I'm sorry if something gets lost in translation, as my professor wrote all questions in portuguese, but what does it mean to ask the probability of finding "state $|\alpha \rangle$ in state $|\beta\rangle$"?. What does it mean and how do I actually calculate it?
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$\begingroup$ But what is the physical significance of it? Is it the probability of finding the particle state $|\beta\rangle$ after measuring state $|\alpha\rangle$? $\endgroup$– Hector FreireSep 20, 2022 at 17:00
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1$\begingroup$ Assuming $|\beta\rangle, |\alpha\rangle$ are normalized, it (presumably) means that if you have a quantum system prepared in a state $|\beta\rangle$ and measure, for example, the observable $|\alpha\rangle\langle \alpha|$, then the probability to obtain the measurement value $1$ equals to $|\langle \alpha|\beta\rangle|^2$. The state after the measurement then is $|\alpha\rangle$. You can in fact measure any observable, as long as $|\alpha\rangle$ is a non-degenerate eigenvector of the corresponding hermitian operator. $\endgroup$– Tobias FünkeSep 20, 2022 at 17:03
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$\begingroup$ ...the aforementioned probability is then the probability to measure the value $a$, where $A|\alpha\rangle = a|\alpha\rangle$ is the eigenvalue to the (non-degenerate) eigenvector $|\alpha\rangle$ of the hermitian operator $A$, corresponding to the observable you measure. $\endgroup$– Tobias FünkeSep 20, 2022 at 17:09
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$\begingroup$ Thanks for the help! $\endgroup$– Hector FreireSep 20, 2022 at 17:13
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$\begingroup$ Well, I think the question can be interpreted so: you measured $|\alpha\rangle$ and measure immediately that; what is the probability that you get $|\beta\rangle$ $\endgroup$– kludgSep 20, 2022 at 17:13
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1 Answer
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The question can be formulated as follows " we prepare a system in state $|\alpha\rangle$ and measure. What is the probability that we measure $|\beta\rangle$?. The answer is obvious: assuming states are normalized, it is square of the norm of overlap the states, $|\langle\alpha|\beta\rangle|^2$.
I think the question is artificially confusing because we don't specify what we prepare and measure. The answer is different from trivial $\{0,1\}$ if we prepare and measure different observables ( in terms of Quantum Information Science, prepare and measure in different bases).
