advanced27 min studyLesson 421 in Physics

Trace distance and state discrimination

Turn a difference of density operators into an optimal binary decision and distinguish measurement limitations from preparation uncertainty.

On this page

What you’ll understand

  • Compute trace distance from the spectrum of a Hermitian difference.
  • Derive the optimal equal-prior discrimination probability.
  • Explain why a fixed measurement may discard distinguishing information.

A source secretly chooses between two preparations and gives you one system. Your task is to name the preparation, accepting that mistakes may be unavoidable. This operational question gives trace distance its meaning. We work with normalized density operators on the same finite-dimensional Hilbert space and allow any positive-operator-valued measurement. The preparation labels, their prior probabilities, and the number of available copies belong to the problem specification; changing any of them changes the discrimination task.

A norm with a decision interpretation

A density operator is positive and has unit trace. For two candidate states define the Hermitian difference Δ=ρ−σ\Delta=\rho-\sigma. Its eigenvalues sum to zero, so the sum of its positive eigenvalues equals the absolute sum of its negative eigenvalues. The trace norm adds the absolute eigenvalues of a Hermitian operator. Including the factor one half makes the resulting distance range from zero for identical states to one for states with orthogonal supports.

A binary measurement has effects MM and I−MI-M, with 0≤M≤I0\le M\le I. If outcome MM means guessing rho, equal priors give a success probability equal to one half plus one half of Tr⁡(MΔ)\operatorname{Tr}(M\Delta). In the eigenbasis of the difference, this quantity is maximized by accepting every positive eigenspace and rejecting every negative one. Zero eigenspaces can be assigned either way without affecting success. This derivation explains the optimal measurement rather than merely naming a norm.

D(ρ,σ)=12∥ρ−σ∥1,Psuccopt=12(1+D(ρ,σ))D(\rho,\sigma)=\frac12\|\rho-\sigma\|_1,\qquad P_{\rm succ}^{\rm opt}=\frac12\bigl(1+D(\rho,\sigma)\bigr)
Here DD and success probability are dimensionless. The success formula assumes one copy, equal prior probabilities, unrestricted measurements, and minimum average error.↔ Scroll wide equations horizontally.

Classical records are one measurement away

If both matrices are diagonal in a common basis, measuring that basis produces classical distributions. Trace distance then becomes total variation distance: sum the absolute differences of probabilities and divide by two. No quantum measurement improves on reading this common eigenbasis. For noncommuting matrices, however, the computational basis can hide their difference. The pure states with opposite x-directed Bloch vectors have identical computational-basis probabilities but are perfectly distinguished by an x-basis measurement.

Unequal priors require a weighted difference pρ−(1−p)σp\rho-(1-p)\sigma. Its trace is generally nonzero, and the optimum success probability is one half of one plus its trace norm. Substituting the equal-prior distance formula into that task is incorrect. Similarly, repeated independent copies require comparing tensor powers rather than multiplying a single-copy success probability. A receiver can sometimes improve discrimination further by a collective measurement across the entire block.

Psuccopt(p)=1+∥pρ−(1−p)σ∥12P_{\rm succ}^{\rm opt}(p)=\frac{1+\|p\rho-(1-p)\sigma\|_1}{2}
The dimensionless number pp is the prior probability of rho, between zero and one. The effects may include always guessing the more probable state.↔ Scroll wide equations horizontally.

Discarding information contracts distance

A quantum channel is a linear, completely positive, trace-preserving map. Applying the same channel to either hypothesis cannot increase their trace distance. Any output measurement could instead be implemented as a measurement on the input followed by that channel, so it cannot outperform the unrestricted input optimum. Partial trace is one example: losing a tensor factor can erase distinguishing correlations even if each retained subsystem remains perfectly well defined.

This conclusion concerns deterministic processing with all outcomes included. Conditioning on a rare success flag can increase the distance between normalized conditional states, but its probability must enter the overall task. It also does not say every particular measurement attains the distance. A detector that measures only population can miss coherence entirely. Distinguish a fundamental impossibility from a restriction imposed by the available apparatus, especially when comparing a theoretical error bound with an experimental readout.

Discriminate diagonal qubit preparations

With equal priors, discriminate ρ=diag⁡(0.8,0.2)\rho=\operatorname{diag}(0.8,0.2) from σ=diag⁡(0.3,0.7)\sigma=\operatorname{diag}(0.3,0.7) using one qubit. Find the optimal rule and success probability.

  1. Subtract the matrices in their common basis. The difference has eigenvalues 0.5 and −0.5, so its trace norm is 1 and its trace distance is 0.5.

  2. Assign computational outcome zero to rho because its probability is larger there. Assign outcome one to sigma. These assignments select the positive and negative eigenspaces of the difference.

  3. Average the conditional successes with their priors: one half times 0.8 plus one half times 0.7 equals 0.75. This agrees with the norm formula and exceeds blind guessing by 0.25.

The result

The optimal measurement reads the common basis, guesses rho on zero and sigma on one, and succeeds with probability 0.75.

Explore the model

Interactive lab

Interactive lab · quantum

A basis can hide perfect distinguishability

For each fresh comparison Reset and set both controls. Compare mixing angle 90 degrees with relative phase 0 degrees against angle 90 and phase 180. Predict equal basis probabilities of one half but opposite Bloch x coordinates, +1 and −1. Inspect at time zero. This is pure-qubit geometry, without a measurement, mixed-state editor, discrimination optimizer, or collapse simulation.

A basis can hide perfect distinguishabilityA 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 forIdentical probabilities in one basis do not imply identical states or zero optimal distinguishability.

Check your understanding

Two equal-prior states have trace distance 0.6. What is their optimal single-copy minimum-error success probability?

Try it yourself

A source chooses ρ=∣0⟩⟨0∣\rho=|0\rangle\langle0| with prior 0.9 and σ=I/2\sigma=I/2 with prior 0.1. Find the minimum-error strategy and success.

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

Give me a hint

Diagonalize the prior-weighted difference, retaining the signs of both entries.

Compare with the solution

The weighted difference is diag⁡(0.85,−0.05)\operatorname{diag}(0.85,-0.05). Guess rho on zero and sigma on one; the success is 0.9(1)+0.1(1/2)=0.950.9(1)+0.1(1/2)=0.95. Equivalently its trace norm is 0.90, giving (1+0.90)/2=0.95(1+0.90)/2=0.95. Always guessing rho would achieve only 0.9. An outcome one is decisive for sigma even though sigma has a small prior probability.

Take it with you

  • Trace distance quantifies the best binary decision advantage for equal priors.
  • The optimal effect selects the positive spectrum of the appropriate weighted difference.
  • A common deterministic quantum channel cannot improve optimal distinguishability.

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