Operation of NOT and Z gate on the qubit in the state $|H\rangle = a|0\rangle + b|1\rangle$?

Question

I have this last question for an assignment and I've been stuck on it for hours.

Pauli operators for a two-level system(qubit), $$ \sigma_x = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \sigma_z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} $$ $\sigma_x$ is called the quantum NOT gate and $\sigma_z$ is the Z gate. We have a qubit in the state $|H \rangle = \alpha |0\rangle + \beta|1\rangle$. Which state is obtained after operating with the NOT gate and which state with the Z gate? What happens if we apply operation twice?

Answer 1

Well, $\sigma_x |H\rangle$ is given by matrix vector multiplication:

$$\pmatrix{ 0 & 1\\ 1 & 0 } \pmatrix{ \alpha\\ \beta} =\pmatrix{\beta\\\alpha} $$

The state is $\beta |0\rangle + \alpha|1\rangle$. Similar for the 2nd operation.

Why is \ket in Latex not working?

Answer 2

Thanks! $\alpha$ and $\beta $ are constants. Shouldn't they remain outside the matrix? For example, in ket notations $ |0\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix}, \quad |1\rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix} $

The operation of NOT gate on these would be something like this

$ \sigma_x |H\rangle = \alpha \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} + \beta \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \alpha \begin{pmatrix} 0 \\ 1 \end{pmatrix} + \beta \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \alpha |1\rangle + \beta |0\rangle $ Am I correct?