Notes - Quantum Information HT24, No-cloning theorem
Flashcards
Suppose that Alice and Bob, seperated over a large distance, share two photons entangled in the Bell state $ \vert \pmb \Phi^+\rangle$. If Alice measures her photon in the direction of $\theta$, Bob’s sate will become
\[\begin{aligned}
|0, \theta\rangle &= \cos \frac \theta 2 |0 \rangle + \sin \frac \theta 2 |1\rangle, \quad \text{or} \\\\
|1, \theta\rangle &= \sin \frac \theta 2 |0 \rangle - \cos \frac \theta 2 |1\rangle
\end{aligned}\]
If Bob owned a device that cloned quantum states, how could this be used to send faster-than-light signals?
Bob could copy his state many times, and different measurements to determine the actual state with some probability.
Can you state the no-cloning theorem for two distinct, non-orthogonal pure states, first in words and then in terms of unitary gates?
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
\[\text{Tr}_ M\left[ U_ {ABM}(\pmb \rho_ A \otimes {}_ {BM}|\beta\rangle\langle \beta|_ {BM}) U^\dagger_ {ABM} \right] = {}_ A|\phi_ i\rangle\langle \phi_ i|_ {A} \otimes {}_B|\phi_ i\rangle \langle \phi_ i |_ B\](where $ \vert \phi _ i\rangle$ is the same as above).
Quickly prove the no cloning theorem, i.e.
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.
Mathematically, 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
\[\text{Tr} _ M\left[ U _ {ABM}(\pmb \rho _ A \otimes {} _ {BM} \vert \beta\rangle\langle \beta \vert _ {BM}) U^\dagger _ {ABM} \right] = {} _ A \vert \phi _ i\rangle\langle \phi _ i \vert _ {A} \otimes \vert \phi _ i\rangle \langle \phi _ i \vert _ B\]
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. Mathematically, 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
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:
\[|\Psi_ i \rangle_ {ABM} := U_ {ABM}(|\phi_ i\rangle_ A \otimes |\beta\rangle_ {BM})\]The cloning condition implies
\[\text{Tr}_ M[|\Psi_ i\rangle \langle \Psi_ i |_ {ABM}] = |\phi_ i\rangle\langle\phi_ i|_ A \otimes |\phi_ i\rangle \langle \phi_ i|_ B\]By the result that says
\[Tr _ B[ \vert \Psi\rangle\langle \Psi \vert ] = \vert \alpha\rangle\langle \alpha \vert\]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
\[\vert \Psi\rangle = \vert \alpha\rangle \otimes \vert \gamma\rangle\]for some unit vector $ \vert \alpha\rangle$. Then
for some unit vector $ \vert \gamma\rangle$.
This implies that $ \vert \Psi _ i\rangle$ has the form
\[|\Psi_ i\rangle_ {ABM} = |\phi_ i\rangle_ A \otimes |\phi_ i \rangle_ B \otimes |\mu_ i\rangle_ M\]Then, we have
\[\begin{aligned} \langle\Psi_ 0|\Psi_ 1\rangle &= \langle \phi_ 0|\phi_ 1\rangle^2 \langle \mu_ 0 | \mu_ 1\rangle \\\\ &= \langle \phi_ 0 | \phi_ 1 \rangle \langle \phi_ 0 | \phi_ 1 \rangle \langle \mu_ 0 | \mu_ 1 \rangle \\\\ &= (\langle \phi_ 0|_ A \otimes \langle \phi_ 0|_ B \otimes \langle \mu_ 0 |_ M)(|\phi_ 1\rangle_ A \otimes |\phi_ 1\rangle B \otimes |\mu_ 1 \rangle_ M) \\\\ &= (\langle \phi_ 0 |_ A \otimes \langle \beta|_ {BM}) U^\dagger_ {ABM} U_ {ABM} |\phi_ 1\rangle_ A \otimes |\beta\rangle_ {BM} \\\\ &= (\langle \phi_ 0|_ A \otimes \langle \beta |_ {BM})(|\phi_ 1\rangle_ A \otimes |\beta\rangle_ {BM}) \\\\ &= \langle \phi_ 0 | \phi_ 1 \rangle\langle \beta|\beta\rangle \\\\ &= \langle \phi_ 0 | \phi_ 1\rangle \end{aligned}\]Then taking modulus on both sides, we get
\[|\langle \phi_ 0 | \phi_ 1 \rangle|^2 |\langle \mu_ 0 | \mu_ 1 \rangle| = |\langle \phi_ 0 | \phi_ 1 \rangle|\]But since $ \vert \mu _ 0\rangle$ and $ \vert \mu _ 1\rangle$ are unit vectors, we have the inequality
\[|\langle \phi_ 0 | \phi_ 1\rangle|^2 \ge |\langle \phi_ 0 | \phi_ 1\rangle|\]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.
The no-cloning theorem states that:
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 $B$, eacho f them in the same state as the input system.
Mathematically, 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
\[\text{Tr}_
M\left[ U_
{ABM}(\pmb \rho _
A \otimes |\beta\rangle\langle \beta|_
{BM}) U^\dagger_
{ABM} \right] = |\phi_
i\rangle\langle \phi_
i|_
{A} \otimes |\phi_
i\rangle \langle \phi_
i |_
B\]
State two ways in which we can still “clone” states despite this result.
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
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
\[Tr_B[|\Psi\rangle\langle \Psi|] = |\alpha\rangle\langle \alpha|\]
for some unit vector $ \vert \alpha\rangle$. What can you say about the the structure of $ \vert \Psi\rangle$?
for some unit vector $ \vert \gamma\rangle$.