Quantum Information HT24, Quantum teleportation
Flashcards
Suppose that Alice wants to send Bob a quantum state. Can you give three reasons why this is difficult, and why?
- The original copy must be destroyed (by the no cloning theorem)
- Classical information cannot reconstruct a quantum state (again by the no cloning theorem, since a classical description could be used to reconstruct the quantum state)
- You would need an infinite amount of bits to describe a quantum state (consider giving all the digits to the complex coordinates)
Suppose that:
- Alice has a qubit system $Q$ in a state $ \vert \phi\rangle$
- Bob has a system $B$
- Alice wants to send Bob her qubit $ \vert \phi\rangle$
- Alice has an additional system $A$ such that $A$ and $B$ are entangled in the Bell state
\[\vert \pmb \Phi^+\rangle := \frac{ \vert 0\rangle \otimes \vert 0\rangle + \vert 1\rangle \otimes \vert 1\rangle}{\sqrt 2}\]
Can you describe a protocol that implements quantum teleportation?
- Alice measures the composite system $QA$ in the Bell basis
- Alice sends the outcome of her measurement to Bob (where we label $\{ \vert \Phi^+\rangle, \vert \Psi^+\rangle, \vert \Psi^-\rangle, \vert \Phi^-\rangle\} = \{ \vert \Phi _ 0\rangle, \vert \Phi _ 1\rangle, \vert \Phi _ 2\rangle, \vert \Phi _ 3\rangle\}$, these are chosen so that $ \vert \Phi _ m\rangle = (U _ m \otimes I) \vert \Phi^+\rangle$)
- Bob applies $I _ {(0)}, X _ {(1)}, Y _ {(2)}, Z _ {(3)}$, the Pauli matrices (labelled by indices corresponding to the outcome of the measurement)
Suppose that:
- Alice has a qubit system $Q$ in a state $ \vert \phi\rangle$
- Bob has a system $B$
- Alice wants to send Bob her qubit $ \vert \phi\rangle$
- Alice has an additional system $A$ such that $A$ and $B$ are entangled in the Bell state
\[\vert \pmb \Phi^+\rangle := \frac{ \vert 0\rangle \otimes \vert 0\rangle + \vert 1\rangle \otimes \vert 1\rangle}{\sqrt 2}\]
Then a protocol that implements quantum teleportation is:
- Alice measures the composite system $QA$ in the Bell basis
- Alice sends the outcome of her measurement to Bob
- Bob applies $U _ m = I _ {(0)}, X _ {(1)}, Y _ {(2)}, Z _ {(3)}$, the Pauli matrices (labelled by indices corresponding to the outcome of the measurement)
Denoting the Bell basis by $\{ \vert \Phi^+\rangle, \vert \Psi^+\rangle, \vert \Psi^-\rangle, \vert \Phi^-\rangle\} = \{ \vert \Phi _ 0\rangle, \vert \Phi _ 1\rangle, \vert \Phi _ 2\rangle, \vert \Phi _ 3\rangle\}$ to give the useful relation for the possible outcomes of Alice’s measurements:
\[\vert \Phi _ m\rangle = (U _ m \otimes I) \vert \Phi^+\rangle\]
Quickly prove that this scheme works.
- Alice measures the composite system $QA$ in the Bell basis
- Alice sends the outcome of her measurement to Bob
- Bob applies $U _ m = I _ {(0)}, X _ {(1)}, Y _ {(2)}, Z _ {(3)}$, the Pauli matrices (labelled by indices corresponding to the outcome of the measurement)
The composite system $QAB$ is initially in the state
\[\vert \Psi\rangle _ {QAB} = \vert \phi\rangle _ Q \otimes \vert \Phi^+\rangle _ {AB}\]After Alice performs her measurement and gets outcome $m$ (corresponding to her state being one of $ \vert \Phi _ m\rangle$, Bob’s system is steered into the state
\[\vert \psi _ m \rangle _ B = \frac{ \vert v _ m \rangle}{ \vert \vert \vert v _ m \rangle \vert \vert }\]where
\[\begin{aligned} \vert v _ m\rangle &= (\langle \Phi _ m \vert _ {QA} \otimes I _ B) ( \vert \phi\rangle _ Q \otimes \vert \Phi^+\rangle _ {AB}) \\\\ &= (\langle \Phi^+ \vert _ {QA} \otimes I _ B)(U^\dagger _ {m, Q}\otimes I _ A \otimes I _ B)( \vert \phi\rangle _ Q \otimes \vert \Phi^+\rangle _ {AB}) \\\\ &= (\langle \Phi^+ \vert _ {QA} \otimes I _ B)(U^\dagger _ m \vert \phi\rangle _ Q \otimes \vert \Phi^+\rangle _ {AB}) \\\\ &= \frac 1 2 \sum _ {k, \ell = 0, 1} (\langle k \vert _ Q \otimes \langle k \vert _ A \otimes I _ B)(U^\dagger _ m \vert \phi\rangle _ Q \otimes \vert \ell\rangle _ A \otimes \vert \ell\rangle _ B) \\\\ &= \frac 1 2 \sum _ {k, \ell = 0, 1} (\langle k \vert U^\dagger _ m \vert \phi\rangle \langle k \vert \ell \rangle \vert \ell\rangle _ B) \\\\ &= \frac 1 2\sum _ {k = 0, 1} \langle k \vert U^\dagger _ m \vert \phi\rangle \vert k\rangle \\\\ &= \frac 1 2\sum _ {k = 0,1} (U^\dagger _ m \vert \phi\rangle) _ k \vert k \rangle \\\\ &= \frac 1 2 U^\dagger _ m \vert \phi\rangle _ B \end{aligned}\]So then Bob’s state is
\[\vert \psi _ m\rangle _ B = U^\dagger _ m \vert \phi\rangle _ B\]Since Alice told Bob the outcome of her measurement, Bob knowns which unitary matrix to apply to his system (namely $U _ m$) in order to put the system into exactly the state $ \vert \phi\rangle _ B$.