Quantum Entanglement

Entangled QuBits are special two-particle states. The Bell states describe the state of maximal entanglement for two entangled QuBits Q_A, Q_B:

\Ket{Q_{A},Q_{B}}=\frac{1}{\sqrt{2}}(\Ket{00}\pm\Ket{11}), \Ket{Q_{A},Q_{B}}=\frac{1}{\sqrt{2}}(\Ket{01}\pm\Ket{10})

Excursus: Two-particle states

In the following, for the explanation of the entanglement of two QuBits, the general representation of a two-particle state is required.

The two-particle state of two non-entangled QuBits can be described by the tensor product (dyadic product) starting from the single-particle states:
$\Ket{a}=\begin{bmatrix} a_{0} \\ a_{1} \end{bmatrix}, \Ket{b}=\begin{bmatrix} b_{0} \\ b_{1} \end{bmatrix}, \Ket{ab}=\Ket{a} \otimes \Ket{b}=\begin{bmatrix} a_{0} \times \begin{bmatrix} b_{0}\\ b_{1} \end{bmatrix} & a_{1} \times \begin{bmatrix} b_{0}\\ b_{1} \end{bmatrix} & \end{bmatrix}=\Biggl[ \begin{bmatrix} a_{0}b_{0}\\ a_{0}b_{1}\\ a_{1}b_{0} \\ a_{1}b_{1} \end{bmatrix}\Biggr]$

A general two-particle state of two QuBits is described by the following state vector:
$\Ket{a}=a_{00}\Ket{00}+a_{01}\Ket{01} +a_{10}\Ket{10} +a_{11}\Ket{11}=\Biggl[\begin{bmatrix} a_{00}\\ a_{01}\\ a_{10}\\ a_{11} \end{bmatrix}\Biggr]$

The normalization requires:
$|a_{00}|^2+|a_{01}|^2+|a_{10}|^2+|a_{11}|^2=1$

In a quantum computer, entanglement between two QuBits can be created by using the Hadamard and CNOT gates:

The Hadamard gate

The Hadamard gate has the task of transferring the states |0> or |1> into a superposition state [1]:

\Ket{0}

\Ket{+}

\Ket{-}

\Ket{1}

This corresponds to a projection into the equatorial plane on the Bloch Sphere. The Hadamard gate transforms the state |0> to |+> or |1> to |->. If the QuBit would be measured in the state |+> or |->, it would transition to the |1> or |0> state with a probability of 50%.

Excursus: Mathematical representation of the Hadamard gate.

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

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

The CNOT gate

Two QuBits are required for this gate: Q_A and Q_B. The QuBit Q_A is called control because the final state of Q_B depends on its state [1]:

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

Excursus: Mathematical representation of the CNOT gate.

The CNOT gate (CNOT operator) can be mathematically expressed by a 4×4 matrix, which transforms the state vector of the two QuBits as follows:

CNOT=\Biggl[\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{bmatrix}\Biggr], \Ket{a}=\Biggl[\begin{bmatrix} a_{00}\\ a_{01}\\ a_{10}\\ a_{11} \end{bmatrix}\Biggr], CNOT\Ket{a}=\Biggl[\begin{bmatrix} a_{00}\\ a_{01}\\ a_{11}\\ a_{10} \end{bmatrix}\Biggr]

The CNOT gate thus swaps the amplitudes of a₁₀ and a₁₁!

Creation of a Bell pair

Now a Bell pair is to be generated with these two gates [1]:

\Ket{Q_A,Q_B}=\frac{1}{\sqrt{2}}(\Ket{00}+\Ket{11})

At the beginning Q_A and Q_B both are in the state |0>. The Hadamard gate changes the state of Q_A to |+>. Therefore the CNOT gate changes the state of Q_B to |1> with a probability of 50% (if the measurement at Q_A at the CNOT gate yields 1). This creates a superposition of |00> and |11>, a Bell state with entanglement!

Consider the following case: If one performs a measurement at Q_A that yields the result 0, the Bell state collapses to |00> and therefore afterwards both Q_Aand Q_Bhave the state |0>. This correlation between possible measurement results is characteristic for entanglement.

Sources

[1] M. Ellerhoff. Mit Quanten Rechnen. ISBN 978-3-658-31221-3

Version: 17.06.2024