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[Answers] Quantum state probabilities and amplitudes - absolute value squared?

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Problem Detail: 

It has just appeared to me that I've overlooked a possibly important fact. When we want to calculate probability $p_x$ of measuring the state $X$ and we know the amplitude of $X$ is $\alpha$, we know that $p_x = |\alpha|^2$. Until today I thought that absolute value was not necessary anyway. However, it brings significant difference if we have complex amplitudes with non-zero imaginary parts.

For example, let's say we have a superposition of 16 states, where four amplitudes are like $\frac{i}{4}$, four like $\frac{-1}{4}$ and eight like $\frac{1}{4}$. If we keep with the recipe I wrote above, then this state is fine, as the sum of all probabilities is 1. However, if we skip the absolute value and compute probabilities like $\alpha^2$, then sum of all probabilities in the above state would be $\frac{1}{2}$ and the state would not be stable.

Is the recipe I wrote above ($p_x = |\alpha|^2$ for $\alpha$ being an amplitude of state $X$) correct? Do we calculate the absolute value of the amplitude before we square it?

Asked By : 3yakuya

Answered By : Yuval Filmus

Using $\alpha^2$ doesn't make sense since you get negative or even non-real probabilities! The correct "recipe" is indeed $|\alpha|^2$. This also meshes well with the fact that the operations you are allowed to apply are unitary.

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Question Source : http://cs.stackexchange.com/questions/37688

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