# How to translate matrix back into Dirac notation?

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 $$\begin{bmatrix} 1 \\ 1 \\ 1 \\ -1 \end{bmatrix}$$ translate into $$|00\rangle + |01\rangle + |10\rangle - |11\rangle$$ states?

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

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.

• Oh, okay, it was way simpler than I expected. Thanks! – Michał Zając Jan 23 '19 at 0:11

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

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

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).