Quantenteleportation

The interleaving of QuBit A and B results in:

\Ket{e}=\frac{1}{\sqrt{2}}(\Ket{00}+\Ket{11})=\frac{1}{\sqrt{2}}(\Ket{0}_{A}\Ket{0}_{B}+\Ket{1}_{A}\Ket{1}_{B})\qquad(18)

Thus, with the QuBit QC whose one-particle state is to be transferred, the total states are as follows (expressed by tensor product):

\Ket{Q_C}\otimes\Ket{e}=\frac{1}{\sqrt{2}}(\alpha\Ket{000}+\alpha\Ket{011}+\beta\Ket{100}+\beta\Ket{111})\qquad(19)

The question arises how QuBit B, which is not connected to QuBit C via CNOT, is represented mathematically. This will be explained using two QuBits, where no operation is performed on one QuBit and a Hadamard transformation is performed on the second QuBit [3]:

To be able to describe the state of this system, the identity matrix I is written down for the QuBit q0. This matrix leaves the state of a QuBit unchanged:

I=\begin{bmatrix}
1 & 0\\
0 & 1
\end{bmatrix}\qquad(20)

The overall state G of this system can thus be described with the matrix representation of H (see equation (12)) as follows:

G=H\otimes I=\begin{bmatrix}
H & I\\
I & H
\end{bmatrix}=\begin{bmatrix}
\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 1 & 0\\
\frac{1}{\sqrt{2}} &-\frac{1}{\sqrt{2}} & 0 &1 \\
1& 0 & \frac{1}{\sqrt{2}} &\frac{1}{\sqrt{2}} \\
0 & 1 & \frac{1}{\sqrt{2}} &-\frac{1}{\sqrt{2}}
\end{bmatrix}\qquad(21)

From the matrix representation of two non-entangled qubits (see above representation for q0 and q1) it follows that in the CNOT operation for the uninvolved QuBit B must be multiplied by the identity matrix I (see equation(20)). In the subsequent Hadamard transformation, two identity matrices are required because QuBit A and B are not involved in the transformation:

(H\otimes I \otimes I)(CNOT \otimes I)(\Ket{Q_C}\otimes \Ket{e})=\\
(H\otimes I \otimes I)(CNOT \otimes I)\frac{1}{\sqrt{2}}(\alpha\Ket{000}+\alpha\Ket{011}+\beta\Ket{100}+\beta\Ket{111})=\\
(H\otimes I \otimes I)\frac{1}{\sqrt{2}}(\alpha\Ket{000}+\alpha\Ket{011}+\beta\Ket{110}+\beta\Ket{101})=\\
\frac{1}{2}(\alpha(\Ket{000}+\Ket{011}+\Ket{100}+\Ket{111})+\beta(\Ket{010}+\Ket{010}-\Ket{110}-\Ket{101}))=\\
\frac{1}{2}(\color{green}{\Ket{00}}\color{black}{(}\color{blue}{\alpha\Ket{0}+\beta\Ket{1}}\color{black}{)+} \color{green}{\Ket{01}}\color{black}{(}\color{blue}{\alpha\Ket{1}+\beta\Ket{0}}\color{black}{)+}\color{green}{\Ket{10}}\color{black}{(}\color{blue}{\alpha\Ket{0}-\beta\Ket{1}}\color{black}{)+}\color{green}{\Ket{11}}\color{black}{(}\color{blue}{\alpha\Ket{1}-\beta\Ket{0}}\color{black}{))}(22)

Alice performs a measurement on her two QuBits (A and C). As a measurement result she receives with equal probability (25%) one of the four green marked states in the equation above. Alice sends this measurement result encoded in two classical bits to Bob. Depending on the measurement result, Bob’s QuBit B assumes the state marked in blue.

In this example, the X gate changes the QuBit state |0〉 to |1〉:

\Ket{0}=\begin{bmatrix}
1\\
0
\end{bmatrix};
X=\begin{pmatrix}
0 & 1\\
1 & 0
\end{pmatrix}\qquad(23)


X\Ket{0}=\begin{pmatrix}
0 & 1\\
1 & 0
\end{pmatrix}\begin{bmatrix}
1\\
0
\end{bmatrix}=\begin{bmatrix}
0\\
1
\end{bmatrix}

The Z gate is also a 2×2 matrix:

\Ket{+}=\frac{1}{\sqrt{2}}\begin{bmatrix}
1\\
1
\end{bmatrix};
Z=\begin{pmatrix}
1 & 0\\
0 & -1
\end{pmatrix}\qquad(24)


Z\Ket{+}=Z=\begin{pmatrix}
1 & 0\\
0 & -1
\end{pmatrix}\frac{1}{\sqrt{2}}\begin{bmatrix}
1\\
1
\end{bmatrix}=\frac{1}{\sqrt{2}}\begin{bmatrix}
1\\
-1\end{bmatrix}=\Ket{-}