# No-cloning theorem with 3 particles

I know how to demonstrate that it is not possible to make a unitary operator so that $|a\rangle|0\rangle$ turns into $|a\rangle|a\rangle$ , but is it possible to have $|a\rangle|0\rangle|0\rangle \rightarrow |a\rangle|a\rangle|c(a)\rangle$ which would allow a kind of cloning thanks to a third particle which would be useless except to preserve unitarity ?

• What is c(a) here? – unsym Aug 21 '13 at 18:16
• I mean that the final state of the third particle can depend on the one of the particle to be cloned – agemO Aug 21 '13 at 18:24
• I think you meant the last line to be $a 0 0 \rightarrow a a c(a)$? So that one particle is cloned and the third particle carries away information. – BebopButUnsteady Aug 21 '13 at 18:39
• @agemO Can you be a bit more explicit on what $c(a)$ is? $c(a)=0$ is a function as well! – Ali Aug 21 '13 at 18:54
• "I think you meant the last line to be a00→aac(a)?" exactly I'm so stupid – agemO Aug 21 '13 at 18:57

This does not work.

Imagine the unitary transformation $|a~0~0\rangle \rightarrow |a~a~c(a)\rangle$ So, we have : $|b~0~0\rangle \rightarrow|b~b~c(b)\rangle$

Here, I suppose that the states $a, b, c(a), c(b)$ are normed.

By unitarity, we must have :

$$\langle b~0~0|a~0~0 \rangle = \langle b~b~c(b)|a~a~c(a) \rangle\tag{1}$$ That is :

$$\langle b|a \rangle = \langle b|a \rangle^2 \langle c(b)|c(a) \rangle\tag{2}$$

Suppose $\langle b|a \rangle \neq 0$, then we have :

$$1 = \langle b|a \rangle \langle c(b)|c(a) \rangle\tag{3}$$

Suppose now we that $|\langle b|a \rangle| < 1$, then we should have : $|\langle c(b)|c(a) \rangle| >1$, which is absurd because of the Cauchy-Schwarz inequality

[EDIT]

If you want to release the condition of normed states for $a, b, c(a), c(b)$, you could always write :

$a = ||a||a',~ b = ||b||b',~ c(a) = ||c(a)||c'(a),~c(b) = ||c(b)||c'(b)$ where now, $a', b', c'(a), c'(b)$ are normed, you will have :

$$\frac{1}{||a||~||b||~||c(a)||~||c(b)||} = \langle b'|a' \rangle \langle c'(b)|c'(a) \rangle\tag{4}$$

If I choose $$|\langle b'|a' \rangle| <\frac{1}{||a||~||b||~||c(a)||~||c(b)||}$$, I must have $|\langle c'(b)|c'(a) \rangle| >1$, which is absurd because of the Cauchy-Schwarz inequality

• It is imply because that it the state are normalized, no need for CS inequality – unsym Aug 22 '13 at 22:35
• Perfect, I was like "⟨c(b)|c(a)⟩ can be anything" but no – agemO Aug 23 '13 at 12:00