8.10: A Simple Teleportation Exercise

This circuit teleports |Ψ> from the top wire to the bottom wire. Measurement on the top wire yields |0> or |1>, the power to which the Z matrix on the bottom wire is raised. As shown below Z⁰ is the identity matrix, in other words do nothing because |Ψ> is already on the bottom wire.

\[ \begin{matrix} | \Psi \rangle & \triangleright & \cdot & H & \triangleright & \text{Measure m = 0 or 1} \\ |0 \rangle & \triangleright & \oplus & ~ & \triangleright & Z^m \rightarrow | \Psi \rangle \end{matrix} \nonumber \]

\[ \begin{matrix} |0 \rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix} & |1 \rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix} & | \Psi \rangle = \begin{pmatrix} \sqrt{ \frac{1}{3}} \\ \sqrt{ \frac{2}{3}} \end{pmatrix} \end{matrix} \nonumber \]

The required quantum gates in matrix format:

\[ \begin{matrix} I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} & H = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} & Z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} & Z^0 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} & \text{CNOT} = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{pmatrix} \end{matrix} \nonumber \]

Form the quantum circuit for teleportation:

\[ \begin{matrix} \text{QC = kronecker(H, I) CNOT)} & \text{QC} = \begin{pmatrix} 0.707 & 0 & 0 & 0.707 \\ 0 & 0.707 & 0.707 & 0 \\ 0.707 & 0 & 0 & -0.707 \\ 0 & 0.707 & -0.707 & 0 \end{pmatrix} & \text{QC} = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 1 & 0 & 0 & -1 \\ 0 & 1 & -1 & 0 \end{pmatrix} \end{matrix} \nonumber \]

Write the initial state as a 4-vector using tensor multiplication.

\[ \begin{pmatrix} \sqrt{ \frac{1}{3}} \\ \sqrt{ \frac{2}{3}} \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} \sqrt{ \frac{1}{3}} \\ 0 \\ \sqrt{ \frac{2}{3}} \\ 0 \end{pmatrix} \nonumber \]

Calculate the output state of the circuit and write it as a superposition of |0> and |1> on the top wire.

\[ \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 1 & 0 & 0 & -1 \\ 0 & 1 & -1 & 0 \end{pmatrix} \begin{pmatrix} \sqrt{ \frac{1}{3}} \\ 0 \\ \sqrt{ \frac{2}{3}} \\ 0 \end{pmatrix} = \frac{1}{ \sqrt{2}} \begin{pmatrix} \sqrt{ \frac{1}{3}} \\ \sqrt{ \frac{2}{3}} \\ \sqrt{ \frac{2}{3}} \\ - \sqrt{ \frac{2}{3}} \end{pmatrix} \nonumber \]

\[ \frac{1}{ \sqrt{2}} \begin{pmatrix} \sqrt{ \frac{1}{3}} \\ \sqrt{ \frac{2}{3}} \\ \sqrt{ \frac{2}{3}} \\ - \sqrt{ \frac{2}{3}} \end{pmatrix} = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \\ 0 \end{pmatrix} \begin{pmatrix} \sqrt{ \frac{1}{3}} \\ \sqrt{ \frac{2}{3}} \end{pmatrix} + \begin{pmatrix} 0 \\ 1 \end{pmatrix} \begin{pmatrix} \sqrt{ \frac{1}{3}} \\ - \sqrt{ \frac{2}{3}} \end{pmatrix} \right] \nonumber \]

It is clear that if |0> is measured on the top wire, |Ψ> is on the bottom wire without further action. If |1> is measured on the top wire the qubit on the bottom wire is converted to |Ψ> by multiplication by Z.

\[ \begin{matrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} \begin{pmatrix} \sqrt{ \frac{1}{3}} \\ \sqrt{ \frac{2}{3}} \end{pmatrix} & ~ & ~ & \begin{pmatrix} 0 \\ 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} \sqrt{ \frac{1}{3}} \\ - \sqrt{ \frac{2}{3}} \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \end{pmatrix} \begin{pmatrix} \sqrt{ \frac{1}{3}} \\ \sqrt{ \frac{2}{3}} \end{pmatrix} \end{matrix} \nonumber \]

Matrix summary:

\[ \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 1 & 0 & 0 & -1 \\ 0 & 1 & -1 & 0 \end{pmatrix} \begin{pmatrix} \sqrt{ \frac{1}{3}} \\ 0 \\ \sqrt{ \frac{2}{3}} \\ 0 \end{pmatrix} = \frac{1}{ \sqrt{2}} \begin{pmatrix} \sqrt{ \frac{1}{3}} \\ \sqrt{ \frac{2}{3}} \\ \sqrt{ \frac{1}{3}} \\ - \sqrt{ \frac{2}{3}} \end{pmatrix} = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \\ 0 \end{pmatrix} \otimes \begin{pmatrix} \sqrt{ \frac{1}{3}} \\ \sqrt{ \frac{2}{3}} \end{pmatrix} + \begin{pmatrix} 0 \\ 1 \end{pmatrix} \otimes \begin{pmatrix} \sqrt{ \frac{1}{3}} \\ - \sqrt{ \frac{2}{3}} \end{pmatrix} \right] \nonumber \]

Algebraic summary:

\[ \begin{matrix} Z = \begin{pmatrix} 0 & \text{to} & 0 \\ 1 & \text{to} & -1 \end{pmatrix} & H = \begin{bmatrix} 0 & \text{to} & \frac{(0 + 1)}{ \sqrt{2}} \\ 1 & \text{to} & \frac{(0-1)}{ \sqrt{2}} \end{bmatrix} & \text{CNOT} = \begin{pmatrix} 00 & \text{to} & 00 \\ 01 & \text{to} & 01 \\ 10 & \text{to} & 11 \\ 11 & \text{to} & 10 \end{pmatrix} \end{matrix} \nonumber \]

\[ \left( \sqrt{ \frac{1}{3}} \left| 0 \left\rangle + \sqrt{ \frac{2}{3}} \right|1 \right\rangle \right) |0 \rangle = \sqrt{ \frac{1}{3}} \left| 00 \right\rangle + \sqrt{ \frac{2}{3}} \left|10 \right\rangle \nonumber \]

\( \text{CNOT}\)

\[ \sqrt{ \frac{1}{3}} \left| 00 \right\rangle + \sqrt{ \frac{2}{3}} \left|11 \right\rangle \nonumber \]

\( \text{H} \otimes \text{I}\)

\[ \frac{1}{ \sqrt{2}} \left[ \sqrt{ \frac{1}{3}} \left( |0 \rangle + |1 \rangle \right) |0 \rangle + \sqrt{ \frac{2}{3}} \left( | 0 \rangle - |1 \rangle \right) |1 \rangle \right] = \frac{1}{ \sqrt{2}} \left[ |0 \rangle \left( \sqrt{ \frac{1}{3}} | 0 \rangle + \sqrt{ \frac{2}{3}} |1 \rangle \right) | 1 \rangle + \left( \sqrt{ \frac{1}{3}} |0 \rangle - \sqrt{ \frac{2}{3}} |1 \rangle \right) \right] \nonumber \]

\( \downarrow\)

\[ \frac{1}{ \sqrt{2}} \left[ |0 \rangle \begin{pmatrix} \sqrt{ \frac{1}{3}} \\ \sqrt{ \frac{2}{3}} \end{pmatrix} + |1 \rangle \begin{pmatrix} \sqrt{ \frac{1}{3}} \\ - \sqrt{ \frac{2}{3}} \end{pmatrix} \right] \xrightarrow{Action} \frac{1}{ \sqrt{2}} \left[ I \begin{pmatrix} \sqrt{ \frac{1}{3}} \\ - \sqrt{ \frac{2}{3}} \end{pmatrix} + Z \begin{pmatrix} \sqrt{ \frac{1}{3}} \\ - \sqrt{ \frac{2}{3}} \end{pmatrix} \right] \nonumber \]

Total matrix operator approach:

m = 0

\[ \text{kronecker}(I,~ Z^0 ) \text{kronecker} \left[ \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix},~I \right] \text{kronecker} (H,~I) CNOT \begin{pmatrix} \sqrt{ \frac{1}{3}} \\ 0 \\ \sqrt{ \frac{2}{3}} \\ 0 \end{pmatrix} = \begin{pmatrix} 0.408 \\ 0.577 \\ 0 \\ 0 \end{pmatrix} = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \\ 0 \end{pmatrix} \begin{pmatrix} \sqrt{ \frac{1}{3}} \\ \sqrt{ \frac{2}{3}} \end{pmatrix} \nonumber \]

m = 1

\[ \text{kronecker}(I,~ Z^1 ) \text{kronecker} \left[ \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix},~I \right] \text{kronecker} (H,~I) CNOT \begin{pmatrix} \sqrt{ \frac{1}{3}} \\ 0 \\ \sqrt{ \frac{2}{3}} \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0.408 \\ 0.577 \end{pmatrix} = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \\ 1 \end{pmatrix} \begin{pmatrix} \sqrt{ \frac{1}{3}} \\ \sqrt{ \frac{2}{3}} \end{pmatrix} \nonumber \]

The measurement projection operators used above in the next to the last step are:

m = 0

\[ \begin{pmatrix} 1 \\ 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \nonumber \]

m = 1

\[ \begin{pmatrix} 0 \\ 1 \end{pmatrix} \begin{pmatrix} 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \nonumber \]