Two-State Cloning Bounds and Heralded Cloning

Exact deterministic cloning fails for two distinct nonorthogonal pure states. The canonical no-cloning article proves that statement and distinguishes cloning from broadcasting. Here the same obstruction becomes quantitative: any deterministic cloner has a nonzero error, while an exact heralded cloner can succeed with a calculable probability below one. Neither result asserts that approximate cloning is useless or that probabilistic cloning is forbidden.

Required background. Fidelity and Trace Distance supplies trace-distance contraction under channels and its pure-state formula; No-Cloning and No-Signaling specifies one common physical map for the declared input family.

A lower bound on deterministic cloning error

Take unit vectors $\psi_0,\psi_1$ and choose phases so that $\langle\psi_0,\psi_1\rangle=s\in(0,1)$. Write $\rho_i=|\psi_i\rangle\langle\psi_i|$. A fixed blank register and any fixed auxiliary preparation can be absorbed into a CPTP map $\mathcal C$ from the input to two output systems. Define its joint-output errors by

$$\epsilon_i=D(\mathcal C(\rho_i),\rho_i^{\otimes2}), \qquad D(\rho,\sigma)=\tfrac12\|\rho-\sigma\|_1.$$

The triangle inequality, followed by contraction, yields

$$\begin{aligned} D(\rho_0^{\otimes2},\rho_1^{\otimes2}) &\le\epsilon_0+\epsilon_1 +D(\mathcal C(\rho_0),\mathcal C(\rho_1))\\ &\le\epsilon_0+\epsilon_1+D(\rho_0,\rho_1). \end{aligned}$$

The two pure-state overlaps are $s$ and $s^2$. Therefore

$$\epsilon_0+\epsilon_1 \ge\sqrt{1-s^4}-\sqrt{1-s^2}>0.$$

In particular, the larger error is at least half the right side. This bound follows directly from distinguishability and makes no assumption of unitary evolution on the two observed outputs alone. An environment or discarded register is already allowed by the channel description.

The bound need not be the optimal achievable cloning error. It applies to the distance from the joint product target, not merely to two marginal fidelities. Correlated approximate outputs can have quite different joint and marginal errors; a statement about one metric is not automatically a statement about another.

Exact cloning conditioned on a success flag

Now allow a measurement record indicating success or failure. A convenient isometric description for equal success probability $p$ is

$$V\psi_i= \sqrt p\,\psi_i^{\otimes2}\otimes e_i\otimes|S\rangle +\sqrt{1-p}\,f_i\otimes|F\rangle,$$

where the flag vectors are orthogonal, and all displayed $e_i,f_i$ are normalized. The failure register can include an arbitrary environment. Inner-product preservation gives

$$s=p s^2\langle e_0,e_1\rangle +(1-p)\langle f_0,f_1\rangle.$$

Taking absolute values gives $s\le ps^2+1-p$, and hence

$$p\le\frac1{1+s}.$$

This is a necessary bound even when the successful branch retains input-dependent information in its environment.

It is also attainable for this two-state, equal-success task. Choose a common success environment $e_0=e_1=e$ and a common normalized failure vector $f_0=f_1=f$. At $p=1/(1+s)$ the proposed outputs have unit norms and mutual inner product $ps^2+1-p=s$, exactly matching the input Gram matrix. Because the two inputs are linearly independent, the linear map defined on their span is an isometry. An enlarged output space permits an isometric extension to the remaining input space, defining a physical channel and its flag measurement.

Thus optimal success is strictly below one for distinct nonorthogonal states. Conditional normalization is allowed only after the flag is observed. The complete instrument, including failures and their probabilities, remains a linear physical operation. Removing the failure branch from the description would change the task.

Limits and related no-go statements

As $s\to0$, the heralded bound tends to one: orthogonal states can be copied deterministically. As distinct states approach coincidence, the bound tends to $1/2$. At exactly $s=1$ the promised family contains only one ray, which can simply be prepared twice with certainty. There is no contradiction: linear independence and the demand to distinguish two distinct target relations disappear at that endpoint.

For mixed states, broadcasting asks only for correct marginals and permits correlations. Its exact finite-dimensional criterion is commutativity, proved in No-Broadcasting Theorem. The two-state pure cloning calculation here should not be substituted for that mixed-state theorem.

Exercise: compare the two tasks at a fixed overlap

Set $s=1/\sqrt2$. Find the lower bound on the worst deterministic joint-output trace-distance error and the optimal equal success probability for exact heralded cloning.

Solution

The error bound is $\max_i\epsilon_i\ge(\sqrt3-\sqrt2)/4$, approximately $0.0795$. The heralded success probability is $1/(1+1/\sqrt2)=2-\sqrt2$, approximately $0.586$. These compare different tasks: the first always returns an approximate output, while the second sometimes reports failure and is exact only on successful trials.

References

• Duan, Lu-Ming, and Guang-Can Guo, “Probabilistic Cloning and Identification of Linearly Independent Quantum States,” Physical Review Letters 80, 4999–5002 (1998), doi:10.1103/PhysRevLett.80.4999 — probabilistic exact cloning and Gram-matrix constraints.
• Watrous, John, The Theory of Quantum Information, Cambridge University Press, 2018, doi:10.1017/9781316848142 — channel distinguishability and trace-distance contraction used in the error estimate.
• Wootters, William K., and Wojciech H. Zurek, “A single quantum cannot be cloned,” Nature 299, 802–803 (1982), doi:10.1038/299802a0 — the original exact cloning obstruction.