Commutativity of Born's rule

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etotheipi
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As far as I am aware, if a system is prepared in the state |r> and we measure some observable, the probability of the result of the measurement being the eigenvalue of some eigenvector |u> representing some other state is |<r|u>|2 or <r|u><u|r>.

On a slightly different note, I also know that a state can be represented in terms of basis vectors e.g. |r> = x|u> + y|d>. I have learned that an interpretation of this expression is that if the system is prepared in this state |r>, the probability of being in either the state |u> or |d> is the modulus square of their corresponding coefficients, where, for example, the coefficient of |u> may be expressed as <u|r>. This is consistent with the first paragraph.

However, if we now want to compute the probability that the system takes the state |r> given that it is prepared in the |u> state, according to Born's rule this is simply |<u|r>|2, or <u|r><r|u>, which is identical to the previous probability.

Would I then be right in saying that if some state |r> can be written as a linear combination of basis vectors as in

|r> = x|u> + y|d>

that the probability of measuring |r> given the system is prepared in the state |u> is identical to that of being prepared in the state |r> and measuring |u>, namely the modulus square of the coefficient which is in this case x? This appears to make physical sense in the example of spin just due to the symmetry of the two situations, but I just want to make sure my reasoning is correct. Thanks a bunch in advance!
 

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Orodruin
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Yes.
 
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