Quantum Information HT24, No-cloning theorem

Bob could copy his state many times, and different measurements to determine the actual state with some probability.

It is impossible to construct a copy machine that takes as input a quantum system $A$ in one of two distinct non-orthogonal states $ \vert \phi _ 0\rangle$ and $ \vert \phi _ 1\rangle$ and deterministically returns as output two systems, $A$ and $B$, each of them in the same state as the input system.

There is no unitary gate $U _ {ABM}$ acting on systems $A, B$ and $M$ and no initial state $ \vert \beta\rangle _ {BM}$ of systems $B$ and $M$ such that

\[\text{Tr} _ M\left[ U _ {ABM}(\pmb \rho _ A \otimes {} _ {BM} \vert \beta\rangle\langle \beta \vert _ {BM}) U^\dagger _ {ABM} \right] = {} _ A \vert \phi _ i\rangle\langle \phi _ i \vert _ {A} \otimes {} _ B \vert \phi _ i\rangle \langle \phi _ i \vert _ B\]

(where $ \vert \phi _ i\rangle$ is the same as above).

We have the initial state $ \vert \phi _ i\rangle _ A \otimes \vert \beta\rangle _ {BM}$ and such a unitary gate would transform into the final state $ \vert \Psi _ i \rangle _ {ABM}$ which will be given by:

\[\vert \Psi _ i \rangle _ {ABM} := U _ {ABM}( \vert \phi _ i\rangle _ A \otimes \vert \beta\rangle _ {BM})\]

The cloning condition implies

\[\text{Tr} _ M[ \vert \Psi _ i\rangle \langle \Psi _ i \vert _ {ABM}] = \vert \phi _ i\rangle\langle\phi _ i \vert _ A \otimes \vert \phi _ i\rangle \langle \phi _ i \vert _ B\]

By the result that says

Suppose that two systems $A$ and $B$ are in the pure state $ \vert \Psi\rangle$ and the marginal state on $A$ is pure so that

\[Tr _ B[ \vert \Psi\rangle\langle \Psi \vert ] = \vert \alpha\rangle\langle \alpha \vert\]

for some unit vector $ \vert \alpha\rangle$. Then

\[\vert \Psi\rangle = \vert \alpha\rangle \otimes \vert \gamma\rangle\]

for some unit vector $ \vert \gamma\rangle$.

This implies that $ \vert \Psi _ i\rangle$ has the form

\[\vert \Psi _ i\rangle _ {ABM} = \vert \phi _ i\rangle _ A \otimes \vert \phi _ i \rangle _ B \otimes \vert \mu _ i\rangle _ M\]

Then, we have

\[\begin{aligned} \langle\Psi _ 0 \vert \Psi _ 1\rangle &= \langle \phi _ 0 \vert \phi _ 1\rangle^2 \langle \mu _ 0 \vert \mu _ 1\rangle \\\\ &= \langle \phi _ 0 \vert \phi _ 1 \rangle \langle \phi _ 0 \vert \phi _ 1 \rangle \langle \mu _ 0 \vert \mu _ 1 \rangle \\\\ &= (\langle \phi _ 0 \vert _ A \otimes \langle \phi _ 0 \vert _ B \otimes \langle \mu _ 0 \vert _ M)( \vert \phi _ 1\rangle _ A \otimes \vert \phi _ 1\rangle B \otimes \vert \mu _ 1 \rangle _ M) \\\\ &= (\langle \phi _ 0 \vert _ A \otimes \langle \beta \vert _ {BM}) U^\dagger _ {ABM} U _ {ABM} \vert \phi _ 1\rangle _ A \otimes \vert \beta\rangle _ {BM} \\\\ &= (\langle \phi _ 0 \vert _ A \otimes \langle \beta \vert _ {BM})( \vert \phi _ 1\rangle _ A \otimes \vert \beta\rangle _ {BM}) \\\\ &= \langle \phi _ 0 \vert \phi _ 1 \rangle\langle \beta \vert \beta\rangle \\\\ &= \langle \phi _ 0 \vert \phi _ 1\rangle \end{aligned}\]

Then taking modulus on both sides, we get

\[\vert \langle \phi _ 0 \vert \phi _ 1 \rangle \vert ^2 \vert \langle \mu _ 0 \vert \mu _ 1 \rangle \vert = \vert \langle \phi _ 0 \vert \phi _ 1 \rangle \vert\]

But since $ \vert \mu _ 0\rangle$ and $ \vert \mu _ 1\rangle$ are unit vectors, we have the inequality

\[\vert \langle \phi _ 0 \vert \phi _ 1\rangle \vert ^2 \ge \vert \langle \phi _ 0 \vert \phi _ 1\rangle \vert\]

But this is actually a contradiction. If $x := \vert \langle \phi _ 0 \vert \phi _ 1\rangle \vert $, then we have $0 \le x \le 1$ and $x^2 \ge x$, which has solutions $x = 0$ or $x = 1$. But if $x = 0$, then these states are orthogonal (which is a contradiction by the assumption), or if $x = 1$, then these represent the same state.

This requires that the states be distinct, non-orthogonal, and pure. Thus this says nothing about

• Cloning orthogonal states (e.g. we copy classical information all the time)
• Approximately cloning states

\[\vert \Psi\rangle = \vert \alpha\rangle \otimes \vert \gamma\rangle\]

for some unit vector $ \vert \gamma\rangle$.