Quantum entanglement

Together, the QuBits can assume four different states, like two ordinary bits.
Entangled QuBits consist of a superposition of two certain states and decay into one of the two states after a measurement. For two entangled QuBits Q_A, Q_B the following four states exist:

$\Ket{Q_{A},Q_{B}}=\frac{1}{\sqrt{2}}(\Ket{00}\pm\Ket{11})\qquad(10)\\ \Ket{Q_{A},Q_{B}}=\frac{1}{\sqrt{2}}(\Ket{01}\pm\Ket{10})\qquad(11)$
The state (10) is also called Bell state or Bell pair.

Excursus: Representation of two QuBits via tensor product

For the explanation of the entanglement of two quanta or QuBits, it is necessary to explain the representation of the state of two QuBits, as well as the Hadamard and CNOT gates.
Two non-entangled QuBits can be represented by the following state vector:

\Ket{a}=a_{00}\Ket{00}+a_{01}\Ket{01} +a_{10}\Ket{10} +a_{11}\Ket{11}=\begin{bmatrix} a_{00}\\ a_{01}\\ a_{10}\\ a_{11} \end{bmatrix}\qquad(6)

For the probabilities, again:

|a_{00}|^2+|a_{01}|^2+|a_{10}|^2+|a_{11}|^2=1\qquad(7)

In general, the state or states of two (or more QuBits) can be described by the tensor product (dyadic product):

\Ket{a}=\begin{bmatrix} a_{0} \\ a_{1} \\ \end{bmatrix} ; \Ket{b}=\begin{bmatrix} b_{0} \\ b_{1} \\ \end{bmatrix}\qquad(8)

\Ket{ab}=\Ket{a} \otimes \Ket{b}=\begin{bmatrix} a_{0} \times \begin{bmatrix} b_{0}\\ b_{1} \end{bmatrix} & \\ a_{1} \times \begin{bmatrix} b_{0}\\ b_{1} \end{bmatrix} & \end{bmatrix}=\begin{bmatrix} a_{0}b_{0}\\ a_{0}b_{1}\\ a_{1}b_{0} \\ a_{1}b_{1} \end{bmatrix}\qquad(9)

Question: How are 2 QuBits connected so that they are interleaved?

Two so-called (circuit) gates are needed for entanglement: The Hadamard and CNOT gates.

The Hadamard gate

The Hadamard gate has the task of transferring the (absolute) state of a QuBit (|0〉 or |1〉) into the superposition state [1]:

\Ket{0}

\Ket{+}

\Ket{-}

\Ket{1}

This corresponds on the Bloch sphere to a projection in the equatorial plane. The Hadamard gate transforms the state
|0〉 to |+〉 (see equation (13)) or |1〉 to |-〉 (see equation (14)). If the QuBit would again be measured into this state (|+〉 or |-〉), it will transition to the |1〉 or |0〉 state with a probability of 50%.

Excursus: Mathematical representation of a Hadamard gate

Mathematically, the Hadamard gate (Hadmard operator) is a 2×2 matrix that transforms the vectors (states) |0〉 or |1〉 into the vector (state) |+〉 or |-〉:

$H=\frac{1}{\sqrt{2}}\begin{bmatrix} 1 &1 \\ 1 & -1 \end{bmatrix}\qquad(12)\\ H\Ket{0}=\Ket{+}=\frac{1}{\sqrt{2}}(\Ket{0}+\Ket{1})\qquad(13)\\ H\Ket{1}=\Ket{-}=\frac{1}{\sqrt{2}}(\Ket{0}-\Ket{1})\qquad(14)$

The CNOT gate

The second gate required for entanglement is the CNOT gate. Two QuBits are required for this gate Q_{A and Q}_B. The QuBit Q_A is called the control bit, because the state of Q_B depends on its state [1]:

The CNOT gate inverts the value of QuBit Q_B only if Q_A has the value 1. Otherwise the value of Q_B remains unchanged.

Excursus: Mathematical representation of a CNOT gate

The CNOT gate (CNOT operator) can be mathematically expressed by a 4×4 matrix, which multiplied by the status vector of two QuBits, gives the following state after the CNOT operation.:

CNOT=\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 \end{bmatrix}\qquad(15)

\Ket{a}=\begin{bmatrix} a_{00}\\ a_{01}\\ a_{10}\\ a_{11} \end{bmatrix}

CNOT\Ket{a}=\begin{bmatrix} a_{00}\\ a_{11}\\ a_{10}\\ a_{01} \end{bmatrix}\qquad(16)

The CNOT gate thus swaps the amplitudes of a₀₁ and a₁₁.

Now a Bell pair or entanglement is to be generated with these two gates (operators). This circuit entangles the two qubits Q_A and Q_B with the initial values 0 [1].:

\Ket{Q_A,Q_B}=\frac{1}{\sqrt{2}}(\Ket{0,0}+\Ket{1,1})

Depending on the measurement result of the control QuBit QA, the QuBit QB is changed or not. After a measurement, the state of both qubits is |0.0〉 or |1.1〉 with a probability of 50%.

Sources

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

Quantum entanglement