Fidelity of quantum states ($F(\rho ,\sigma )$)

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]

In the unsquared convention, $F({\sum }_x{p}_x{\rho }_x,{\sum }_x{p}_x{\sigma }_x)\ge {\sum }_x{p}_{x}F({\rho }_x,{\sigma }_x)$. Use Uhlmann's theorem to choose component purifications with nonnegative maximizing overlaps, and then construct matching flagged purifications of a quantum ensemble. Their overlap is the weighted sum on the right; the fidelity of the mixtures maximizes over all such overlaps.

Discarding a subsystem cannot decrease unsquared quantum fidelity: $F({\rho }_A,{\sigma }_A)\ge F({\rho }_{AB},{\sigma }_{AB})$. Maximizing purifications of the joint states are also candidates for purifications of the marginals, with the discarded subsystem included in the reference. Uhlmann's theorem then proves the inequality. Information loss makes the two states at least as similar.

If $M$ is a positive contraction and $p=⟨\psi |M^2|\psi ⟩>0$, the normalized successful post-measurement state is $M|\psi ⟩/\sqrt{p}$. Since $M^2\le M$,A likely outcome therefore preserves a pure input well in quantum fidelity when its measurement operator is positive. Positivity excludes a hidden unitary rotation that could otherwise change the state even for an outcome of probability one.

For normalized pure states $\rho =|\psi ⟩⟨\psi |$ and $\sigma =|\phi ⟩⟨\phi |$, the trace distance and quantum fidelity satisfyThe last two expressions distinguish the unsquared quantum fidelity from the squared quantum fidelity. Identical pure states give $D=0$, and orthogonal pure states give $D=1$. The equality need not hold for mixed states; the Fuchs-van de Graaf inequalities give bounds in that case.

Brief proof: for two distinct pure states, their two-dimensional span contains the nonzero part of $\rho -\sigma$. This restriction has trace zero and determinant $-(1-F^2)$, hence eigenvalues $\pm \sqrt{1-F^2}$. Its trace norm is twice the positive eigenvalue, giving trace distance $D=\sqrt{1-F^2}$. For identical pure states, $\rho -\sigma =0$ directly.

Compare the qubit pure statewith $|0⟩$, using the computational basis. Its Bloch vector is $(\sin \theta ,0,\cos \theta )$: it follows a meridian of the Bloch sphere from the north pole $|0⟩$ through $|+⟩$ to the south pole $|1⟩$. The inner product with $|0⟩$ is $\cos (\theta /2)$, nonnegative on this interval. Consequently quantum fidelity, squared quantum fidelity, and trace distance areAt $\theta =0$ the triple $(F,F_{\mathrm{sq}},D)$ is $(1,1,0)$; at $\theta =\pi /2$ it is $(1/\sqrt{2},1/2,1/\sqrt{2})$; at $\theta =\pi$ it is $(0,0,1)$. The angle between the Bloch vectors is $\theta$, while the quantum fidelity uses its half-angle. The pure-state trace distance and fidelity identity becomes the familiar relation ${\cos }^2(\theta /2)+{\sin }^2(\theta /2)=1$.

For equal prior probabilities, the Holevo–Helstrom theorem gives optimal quantum state discrimination success $P_{\mathrm{succ}}=(1+D)/2$. Along this meridian it rises from $1/2$ to one, with $(1+1/\sqrt{2})/2\approx 0.8536$ at $|+⟩$. This connects the geometric trace distance to the best possible decision from a single preparation; it is a different measurement task from the target-acceptance probability defining squared quantum fidelity.

// Target: quantum-information-theory.bigb

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

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.

A POVM $\{E_i\}$ gives outcome probability distributions $p_i=\mathrm{Tr}(\rho E_i)$ and $q_i=\mathrm{Tr}(\sigma E_i)$ by the Born rule. Their classical overlap, the Bhattacharyya coefficient, is $B(p,q)={\sum }_i\sqrt{p_i{q}_i}$. Fuchs and Caves showed that minimizing this overlap over all POVMs gives the unsquared quantum fidelity:The same POVM is applied to either possible input quantum state. Each measurement in quantum mechanics has $B(p,q)\ge F$; some measurement attains equality, including when the density operators have zero eigenvalues. Thus quantum fidelity quantifies the smallest classical overlap that observation can reveal. For fidelity of commuting mixed states, measurement in a common eigenbasis attains the minimum. This operational interpretation compares statistics without assigning a probability that two mixed states are equal. ^{[ref] [ref]}

For all finite-dimensional density operators, including pure states and mixed states, the unsquared quantum fidelity and trace distance obeyIn the squared quantum fidelity convention this is $1-\sqrt{F_{\mathrm{sq}}}\le D\le \sqrt{1-F_{\mathrm{sq}}}$. The bounds translate a high quantum fidelity into a small operationally meaningful trace distance. If $F\ge 1-\epsilon$ with $0\le \epsilon \le 1$, then $D\le \sqrt{2\epsilon -{\epsilon }^2}$; if $D\le \delta$, then $F\ge 1-\delta$. The upper bound is attained by the pure-state trace distance and fidelity identity. For the fidelity of commuting mixed states example, the bounds read $0.0652\lesssim 0.3\lesssim 0.3551$. ^[ref]

This is the square of the unsquared quantum fidelity, another common convention for fidelity of quantum states. If one state is the normalized pure state $|\psi ⟩⟨\psi |$, thenBy the Born rule, the squared quantum fidelity here is the probability that a measurement in quantum mechanics with acceptance orthogonal projector $|\psi ⟩⟨\psi |$ accepts $\sigma$. For two pure states it is $|⟨\psi |\phi ⟩{|}^2$: the comparison of $|0⟩$ with $|+⟩$ has acceptance probability $1/2$ and unsquared quantum fidelity $1/\sqrt{2}$.

For two general mixed states, neither quantum fidelity nor squared quantum fidelity is a probability that the states are equal. Equality of two specified density operators is a definite mathematical condition. The measurement characterization of quantum fidelity instead describes overlap between outcome probability distributions. The ensemble average used in reliable quantum source compression is normally the average squared quantum fidelity with the emitted pure state; the convention must be stated when comparing formulas.