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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.

Let distinct nonorthogonal pure states have overlap magnitude $0<c<1$, and let one quantum channel take $n$ identical inputs to $m$ outputs. Let ${ϵ}_u,{ϵ}_v$ be the respective probabilities that a test for the ideal $m$-copy pure state fails. Trace distance contraction, its triangle inequality and the pure-target upper bound on trace distance give $\sqrt{1-c^{2m}}\le \sqrt{{ϵ}_u}+\sqrt{1-c^{2n}}+\sqrt{{ϵ}_v}$. Squaring with the Cauchy-Schwarz inequality gives the displayed strategy-independent positive bound. It establishes both a worst-case bound and an equal-prior average bound. Averaging rotated pairs establishes a positive Haar-average bound for unknown qubit inputs, without assuming independent clone errors.

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.

Let $P_{\varphi }=|\varphi ⟩⟨\varphi |$. An arbitrary deterministic strategy, including auxiliary systems, intermediate measurements, adaptive actions and discarded records, is a quantum channel $\mathcal{E}$ from the two input qubits to the three returned qubits. Its output is $R_{\varphi }=\mathcal{E}(P_{\varphi }^{\otimes 2})$. The three local tests commute, and their joint all-pass projector is $P_{\varphi }^{\otimes 3}$. Thusis exactly the probability of at least one failed test. No independence of the three returned qubits is assumed.

We prove a uniform positive obstruction using trace distance, $D(\rho ,\sigma )=\Vert \rho -\sigma {\Vert }_1/2$. Two normalized pure kets obeyIndeed their projector difference is supported on their two-dimensional span, has trace zero, and has eigenvalues $\pm \sqrt{1-|⟨u|v⟩{|}^2}$.

Two further estimates will cover even mixed outputs. First, trace distance contracts under a quantum channel. To see this directly, the positive spectral projector of the trace-zero difference gives $D(\rho ,\sigma )={\max }_{0\le A\le I}\mathrm{Tr}[A(\rho -\sigma )]$. The adjoint of a channel is positive and unital, so it takes any output effect into an input effect. Maximizing over this subset cannot exceed the input maximum, proving contraction.

Second, for a mixed state $R={\sum }_j{q}_j|u_j⟩⟨u_j|$ and a pure target $P=|v⟩⟨v|$, the trace-norm triangle inequality and weighted Cauchy-Schwarz inequality give the pure-target upper bound on trace distance:In particular $D(R_{\varphi },P_{\varphi }^{\otimes 3})\le \sqrt{{ϵ}_{\varphi }}$.

Now take $|{\varphi }_0⟩=|0⟩$ and $|{\varphi }_1⟩=|+⟩=(|0⟩+|1⟩)/\sqrt{2}$. The two-copy input distance is $D_2=\sqrt{1-1/4}=\sqrt{3}/2$, whereas the ideal three-copy distance is $D_3=\sqrt{1-1/8}=\sqrt{7/8}$. Writing $R_j=R_{{\varphi }_j}$, contraction and the triangle inequality yieldLet $\delta =D_3-D_2>0$. Then $\sqrt{{ϵ}_0}+\sqrt{{ϵ}_1}\ge \delta$, and squaring givesThis is the copy-number trace-distance obstruction. It applies to every Bob strategy, irrespective of its internal complexity. If Alice chooses these two possible inputs with equal probabilities, the left side is her unconditional failure probability. For every device, at least one of the two inputs also has failure probability at least ${\delta }^2/4$, establishing the worst-case guarantee.

The same positive bound holds if Alice chooses a uniformly random pure-qubit state. Apply the preceding inequality to $U|0⟩$ and $U|+⟩$ for every single-qubit unitary $U$, since their overlap is unchanged. Average over normalized Haar measure. Both rotated kets have the same uniform distribution on the Bloch sphere, so each of the two average failures is the uniform-input average. Consequently that average is also at least ${\delta }^2/4$. A constant satisfying the strict requested inequality in either this average or the worst-case interpretation isThe proven bound is $2p$, hence strictly greater than $p$. This proof establishes a quantitative gap, rather than relying only on the impossibility of exact quantum cloning.

There is a necessary quantifier qualification because the PDF does not specify Alice's input distribution. A positive lower bound for every fixed state against every strategy would be false: a device hardwired to return $|000⟩$ never fails when the input is $|0{⟩}^{\otimes 2}$. The unknown-state assertion is therefore meaningful as a worst-case guarantee over possible inputs, or as an average for a specified hidden input ensemble. Both the equal-prior example and the uniform-input version have been proved here, with a constant independent of Bob's strategy.