Quantum computers are built on top of single-qubit and 2-qubit operators. In the last two articles, we covered few single-qubit gates, and especially the Hadamard gate which puts a qubit into superposition.

Here, we will explore the 2-qubit operators and more precisely we will look at putting qubits into **entanglement** with **the C-NOT gate**.

##### Classical XOR gate

The Controlled-Not gate (CNOT) is analogous to the XOR gate (Exclusive OR) in classical computing. We have already presented in our previous article Introduction to quantum logic gates the **XOR gate** table truth: it gives a true (1 or HIGH) output when the number of true inputs is odd.

##### Quantum C-NOT gate

The quantum CNOT gate has two inputs, and thus two outputs. **The target input is negated only if the control input is set to 1**. If the control input is 0, the gate has no effect. The control qubit is not changed by the gate.

Below is a snapshot of both classical and quantum diagrams. We verify easily that they mirror each other: the quantum Target output column matches the y+x column of the classical XOR gate.

As we already know, each gate/operator can be expressed as a matrix. As the C-NOT gate takes two qubits as inputs and two qubits as output it will be a **4x4 matrix**.

There is a useful technique to transform a truth table to a matrix. starting at row 0 column 0, you label the columns and rows consecutively in binary, from 00 to 11 for example. You then place a 1 in a cell if the input maps to the output; 0 otherwise.

For example, in our case, as |11> maps to |10> we set the matrix's value of 4th line, 3th column to 1.

That's it, you're left with a matrix for your gate.

Let's try to apply the CNOT-gate by example to the |00> state, by multiplying the CNOT matrix to the basis state vector

We observe that it is the expected result ;-)

If we now try to apply the CNOT gate to the |10> state, by definition as the control qubit is |1>, the second qubit should be flipped from |0> to |1>. Let's observe it:

As expected, we get |11>.

**Remark 1**: More generally, the inputs are allowed to be a linear superposition of {|0>,|1 >}. The CNOT gate transforms the quantum state:

a|00> +b|01 +c|10 +d|11>

into

a|00> +b|01> +c|11> +d|10>

**Remark 2**: The first experimental realization of a CNOT gate was accomplished in 1995 with a single Beryllium ion.

##### Bell State

Let's try now to prepare a two-qubit system by example in |00> state, and then:

- apply **the Hadamard gate** to the first one so we get a 50/50 superposition state as detailed in Introduction to quantum logic gates

- then apply **the C-NOT gate** with the second qubit |0> acting as the control qubit

In the first step, the Hadamard gate transforms |00> in the superposition state (H|0>)|0> as per below:

This will then act as **a control input to the CNOT gate**, which only inverts the target (the second qubit) when the control (the first qubit) is 1. Thus, the CNOT gate transforms the second qubit as follows:

In the first line, we used linearity, and in the second line we used the definition of the CNOT gate.

The result is one **entangled state**, the one we have taken in example in our articles Expectation value of an entangled state and Correlation in entangled state

**Remark 1**: we remind here what this entanglement means: the qubit held by Alice (subscript "A") can be 0 as well as 1. If Alice measured her qubit in the standard basis, the outcome would be perfectly random, either possibility 0 or 1 having probability 1/2. But if Bob (subscript "B") then measured his qubit, the outcome would be the same as the one Alice got. So, if Bob measured, he would also get a random outcome on first sight, but if Alice and Bob communicated, they would find out that, although their outcomes seemed random, they are perfectly correlated.

**Remark 2**: this state represents one of the **four maximally entangled two-qubit Bell states**. The different ways to create different Bell states |β_{00}>, |β_{01}>, |β_{10}>, |β_{11}> is to apply the precedent two gates to respectively each initial two qubit input |00>, |01>, |10> and |11>.

Let's detail the creation of the three remaining Bell states

Applying the CNOT gate to this state gives:

Let's focus on the Bell state |β_{10}> which takes |10> as input state:

Finally let's consider the last Bell state |β_{11}> which takes |11> as input state:

Finally below are the different ways to create different Bell states: