No-cloning theorem with $3$ particles

Question

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?

Answer 1

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

Answer 2

Let's consider the most generic situation and ask the question:

Does a linear unitary operator $U$ exist such that $U\vert \psi\rangle_1\otimes\vert\alpha\rangle_2=e^{if(\psi,\alpha)}\vert\psi\rangle_1\otimes\big(\vert \psi\rangle\otimes\vert\xi_\psi\rangle\big)_2$, $\forall\vert\psi\rangle\in\mathcal{H}_1$?

Notice that this includes the case of the substrate state using any number of particles that can also be entangled among themselves. Allowing for any kind of entanglement that one might want in the substrate state helps address questions along the lines of Does the success of quantum teleportation means quantum cloning is possible too?. Also, the post-cloning state has the room for storing whatever unwanted stuff one might have at the end of the process in addition to the cloned copy. So, as claimed, this is the most generic form of the cloning procedure that one can imagine.

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Now, the proof that such an operator cannot be constructed is pretty much just a copy of the proof provided by @Trimok in their elegant answer. I just wanted to emphasize that the same proof can be carried out for the most generic cloning claim -- not just for a cloning claim involving two particles in the substrate state.

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If such a postulated operator were to exist, the following should hold:

• $U\vert \psi\rangle_1\otimes\vert\alpha\rangle_2=e^{if(\psi,\alpha)}\vert\psi\rangle_1\otimes\big(\vert \psi\rangle\otimes\vert\xi_\psi\rangle\big)_2$
• $U\vert \phi\rangle_1\otimes\vert\alpha\rangle_2=e^{if(\phi,\alpha)}\vert\phi\rangle_1\otimes\big(\vert \psi\rangle\otimes\vert\xi_\phi\rangle\big)_2$

This implies, from unitarity, that

$$\langle\psi\vert\phi\rangle=e^{-if(\psi,\alpha)+if(\phi,\alpha)}(\langle\psi\vert\phi\rangle)^2\langle\xi_\psi\vert\xi_\phi\rangle$$

Thus, if $\langle\psi\vert\phi\rangle\neq 0$,

$$1=e^{-if(\psi,\alpha)+if(\phi,\alpha)}\langle\psi\vert\phi\rangle\langle\xi_\psi\vert\xi_\phi\rangle\implies\vert\langle\xi_\psi\vert\xi_\phi\rangle\vert^2=\frac{1}{\vert\langle\psi\vert\phi\rangle\vert^2}\geq1$$

which can be consistent only if $\vert\psi\rangle=\vert\phi\rangle$ and inconsistent otherwise. Thus, for generic states, $\vert\psi\rangle,\vert\phi\rangle$, unitarity prohibits our proposed operator.