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The absolute value of a linear operator $L$ is the positive square root of an operator $L^{†}L$. Its eigenvalues are the singular values of $L$, its trace is the trace norm of $L$, and it appears in the polar decomposition of a bounded operator $L=U|L|$.

When two density operators are commuting elements, choose a common orthonormal eigenbasis and write $\rho =\mathrm{diag}(p_i)$, $\sigma =\mathrm{diag}(q_i)$. Their quantum fidelity becomes the classical overlap of the two probability distributions, while their trace distance becomes total variation distance:This follows by computing the positive square root of an operator entry by entry in the common eigenbasis. For the qubit mixed states $\rho =\mathrm{diag}(0.9,0.1)$ and $\sigma =\mathrm{diag}(0.6,0.4)$,The two contributions to quantum fidelity are about $0.734847$ and $0.2$. In contrast, the chance that two independent computational basis measurements, one on each preparation, print the same label is $0.9(0.6)+0.1(0.4)=0.58$. That particular event has a specified probability; neither $F$ nor $F_{\mathrm{sq}}$ is that probability.

Here trace distance $0.3$ means an equal-prior quantum state discrimination success of $0.65$ by the Holevo–Helstrom theorem. Measuring in the common computational basis and guessing $\rho$ on outcome zero and $\sigma$ on outcome one achieves $[0.9+0.4]/2=0.65$. Also $\sqrt{1-F^2}\approx 0.3551>D$: the pure-state trace distance and fidelity identity is an equality for pure states, rather than a formula for arbitrary mixed states.

For density operators $\rho ,\sigma$ on the same finite-dimensional Hilbert space, quantum fidelity measures how close their quantum states are. This article uses the unsquared convention:Each square root is the positive square root of an operator, and $\Vert \cdot {\Vert }_1$ is the trace norm. Quantum fidelity is symmetric in its inputs; it is one exactly when $\rho =\sigma$, and zero exactly when $\rho \sigma =0$, meaning that the support of a positive operator for one state is orthogonal to that for the other. Larger quantum fidelity means greater similarity. The separate squared quantum fidelity is $F_{\mathrm{sq}}=F^2$. ^[ref]

For two normalized pure states, quantum fidelity is the magnitude of their inner product, $F=|⟨\psi |\phi ⟩|$. Thus $|0⟩$ compared with itself has $F=1$, while $|0⟩$ and $|1⟩$ have $F=0$. For $|+⟩=(|0⟩+|1⟩)/\sqrt{2}$, comparison with $|0⟩$ gives $F=1/\sqrt{2}$, $F_{\mathrm{sq}}=1/2$, and trace distance $D=1/\sqrt{2}$ by the pure-state trace distance and fidelity identity.

Why compare density operators rather than their entries? A change of coordinates should not make the same physical preparations look more or less alike. Unitary invariance of trace distance and fidelity makes quantum fidelity independent of a common basis choice. Multiplying a pure state vector by a global phase leaves its density operator unchanged, so its quantum fidelity with the original state is one. These features make quantum fidelity useful for checking whether quantum channels, state preparations, and Schumacher compression preserve a target quantum state.

The connection to Uhlmann's theorem supplies another beautiful interpretation: quantum fidelity is the largest absolute inner product between pure vectors obtained from a purification of a density operator for each state, using a sufficiently large common reference space. Uhlmann's 1976 paper extended transition probability to algebraic states, including those of a C-star algebra; for density operators its transition probability corresponds to our squared quantum fidelity. The measurement characterization of quantum fidelity gives a complementary description directly in terms of observable outcome statistics. ^[ref]

Choose an orthonormal basis of eigenvectors of the positive-definite operator $\alpha$ using the proved spectral theorem. Its eigenvalues ${\lambda }_j$ are strictly positive, because $⟨\alpha e_j,e_j⟩={\lambda }_j>0$. Define $\beta e_j=\sqrt{{\lambda }_j}{e}_j$. It is a positive-definite Hermitian operator and ${\beta }^2=\alpha$, establishing existence of the positive square root of an operator.

For uniqueness, let $B$ be any positive-definite Hermitian square root. The identity $B^2=\alpha$ implies $B\alpha =B^3=\alpha B$, so $B$ preserves each eigenspace $E_{\lambda }$ of $\alpha$. Its restriction to that space is Hermitian and positive-definite. Diagonalize that restriction by the same spectral theorem. Every resulting eigenvalue $\mu$ satisfies $\mu >0$ and ${\mu }^2=\lambda$, hence $\mu =\sqrt{\lambda }$. Therefore $B$ acts as $\sqrt{\lambda }I$ on the entire eigenspace, even when $\lambda$ is repeated. These eigenspaces span the space, so the positive-definite Hermitian square root is unique, with eigenvalues $\sqrt{{\lambda }_j}$.

For a positive operator $A$, the spectral theorem writes $A={\sum }_j{a}_j|j⟩⟨j|$ with $a_j\ge 0$. Hence its unique positive square root of an operator is $\sqrt{A}={\sum }_j\sqrt{a_j}|j⟩⟨j|$. Conjugating this spectral decomposition shows $\sqrt{UA{U}^{†}}=U\sqrt{A}{U}^{†}$ for a unitary operator $U$.

Let $\Delta ={\rho }_1-{\rho }_2$ and ${\rho }_i^{\prime }=U{\rho }_i{U}^{†}$. Then ${\Delta }^{\prime }=U\Delta U^{†}$ and $({\Delta }^{\prime }{)}^{†}{\Delta }^{\prime }=U{\Delta }^{†}\Delta U^{†}$. The square-root covariance and the cyclic property of the trace giveFor the quantum fidelity, covariance first gives $({\rho }_1^{\prime }{)}^{1/2}=U{\rho }_1^{1/2}{U}^{†}$. ThereforeApply the same square-root identity and the cyclic property of the trace to obtainThus unitary invariance of trace distance and fidelity givesThe same $U$ acts on both quantum states: these quantities depend on the relative distinguishability of the quantum states, not on a choice of basis.

A bounded self-adjoint operator $T$ is a positive operator when $⟨T\xi ,\xi ⟩\ge 0$ for every $\xi$. Equivalently its quadratic form is nonnegative. We construct its positive square root of an operator directly and prove uniqueness.

If $T=0$ its positive square root is zero. Otherwise set $r=\Vert T\Vert$, $B=T/r$, and $C=I-B$. Then $0\le B\le I$ and $C$ is positive. It is also a contraction. Indeed, the Cauchy-Schwarz inequality for the positive form $⟨B\xi ,\eta ⟩$ gives $\Vert B\xi {\Vert }^2\le ⟨B\xi ,\xi ⟩$, and henceThe scalar binomial series has the formThe last equality follows by taking $z\uparrow 1$ in the nonnegative series. Thereforeconverges in operator norm. Every $C^n$ is a positive contraction: its even powers are squares, its odd powers have form $C^{k}C{C}^k$, and their norms are at most one. Each partial sum for $S/\sqrt{r}$ is consequently positive, since it is at least $(1-{\sum }_{n\le N}{c}_n)I$. The limit is positive. Multiplying the absolutely convergent series and using the scalar coefficient identity gives $S^2=r(I-C)=T$. This construction also shows that $S$ is a norm limit of polynomials in $T$ and thus commutes with every operator commuting with $T$.

Now suppose $R$ is another positive square root. It commutes with $T=R^2$, hence with $S$. ThusThe difference vanishes on the range of $R+S$ and on its closure. It also vanishes on its kernel: if $(R+S)\xi =0$, the fact that $R$ and $S$ are positive operators gives $⟨R\xi ,\xi ⟩=⟨S\xi ,\xi ⟩=0$, and the Cauchy-Schwarz inequality for their positive forms implies $R\xi =S\xi =0$. Since $R+S$ is self-adjoint, the orthogonal sum of its kernel and closed range is all of $H$. Therefore $R=S$, proving

For a bounded self-adjoint $T$, nonnegative spectrum is equivalent to $⟨Tx,x⟩\ge 0$. Here is a useful elementary justification of the nontrivial direction. Let $a={\inf }_{\Vert x\Vert =1}⟨Tx,x⟩$ and $S=T-aI$, so $S$ has nonnegative quadratic form. Cauchy-Schwarz for that positive form gives $\Vert Sx{\Vert }^2\le \Vert S\Vert ⟨Sx,x⟩$. Taking unit vectors approaching the infimum makes $S{x}_j\to 0$, hence $a\in \sigma (T)$. Thus nonnegative spectrum forces $a\ge 0$. The reverse direction follows from the lower bound and adjoint argument for $T-\lambda$ when $\lambda <0$, together with reality of the self-adjoint spectrum.

If $T=0$ its positive square root is zero. Otherwise set $M=\Vert T\Vert$ and $C=I-T/M$, so $0\le C\le I$ and $\Vert C\Vert \le 1$. Write the scalar binomial seriesThe last equality follows by taking $z\uparrow 1$ in this positive-coefficient series. Consequentlyconverges absolutely in operator norm. Each partial sum is positive, because $0\le C^n\le I$ and the sum of its coefficients is at most one. The limit is positive and self-adjoint. Multiplying the absolutely convergent series gives $B^2=M(I-C)=T$.

For uniqueness let $R$ be another positive square root. It commutes with $T=R^2$, hence with the polynomial-norm-limit $B$. Thus $(R-B)(R+B)=0$. The kernel of $R+B$ is $\ker R\cap \ker B$: zero quadratic form for a positive operator implies it annihilates the vector, by positive-form Cauchy-Schwarz. The difference $R-B$ vanishes there and on $\mathrm{ran}(R+B)$; the latter is dense in the kernel complement because $R+B$ is self-adjoint. Therefore $R=B$. The positive square root of an operator exists and is unique.

For positive operators, the unique positive square root of an operator respects tensor products: $\sqrt{A\otimes B}=\sqrt{A}\otimes \sqrt{B}$. To verify this, diagonalize $A,B$; the product eigenvectors have eigenvalues $a_i{b}_j$, whose square roots are $\sqrt{a_i}\sqrt{b_j}$. Hence the operator inside the outer square root in the fidelity factors asTaking its positive square root and using $\mathrm{Tr}(C\otimes D)=\mathrm{Tr}C\mathrm{Tr}D$ gives $\boxed{F(\rho \otimes \sigma ,{\rho }^{\prime }\otimes {\sigma }^{\prime })=F(\rho ,{\rho }^{\prime })F(\sigma ,{\sigma }^{\prime })}$.

The absolute value of an operator, trace norm, and operator norm are respectivelyThe positive square root of an operator makes $|L|$ well defined. Its eigenvalues $s_j\ge 0$ are the singular values of $L$, so $\Vert L{\Vert }_1={\sum }_j{s}_j$ and $\Vert L{\Vert }_{\mathrm{op}}={\max }_j{s}_j$.

Use the polar decomposition of a bounded operator $L=U|L|$, with the given unitary operator $U$. In an orthonormal eigenbasis $e_j$ for $|L|$,The operator norm assumption and unitarity imply $|⟨e_j|ZU|e_j⟩|\le \Vert Z{\Vert }_{\mathrm{op}}\le 1$. The triangle inequality therefore yields the required trace estimate:Choosing $Z=U^{†}$ attains equality, which also gives the finite-dimensional trace duality formula $\Vert L{\Vert }_1={\max }_{\Vert Z{\Vert }_{\mathrm{op}}\le 1}|\mathrm{Tr}(ZL)|$. A singular $L$ causes no difficulty: in a finite-dimensional square space the polar factor can be extended to a unitary operator on the complementary subspaces.

With $|\Omega {⟩}_{RA}=n^{-1/2}{\sum }_i|i{⟩}_R|i{⟩}_A$,The partial trace over $R$ isIts norm is $\mathrm{Tr}{\rho }_A=1$. The positive square root of an operator $\sqrt{{\rho }_A}$ is unique, proving the claim.

The two rank-one operators have orthogonal ranges. Their Hilbert-Schmidt inner product is zero, while $\Vert C_i{\Vert }_{\mathrm{HS}}={\lambda }_i$, so the Hilbert-Schmidt distance between covariance operators isFor each $C_i$, the positive square root of an operator is $C_i^{1/2}=\sqrt{{\lambda }_i}{e}_i\otimes e_i$. Orthogonality therefore gives the square-root distance between covariance operatorsFinally $C_2^{1/2}{C}_1^{1/2}=0$, so every singular value in the Procrustes cross-term vanishes. Hence

A bounded Hermitian operator $T$ on a Hilbert space is positive, also called a positive semidefinite operator, whenThe value must be real as well as nonnegative. This permits zero values on nonzero vectors: the zero linear operator is positive. For a bounded Hermitian operator, positivity is equivalent to its spectrum being contained in $[0,\infty )$; in finite dimension, this is the condition that all eigenvalues are nonnegative.^[ref]

The Hermitian operator diagram separates three cases. $D=\mathrm{diag}(1,-1)$ is self-adjoint but not positive, because $q_D((0,1{)}^{\mathrm{T}})=-1$. The orthogonal projection $P=\mathrm{diag}(1,0)$ is positive semidefinite, with $q_P(x)=|x_1{|}^2$, but is not strictly positive on every nonzero vector. The complex Hermitian matrix $H=\left(\begin{array}{cc}2& i\\ -i& 2\end{array}\right)$ is positive definite because $q_H(a{e}_1+b{e}_3)=|a{|}^2+3|b{|}^2>0$ whenever $(a,b)\ne (0,0)$. For real symmetric matrices, MIT's lecture 27 relates these tests to the geometry of quadratic forms and their minima.^[ref]

A fundamental source of positive operators is $T^{\ast }T$ for a bounded linear operator $T:H\to K$:For the complex matrix $T=\left(\begin{array}{cc}1& i\\ 2& 1-i\end{array}\right)$ used above,Its eigenvalues are $4-\sqrt{6}$ and $4+\sqrt{6}$, both positive. Positivity organizes the squared error minimized by a least-squares solution: the same $T^{\ast }T$ occurs in the normal equation for a linear inverse problem. The positive square root of an operator gives the converse squared-length description, $q_T(x)=\Vert T^{1/2}x{\Vert }^2$ for every bounded positive operator $T$.

On a real Hilbert space, nonnegativity of $⟨Tx,x⟩$ alone does not imply self-adjointness, so the Hermitian operator hypothesis is part of the definition. On a complex Hilbert space, real-valuedness of all these expressions already forces self-adjointness by the polarization identity.

Strict positivity and a uniform positive lower bound also differ in infinite dimension. On the l2 sequence space, the bounded Hermitian operator $(Tx{)}_n=x_n/n$ satisfiesConsequently no $\gamma >0$ satisfies $q_T(x)\ge \gamma \Vert x{\Vert }^2$ for every $x$. Compare the positive definite symmetric operator and uniformly positive definite symmetric operator conditions.

For any unitary $U$, the unique positive square root of an operator obeys $\sqrt{UA{U}^{†}}=U\sqrt{A}{U}^{†}$. Conjugating the positive operators appearing in trace distance and quantum fidelity therefore conjugates their square roots. Cyclicity of the trace removes the unitary factors, proving invariance of both quantities under simultaneous unitary transformations of the states.

// Target: quantum-information-theory.bigb