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I started a "principles of quantum mechanics" course this year, and already have a problem with successive measurements (using Dirac notation). The question is about incompatible observables A and B. An experiment is performed to measure observable A, then observable B, and then A again, where the eigenvalues A1 and A2 can then be retrieved each with their own probability.

To find the probability of getting the value of observable A to be A1 in the end, I literally draw a probability tree, and count and sum the probabilities along the required branch. However, I would like to know how this is done using Dirac's notation (for more complex successive measurements, for example).

Here is the exercise:

The observables A and B are represented by operators $ \hat A $ and $ \hat B $ with eigenfunctions { u$_i$(x) } and { v$_i$(x) } respectively, such that

v$_1$(x) = { $\sqrt 3 u_1(x) + u_2(x)$ } / 2

v$_2$(x) = { $u_1(x) − \sqrt 3 u_2(x)$ } / 2

A certain system is subjected to three successive measurements:

(i) a measurement of A

(ii) a measurement of B

(iii) another measurement of A

Show that if measurement (i) yields the value A1 there is a probability of 5/8 that (iii) will yield A1 and a probability of 3/8 that it will yield A2.

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closed as off-topic by ACuriousMind, John Rennie, user36790, Kyle Kanos, HDE 226868 Oct 1 '15 at 1:45

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    $\begingroup$ but have no idea how to use LaTeX. To be straight with you, it's not going to help you get an answer saying that, MathJax is just a markup notation, it's not complicated at all meta.math.stackexchange.com/questions/5020/…. The best of luck with it anyway $\endgroup$ – user81619 Sep 30 '15 at 16:16
  • $\begingroup$ Thanks for that, I was looking for the link. Added the exact question/exercise $\endgroup$ – Darbininkai Broliai Sep 30 '15 at 16:34
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If the first measurement yields the value $A_1$ with certainty, this means the initial state has collapsed into $u_1$ after the first observation. In particular one has, inverting the above back: $$ |u_1\rangle =\frac{\sqrt{3}}{2}|v_1\rangle + \frac{1}{2}|v_2\rangle. $$ Now a measurement of the observable $B$ must be performed and then one more measurement of $A$. The probability of getting again $A_1$ is, by Bayes' theorem $$ P(A_1 | B) = P(v_1)P(A_1|v_1) + P(v_2)P(A_1|v_2). $$ Computing each contribution starting from the components coefficients that you have in $|u_1\rangle$ we find $$ P(A_1 | B) = {\left(\frac{\sqrt{3}}{2}\right)}^2\cdot{\left(\frac{\sqrt{3}}{2}\right)}^2 + {\left(\frac{1}{2}\right)}^2\cdot {\left(\frac{1}{2}\right)}^2 = \frac{5}{8}. $$

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