How does one interpret intuitively the CNOT gate?

Question

How does one interpret the CNOT gate? The CNOT gate takes a separable state and turns into an entangled state. The oracle in the Deutsch algorithm does the same thing. But how does one understand this intuitively?

Answer 1

The CNOT gate is two-qubit operation, where the first qubit is usually referred to as the control qubit and the second qubit as the target qubit.

• Case 1: The CNOT gate flips the second qubit (the target qubit) if and only if the first qubit (the control qubit) is 1
• Case 2: The CNOT leaves the target qubit unchanged when the control qubit is in state 0

That's the logical table for getting a better understanding:

Note: Any quantum circuit can be simulated to an arbitrary degree of accuracy using a combination of CNOT gates and single qubit rotations.

Answer 2

@DEX7RA's answer is really good. I just want to add another way of understadning the $CNOT$ gate, and any other gate, intuitively from its matrix representation. To do this, first let's remember the four basis states of a state space of two qubits:

$$ |00\rangle = \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}, \; |01\rangle = \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix}, \; |10\rangle = \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix}, \; \text{and} \; |11\rangle = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix} $$

Now, the matrix representation of the $CNOT$ gate is:

$$ \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{pmatrix} $$

When you multiply the $CNOT$ matrix by any of the basis states, you are pulling out one of the columns of the matrix. The column you pull out depends on the index where the 1 is located in the state vector. So for $|00\rangle$ and $|01\rangle$, you pick the first and second columns respectively, and you can see that the state doesn't change. (We are considering the leftmost qubit as the control.)

For $|10\rangle$, you pull out the third column that is $\begin{pmatrix} 0 & 0 &0 &1 \end{pmatrix}^T=|11\rangle$. As you can see, the target qubit was flipped becuase the control qubit was set to 1. Similarly, the state $|11\rangle$ is transformed into $|10\rangle$.

You can apply this same concept to any other matrix and with any other input state, even one in superposition. Consider, for example, the following state:

$$ |\psi\rangle = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 0 \\ 1 \\ 0 \end{pmatrix} $$

Multiplying the $CNOT$ matrix by this state, you pull out columns 1 and 3, both scaled by a factor of $\frac{1}{\sqrt{2}}$. So, you get the state:

$$ CNOT|\psi\rangle = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix} +\frac{1}{\sqrt{2}}\begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix} = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 0 \\ 0 \\ 1 \end{pmatrix} $$

Which is entangled ;)

Answer 3

In addition to the answers posted above, a different way to understand the operation on the target (the control bit always remain the same) is by adding the target and control modulo 2.

For instance, suppose we have

• Before: control $\vert 0 \rangle$ and target $\vert 0 \rangle$. After: control $\vert 0 \rangle$ and target $\vert 0 \oplus 0 \rangle = \vert 0 \rangle$
• Before: control $\vert 0 \rangle$ and target $\vert 1 \rangle$. After: control $\vert 0 \rangle$ and target $\vert 0 \oplus 1 \rangle = \vert 1 \rangle$
• Before: control $\vert 1 \rangle$ and target $\vert 1 \rangle$. After: control $\vert 1 \rangle$ and target $\vert 1 \oplus 1 \rangle = \vert 0 \rangle$

Note that the control bit always remains the same. Hope that helps :)