Basic Quantum Computing - Teleportation

In super dense coding, we used a qubit to send 2 bits of classical information, quantum teleportation is a way of using classical bits to send a qubit.

The Setup

Once again, we have Alice and Bob, they meet up and create a pair of qubits in the entangled state:

\[\frac{\ket{0}_A\ket{0}_B + \ket{1}_A\ket{1}_B}{\sqrt{2}}\]

They then go their seperate ways with one qubit each.

Alice sends a qubit to Bob

Once day, Alice wants to send a qubit to Bob.

Alice starts with the qubit

\[(a\ket{0} + b\ket{1})\]

So the combined system of three qubits is given by

\[(a\ket{0} + b\ket{1}) \otimes \frac{\ket{0}_A\ket{0}_B + \ket{1}_A\ket{1}_B}{\sqrt{2}}\] \[=\frac{a\ket{00}_A\ket{0}_B + a\ket{01}_A\ket{1}_B + b\ket{10}_A\ket{0}_B + b\ket{11}_A\ket{1}_B}{\sqrt{2}}\]

Alice applies a CNOT gate on her two qubits. This changes her second qubit to a 1 if her first qubit is a 0

\[=\frac{a\ket{00}_A\ket{0}_B + a\ket{01}_A\ket{1}_B + b\ket{11}_A\ket{0}_B + b\ket{10}_A\ket{1}_B}{\sqrt{2}}\]

giving followed by a Hadamard gate on the first qubit

\[=\frac{a\ket{00}_A\ket{0}_B}{2} + \frac{a\ket{10}_A\ket{0}_B}{2} + \frac{a\ket{01}_A\ket{1}_B}{2} + \frac{a\ket{11}_A\ket{1}_B}{2}\] \[+\frac{b\ket{01}_A\ket{0}_B}{2} - \frac{b\ket{11}_A\ket{0}_B}{2} + \frac{b\ket{00}_A\ket{1}_B}{2} - \frac{b\ket{10}_A\ket{1}_B}{2}\]

This leaves the state:

\[\frac{1}{2}(\ket{00}_A(a\ket{0}_B + b\ket{1}_B) + \ket{01}_A(b\ket{0}_B + a\ket{1}_B)\] \[+ \ket{10}_A(a\ket{0}_B - b\ket{1}_B)) + \ket{11}_A(b\ket{0} _B-a\ket{1}_B)\]

Alice then measures her two qubits. She’ll get one of 4 results: 00, 01, 10 and 11.

Bob receives the qubit

She then send her results to Bob clasically.

Based on what Alice sends to him, Bob performs a unitary operation on his qubit:

If Bob receives 00 from Alice then the system is in the state:

\[\ket{00}_A(a\ket{0}_B + b\ket{1}_B)\]

So Bob has to do nothing (or does the identity transformation)

If Bob receives 01 from Alice then the system is in the state:

\[\ket{01}_A(b\ket{0}_B + a\ket{1}_B)\]

So Bob has to swap the co effients around so he applies the X gate

\[X = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\]

Which leaves him with

\[\ket{01}_A(a\ket{0}_B + b\ket{1}_B)\]

If Bob receives 10 from Alice then the system is in the state:

\[\ket{10}_A(a\ket{0}_B - b\ket{1}_B))\]

So Bob has to change -b to b so he applies the Z gate

\[Z = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}\]

This gives the final state as

\[\ket{10}_A(a\ket{0}_B + b\ket{1}_B))\]

Finally, if Bob receives 11 from Alice then the system is in the state:

\[\ket{11}_A(b\ket{0} _B-a\ket{1}_B)\]

Then he applies the following gate

\[iY = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}\]

Leaving

\[\ket{11}_A(a\ket{0} _B+b\ket{1}_B)\]

This means no matter what Alice sends to him, at the end the state of Bob’s system is

\[a\ket{0} _B+b\ket{1}_B\]

which is the qubit Alice wanted to send him.

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