No-cloning theorem does not seem precise

As per wikipedia, no-cloning theorem states that it is impossible to create an identical copy of an arbitrary unknown quantum state.

But from which distribution is this unknown quantum state sampled from? What does the counterfeiter know about this distribution? What is the maximum probability that a counterfeiter could succeed?

Let's say if we randomly pick one of the 2 quantum states $$|\psi_0\big>$$ and $$|\psi_1\big>$$ and give it to a counterfeiter (the states $$|\psi_0\big>$$ and $$|\psi_1\big>$$ are fixed ahead of time and are known to the counterfeiter) . The counterfeiter could measure the given state in $$|\psi_0\big>$$ and $$|\psi_0^\perp\big>$$ basis and forge with non-zero probability.

I believe success probability of a counterfeiter depends on the distribution on which the unknown state is sampled from. I would like to know for what distribution of states the no cloning theorem is applicable.

• In your question, what are $|\psi_0\rangle$ and $|\psi_1\rangle$? Are they known to be one of $\{|0\rangle,|1\rangle,|+\rangle,|-\rangle\}$? What do we know about $\langle\psi_0|\psi_1\rangle$? Jul 25, 2019 at 19:26
• If $|\psi_0\rangle$ were $|1\rangle$ and $|\psi_1\rangle$ were $|+\rangle$, and the counterfeiter (Eve) were given $|\psi_b\rangle$ for a random coin $b\in\{0,1\}$ unknown to her, she could not "forge with non-zero probability," correct? Jul 25, 2019 at 21:20
• Or are you saying that $|\psi_0\rangle=\alpha_0|0\rangle+\beta_0|1\rangle$, and $|\psi_1\rangle=\alpha_1|0\rangle+\beta_1|1\rangle$ with $\alpha_0^2$ and $\alpha_1^2$ chosen uniformly at random from $[0,1]$? That seems like @AHusain's answer already addresses this. Jul 25, 2019 at 21:27
• If the measurement output in the computational basis is $0$ then Eve knows that $b=1$. If the measurement output is $1$ in the computational basis, then she can guess, with 50% success, that $b=0$. However, she can can do better than 50%! For example, please see the answers in this question, which may be of some help. Jul 25, 2019 at 22:19
• The no-cloning theorem isn't about the kinds of operations you're talking about. Once you're allowing the operation to fail or to destroy the input state, you've left the no-cloning theorem's domain of applicability. The no-cloning theorem is a theorem about perfectly reliable cloning operations that work 100% of the time and always leave the input in exactly the state it started in. Jul 26, 2019 at 4:14

The no-cloning theorem itself can be stated very precisely.

Given an unknown pure state $$|\psi\rangle$$ that is drawn from a distribution $$\{p_i,|\phi_i\rangle\}$$ (known to the counterfeiter), it is impossible to create a perfect clone with unit probability unless all the states are orthogonal, $$\langle\phi_i|\phi_j\rangle=\delta_{ij}\ \forall\ i,j$$.

Of course, as you say, there are strategies that either (i) succeed perfectly but with some probability of success less than 1, or (ii) always succeed but create the target state with some finite accuracy, as measured by the fidelity (F<1). In my experience, (ii) is the more common case.

What these parameters are is highly dependant upon the distribution that the states are drawn from, and there's no universal answer. The sort of cases you will find that have been explicitly calculated include:

1. Universal cloning, where the distribution is uniform over all possible pure states of a fixed Hilbert space dimension.
2. Equatorial cloning (this is my name for it, many in the literature use 'phase covariant cloning' but I find that misleading). The system to be cloned is a qubit, equally likely to be any pure state on a great circle (usually the equator) of the Bloch sphere, i.e. $$(|0\rangle+e^{i\phi}|1\rangle)/\sqrt{2}$$ for any $$\phi\in[0,2\pi)$$.
3. Phase covariant cloning (again, my name for it). A qubit state $$\cos\theta|0\rangle+\sin\theta e^{i\phi}|1\rangle$$ for some arbitrary distribution $$f(\theta)$$, but still uniform over $$\phi$$. In other words, the distribution is invariant under rotations around the z axis of the Bloch sphere.
4. A discrete pair of non-orthogonal states.

You can derive these success parameters for arbitrary numbers of clones, and you can even demand that different clones have different fidelities, and derive the optimal trade-off between them (full disclosure: I've worked on that quite a bit).

I should probably emphasise that what you want from cloning is highly dependant upon application. In a lot of theory, it is the simple fact that perfect cloning is impossible that is useful within other proofs. Figures of merit are only really relevant if you have a concrete scenario where you want to implement approximate cloning. I'm not aware of many. Note that, for example, it's not so useful in security proofs of e.g. quantum key distribution, because that might allow you to bound how well a particular class of strategies perform, but how do you know there aren't other strategies not included in your analysis?

To keep the problem small, let's say 1 qubit.

In the original statement $$| \psi \rangle$$ could be any state $$\alpha | 0 \rangle + \beta | 1 \rangle$$ for whatever $$\alpha$$ and $$\beta$$ produce a well defined state.

Quantifying over all so not a probabilistic statement. Say you uniformly draw over all possible states. It is impossible to guarantee that you have gotten $$| \psi \rangle \otimes | \psi \rangle$$.

However if you give a probability distribution with smaller support, then it does become possible. Let's say you know the state is either $$|0\rangle$$ or $$|1\rangle$$ with probabilities $$p$$ and $$1-p$$ respectively. Then you could guarantee that you have gotten $$| \psi \rangle \otimes | \psi \rangle$$. You just apply a $$CNOT$$ on $$| \psi \rangle \otimes | 0 \rangle$$.

Further example, suppose the state is either $$|0\rangle$$, $$|1\rangle$$, $$| + \rangle$$ or $$|-\rangle$$ with probabilities $$p-\epsilon_1$$, $$1-p-\epsilon_2$$, $$\epsilon_1$$ and $$\epsilon_2$$ respectively. Then you could use the same gate and it would still be likely to succeed because $$\epsilon_1$$ and $$\epsilon_2$$ are so small.

You could make a statement quantified like $$\forall | \psi \rangle$$ in the support of a given probability distribution. Then it would depend on the distribution. If it was the first one, supported only on 2 states, then it would be possible to get $$| \psi \rangle \otimes | \psi \rangle$$. But if it was the second it would be impossible to guarantee that result even if it would be quite likely.

• "Say you uniformly draw over all possible states." This isn't possible as there's no uniform distribution over $\mathbb{R}$. Jul 26, 2019 at 8:43
• @otah007 "Say you uniformly draw over all possible states": The set of states is compact, as the probabilities for all outcomes have to sum to 1 Jul 26, 2019 at 12:00