How to translate matrix back into Dirac notation?

Question

In Circuit composition and entangled states section of Wikipedia's article on Quantum logic gates the final result of a combined Hadamard gate and identity gate on $|\Phi^{+}\rangle$ state is: $$ M \frac{|00\rangle + |11\rangle}{\sqrt{2}} = \frac{1}{2} \begin{bmatrix} 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ 1 & 0 & -1 & 0 \\ 0 & 1 & 0 & -1\end{bmatrix} \begin{bmatrix}1 \\ 0 \\ 0 \\ 1\end{bmatrix} = \frac{1}{2} \begin{bmatrix} 1 \\ 1 \\ 1 \\ -1 \end{bmatrix} = \frac{|00\rangle + |01\rangle + |10\rangle - |11\rangle}{2}$$

How exactly does the $(1,1,1,-1)^T$ translate into $|00\rangle + |01\rangle + |10\rangle - |11\rangle$ states? I had no problems translating $|00\rangle$ into $(1,0,0,0)^T$ and $|11\rangle$ into $(0,0,0,1)^T$ but I'm not exactly sure how to reverse the process in this example.

Answer 1

First read about the standard representation of qubit systems of and the basics of bra-ket notation.

I had no problems translating $|00\rangle$ into $\begin{bmatrix}1 & 0 & 0 & 0\end{bmatrix}^T$ and $|11\rangle$ into $\begin{bmatrix}0 & 0 & 0 & 1\end{bmatrix}^T$ but I'm not exactly sure how do you reverse the process in this example.

Cool. Then you should also be able to understand that $|00\rangle + |11\rangle \equiv \begin{bmatrix}1 & 0 & 0 & 0\end{bmatrix}^T + \begin{bmatrix}0 & 0 & 0 & 1\end{bmatrix}^T = \begin{bmatrix}1 & 0 & 0 & 1\end{bmatrix}^T$. Yes? Read on.

How exactly does the $\begin{bmatrix} 1 & 1 & 1 & -1 \end{bmatrix}^T$ translate into $|00\rangle + |01\rangle + |10\rangle - |11\rangle$ states?

The standard basis states of a $2$-qubit sytem $|00\rangle,|01\rangle,|10\rangle,|11\rangle$ i.e. standard basis elements of $\Bbb C^2\times \Bbb C^2$ can be mapped to the four $4\times 1$ column vectors $$\begin{bmatrix} 1 & 0 & 0 & 0 \end{bmatrix}^T, \begin{bmatrix} 0 & 1 & 0 & 0 \end{bmatrix}^T,\begin{bmatrix} 0 & 0 & 1 & 0 \end{bmatrix}^T \& \ \begin{bmatrix} 0 & 0 & 0 & 1 \end{bmatrix}^T.$$ This is essentially an isomorphism from $\Bbb C^2\times \Bbb C^2$ to $\Bbb R^4$.

You say that you already know the following mappings:

$$\begin{bmatrix} 1 & 0 & 0 & 0 \end{bmatrix}^T \to |00\rangle$$ $$\begin{bmatrix} 0 & 1 & 0 & 0 \end{bmatrix}^T \to |01\rangle$$ $$\begin{bmatrix} 0 & 0 & 1 & 0 \end{bmatrix}^T \to |10\rangle$$ $$\begin{bmatrix} 0 & 0 & 0 & 1 \end{bmatrix}^T \to |11\rangle$$

Now you simply need to expand the column vector $\begin{bmatrix} 1 & 1 & 1 & -1 \end{bmatrix}^T$ in terms of its basis elements, as follows:

$$\begin{bmatrix} 1 & 1 & 1 & -1 \end{bmatrix}^T$$ $$= \begin{bmatrix} 1 & 0 & 0 & 0 \end{bmatrix}^T + \begin{bmatrix} 0 & 1 & 0 & 0 \end{bmatrix}^T + \begin{bmatrix} 0 & 0 & 1 & 0 \end{bmatrix}^T + \begin{bmatrix} 0 & 0 & 0 & -1 \end{bmatrix}^T$$

$$= \begin{bmatrix} 1 & 0 & 0 & 0 \end{bmatrix}^T + \begin{bmatrix} 0 & 1 & 0 & 0 \end{bmatrix}^T + \begin{bmatrix} 0 & 0 & 1 & 0 \end{bmatrix}^T - \begin{bmatrix} 0 & 0 & 0 & 1 \end{bmatrix}^T$$

$$\implies \begin{bmatrix} 1 & 1 & 1 & -1 \end{bmatrix}^T \equiv |00\rangle + |01\rangle + |10\rangle - |11\rangle$$

In case you can't understand the last couple of steps, review Matrix addition. And you're done!

^{Note: $T$ stands for transpose.}

Answer 2

This is using vector addition.

$|00\rangle = \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}$

$|01\rangle = \begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \end{bmatrix}$

so:

$|00\rangle + |01\rangle = \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} + \begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 0 \\ 0 \end{bmatrix}$

by the rules of vector addition. Adding in the other two terms:

$|00\rangle + |01\rangle + |10\rangle - |11\rangle = \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} + \begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \end{bmatrix} + \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix} - \begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 1 \\ -1 \end{bmatrix}$

The reason you might want to write a state vector this way is to see how measurement of one qbit affects the other, for example (see here).