Quantenteleportation

The interleaving of QuBit A and B results in:

<br /> \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)<br />

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

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

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:
<br /> I=\begin{bmatrix}<br /> 1 & 0\\<br /> 0 & 1<br /> \end{bmatrix}\qquad(20)<br />
The overall state G of this system can thus be described with the matrix representation of H (see equation (12)) as follows:
<br /> G=H\otimes I=\begin{bmatrix}<br /> H & I\\<br /> I & H<br /> \end{bmatrix}=\begin{bmatrix}<br /> \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 1 & 0\\<br /> \frac{1}{\sqrt{2}} &-\frac{1}{\sqrt{2}} & 0 &1 \\<br /> 1& 0 & \frac{1}{\sqrt{2}} &\frac{1}{\sqrt{2}} \\<br /> 0 & 1 & \frac{1}{\sqrt{2}} &-\frac{1}{\sqrt{2}}<br /> \end{bmatrix}\qquad(21)<br />

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:
<br /> (H\otimes I \otimes I)(CNOT \otimes I)(\Ket{Q_C}\otimes \Ket{e})=\\<br /> (H\otimes I \otimes I)(CNOT \otimes I)\frac{1}{\sqrt{2}}(\alpha\Ket{000}+\alpha\Ket{011}+\beta\Ket{100}+\beta\Ket{111})=\\<br /> (H\otimes I \otimes I)\frac{1}{\sqrt{2}}(\alpha\Ket{000}+\alpha\Ket{011}+\beta\Ket{110}+\beta\Ket{101})=\\<br /> \frac{1}{2}(\alpha(\Ket{000}+\Ket{011}+\Ket{100}+\Ket{111})+\beta(\Ket{010}+\Ket{010}-\Ket{110}-\Ket{101}))=\\<br /> \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)<br />

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〉:
<br /> \Ket{0}=\begin{bmatrix}<br /> 1\\<br /> 0<br /> \end{bmatrix};<br /> X=\begin{pmatrix}<br /> 0 & 1\\<br /> 1 & 0<br /> \end{pmatrix}\qquad(23)<br />
<br /> X\Ket{0}=\begin{pmatrix}<br /> 0 & 1\\<br /> 1 & 0<br /> \end{pmatrix}\begin{bmatrix}<br /> 1\\<br /> 0<br /> \end{bmatrix}=\begin{bmatrix}<br /> 0\\<br /> 1<br /> \end{bmatrix}<br />

The Z gate is also a 2×2 matrix:
<br /> \Ket{+}=\frac{1}{\sqrt{2}}\begin{bmatrix}<br /> 1\\<br /> 1<br /> \end{bmatrix};<br /> Z=\begin{pmatrix}<br /> 1 & 0\\<br /> 0 & -1<br /> \end{pmatrix}\qquad(24)<br />
<br /> Z\Ket{+}=Z=\begin{pmatrix}<br /> 1 & 0\\<br /> 0 & -1<br /> \end{pmatrix}\frac{1}{\sqrt{2}}\begin{bmatrix}<br /> 1\\<br /> 1<br /> \end{bmatrix}=\frac{1}{\sqrt{2}}\begin{bmatrix}<br /> 1\\<br /> -1\end{bmatrix}=\Ket{-}<br />