In measurement in the computational basis, I was being told that it is a way to extract information from a qubit, and it outputs a classical bit.

For the quantum state $\alpha |0\rangle + \beta |1\rangle$, the possible outputs are:

$0$, with probability $|\alpha|^2$ and

$1$, with probability $|\beta|^2$.

Why are the probabilities $|\alpha|^2$ and $|\beta|^2$? I assume this is related to the figure below? (where $\alpha$ being the possibilities for it to fall in the x-axis and $\beta$ for the y-axis?)

  • 3
    $\begingroup$ Maybe you need to read some basic books about Quantum Mechanics. For example, maybe nielsen's chapter 2 is enough for you. In quantum mechanics, the probability is always connected to the $\mid\cdot\mid^2$. It's a basic assumption. $\endgroup$
    – narip
    Commented May 29, 2021 at 9:21
  • 1
    $\begingroup$ @narip Sure, thanks for the recommendation. I'll take a look. $\endgroup$ Commented May 29, 2021 at 9:29

1 Answer 1


It is postulate or axiom of quantum mechanics that if a state $|\psi\rangle $ that is a linear superposition of eigenstates $\{ |e_i\rangle\}$ of some observable, $$ |\psi \rangle = \sum_i \alpha_i |e_i\rangle $$ then upon making measurement with respect to this observable, the state is observed in the state $|e_i\rangle$ with probability $|a_i|^2$. That is, $P(|e_i\rangle) = |a_i|^2$. Also note that $\sum |\alpha_i|^2 = 1$ is a necessary condition.

In quantum computing, when we talk about measurement, it usually correspond to measuring in the computational or Z basis. So for a single qubit system, your observable would be the single Pauli Z matrix, $Z = \begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}$, which has eigenvectors of $|0\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix} $ and $|1\rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix} $.

Now, your state is $|\psi \rangle = 0.6|0\rangle + 0.8|1\rangle $ is already written as the linear combination of the eigenstates of the Pauli Z matrix. And thus, the probability to observe $|0\rangle$ is $|0.6|^2$, that is, $P(|0\rangle ) = |0.6|^2$. Similarly, $P(|1\rangle) = |0.8|^2$.


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