Continuous-Gate Universality and Exact Synthesis

Taking $\lambda =1$ gives $U=1\cdot U$, so $U{\sim }_{\mathrm{p}\mathrm{h}\mathrm{a}\mathrm{s}\mathrm{e}}U$.

If $U=\lambda V$ and $\mathrm{\mid }\lambda \mathrm{\mid }=1$, then $V=\bar{\lambda }U$ because $\lambda \bar{\lambda }=1$; also $\mathrm{\mid }\bar{\lambda }\mathrm{\mid }=1$. Hence $U{\sim }_{\mathrm{p}\mathrm{h}\mathrm{a}\mathrm{s}\mathrm{e}}V$ implies $V{\sim }_{\mathrm{p}\mathrm{h}\mathrm{a}\mathrm{s}\mathrm{e}}U$.

If $U=\lambda V$ and $V=\mu W$, then $U=(\lambda \mu )W$ and $\mathrm{\mid }\lambda \mu \mathrm{\mid }=1$. Thus $U{\sim }_{\mathrm{p}\mathrm{h}\mathrm{a}\mathrm{s}\mathrm{e}}V$ and $V{\sim }_{\mathrm{p}\mathrm{h}\mathrm{a}\mathrm{s}\mathrm{e}}W$ imply $U{\sim }_{\mathrm{p}\mathrm{h}\mathrm{a}\mathrm{s}\mathrm{e}}W$. This proves transitivity; together the three properties establish an equivalence relation.

The $1\times 1$ matrices $[1]$ and $[-1]$ are phase-equivalent by the scalar $-1$, but they are not equal because their entries differ. Thus phase equivalence is distinct from exact equality.

The product $Q:=U_0{U}_1^{\ast }$ is unitary by [F3] and hence normal. By [F1], choose a unitary $V$ and diagonal unitary $D$ with $Q=V{D}^2{V}^{\ast }$: diagonalize $Q$, write each eigenvalue as $e^{i{t}_j}$ by [F2], and put $e^{i{t}_j\mathrm{/}2}$ on the corresponding diagonal of $D$.

Set $W:=D^{\ast }{V}^{\ast }{U}_0$. Then $VDW=U_0$. Also $V{D}^{\ast }W=V{D}^{-2}{V}^{\ast }{U}_0$. Taking adjoints in $U_0{U}_1^{\ast }=V{D}^2{V}^{\ast }$ gives $U_1{U}_0^{\ast }=V{D}^{-2}{V}^{\ast }$, so $V{D}^{-2}{V}^{\ast }{U}_0=U_1$. Thus the two diagonal blocks in the displayed factorization are respectively $U_0$ and $U_1$.

The factors $V,D,W$ are unitary, and the two block products agree with $U_0,U_1$; hence the asserted exact block factorization holds, including repeated eigenvalues. Only a finite orthonormal eigenbasis and finitely many square roots are chosen.

A skew-Hermitian matrix has purely imaginary diagonal entries, giving $N$ real parameters. For each $i<j$, its $(i,j)$ entry is an arbitrary complex number and the $(j,i)$ entry is then forced to be its negative conjugate, giving two real parameters. Thus $\mathfrak{u}(N)$ has real dimension $N+2\left(\genfrac{}{}{0px}{}{N}{2}\right)=N^2$.

If $(I-X)v=0$ for $X^{\ast }=-X$, then $Xv=v$. But $v^{\ast }Xv$ is purely imaginary because $X^{\ast }=-X$, while this equality gives $v^{\ast }Xv=v^{\ast }v>0$ for nonzero $v$. Hence $I-X$ is invertible. Direct multiplication, using that $I-X$ and $I+X$ commute, shows $(I+X)(I-X{)}^{-1}$ is unitary.

On the open set of unitary matrices $Q$ near $I$ where $Q+I$ is invertible, the inverse formula is $X=(Q-I)(Q+I{)}^{-1}$. The factors are polynomials in $Q$, so they commute; using $Q^{\ast }=Q^{-1}$ gives $X^{\ast }=-X$. The forward and inverse formulas are smooth rational matrix maps, and therefore define a chart of dimension $N^2$ near $I$.

Left translation by any fixed $Q_0\in U(N)$ sends a Cayley neighborhood of $I$ to a chart neighborhood of $Q_0$, so these translates cover $U(N)$. The group is closed, since $Q^{\ast }Q=I$ is a closed matrix equation, and bounded, since every column has norm one. Identifying $M_N(\mathbb{C})$ with ${\mathbb{R}}^{2N^2}$, [F1] makes $U(N)$ compact. Its subspace topology is Hausdorff and second-countable: the Euclidean metric restricts to a separating metric, and intersections with rational Euclidean balls form a countable base. On overlaps, translated Cayley transition maps are compositions of the forward and inverse rational formulas from step 3.1 with fixed matrix multiplication; matrix inversion is smooth on its open domain, so the transitions are smooth. Thus a finite subcover of these neighborhoods is a finite smooth atlas on a topological $N^2$-manifold, and [F2] supplies its smooth structure.

The Frobenius inner product is left invariant because $\mathrm{Re}\mathrm{tr}((Q_{0}A{)}^{\ast }{Q}_{0}B)=\mathrm{Re}\mathrm{tr}(A^{\ast }B)$. By [F3] it defines a Riemannian metric; [F4] and [F5] give its global positive smooth volume density. In each of the finitely many charts from step 4.1, take all rational coordinate balls whose closed balls lie in the chart domain. These balls form a countable cover without making pointwise choices. Compactness gives a finite subcover. On each selected closed ball the smooth density coefficient is bounded, so its integral is finite; every nonempty coordinate ball has positive integral. The total smooth volume is therefore finite and positive, so it can be normalized.

Under [A1], [F6] and [F7] make the associated Borel measure a regular left-invariant Radon measure; the normalization from step 5.1 makes it a probability. In each chart the density coefficient is smooth and strictly positive, so a Borel set is ${\mu }_F$-null exactly when its coordinate image is Lebesgue-null.

Under [A2], this normalized left-invariant probability on the compact group agrees with normalized Haar probability by uniqueness in [F8]. This proves only the optional Haar branch; the volume-null statement was proved in step 6.1.

Apply [F1] to $A$ to obtain unitary $L_0,R_0$ and $C_0=\mathrm{diag}(c_1,\dots ,c_k)$ with $A=L_0{C}_0{R}_0^{\ast }$ and $c_i\ge 0$. Since $A$ is the compression of the unitary $U$ to the first block, $\mathrm{\parallel }Av\mathrm{\parallel }\le \mathrm{\parallel }v\mathrm{\parallel }$ for every $v$, so each singular value satisfies $c_i\le 1$. Put $S_0=\mathrm{diag}(s_i)$ with $s_i=\sqrt{1-c_i^2}$.

The first block row of $U{U}^{\ast }=I$ gives $A{A}^{\ast }+B{B}^{\ast }=I$, hence $(L_0^{\ast }B)(L_0^{\ast }B{)}^{\ast }=S_0^2$. For every $i$ with $s_i>0$, the row $s_i^{-1}{e}_i^{\ast }{L}_0^{\ast }B$ has norm one, and these rows are mutually orthogonal by the displayed Gram identity. Extend them to an orthonormal basis using [F3], and choose the signs of the completed rows so that a unitary $W$ satisfies $L_0^{\ast }B=-S_0{W}^{\ast }$.

The first block column of $U^{\ast }U=I$ gives $A^{\ast }A+C^{\ast }C=I$, hence $(C{R}_0{)}^{\ast }(C{R}_0)=S_0^2$. For every $i$ with $s_i>0$, the column $s_i^{-1}C{R}_0{e}_i$ has norm one, and these columns are mutually orthogonal. Extend them to an orthonormal basis using [F3]; the resulting unitary $Z$ satisfies $C{R}_0=Z{S}_0$. Therefore $\tilde{U}:=\mathrm{diag}(L_0^{\ast },Z^{\ast })U\mathrm{diag}(R_0,W)=\left(\begin{array}{cc}{C}_0& -S_0\\ S_0& E\end{array}\right)$ for some $E$.

The upper-right block of $\tilde{U}{\tilde{U}}^{\ast }=I$ gives $S_0{E}^{\ast }=C_0{S}_0$. Thus, for each $i$ with $s_i>0$, column $i$ of $E$ is $c_i{e}_i$. The lower-right blocks of $\tilde{U}{\tilde{U}}^{\ast }=I$ and ${\tilde{U}}^{\ast }\tilde{U}=I$ give $E{E}^{\ast }=E^{\ast }E=C_0^2$. For such $i$, the prescribed entry $E_{ii}=c_i$ already contributes $c_i^2$ to row $i$'s squared norm, so $E{E}^{\ast }=C_0^2$ forces every other entry in that row to vanish. Thus both the rows and columns indexed by $P=\{i:s_i>0\}$ have no cross-block entries. On $K=\{i:s_i=0\}$ one has $c_i=1$; the $K\times K$ block of $E$ has orthonormal rows and columns and is therefore unitary. This also covers $c_i=0$, where the corresponding row and column are zero, and $c_i=1$, where the index belongs to $K$.

Let $Q$ be that unitary $K\times K$ block and replace $Z$ by $Z^{\mathrm{\prime }}=Z\mathrm{diag}(I,Q)$. Since $Q$ acts only where $S_0=0$ and $C_0=I$, this leaves $C{R}_0=Z^{\mathrm{\prime }}{S}_0$ unchanged and changes the lower-right block to $C_0$. Hence $U=\mathrm{diag}(L_0,Z^{\mathrm{\prime }})\left(\begin{array}{cc}{C}_0& -S_0\\ S_0& C_0\end{array}\right)\mathrm{diag}(R_0^{\ast },W^{\ast })\mathrm{.}$ Both outside factors are unitary, and $C_0^2+S_0^2=I$ by definition. The argument includes $c_i=0$, $c_i=1$, repeated values, and empty positive or zero supports.

Put $S=\mathrm{span}\{\mathrm{\mid }x⟩,\mathrm{\mid }y⟩\}$. By [F3], the selected pair is orthonormal. By [F1], $H=S\oplus S^{\perp }$, and relative to this sum the defined operator is $V\oplus I_{S^{\perp }}$.

The direct sum preserves norms because $V$ is unitary on $S$ and the identity preserves norms on $S^{\perp }$. Its inverse is $V^{\ast }\oplus I_{S^{\perp }}$, so it is an invertible isometry and hence unitary by [F2]. The scalar phase in $V$ is part of this exact direct-sum matrix.

If $b\mathrm{\ne }0$, then $h>0$ and direct multiplication gives $G^{\ast }G=I_2$ and $G(a,b{)}^T=(h,0{)}^T$. Thus the row operation is defined and left multiplication by its coordinate embedding preserves unitarity of $M$. If $b=0$, skipping the operation leaves the desired entry zero.

Inductively, before processing column $j$, columns $0,\dots ,j-1$ have zero entries below their diagonal positions. Each row pair used for column $j$ has indices at least $j$, so both entries in every earlier column are zero and the row operation preserves those zeros. Descending $r$ therefore makes column $j$ zero below its diagonal without undoing a zero already made in that column.

After the finite loops, $D=G_s\cdots G_{1}U$ is upper triangular and unitary. Its first column has only its first entry nonzero, which has modulus one. Orthogonality with every later column forces the first entry of each later column to be zero. Removing the first row and column leaves an upper-triangular unitary block; induction proves that $D$ is diagonal with unit-modulus diagonal entries.

Rearranging $D=G_s\cdots G_{1}U$ gives $U=G_1^{\ast }\cdots G_s^{\ast }D$ exactly. The finite loops have ${\sum }_{j=0}^{N-2}(N-j-1)=N(N-1)\mathrm{/}2$ iterations (zero for $N=1$), with at most one row elimination per iteration, so they terminate with at most that many eliminations and at most $N$ diagonal phases; the zero tests skip any already-zero entry. Under the convention that a matrix product acts on column states from right to left, the same identity orders the actions as $D$ first and then $G_s^{\ast },\dots ,G_1^{\ast }$.

Direct multiplication of the matrices in [F8] gives $Y^2=Z^2=I_2$. Grouping the even and odd terms of the exponential series in [F7] then gives $R_z(t)=\mathrm{diag}(e^{-it\mathrm{/}2},e^{it\mathrm{/}2})$ and $R_y(t)=\left(\begin{array}{cc}\cos (t\mathrm{/}2)& -\sin (t\mathrm{/}2)\\ \sin (t\mathrm{/}2)& \cos (t\mathrm{/}2)\end{array}\right)$. Thus the matrices used in the Statement match the QC7 convention exactly.

The Leibniz formula [F4] gives $\det (\bar{U})=\overline{\det U}$; by [F5], $\det (U^{\ast })=\det ({\bar{U}}^{\mathsf{T}})=\overline{\det U}$. Thus [F6] and $U^{\ast }U=I$ give $\mathrm{\mid }\det U{\mathrm{\mid }}^2=\det (U^{\ast }U)=1$.

By [F1], choose $\delta \in \mathbb{R}$ with $\det U=e^{2i\delta }$ and put $W=e^{-i\delta }U$. The scalar has modulus one and [F4],[F6] give $\det W=e^{-2i\delta }\det U=1$, so $W$ is unitary with determinant one.

Write the first column of $W$ as $(a,c{)}^T$. Its norm is one, so $(\mathrm{\mid }a\mathrm{\mid },\mathrm{\mid }c\mathrm{\mid })$ is on the unit circle in the closed first quadrant. By [F2], choose $t\in [0,2\pi )$ with $\mathrm{\mid }a\mathrm{\mid }=\cos t$ and $\mathrm{\mid }c\mathrm{\mid }=\sin t$. By [F3], sine is positive on $(0,\pi )$: it increases from $0$ to $1$ on $[0,\pi \mathrm{/}2]$ and decreases to $0$ on $[\pi \mathrm{/}2,\pi ]$. Its $\pi$-shift makes it negative on $(\pi ,2\pi )$, so $\sin t\ge 0$ forces $t\in [0,\pi ]$. Cosine decreases strictly on $[0,\pi ]$ and vanishes at $\pi \mathrm{/}2$, so $\cos t\ge 0$ then forces $t\in [0,\pi \mathrm{/}2]$. Set $\theta =2t$. This also covers either zero coordinate.

The unit vector orthogonal to $(a,c{)}^T$ is a unit scalar multiple of $(-\bar{c},\bar{a}{)}^T$. If its scalar is $s$, the determinant of the matrix with these columns is $s(\mathrm{\mid }a{\mathrm{\mid }}^2+\mathrm{\mid }c{\mathrm{\mid }}^2)=s$; since $\det W=1$, the second column is exactly $(-\bar{c},\bar{a}{)}^T$. Choose phases $\gamma ,\eta$ so $a=e^{-i\gamma }\cos t$ and $c=e^{i\eta }\sin t$; when one modulus is zero choose its phase arbitrarily.

Put $\alpha =\gamma +\eta$ and $\beta =\gamma -\eta$. Direct multiplication of the matrices obtained in step 1.1 gives $R_z(\alpha )R_y(\theta )R_z(\beta )=\left(\begin{array}{cc}{e}^{-i\gamma }\cos t& -e^{-i\eta }\sin t\\ e^{i\eta }\sin t& e^{i\gamma }\cos t\end{array}\right)=W\mathrm{.}$ The phase choices in step 4.1 make this exactly the two columns of $W$, including the cases $t=0$ and $t=\pi \mathrm{/}2$.

Restoring the scalar from step 2.1 proves the formula for arbitrary $U$. If $\det U=1$, choose $\delta =0$ in step 2.1, so $W=U$, and apply steps 3.1–5.1.

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.

The diagonal blocks of $C(\lambda U)$ are $I_H$ and $\lambda U$. Multiplying the blocks of $(P_{\lambda }\otimes I_H)C(U)$ gives $I_H$ and $\lambda U$, so the two full matrices are equal exactly.

Under the whole-output update specified in the Given, $(\lambda I)\rho (\lambda I{)}^{\ast }=\lambda \bar{\lambda }\rho =\rho$ because $\mathrm{\mid }\lambda \mathrm{\mid }=1$. Hence the whole-output scalar leaves the density operator unchanged.

By [F2], $U$ and $\lambda U$ are phase-equivalent as target matrices. If $\lambda \mathrm{\ne }1$, $P_{\lambda }$ is not a scalar multiple of the identity: its diagonal entries are $1$ and $\lambda$. Thus step 1.1 places a phase on only the active control block. Left multiplication by $P_{\bar{\lambda }}\otimes I_H$ changes the blocks back to $I_H,U$, so it restores $C(U)$ exactly. For $\lambda =1$ no correction is needed.

By [F4], each defined measurement weight is $\mathrm{tr}(\rho E_a)$. The equality of density operators in step 1.2 makes these weights identical for every outcome $a$. Under the normalized-law conclusion [F5], the two probability distributions therefore agree. The global scalar in step 1.2 preserves the whole measurement law, while the controlled target scalar in step 1.1 does not become a whole-output scalar.