There are a couple of posts on this question, but I think they are not satisfactory. The question is Nielsen and Chuang's QCQI, Exercise 1.2, page 57, which asks "Explain how a device which, upon input of two non-orthogonal states correctly identified the state, could be used to build a device which cloned the states..." The answer offered here is that if you knew the state you could synthesize a copy of it. That sounds convincing but I notice that they do not use the condition that the states are not orthogonal. Clearly, orthogonal states can be distinguished and yet doing so cannot lead to cloning. So the posted answer seems to be incomplete or incorrect.

If was assume that the hypothetical circuit that distinguishes non-orthogonal states outputs a qubit $|0\rangle$ for the first case and $|1\rangle$ for the second case, then I think it is easy to show that the circuit would not be unitary. So in a sense this is an answer to the exercise, but not exactly since it doesn't mention cloning.


1 Answer 1


TL;DR: The assumption of non-orthogonality is implicitly used by the linked answer. It is needed due to a "loophole" in the no-cloning theorem that allows cloning of known orthogonal states.

Universal cloner is prohibited

The no-cloning theorem is the statement that there is no unitary $U$ such that

$$ U|\gamma\rangle|b\rangle = |\gamma\rangle|\gamma\rangle\tag1 $$

for all states $|\gamma\rangle$ and a fixed state $|b\rangle$.

The state $|\gamma\rangle$ represents the input state to be cloned and thus can be anything and the state $|b\rangle$ represents a blank state to be overwritten in the process of cloning the input state. See also box $12.1$ on page $532$ in Nielsen & Chuang. It is instructive to see the proof of the theorem.

Proof Suppose unitary $U$ exists. Let $|u\rangle$ and $|v\rangle$ be two normalized distinct non-orthogonal quantum states. Their inner product is

$$ x=\langle u|v\rangle = \langle u|v\rangle\langle b|b\rangle = \langle u|\langle b|U^\dagger U|v\rangle|b\rangle = \langle u|v\rangle\langle u|v\rangle=x^2. $$

Thus, $x=0$ or $x=1$ contradicting the assumption that $|u\rangle$ and $|v\rangle$ are distinct and non-orthogonal. $\square$

Loophole: Known orthogonal states may be cloned

The key observation to make about the proof is that it uses the requirement that the cloner is supposed to work for all inputs to apply it to non-orthogonal inputs which then leads to a contradiction. However, we avoid the contradiction if the states are orthogonal!

Thus, a weak cloner that satisfies $(1)$ when $|\gamma\rangle$ is one of two known orthogonal states $|\phi\rangle$ and $|\psi\rangle$ is not prohibited. In fact, it is easy to check that any unitary extension of the partial isometry

$$ V = |\phi\rangle|\phi\rangle\langle\phi|\langle b| +|\psi\rangle|\psi\rangle\langle \psi|\langle b|\tag2 $$

achieves cloning for $|\phi\rangle$ and $|\psi\rangle$.

Mathematical intuition for the loophole

This is an instance of a general property of linear functions wherein we are free to dictate their action on a basis, but once the action on the basis is chosen the action on all other vectors is fixed by linearity. The essence of the no-cloning theorem is that the linear extension of cloning the basis vectors fails to act as cloning on all other vectors.

Physical intuition for the loophole

The loophole is not surprising when we look at the no-cloning theorem from the perspective of the correspondence principle, i.e. the common sense requirement that since our macroscopic world is classical there must exist conditions under which quantum mechanics reproduces classical physics. It turns out that known orthogonal quantum states essentially encode classical information and therefore exhibit all its characteristic properties such as reliable distinguishability and clonability.

No-cloning theorem and distinguishability

Returning to the question, we find that the existence of the above loophole in the no-cloning theorem explains the need for the assumption that $|\psi\rangle$ and $|\phi\rangle$ are not orthogonal.

That sounds convincing but I notice that they do not use the condition that the states are not orthogonal.

The non-orthogonality assumption is needed to arrive at a contradiction, because, contrary to what is stated in the question, the no-cloning theorem does not actually rule out a cloner for known orthogonal states.

Clearly, orthogonal states can be distinguished and yet doing so cannot lead to cloning.

This is not true. Known orthogonal states can be cloned at will. See $(2)$ above for an explicit formula for the cloner.

  • $\begingroup$ How did you get V? $\endgroup$
    – morpheus
    Oct 11, 2022 at 1:11

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