# Quantenteleportation

## Quantum teleportation

The quantum teleportation is shown schematically and as a circuit diagram consisting of CNOT, Hadamard and XZ gates in the following two sketches [1,2].:

The goal of quantum teleportation is to transfer the state of QuBit C to another QuBit B. Due to the No-Cloning Theorem, it is not possible to transfer the state of one QuBit C to another without changing the state of QuBit C. The state of QuBit C is initially Q_{C}:

\Ket{Q_C}=\alpha\Ket{0}+\beta\Ket{1}\qquad(17)

Three QuBits are required for execution: One QuBit has the state to be transferred, the other two interleaved QuBits (A and B) are used to transfer the state from QuBit C to QuBit B.

### Sequence of a quantum teleportation

- Two QuBits A and B are interleaved with each other. Bob keeps the QuBit B and sends the second QuBit to Alice.
- Alice concatenates her QuBit C with the received QuBit A via a CNOT gate and performs a Hadamard transformation
- Alice measures the state of QuBit C and A
- Alice communicates the result of the measurements- encoded by two bits- via a classical communication channel
- Depending on the measurement result, Bob performs transformations on his QuBit B (called XZ or Pauli transformation) so that his QuBit B has the same state as QuBit C.

First, qubit A and B are interleaved with each other. Then QuBit A (Alice) is connected to QuBit C via a CNOT gate. QuBit C is then brought into the superposition state with a Hadamard gate (see circuit diagram).

## Excursus: Entanglement and superposition of qubits A, B and C

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)

## Excursus: Matrix representation of two non-entangled qubits

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)

## Excursus: Mathematical description of quantum entanglement

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.

Alice performs a measurement on her two QuBits (A and C). She obtains one of the four measurement results in the following table with equal probability (25%):

Measurement result (Alice) | QuBit B state (Bob) |

|00〉 | α|0〉+β|1〉 |

|01〉 | α|1〉+β|0〉 |

|10〉 | α|0〉-β|1〉 |

|11〉 | α|1〉-β|0〉 |

*Note: the states in which Bob’s QuBit is located result from the mathematical descriptions of the Hadamard and CNOT gates and their connections with the three QuBits (see circuit diagram and digressions on quantum teleportation).*

### X and Z gates

From the above table it can be seen that for the measurement state |00〉 Bob’s QuBit Q_{B} is already in the state of the QuBit Q_{C} to be transmitted (compare equation (17)). For the three other measurement states, an adjustment of Q_{B} is required so that Q_{B} has the same state as the QuBit Q_{C}. So-called X and Z gates are required for this.

The X gate inverts the state from |0〉 to |1〉 and vice versa [1]:

The Z gate results in a change of state |+〉 to |-〉 or |-〉 to |+〉 [1]:

By using these gates Bob can bring his QuBit Q_{B} into the state of Q_{C}.

The following table shows, depending on the measurement result, the gates to be used, which Bob must apply to his qubit Q_{B}:

Measurement result (Alice) | required gates |

|00〉 | none |

|01〉 | X |

|10〉 | Z |

|11〉 | ZX |

The transformation of a QuBit with X and Z gates is also called **Pauli transformation**.

## Excursus: Mathematical representation of X and Z gates

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{-}

## Sources

[1]: M. Ellerhoff, Mit Quanten Rechnen. Wiesbaden: Springer Spektrum , ISBN 978-3-658-31221-3

[2]: P. Kaufmann, S. Naegele-Jackson, II. Quantenrevolution – die Welt der Qubits: DFN-Mitteilungen Ausgabe 99 Juni 20/21 Seite 22

[3] https://qiskit.org/textbook/preface.html