Quantum Teleportation

Let say Alice wants to pass her qubit |\psi\rangle=a|0\rangle+b|1\rangle to Bob. By no cloning theorem, we know that Alice cannot duplicate her qubit to Bob. However, she can “teleport” her qubit to Bob if they share a Bell state \frac{1}{\sqrt{2}}(|00\rangle+|11\rangle).

First, Alice will apply CNOT gate from |\psi\rangle to her qubit in the shared Bell state. This will result in a combined state of

\frac{1}{\sqrt{2}}(a|000\rangle+b|110\rangle+a|011\rangle+b|101\rangle),

where three bits are respectively the state she wants to teleport, her qubit in the Bell state, and Bob’s qubit in the Bell state. Now, if she measures her qubit in the Bell state with the standard basis (|0\rangle and |1\rangle), the joint state will become

a|00\rangle+b|11\rangle if the measured qubit is |0\rangle

and

a|01\rangle+b|10\rangle if the measured qubit is |1\rangle.

Now, Alice may call Bob on the phone and tell him what she has measured. If it is the latter case, Bob can apply the bit flip gate to his qubit and result in the state in the first case. So after this operation, they will always share the quantum state

a|00\rangle+b|11\rangle.

Now if Alice applies Hadamard gate to her remaining qubit, it results in

\frac{1}{\sqrt{2}}(a|00\rangle+a|10\rangle+b|01\rangle-b|11\rangle) and if she measures it with the standard basis, the remaining state (i.e., Bob’s qubit) becomes

a|0\rangle+b|1\rangle if her measurement is |0\rangle and

a|0\rangle-b|1\rangle if her measurement is |1\rangle.

At this time, she can call Bob again and tell him the result, if it is the former case, Bob gets the original state |\psi\rangle precisely. For the latter case, he can further apply the phase shift gate to his qubit to recover |\psi\rangle.

 

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