# Why are the probabilities $|\alpha|^2$ and $|\beta|^2$ when measuring in the computational basis?

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?)

• 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. Commented May 29, 2021 at 9:21
• @narip Sure, thanks for the recommendation. I'll take a look. Commented May 29, 2021 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, $$P(|1\rangle) = |0.8|^2$$.