advanced27 min studyLesson 422 in Physics

Fidelity and Bures distance

Separate squared fidelity from root fidelity and connect overlaps, purification geometry, and operational distance bounds.

On this page

What you’ll understand

  • State both common fidelity conventions explicitly.
  • Calculate fidelity and Bures distance in tractable examples.
  • Use distance inequalities without treating a bound as an equality.

When a device is intended to prepare a target state, an overlap is often easier to interpret than an optimal decision rule. Fidelity supplies that overlap, including for mixed states. Unfortunately, two different quantities share the name fidelity in the literature. We reserve capital F for the squared convention and lowercase f for its positive square root. Declaring this choice prevents a square-root error from propagating into distance bounds, error budgets, and comparisons between experiments.

Define the square roots before taking a trace

For positive density matrices, the square root means the unique positive matrix square root obtained spectrally. The operator between the outer roots in the definition is positive, even when the two states do not commute. The equivalent trace-norm expression is useful computationally because it involves singular values. Ordinary matrix multiplication cannot usually replace these square roots. In particular, the Hilbert–Schmidt overlap of two mixed states is generally neither fidelity convention.

If one state is a pure projector, squared fidelity becomes the probability of accepting that projector when measuring the other state. If both are pure, it is the squared absolute inner product. A global phase cancels, whereas a relative phase can change the overlap. If the states commute, calculate root fidelity by summing square roots of matched classical probabilities, then square that entire sum. Squaring each term before summing would remove the cross terms and give a different quantity.

f(ρ,σ)=Tr⁡ρ σρ=∥ρσ∥1,F=f2f(\rho,\sigma)=\operatorname{Tr}\sqrt{\sqrt\rho\,\sigma\sqrt\rho}=\|\sqrt\rho\sqrt\sigma\|_1,\qquad F=f^2
Both conventions are dimensionless and lie in [0,1]. This module calls F squared fidelity and f root fidelity; some primary papers call f fidelity.↔ Scroll wide equations horizontally.

Purifications explain the geometry

Choose a reference factor large enough to purify either state. A purification is a normalized vector on system tensor reference whose partial trace over the reference gives the specified state. Uhlmann's theorem says root fidelity is the largest absolute overlap between such purifications. Optimization over the unused reference is essential: two arbitrary purifications need not achieve it. This is a theorem about states on a declared tensor product, rather than an assertion that a mixed state secretly has one preferred wavefunction.

The Bures distance used here is the square root of twice one minus root fidelity. The related Bures angle is the inverse cosine of root fidelity; it is a different function. For normalized states, zero Bures distance means equality, and orthogonal supports give distance square root of two. The formula has no spatial units. It measures separation in state space, so interpreting it as the physical travel distance of a particle or as elapsed evolution time would require additional structure.

dB(ρ,σ)=2−2f(ρ,σ),1−F≤D(ρ,σ)≤1−Fd_B(\rho,\sigma)=\sqrt{2-2f(\rho,\sigma)},\qquad 1-\sqrt F\le D(\rho,\sigma)\le\sqrt{1-F}
The second relation gives the Fuchs–van de Graaf bounds using squared F and the half-trace-norm D. All quantities refer to the same normalized finite-dimensional states.↔ Scroll wide equations horizontally.

A useful overlap is not an entire error model

A deterministic channel cannot decrease fidelity between two input hypotheses: processing can make states harder to distinguish. Unitary processing preserves it. This direction agrees with trace-distance contraction, although one measure increases and the other decreases. For two pure states the upper trace-distance bound is an equality. Mixed states generally require the inequality, so obtaining a single fidelity number does not determine the exact discrimination error.

State fidelity also differs from a guarantee about a quantum device on every input. A gate can reproduce one test state perfectly and fail badly on another, or fail on correlations with a reference. Average fidelity, worst-case fidelity, and entanglement-sensitive channel metrics answer different questions. Before comparing quoted values, identify the target, input ensemble, convention, and optimization. An uncertainty in estimating fidelity is another issue entirely; the exact algebraic definitions do not themselves provide statistical confidence intervals.

A pure-state overlap with exact radicals

Compare ∣0⟩|0\rangle with ∣ψ⟩=(3∣0⟩+∣1⟩)/2|\psi\rangle=(\sqrt3|0\rangle+|1\rangle)/2. Compute F, f, Bures distance, and trace distance.

  1. Read the normalized zero-basis amplitude as 3/2\sqrt3/2. Its squared modulus gives F=3/4F=3/4, while root fidelity is f=3/2f=\sqrt3/2. The two numbers are deliberately different.

  2. Insert root fidelity into the Bures expression: dB=2−3≈0.517638d_B=\sqrt{2-\sqrt3}\approx0.517638. Using F in place of f would incorrectly give a different distance.

  3. Because both states are pure, D=1−F=1/2D=\sqrt{1-F}=1/2. The lower bound is 1−3/2≈0.1339751-\sqrt3/2\approx0.133975, so the computed distance lies within both bounds.

The result

Squared fidelity is 0.75, root fidelity about 0.866025, Bures distance about 0.517638, and trace distance 0.5.

Explore the model

Interactive lab

Interactive lab · quantum

Read a pure target overlap

Reset and set mixing angle to 60 degrees and relative phase to 0 degrees. Predict zero-basis probability 0.75, which equals squared fidelity with the pure target zero. Repeat after Reset with both controls set to 60 and 90 degrees; the same target fidelity remains 0.75 although Bloch x changes to zero. The instrument displays a pure qubit only; it computes no general mixed-state fidelity or Bures distance.

Read a pure target overlapA normalized two-level pure state, not an interacting field simulation. Playback adds one radian per second to the relative phase; basis probabilities stay fixed. Current readings: Probability |0⟩: 71.1%; Probability |1⟩: 28.9%; Bloch (x, y, z): 0.78, 0.45, 0.42. All parameters are adjustable with the labeled controls below.xyz|0⟩|1⟩Basis probabilitiesP(0)71.1%P(1)28.9%
Bloch vector / P(0)P(1)Phase latitude
Probability |0⟩
71.1%
Probability |1⟩
28.9%
Bloch (x, y, z)
0.78, 0.45, 0.42
t = 0 s

A normalized two-level pure state, not an interacting field simulation. Playback adds one radian per second to the relative phase; basis probabilities stay fixed. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA phase can change a state while leaving its overlap with a particular basis target unchanged.

Check your understanding

For commuting states with probabilities (1,0) and (1/4,3/4), which pair gives squared fidelity F and root fidelity f?

Try it yourself

For ρ=I/2\rho=I/2 and σ=diag⁡(0.8,0.2)\sigma=\operatorname{diag}(0.8,0.2), find F, f, and D. Check whether the pure-state equality for D holds.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Use the common eigenbasis for both the square-root overlap and the absolute probability differences.

Compare with the solution

Root fidelity is 0.4+0.1=3/10≈0.948683\sqrt{0.4}+\sqrt{0.1}=3/\sqrt{10}\approx0.948683, so F=0.9F=0.9. Trace distance is (0.3+0.3)/2=0.3(0.3+0.3)/2=0.3. The bounds are 1−3/10≤0.3≤0.11-3/\sqrt{10}\le0.3\le\sqrt{0.1}. The upper value is about 0.316228, not 0.3, showing why the pure-state equality cannot be assumed for a mixed pair.

Take it with you

  • Define the fidelity convention before using a numerical result or inequality.
  • Bures distance depends on root fidelity, and its angle is a different quantity.
  • One successful state preparation does not establish a worst-case channel guarantee.

Read further

Original sources and deeper explanations to follow at your own pace.

Optional progress, saved in this browser when available. Revisit any lesson at any time.

Lesson not marked complete.