Trace Distance, Fidelity, and Information Limits

The map $U\mapsto U^{\ast }U$ is continuous because each entry is a finite sum of products of real-coordinate polynomial expressions from [F2]. The singleton $\{I_d\}$ is closed in the finite-dimensional Euclidean matrix space, so its inverse image $U(d)$ is closed. If $U=(u_{ij})\in U(d)$, each column has squared Euclidean norm ${\sum }_i\mathrm{\mid }{u}_{ij}{\mathrm{\mid }}^2=1$; hence every real and imaginary entry lies in $[-1,1]$. Thus $U(d)$ is bounded in ${\mathbb{R}}^{2d^2}$.

By Heine–Borel and the agreement of metric and topological compactness in [F1], the closed bounded set $U(d)$ from step 1.1 is compact in its Euclidean subspace topology. Distinct points of $U(d)$ have disjoint sufficiently small Euclidean balls, so its subspace topology is Hausdorff by [F5]. It is nonempty because $I_d\in U(d)$.

By [F3], each $U\in U(d)$ has inverse $U^{\ast }$, so the defining set is closed under inverses. If $U,V\in U(d)$, then $(UV{)}^{\ast }(UV)=V^{\ast }(U^{\ast }U)V=V^{\ast }V=I_d$, so it is closed under multiplication; associativity and the identity follow from [F4]. Matrix multiplication is continuous by the finite coordinate formulas of step 1.1, and inversion is $U\mapsto U^{\ast }$, which is continuous by [F2]. Thus the group with its subspace topology is a topological group by [F5].

Diagonalize $X$ and let $M\ge 0$ be its largest eigenvalue; if $X=0$, take $M=0$. Diagonalize $\sigma$ and let $m>0$ be its smallest eigenvalue, which is positive by faithfulness and [F2]. For every $v\in H$, these spectral bounds give $⟨Xv,v⟩\le M\mathrm{\parallel }v{\mathrm{\parallel }}^2$ and $⟨\sigma v,v⟩\ge m\mathrm{\parallel }v{\mathrm{\parallel }}^2$.

Set $c=M\mathrm{/}m$. Then $c$ is finite and nonnegative, and for every $v$ step 1.1 gives $⟨(c\sigma -X)v,v⟩\ge (cm-M)\mathrm{\parallel }v{\mathrm{\parallel }}^2=0$. By [F3], $c\sigma -X\ge 0$, which is exactly $X\le c\sigma$. This also covers $X=0$ with $c=0$.

Put $A=\sqrt{\rho }\sqrt{\sigma }$. Since $A^{\ast }=\sqrt{\sigma }\sqrt{\rho }$, we have $A^{\ast }A=\sqrt{\sigma }\rho \sqrt{\sigma }$ and $A{A}^{\ast }=\sqrt{\rho }\sigma \sqrt{\rho }$. Both are non-negative. By [L1], the eigenvalues of $A^{\ast }A$ and $A{A}^{\ast }$ are the squares of the singular values of $A$; because $A$ is an endomorphism, the SVD lists the same $d$ singular values on both sides, with zero multiplicity included.

The unique non-negative square roots of $A^{\ast }A$ and $A{A}^{\ast }$ therefore have the same eigenvalue list, namely the singular values of $A$. Their traces both equal the sum of those singular values, which is $\mathrm{\parallel }A{\mathrm{\parallel }}_1$ by [L1]. Substituting $A{A}^{\ast }=\sqrt{\rho }\sigma \sqrt{\rho }$ proves the stated identity and, by [F1], its equivalent root-fidelity expression.

If either $A$ or $B$ is zero, or a domain or codomain factor is zero-dimensional, then $A\otimes B=0$ and at least one factor trace norm is zero, so the formula holds. Otherwise choose SVDs $Az={\sum }_{i=1}^r{s}_i⟨z,v_i⟩u_i$ and $Bw={\sum }_{j=1}^q{t}_j⟨w,q_j⟩p_j$ as in [F2], with all $s_i,t_j>0$.

By [F3], on elementary tensors $(A\otimes B)(z\otimes w)={\sum }_{i=1}^r{\sum }_{j=1}^q{s}_i{t}_j⟨z\otimes w,v_i\otimes q_j⟩u_i\otimes p_j\mathrm{.}$ The families $(v_i\otimes q_j{)}_{i,j}$ and $(u_i\otimes p_j{)}_{i,j}$ are orthonormal because their pairings are products of the factor pairings. Since elementary tensors span, this is a singular-value decomposition of $A\otimes B$ with positive singular values exactly $s_i{t}_j$. Thus [F1] gives $\mathrm{\parallel }A\otimes B{\mathrm{\parallel }}_1={\sum }_{i,j}{s}_i{t}_j=\left({\sum }_i{s}_i\right)\left({\sum }_j{t}_j\right)=\mathrm{\parallel }A{\mathrm{\parallel }}_1\mathrm{\parallel }B{\mathrm{\parallel }}_1\mathrm{.}$

Let $Q=I-P_S$. By [F1], the range of ${\mathrm{Tr}}_{2}Y$ lies in $S$, so $Q{\mathrm{Tr}}_{2}Y=0$ and $\mathrm{tr}(Q{\mathrm{Tr}}_{2}Y)=0$. By [L1], $\mathrm{tr}((Q\otimes I)Y)=0$. Cyclicity of trace rewrites this as $\mathrm{tr}(Y^{1\mathrm{/}2}(Q\otimes I)Y^{1\mathrm{/}2})=0$. Since [L2] gives $(Q\otimes I{)}^{\ast }=(Q\otimes I{)}^2=Q\otimes I$, the operator inside the trace is $Z^{\ast }Z\ge 0$ for $Z=(Q\otimes I)Y^{1\mathrm{/}2}$; [F2] makes it zero, hence $Z=0$ because $⟨Z^{\ast }Zv,v⟩=\mathrm{\parallel }Zv{\mathrm{\parallel }}^2$. Thus $(Q\otimes I)Y=Y(Q\otimes I)=0$, and $Y=(P_S\otimes I)Y(P_S\otimes I)$.

Applying the same argument to the second marginal with the projection $I\otimes Q$ gives $Y=(I\otimes P_S)Y(I\otimes P_S)$.

The two commuting projections $P_S\otimes I$ and $I\otimes P_S$ have product $P_S\otimes P_S$ by [L3]. Combining step 1.1 and step 2.1 yields $Y=(P_S\otimes P_S)Y(P_S\otimes P_S)$, so the range and hence the support of $Y$ lies in $S\otimes S$.

Put $c=\mathrm{\mid }⟨u\mathrm{\mid }v{⟩}_D\mathrm{\mid }\in [0,1]$ by [F2] and the unit norms. If $c=1$, equality in [F2] makes $v$ a unit scalar multiple of $u$, so $P_u=P_v$ and both sides are zero. Suppose $c<1$. Multiplying $v$ by a unit scalar does not change $P_v$ and lets us assume $⟨v,u{⟩}_{\mathrm{r}\mathrm{o}\mathrm{o}\mathrm{t}}=c$: if $a=⟨v,u{⟩}_{\mathrm{r}\mathrm{o}\mathrm{o}\mathrm{t}}\mathrm{\ne }0$, multiply $v$ by $\bar{a}\mathrm{/}\mathrm{\mid }a\mathrm{\mid }$. Define $w=(v-cu)\mathrm{/}\sqrt{1-c^2}$. Direct expansion using [F1] gives $⟨w,u{⟩}_{\mathrm{r}\mathrm{o}\mathrm{o}\mathrm{t}}=0$ and $\mathrm{\parallel }w\mathrm{\parallel }=1$, so $(u,w)$ is orthonormal and $v=cu+\sqrt{1-c^2}w$.

In the orthonormal basis $(u,w)$ of their span, $P_u-P_v$ has matrix $\left(\begin{array}{cc}1-c^2& -c\sqrt{1-c^2}\\ -c\sqrt{1-c^2}& -(1-c^2)\end{array}\right)$. Its trace is zero and determinant is $-(1-c^2)$, so its two eigenvalues are $\sqrt{1-c^2}$ and $-\sqrt{1-c^2}$; it is zero on the orthogonal complement. By [L2], $\mathrm{\parallel }{P}_u-P_v{\mathrm{\parallel }}_1=2\sqrt{1-c^2}$.

Applying [F3] to step 2.1 gives $D(P_u,P_v)=\sqrt{1-c^2}$; together with the $c=1$ case in step 1.1 this proves the formula for every pair of unit vectors.

For $k\ge 1$, ${\mathrm{\Pi }}^{\ast }=\frac{1}{k!}{\sum }_{\pi }{U}_{\pi }^{\ast }=\frac{1}{k!}{\sum }_{\pi }{U}_{{\pi }^{-1}}=\mathrm{\Pi }$ by [L1] and reindexing by inversion. Also ${\mathrm{\Pi }}^2=\frac{1}{(k!{)}^2}{\sum }_{\pi ,\sigma }{U}_{\sigma \circ \pi }=\frac{1}{(k!{)}^2}{\sum }_{\tau \in S_k}k!U_{\tau }=\mathrm{\Pi },$ using [F1]. Thus $\mathrm{\Pi }$ is an orthogonal projection.

For every $\tau \in S_k$, $U_{\tau }\mathrm{\Pi }=\frac{1}{k!}{\sum }_{\pi }{U}_{\pi \circ \tau }=\mathrm{\Pi }$ by [L1] and reindexing by right multiplication. Hence every vector in $\mathrm{ran}\mathrm{\Pi }$ is invariant. Conversely, if $U_{\pi }v=v$ for all $\pi$, then $\mathrm{\Pi }v=\frac{1}{k!}{\sum }_{\pi }v=v$. Therefore $\mathrm{ran}\mathrm{\Pi }={\mathrm{Sym}}^k(H)$.

For $k=0$, $S_0$ has one element and the defining empty tensor power is the one-dimensional tensor unit, so the stated average is $I$ and its range is the whole space.

For fixed $r$, let $M_r$ be the maximum of $\mathrm{\parallel }(\mathrm{\Phi }\otimes {\mathrm{id}}_r)(R_{u,v}){\mathrm{\parallel }}_1$ over unit vectors $u,v\in H\otimes {\mathbb{C}}^r$. Their unit spheres are compact by [F1]. In orthonormal coordinates each entry of $R_{u,v}$ is a finite product of coordinates of $u$ and conjugates of coordinates of $v$; applying the finite linear map $\mathrm{\Phi }\otimes {\mathrm{id}}_r$ remains continuous, and [L4] makes the output trace norm continuous. Hence the maximum exists. By [L1], every $X$ with $\mathrm{\parallel }X{\mathrm{\parallel }}_1\le 1$ has an SVD $X={\sum }_j{s}_j{R}_{u_j,v_j}$ with unit $u_j,v_j$, $s_j\ge 0$, and ${\sum }_j{s}_j\le 1$. Linearity and the trace-norm triangle inequality give $\mathrm{\parallel }(\mathrm{\Phi }\otimes {\mathrm{id}}_r)(X){\mathrm{\parallel }}_1\le {\sum }_j{s}_j{M}_r\le M_r$. For unit $u,v$, the adjoint calculation gives $R_{u,v}^{\ast }{R}_{u,v}=R_{v,v}$, a rank-one orthogonal projection, so its singular values are one $1$ and the remaining zeros; hence $\mathrm{\parallel }{R}_{u,v}{\mathrm{\parallel }}_1=1$. Thus the induced norm equals $M_r$ and is attained.

Now take any finite ancilla $Z$ and unit vectors $x,y\in H\otimes Z$. By [L3], their Schmidt ranks $a,b$ are at most $\min (d,\dim Z)$. Let $m=\max (a,b)$ and pad each Schmidt list to an orthonormal $m$-tuple, obtaining $x^{\mathrm{\prime }},y^{\mathrm{\prime }}\in H\otimes {\mathbb{C}}^m$ by replacing the respective ancilla Schmidt vectors by the standard basis vectors and retaining the Schmidt coefficients and $H$-vectors. Define $V_x,V_y:{\mathbb{C}}^m\to Z$ to send that basis to the respective padded orthonormal ancilla families; thus $x=(I_H\otimes V_x)x^{\mathrm{\prime }}$ and $y=(I_H\otimes V_y)y^{\mathrm{\prime }}$. The tensor-map definition and product pairing show each map $I\otimes V_x$ and $I\otimes V_y$ is an isometry. Factor permutations are unitary because they carry an orthonormal product basis bijectively to another such basis. Thus $(\mathrm{\Phi }\otimes {\mathrm{id}}_Z)(R_{x,y})=(I_K\otimes V_x)(\mathrm{\Phi }\otimes {\mathrm{id}}_m)(R_{x^{\mathrm{\prime }},y^{\mathrm{\prime }}})(I_K\otimes V_y{)}^{\ast }\mathrm{.}$ To verify the equality, expand operators in elementary tensors $A\otimes B$: $\mathrm{\Phi }\otimes \mathrm{id}$ sends this to $\mathrm{\Phi }(A)\otimes B$, and the two isometries send each ancilla matrix unit $R_{e_i,e_j}$ to $R_{z_i,w_j}$. For isometries $U,V$, $(U{A}_0{V}^{\ast }{)}^{\ast }(U{A}_0{V}^{\ast })=V{A}_0^{\ast }{A}_0{V}^{\ast }$. The positive eigenvalues of this operator are those of $A_0^{\ast }{A}_0$: an eigenvector $z$ with positive eigenvalue maps to $Vz$, and any eigenvector with positive eigenvalue must lie in the range of $V$ because the operator vanishes on its orthogonal complement. Hence left/right isometric embeddings preserve the nonzero singular values and trace norm. Embedding ${\mathbb{C}}^m$ isometrically into ${\mathbb{C}}^d$ on both sides is the same case, so the original output has trace norm at most $\mathrm{\parallel }\mathrm{\Phi }\otimes {\mathrm{id}}_d{\mathrm{\parallel }}_{1\to 1}$.

By step 1.1, the induced norm for each $Z$ is attained by a rank-one unit input, so step 1.2 gives $\mathrm{\parallel }\mathrm{\Phi }\otimes {\mathrm{id}}_Z{\mathrm{\parallel }}_{1\to 1}\le \mathrm{\parallel }\mathrm{\Phi }\otimes {\mathrm{id}}_d{\mathrm{\parallel }}_{1\to 1}$. Taking the supremum over all finite $Z$ gives $\mathrm{\parallel }\mathrm{\Phi }{\mathrm{\parallel }}_{\diamond }\le \mathrm{\parallel }\mathrm{\Phi }\otimes {\mathrm{id}}_d{\mathrm{\parallel }}_{1\to 1}$. The reverse inequality holds because $r=d$ is one of the ancilla dimensions in the supremum. Step 1.1 at $r=d$ supplies an attaining rank-one operator.

Write a binary POVM as $(E,I-E)$ with $0\le E\le I$. Its equal-prior success probability is $\frac{1}{2}\mathrm{tr}(E\rho )+\frac{1}{2}\mathrm{tr}((I-E)\sigma )=\frac{1}{2}+\frac{1}{2}\mathrm{tr}(E\mathrm{\Delta }),\mathrm{\Delta }=\rho -\sigma ,$ using $\mathrm{tr}\sigma =1$.

For any such effect, [L2] gives $\mathrm{tr}(E\mathrm{\Delta })=\mathrm{tr}(E{\mathrm{\Delta }}_{+})-\mathrm{tr}(E{\mathrm{\Delta }}_{-})\le \mathrm{tr}{\mathrm{\Delta }}_{+}$: the second trace is nonnegative by [L3], and $\mathrm{tr}((I-E){\mathrm{\Delta }}_{+})\ge 0$ by [L3] bounds the first by $\mathrm{tr}{\mathrm{\Delta }}_{+}$. Thus [L2] implies $\mathrm{tr}(E\mathrm{\Delta })\le D(\rho ,\sigma )$.

Let $E_{+}$ be the orthogonal projection onto the positive eigenspaces of $\mathrm{\Delta }$. Then $E_{+}\mathrm{\Delta }={\mathrm{\Delta }}_{+}$, so $\mathrm{tr}(E_{+}\mathrm{\Delta })=\mathrm{tr}{\mathrm{\Delta }}_{+}=D(\rho ,\sigma )$ by [L2]. The binary projective measurement $(E_{+},I-E_{+})$ therefore attains the upper bound from step 1.2.

Substituting the maximum $\mathrm{tr}(E\mathrm{\Delta })=D$ into the success formula of step 1.1 gives $\frac{1}{2}+\frac{1}{2}D=\frac{1}{2}(1+D)$, as claimed.

Let $\mathrm{\Delta }=\rho -\sigma ={\mathrm{\Delta }}_{+}-{\mathrm{\Delta }}_{-}$. Since $\mathrm{tr}\mathrm{\Delta }=0$, [L1] gives $\mathrm{tr}{\mathrm{\Delta }}_{+}=\mathrm{tr}{\mathrm{\Delta }}_{-}=D\ge 0$, proving the lower bound.

If $\mathrm{supp}\rho \perp \mathrm{supp}\sigma$, then $\rho \sigma =0$ and $\mathrm{\Delta }$ acts as $\rho$ on $\mathrm{supp}\rho$, as $-\sigma$ on $\mathrm{supp}\sigma$, and as zero on the orthogonal complement. Hence $\mathrm{\parallel }\mathrm{\Delta }{\mathrm{\parallel }}_1=\mathrm{tr}\rho +\mathrm{tr}\sigma =2$ and $D=1$.

Let $P_{+}$ project onto the strictly positive eigenspaces of $\mathrm{\Delta }$. Then $0\le P_{+}\le I$ and $D=\mathrm{tr}(P_{+}\mathrm{\Delta })=\mathrm{tr}(P_{+}\rho )-\mathrm{tr}(P_{+}\sigma )\le \mathrm{tr}(P_{+}\rho )\le 1$. The first inequality uses $\mathrm{tr}(P_{+}\sigma )\ge 0$ by [L2], and the last follows from $\mathrm{tr}((I-P_{+})\rho )\ge 0$ by [L2] and $\mathrm{tr}\rho =1$.

If $D=0$, [F1] gives $\mathrm{\parallel }\rho -\sigma {\mathrm{\parallel }}_1=0$, so definiteness of the trace norm gives $\rho =\sigma$. Conversely equal states have distance zero by [F1].

Conversely, if $D=1$, then both inequalities in step 2.1 are equalities: $\mathrm{tr}(P_{+}\rho )=1$ and $\mathrm{tr}(P_{+}\sigma )=0$. Thus $\mathrm{tr}((I-P_{+})\rho )=0$ and $\mathrm{tr}(P_{+}\sigma )=0$. Applying [F3] to each positive operator gives $(I-P_{+})\sqrt{\rho }=0$ and $P_{+}\sqrt{\sigma }=0$, so $\mathrm{supp}\rho \subseteq \mathrm{ran}{P}_{+}$ and $\mathrm{supp}\sigma \subseteq \ker P_{+}$. These subspaces are orthogonal.

If $k=0$, the tensor unit is one-dimensional and the action is trivial, so the representation is irreducible. Suppose $k\ge 1$. Let $e_1,\dots ,e_d$ be the standard orthonormal basis of ${\mathbb{C}}^d$. For each multi-index $\alpha =({\alpha }_1,\dots ,{\alpha }_d)\in {\mathbb{N}}^d$ with ${\sum }_j{\alpha }_j=k$, let ${\mathrm{\Omega }}_{\alpha }$ be the set of words of length $k$ in which $j$ occurs ${\alpha }_j$ times, and set $v_{\alpha }=\mathrm{\mid }{\mathrm{\Omega }}_{\alpha }{\mathrm{\mid }}^{-1\mathrm{/}2}{\sum }_{(i_1,\dots ,i_k)\in {\mathrm{\Omega }}_{\alpha }}{e}_{i_1}\otimes \cdots \otimes e_{i_k}$. The sets ${\mathrm{\Omega }}_{\alpha }$ partition the elementary tensor basis, so the $v_{\alpha }$ are orthonormal. A vector fixed by all permutations has equal coefficients on each permutation orbit ${\mathrm{\Omega }}_{\alpha }$; hence these vectors span exactly the symmetric subspace in [F2].

Put $t=(2k+1{)}^{-1}$ and $D=\mathrm{diag}(e^{it},e^{i{t}^2},\dots ,e^{i{t}^d})\in U(d)$. Each $v_{\alpha }$ is an eigenvector of $D^{\otimes k}$ with eigenvalue ${\lambda }_{\alpha }=e^{i{\sum }_j{\alpha }_j{t}^j}$. These eigenvalues are distinct: for $\alpha \mathrm{\ne }\beta$, let $\mathrm{\ell }$ be the first index with $c_{\mathrm{\ell }}:={\alpha }_{\mathrm{\ell }}-{\beta }_{\mathrm{\ell }}\mathrm{\ne }0$. Then $\mathrm{\mid }{c}_{\mathrm{\ell }}\mathrm{\mid }{t}^{\mathrm{\ell }}\ge t^{\mathrm{\ell }}$, while $\mathrm{\mid }{\sum }_{j>\mathrm{\ell }}{c}_j{t}^j\mathrm{\mid }\le k{t}^{\mathrm{\ell }+1}\mathrm{/}(1-t)=t^{\mathrm{\ell }}\mathrm{/}2$; thus the phase difference is nonzero. Also $\mathrm{\mid }{\sum }_j{c}_j{t}^j\mathrm{\mid }\le kt\mathrm{/}(1-t)=1\mathrm{/}2<2\pi$, so the phase difference is not a nonzero multiple of $2\pi$.

Let $W\mathrm{\ne }\{0\}$ be a subspace invariant under every $U^{\otimes k}$, and choose $0\mathrm{\ne }w={\sum }_{\alpha }{c}_{\alpha }{v}_{\alpha }\in W$. At least one coefficient $c_{\alpha }$ is nonzero. Because the eigenvalues in step 1.2 are distinct, the interpolation polynomial $p_{\alpha }(z)={\prod }_{\beta \mathrm{\ne }\alpha }(z-{\lambda }_{\beta })\mathrm{/}({\lambda }_{\alpha }-{\lambda }_{\beta })$ satisfies $p_{\alpha }(D^{\otimes k})w=c_{\alpha }{v}_{\alpha }$. Invariance under $D^{\otimes k}$ and linearity imply $v_{\alpha }\in W$. If $d=1$, step 1.1 has only the single multi-index $(k)$, so this already gives $W={\mathrm{Sym}}^k(\mathbb{C})$. Hence assume $d\ge 2$ in the transfer argument below.

Fix indices $i\mathrm{\ne }j$ with ${\alpha }_i>0$, let $q>0$ be sufficiently small, and set $c=(1+q^2{)}^{-1\mathrm{/}2}$ and $s=q(1+q^2{)}^{-1\mathrm{/}2}$, so $c^2+s^2=1$. The unitary rotation sending $e_i\mapsto c{e}_i+s{e}_j$, $e_j\mapsto -s{e}_i+c{e}_j$ and fixing all other $e_{\mathrm{\ell }}$ sends $v_{\alpha }$ into the symmetric subspace by [L1]. The coefficient of $v_{\beta }$ for $\beta =\alpha -e_i+e_j$ is the positive factor $\sqrt{\mathrm{\mid }{\mathrm{\Omega }}_{\beta }\mathrm{\mid }\mathrm{/}\mathrm{\mid }{\mathrm{\Omega }}_{\alpha }\mathrm{\mid }}$ times $s{c}^{{\alpha }_i+{\alpha }_j-1}P(q^2),P(z)={\sum }_{r=0}^{\min ({\alpha }_i-1,{\alpha }_j)}(-1{)}^r\left(\genfrac{}{}{0px}{}{{\alpha }_i-1}{r}\right)\left(\genfrac{}{}{0px}{}{{\alpha }_j+1}{r+1}\right)z^r\mathrm{.}$ This follows by counting which $r$ output $i$-positions came from $j$ and which $r+1$ output $j$-positions came from $i$ in the tensor expansion. Since $P(0)={\alpha }_j+1>0$ and $P$ is a polynomial, choose $q>0$ small enough that $P(q^2)\mathrm{\ne }0$. The rotated vector lies in $W$ by invariance; applying the interpolation projection from step 2.1 for weight $\beta$ isolates a nonzero multiple of $v_{\beta }$, so $v_{\beta }\in W$.