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
    May 29 '21 at 9:21
  • 1
    $\begingroup$ @narip Sure, thanks for the recommendation. I'll take a look. $\endgroup$ May 29 '21 at 9:29

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, $P(|1\rangle) = |0.8|^2$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.