mjhjmtu

Can anyone explain how the CNOT matrix below is a valid presentation for the four-qubit states that follow after?

|0 0> -> |0 0> 
|0 1> -> |0 1>
|1 0> -> |1 1>
|1 1> -> |1 0>

Source: Wikipedia

The one concept that I think would really help you is knowing how to turn those 4 states, $|00rangle, |01rangle, |10rangle, |11rangle$, into vectors, so that you can do the matrix multiplication.

Let me show you.

$$
beginalign
|00rangle = beginbmatrix 1 \ 0 \ 0 \ 0 endbmatrix,|01rangle = beginbmatrix 0 \ 1 \ 0 \ 0 endbmatrix, |10rangle = beginbmatrix 0 \ 0 \ 1 \ 0 endbmatrix, |11rangle = beginbmatrix 0 \ 0 \ 0 \ 1 endbmatrix
endalign
$$

Now if you do the matrix multplication: $rmCNOT times |00rangle$

You will see that you will get exactly what you said, which is $|00rangle$, and the same is true for the rest of them!

This is using the convention that $|0rangle = beginbmatrix 1 \ 0 endbmatrix$ and $|1rangle = beginbmatrix 0 \ 1 endbmatrix$, and $|abrangle = |arangle otimes |brangle$ where $otimes$ is the left Kronecker product.