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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 $2$ ?

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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!)

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    $\begingroup$ So plz me an example that why using square has the advantages. $\endgroup$
    – Hamit
    Oct 9, 2015 at 19:50
  • $\begingroup$ Mathematical convenience is one advantage: the machinery of Hilbert space is nicely set up to deal with quadratic forms. But the main advantage is that it seems to describe reality, in the sense that it agrees quite well with the results of experiments. $\endgroup$ Oct 9, 2015 at 19:55
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    $\begingroup$ Stern-Gerlach experiment and various variations on it. $\endgroup$ Oct 9, 2015 at 20:07
  • $\begingroup$ @HamedBaghalGhaffari: Whether or not it's an advantage is irrelevant; the reason we use the squares is that this is how the world works; the laws of physics, which are found from experiment, say so. $\endgroup$
    – Javier
    Oct 9, 2015 at 20:11
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    $\begingroup$ I was just about to suggest the Stern-Gerlach experiment. Have a look at this MIT lecture, starting at 39:00, as marked: youtu.be/AX9769eQV24?t=39m1s $\endgroup$
    – Simon S
    Oct 9, 2015 at 20:14
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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.

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    $\begingroup$ Sorry! But then $(\sqrt{2})^2+(\sqrt{2})^2=4\neq 1$ so your answer is not working and also what do you exactly mean by adding two qubits? $\endgroup$
    – H.W.
    Oct 19, 2015 at 17:50
  • $\begingroup$ You are correct in that the examples i showed of adding cubits doesn't result in a normalized vector. The point was to just show destructive and constructive interference though. If you want a better example, pass the vector $1/\sqrt{2}(|0\rangle+|1\rangle)$ through a Hadamard gate, and then pass the vector $1/\sqrt{2}(|0\rangle-|1\rangle)$ through and observe the difference. The Hadamard gate is the matrix: $$1/\sqrt{2}\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}$$ $\endgroup$
    – Alan Wolfe
    Oct 19, 2015 at 18:34

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