# Why does a quantum cloner imply superluminal communication

I am wondering why superluminal communication would be possible if a quantum cloner would exist?

The common argument (FLASH) goes as follows: Alice and Bob share the Bell state $$|\psi^-\rangle = \frac{1}{\sqrt 2}(|01\rangle - |10\rangle)$$ Now if Bob could make $2N$ copies of his qubit, he could find out in which basis Alice has measured before a signal with this information could reach him.

Let's consider Bob makes $2N$ copies of his qubit: $$\frac{1}{\sqrt 2}(|01\rangle - |10\rangle) \quad \longrightarrow \quad \frac{1}{\sqrt 2}(|0\underbrace{1...1}_{2N}\rangle - |1\underbrace{0...0}_{2N}\rangle)$$ Now if Alice measures in the z-basis $\{|0\rangle,|1\rangle\}$ Bob will be left with either $|1...1\rangle$ or $|0...0\rangle$. So every qubit is in a pure state: either $|1\rangle$ or $|0\rangle$. So he now measures $N$ qubits in the z-basis and $N$ in the x-basis $\{|+\rangle,|-\rangle\}$. Since he will get $N$ times the same result for the measurements in the z-basis he concludes that Alice has measured in the z-basis.

However, if Alice measures in the x-basis Bob will be left with either $$\frac{1}{\sqrt 2}(|1...1\rangle - |0...0\rangle)$$ or $$\frac{1}{\sqrt 2}(|1...1\rangle + |0...0\rangle)$$ Thus all of his qubits are in a completely mixed state $\frac 1 2 (|0\rangle\langle 0| + |1\rangle\langle 1|) = \frac 1 2 (|+\rangle\langle +| + |-\rangle\langle -|)$. So if Bob now measures $N$ qubits in the z-basis and $N$ in the x-basis he will never get $N$ times the same result for one particular basis. So he can conclude nothing.

So what am I doing wrong?

• That's not cloning. What you're doing with the EPR state is an allowed operation in quantum mechanics. Cloning is an operation on a pure state that takes $|0\rangle \rightarrow |0\rangle|0\rangle$, $|1\rangle \rightarrow |1\rangle|1\rangle$, $|+\rangle \rightarrow |+\rangle|+\rangle$, $|-\rangle \rightarrow |-\rangle|-\rangle$. – Peter Shor Mar 6 '15 at 19:54

The things are simpler than you say. Though, let me work with the polarization singlet of photons,

$$|S\rangle = \frac {|+\rangle |+\rangle + |-\rangle |-\rangle}{\sqrt {2}}.$$

Let Bob prepare 4 sets of polarized photons:

(1) $|+\rangle$, (2) $|-\rangle$, (3) $|u\rangle = \frac {|+\rangle + |-\rangle}{\sqrt{2}}$, (4) $|v\rangle = \frac {|+\rangle - |-\rangle}{\sqrt{2}}$.

A) Assume that Alice measures in the base $\{ |+\rangle, |-\rangle \}$. At a certain trial, she obtains $|+\rangle$. Bob's photon, therefore, will be polarized $|+\rangle$, and he makes $4N$ copies. Each copy he brings to a beam-splitter.

(a) Each copy from the first set of $N$ he meats with a photon from the set $|+\rangle$. All the trials will give that both photons go either to the one side of the BS, or to the other. In no case will he get one photon to one side and the other to the other side.

(b) Each copy from the 2nd set of $N$ he meats with a photon from the set $|-\rangle$. Part of the trials will give that both photons go either to the one side of the BS, and part of the trials will give one photon to one side and the other to the other side.

(c) Each copy from the 3rd set of $N$ he meats with a photon from the set $|u\rangle$. Part of the trials will give that both photons go either to the one side of the BS, and part of the trials will give one photon to one side and the other to the other side.

(d) Each copy from the 4th set of $N$ he meats with a photon from the set $|v\rangle$. Part of the trials will give that both photons go either to the one side of the BS, and part of the trials will give one photon to one side and the other to the other side.

B) Assume that Alice measures in the base $\{ |u\rangle, |v\rangle \}$, and in a certain trial she gets $|v\rangle$. So, Bob gets the same. Again he makes $4N$ copies, and brings copy to the beam-splitter, and meets the copy with photons according to the protocol (a) thru (d).

The only case in which he gets that both photons go to the same side of the BS, is (d).

Thus, Bob could know in what basis Alice measured.