# Question about the no-cloning theorem

1. Nov 24, 2014

### jk22

In this theorem we have a unitary transformation $U|a>|b>=|a>|a>$

But isnt it obvious that this is a rotation on a subspace but this rotation should depend on both |a> and |b> ?

With this dependence it seems to me the conclusion cannot be reached since the unitarity is U(a,b)U(a,b)^+=1 but U(c,b)U(a,b)^+ is not forcedly 1.

2. Nov 24, 2014

### VantagePoint72

It's not really clear to me what you're asking. The point of the no cloning theorem is that the unitary you give does not exist. Is that the conclusion you're saying cannot be reached? Can you be more specific and quote the proof you're reading and point us to the particular part of the proof you're not following?

In any case, a "rotation" that depends on the thing you're "rotating" is not a rotation at all, so that part of your comment doesn't track.

If we consider the simple case of cloning a single qubit with no working bits then the argument is pretty simple. We have that $U|0\rangle |0\rangle = |0\rangle |0\rangle$ and $U|1\rangle |0\rangle = |1\rangle |1\rangle$ (it is not necessary to assume anything about what the proposed unitary does when the target qubit is non-zero). Then if $|\psi\rangle = \alpha |0\rangle | + \beta |1\rangle$, $|\alpha|^2 + |\beta|^2 = 1$, we have by linearity that
$U|\psi\rangle|0\rangle = \alpha U|0\rangle |0\rangle + \beta U |1\rangle |0\rangle = \alpha |0\rangle |0\rangle + \beta |1\rangle |1\rangle \neq |\psi\rangle|\psi\rangle$ for all $\alpha$ and $\beta$.

With a bit more work, the argument can be generalized to arbitrary states and with an arbitrarily-big ancilla register for the cloning operator to use as working space. More general proofs use the fact that a unitary operator must, by definition, preserve the inner product between all pairs of states and show that any proposed cloning operator cannot preserve inner products between non-parallel, non-orthogonal states.

Last edited: Nov 24, 2014
3. Nov 25, 2014

### jk22

U cannot be linear in fact it is not even an application since :

U|2a>|b>=|2a>|2a>=4|a>|a>=U2|a>|b>
=U|a>|2b>=|a>|a>

4. Nov 25, 2014

### jk22

Another proof that such an operation cannot exist

5. Nov 25, 2014

### jk22

To be complete only the null state can be cloned since it gives again a null state whicj is an impossible event.