Theorem: Single-copy and asymptotic detection

For independent identically prepared outputs of the fixed interface:

1. With equal priors, the optimal single-copy success probability is $\frac{1}{2}(1+\delta )$.

2. The optimal symmetric error exponent is

   $${\xi }_C=-\log \underset{0\le s\le 1}{\min }\mathrm{tr}({\varrho }^s{\varsigma }^{1-s}),$$

   with endpoint powers interpreted as support projections. It satisfies

   $$-\frac{1}{2}\log (1-{\delta }^2)\le {\xi }_C\le -\log (1-\delta ).$$

3. At fixed type-I error in $(0,1)$, the Stein exponent, with $\varrho$ as the null state, is $D(\varrho \Vert \varsigma )$, and

   $$2{\delta }^2\le D(\varrho \Vert \varsigma )\le \log k\le |J|\log 2.$$

Thus the two models are distinguishable at a positive exponential rate through this interface if and only if $\delta >0$. The Chernoff exponent has a two-sided bound using $\delta$ alone; the Stein upper bound requires additional information, here the pinching-sector count.

Proof. The first assertion is Helstrom optimization.

The quantum Chernoff theorem identifies the symmetric exponent Audenaert2007,NussbaumSzkola2009. Put $q_{\ast }={\min }_s\mathrm{tr}({\varrho }^s{\varsigma }^{1-s})$. The standard trace inequality used in its proof gives $q_{\ast }\ge 1-\delta$. Also,

where the last inequality is Fuchs–van de Graaf FvdG1999. Taking negative logarithms proves (eq:chernoff).

The quantum Stein theorem gives the asymmetric exponent HiaiPetz1991,OgawaNagaoka2000. Quantum Pinsker gives $D\ge \frac{1}{2}\Vert \varrho -\varsigma {\Vert }_1^2=2{\delta }^2$ Watrous2018. Lemma lem:domination gives the upper bound and ensures the required support inclusion. If $\delta =0$, the output states are identical at every sample size. If $\delta >0$, the lower bounds are positive.