=Z^2=I_2$, and the matrices in [F4] are traceless and Hermitian. In the magic basis let $J_{ij}$ be the real skew-symmetric generator with $J_{ij}{e}_i=e_j$, $J_{ij}{e}_j=-e_i$, and zero on the other basis vectors. Direct action on the four displayed vectors gives the six local generators $iX\otimes I\mapsto J_{13}+J_{24},iY\otimes I\mapsto J_{14}-J_{23},iZ\otimes I\mapsto J_{12}-J_{34},$ $I\otimes iX\mapsto J_{13}-J_{24},I\otimes iY\mapsto -J_{14}-J_{23},I\otimes iZ\mapsto J_{12}+J_{34}\mathrm{.}$ These are a basis of the six-dimensional real space $\mathfrak{s}\mathfrak{o}(4)$ of skew-symmetric matrices. Thus every $J_{ij}$ is the magic-basis image of some real linear combination $A\otimes I+I\otimes B$ of local generators, where $A,B$ are traceless skew-Hermitian $2\times 2$ matrices. The two summands commute, so its exponential is $e^A\otimes e^B$. A traceless skew-Hermitian $2\times 2$ matrix has eigenvalues $ir,-ir$ for some real $r$, so $e^A,e^B\in SU(2)$. Hence each plane rotation $e^{t{J}_{ij}}$ is a local image. Conversely, [F1] expresses every local $SU(2)$ factor as a product of $R_z$ and $R_y$ rotations; their magic-basis images are exponentials of the displayed real skew generators and hence real orthogonal. Finally, Givens elimination writes every $R\in SO(4)$ as a finite product of plane rotations: for a unit first column, successively rotate coordinate pairs to zero its last three entries and make the first entry $1$; orthogonality then leaves a lower-right $SO(3)$ block, to which the same procedure applies recursively. Therefore the local image is all of $SO(4)$, and in particular $R_{0}O$ and $O^T$ are local images.

In the same basis, direct action of the Pauli products gives the diagonal sign lists $(+,-,+,-)$, $(-,+,+,-)$, and $(+,+,-,-)$ for $X\otimes X$, $Y\otimes Y$, and $Z\otimes Z$, respectively. These mutually orthogonal trace-zero lists form a basis of the three-dimensional real trace-zero diagonal subspace. Since $D_0$ is diagonal unitary with determinant one, choose arguments for its four diagonal entries; their sum is a multiple of $2\pi$, so adjusting one argument by that multiple gives phases summing exactly to zero. Therefore those phases are a real linear combination of the three sign lists, and $D_0=\exp (i(aX\otimes X+bY\otimes Y+cZ\otimes Z))$ in the magic basis. Conjugating $\widehat{W}=R_{0}O{D}_0{O}^T$ back by $Q$, using the local representations from step 4.1, and restoring $e^{i\delta }$ gives the stated exact factorization.