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For two pure states of overlap $0<c<1$, a quantum channel attempts to turn $M$ copies into $N>M$ copies. Let $e_i$ be the failure probability of the all-copies projector test on input $i$. The pure-state trace distance and fidelity identity, trace-distance contraction under quantum channels, pure-target upper bound on trace distance and triangle inequality give the displayed strictly positive gap $\Delta$. Hence ${\max }_i{e}_i\ge {\Delta }^2/4$, and for positive priors ${\pi }_0+{\pi }_1=1$, weighted Cauchy-Schwarz inequality gives ${\pi }_0{e}_0+{\pi }_1{e}_1\ge {\pi }_0{\pi }_1{\Delta }^2$. No positive state-by-state bound holds: preparing one candidate perfectly makes its own conditional failure probability zero.

There is a necessary distinction between a failure bound averaged over the unknown state, a worst-case bound, and a separate bound for each candidate state. The last interpretation is false: a forger can discard the inputs and always prepare $|{\psi }_0{⟩}^{\otimes N}$. If the actual state is $|{\psi }_0⟩$, every authenticity test passes with probability one. The meaningful uniform statement concerns worst-case failure, or average failure under any fixed prior giving both states positive probability. The PDF does not specify that prior; the argument below gives an explicit bound for either interpretation.

Put $c=|⟨{\psi }_0|{\psi }_1⟩|\in (0,1)$. Any unconditional strategy producing $N$ output qubits, including ancillas, adaptive measurement in quantum measurements and correlated outputs, is a quantum channel $\mathcal{E}$. Its inputs and ideal outputs are the density operatorsThe projective measurements on the different banknotes commute, and their all-pass effect is exactly $P_i$. Therefore the probability of at least one rejection, conditional on input $i$, isThis remains true for entangled outputs; no independence of their individual test outcomes is assumed.

Use trace distance $D(\rho ,\sigma )=\frac{1}{2}\Vert \rho -\sigma {\Vert }_1$. The pure-state trace distance and fidelity identity givesFor completeness, two pure-state projectors differ by a matrix with nonzero eigenvalues $\pm \sqrt{1-|⟨u|v⟩{|}^2}$, which proves this identity. Also the pure-target upper bound on trace distance gives $D({\rho }_i,P_i)\le \sqrt{e_i}$. To see the mixed-state step, write ${\rho }_i={\sum }_j{q}_j|u_j⟩⟨u_j|$ and use convexity of the trace norm followed by the Cauchy-Schwarz inequality:Finally, trace-distance contraction under quantum channels gives $D({\rho }_0,{\rho }_1)\le D({\sigma }_0,{\sigma }_1)$. This follows from $D(\rho ,\sigma )={\max }_{0\le E\le I}\mathrm{Tr}[E(\rho -\sigma )]$: the adjoint channel is positive and unital, so its image of any effect $E$ is still an effect.

The triangle inequality now yieldsThus the cloning failure bound from trace-distance contraction isand in particular $\max (e_0,e_1)\ge {\Delta }^2/4$. For positive priors ${\pi }_0+{\pi }_1=1$, another Cauchy-Schwarz inequality givesso the average rejection probability satisfiesFor equal priors, one may take $p_0={\Delta }^2/8$; the same choice is a strict lower bound on worst-case rejection. This proves a quantitative version of the no-cloning theorem for the bank's all-pass test. It concerns an unconditional attempt: heralded success cannot be postselected while omitting failed attempts from the accounting.

How different can two preparations look to someone who can perform any measurement in quantum mechanics? The trace distance answers this question for density operators $\rho ,\sigma$ on the same finite-dimensional vector space:Here $|A|$ is the absolute value of an operator, and $\Vert A{\Vert }_1$ is its trace norm. Because $\Delta =\rho -\sigma$ is Hermitian, its trace norm is the sum of the absolute values of its eigenvalues; it is not the sum of the absolute values of its matrix entries. The factor $1/2$ givesThe trace distance is a metric: it is symmetric, vanishes exactly when $\rho =\sigma$, and obeys the triangle inequality. Identical preparations have trace distance zero; preparations with orthogonal occupied eigenspaces have trace distance one and can be identified perfectly in a single measurement in quantum mechanics.

Operationally, the trace distance is the largest total variation distance between the outcome probability distributions obtainable by a POVM. Equivalently, the variational characterization of trace distance says it is the largest possible probability gap for one event in a binary measurement in quantum mechanics. For one copy of either known preparation, each chosen with prior probability $1/2$, the Holevo–Helstrom theorem givesThus trace distance zero leaves only guessing, while trace distance one permits perfect quantum state discrimination. The optimum ranges over measurements; different measurement bases can expose different amounts of distinguishability. ^[ref]

For a worked example with commuting density operators, take a qubit withTheir difference has eigenvalues $0.5,-0.5$, so $D=(|0.5|+|-0.5|)/2=0.5$. Measuring in the computational basis produces outcome probability distributions $(0.8,0.2)$ and $(0.3,0.7)$, whose total variation distance is also $0.5$. Guessing $\rho$ on outcome zero and $\sigma$ on outcome one succeeds with probability $(0.8+0.7)/2=0.75$. A projective measurement in the Hadamard basis instead produces $(0.5,0.5)$ for both preparations and gives no discrimination advantage.

The trace-distance contraction under quantum channels ensures that a common quantum channel, including the partial trace, cannot increase the trace distance. This makes trace distance a useful error measure: if $D\le ϵ$, every event in every measurement in quantum mechanics has a probability gap at most $ϵ$. Losing access to a subsystem can hide distinguishability, but cannot create more of it.