# No-cloning theorem and distinguishing between two non-orthogonal quantum states revisited revisited

There are many posts to this question from Nielson and Chuang's Quantum Computation and Quantum Information Exercise 1.2 page 57. It is required to prove that if a hypothetical device exists, which could distinguish between two non-orthogonal quantum states $$|\psi\rangle$$ or $$|\phi\rangle$$ and correctly output the answer, we could create a device that cloned the states $$|\psi\rangle$$ and $$|\phi\rangle$$ , contradicting the no-cloning theorem.

I propose an algorithm that will clone any state.

1. Take $$|\psi\rangle$$ and $$|\psi'\rangle$$ as basis, with $$|\psi'\rangle$$ being orthogonal to $$|\psi\rangle$$ .
2. Measure a new qubit with this basis.
3. If the result is $$|\psi\rangle$$ , then we are done. If the result is $$|\psi'\rangle$$ , then use the X gate to get $$|\psi\rangle$$ as the result.
4. Repeat this for all other qubits taking those qubits and their orthogonal qubit as the basis.

I suspect that changing the basis is a problem, but I am not able to pinpoint what exactly is the problem. I am quite new to quantum computing and even anything quantum in general. So, I require easy-to-understand explanations.

In other answers, cloning $$|\psi\rangle$$ is considered trivial if we identify which one of the qubits is $$|\psi\rangle$$ . But this is not trivial to me, we only know that this qubit is $$|\psi\rangle$$ and not $$|\phi\rangle$$ but we didn't know the states of $$|\psi\rangle$$ and $$|\phi\rangle$$ from the beginning. We only know that out of two unknown non-orthogonal states, our test qubit is $$|\psi\rangle$$ , and $$|\phi\rangle$$ is the other one. So how can we clone a qubit if nothing is known about it?

The crucial point about the no-cloning theorem is (as you already suspect) the basis choice.

You can always find a specific cloning machine for a specific basis, but you cannot find a general cloning machine that accepts any state and clones it.

This is probably best understood via examples and this topic has been discussed before. Please have a look here or here.

## Edit: a simple example

The simplest version of the argument that I know of is this:

Let's say you have a machine $$M$$ that is able to clone the states $$|0\rangle$$ and $$|1\rangle$$. This could be written like this:

$$|0\rangle \overset{M}{\to} |0\rangle |0\rangle$$ $$|1\rangle \overset{M}{\to} |1\rangle |1\rangle$$

This mapping property fully determines how that machines acts on the state $$|+\rangle = \frac{1}{\sqrt 2} (|0\rangle + |1\rangle)$$ due to the linearity of quantum mechanics:

$$|+\rangle = \frac{1}{\sqrt 2} (|0\rangle + |1\rangle) \overset{M}{\to} \frac{1}{\sqrt 2}(|0\rangle |0\rangle + |1\rangle |1\rangle) \neq |+\rangle |+\rangle$$

So, the machine fails to clone the $$|+\rangle$$ state. In fact, it will fail to clone any state other than $$|0\rangle$$ or $$|1\rangle$$.

To conclude, you can define a machine that is able to clone states of a given basis. But by that choice you restrict that machine from cloning any other states.

• I am confused about why the basis matters, you clone on a specific basis 1, and then the qubit is sent into a circuit with a different basis 2, this time the cloned state is useful as although we knew its state in the basis 1, its state in basis 2 is a superposition. Keep in mind that we don't know how basis 1 and 2 related to each other. Commented Jun 22 at 4:34
• I have edited my answer to include a simple example. Commented Jun 24 at 11:25

I think what you asking about is the following setting: you are given a qubit which is either $$|\psi\rangle$$ or $$|\psi'\rangle$$ with are not orthogonal. You know what those two possible states are, but not which it is. You have to clone it.

You start by measuring in a basis that prijects onto $$|\psi\rangle$$. If you get the other answer, you know you did not have $$|\psi\rangle$$, so you produce multiple copies of $$|\psi'\rangle$$. That part is no problem.

The problem is what happens if you get the answer corresponding to $$|\psi\rangle$$? Since $$|\psi\rangle$$ and $$|\psi'\rangle$$ are not orthogonal, there's some chance that the original state was $$|\psi'\rangle$$, you simply don't know. Even worse, now, whatever the state was initially, it's $$|\psi\rangle$$. So all information about what the state was has been lost. You cannot tell what the state was, and you have no hope to clone it.