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In quantum mechanics the possible outcomes of a measurement of a quantity corresponding to a linear hermitian operator $A$ are only the eigenvalues $a_i$ of $A$ (see e.g. http://wwwitp.physik.tu-berlin.de/brandes/public_html/qm/umist_qm/node65.html). The expectation value of $A$ is computed as $$ \langle A \rangle_\psi = \sum_j a_j |\langle \psi | \phi_j \rangle|^2 $$ where $|\langle \psi | \phi_j \rangle|^2$ is called the transition probability (see https://en.wikipedia.org/wiki/Expectation_value_(quantum_mechanics)).

In statistics a random variable $X$ is said to be a set of possible values $x_i$ from a random experiment (see e.g. https://www.mathsisfun.com/data/random-variables.html). The expected value of $X$ is computed as $$ \operatorname{E}[X] = \sum_{i=1}^\infty x_i\, p_i, $$ where $p_i$ are the probabilities of the $x_i$ (see https://en.wikipedia.org/wiki/Expected_value).

So can one say, that the set of expectation values $\{a_j\}$ with transition probabilities $|\langle \psi | \phi_j \rangle|^2$ in QM corresponds to a random variable $X = \{x_i\}$ with probabilities $p_i$ in statistics?

asmaier
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    I'm not sure how this is different from your other question, and how this is not also answered by the duplicate linked there. "Having an eigenvalue $\phi_j$" is precisely the type of proposition that ValterMoretti considers in his answer. – ACuriousMind Feb 06 '16 at 21:27
  • The term "eigenvalue" does not appear in the answer of ValterMoretti you link to, so I think my question is different. – asmaier Feb 06 '16 at 21:35
  • Look at the proposition $P$: "The $x$-value of the spin is $1/2$" - that is the statement that the observable $\sigma_x$ takes a certain eigenvalue. He doesn't explicitly say it, but most propositions you will consider in practice are of course statements about the values of observables, and hence eigenvalues. – ACuriousMind Feb 06 '16 at 21:42
  • But ValterMoretti is talking about measurements of two observables (which as I understand don't commute in QM in general). That is not what I'm asking about. My question is much simpler and only concerns the measurement of a single observable. – asmaier Feb 06 '16 at 21:52
  • If you only have one observable, then everything commutes (the projectors onto the eigenspaces commute because their product is zero), and you indeed get back classical/commutative probability. I maintain this is all in the answer - you just have to read it. – ACuriousMind Feb 06 '16 at 22:13
  • Sorry, I overlooked this clear statement in the wall of text by Moretti: "As a matter of fact, a maximal set of pairwise commuting projectors has formal properties identical to those of classical logic: is a Boolean σ-algebra." – asmaier Feb 07 '16 at 09:42

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