Why do we use square in measuring a qubit with probability?

Question

A superposition is as follow: $$ \vert\psi\vert = \alpha\vert 0\rangle + \beta\vert 1\rangle. $$

When we measure a qubit we get either the result 0, with probability $\vert \alpha\vert^{2},$ or the result 1, with probability $\vert \beta\vert^{2}.$ Blockquote

Why does the $\alpha$ have the power $2$ ?

Answer 1

Depending on how you are setting up your quantum states, it is usually taken as an axiom of the system that if a state $|\psi\rangle$ is a linear superposition of eigenstates $\{ |e_n\rangle \}$ of some observable where

$$|\psi\rangle = \sum_n \alpha_n |e_i\rangle \ , \ \alpha_n \in \mathbb C \text{ and the coefficients are normalized: } \sum_n |\alpha_n|^2 = 1$$

then when we make a measurement with respect to that observable, the state is observed in state $|e_i\rangle$ with probability $|\alpha_i|^2$. Whatever else the probability is, it cannot be $\alpha_i$, as that is complex. $|\alpha_i|$ is a possible choice. But it will turn out that the mathematics and physics of the model are much more fruitful if we say the probability is $|\alpha_i|^2$.

In other words, it's a model which has made an arbitrary choice about how to interpret it and use it. And it turns out it's a very successful model.

(To be convinced of that last point, keep on using the model in your course or elsewhere!)

Answer 2

The reason amplitudes are used, which square to probabilities, instead of using probabilities themselves, is that it lets you have the same probability values with different phases.

This is how we can have both of these amplitude vectors represent a 50% chance of a qubit being true or false: $$1/\sqrt{2}(|0\rangle+|1\rangle)$$ $$1/\sqrt{2}(|0\rangle-|1\rangle)$$

Which is written like this when not using the ket notation:

$$[1/\sqrt{2}, 1/\sqrt{2}]$$ $$[1/\sqrt{2}, -1/\sqrt{2}]$$

Why that is important is because it lets us change the phase without affecting probability. When we combine values, depending on phase, they will either add together, or cancel each other out.

Here they add together: $$[1/\sqrt{2}, 1/\sqrt{2}] + [1/\sqrt{2}, 1/\sqrt{2}] = [\sqrt{2}, \sqrt{2}]$$

And here they cancel out: $$[1/\sqrt{2}, 1/\sqrt{2}] + [1/\sqrt{2}, -1/\sqrt{2}] = [\sqrt{2}, 0]$$

This allows deconstructive interference to happen, which is observable in the real world with experimentation, but is also one of the things that makes quantum computing powerful.