For example, given two spin-1/2 particles in basis $|++\rangle,|+-\rangle,|-+\rangle,|--\rangle$. If the first measurement of spin is carried out and gives $1/2$ for particle one, what is the probability of getting $1/2$ for particle 2?

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    $\begingroup$ Well, that will depend on what state the spins are in, won't it? $\endgroup$ Commented Oct 18, 2017 at 17:23
  • $\begingroup$ Are you taking about conventional QM, or this is some kind of a modification? In conventional formalism measurements are independent, as far as I am concerned. Hence, conditional probability is not even introduced. $\endgroup$
    – MsTais
    Commented Oct 18, 2017 at 17:35
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    $\begingroup$ @MsTais That is incorrect. If you have a state $\left \lvert \uparrow \uparrow \right \rangle + \left \lvert \downarrow \downarrow \right \rangle$ then of course the probabilities are correlated. $\endgroup$
    – DanielSank
    Commented Oct 18, 2017 at 17:41

2 Answers 2


So just to take this really really low-level, the definition of conditional probability is $$\operatorname{Pr}(A|B) = {\operatorname{Pr}(A\cap B)\over\operatorname{Pr}(B)}.$$To get these probabilities we need indicator observables, observables which are $1$ if and only if the particle is in an "acceptable" state, or zero otherwise. For example the indicator observable for detecting if a harmonic oscillator is in one of its lowest three states is $|0\rangle\langle 0| + |1\rangle\langle 1| + |2\rangle\langle 2|.$

If we are dealing with two intimately connected systems then the indicator $\hat P_{A\cap B}$ will not have an obvious connection to $\hat P_A$ or $\hat P_B$ and you will just have to list out all of the basis states which satisfy both of the properties that you desire. But for separable systems like you're describing now, we have the nice property that $\hat P_{A\cap B} = \hat P_A \otimes \hat P_B$ with the tensor product $\otimes$ in there.

So then the above equation will just become,$$\operatorname{Pr}(A|B) = {\langle\hat P_A\otimes \hat P_B\rangle\over\langle\hat 1\otimes\hat P_B\rangle}.$$

  • $\begingroup$ Nice, thanks! But what's the intuition behind $\hat{I} \otimes \hat{P}_B$? I get that it has to be in the two-particle hilbert space, but I would had never got it myself. $\endgroup$ Commented Oct 18, 2017 at 21:11
  • $\begingroup$ Any observable $\hat 0$ which acts only on the $B$-subspace of the Hilbert space is expressed in terms of the bigger space as $\hat 1\otimes\hat O.$ Or you can just expand it out, for the state $|b\rangle$ for the second subsystem the compatible states are $\sum_n |nb\rangle\langle nb| = \hat 1\otimes |b\rangle\langle b|.$ $\endgroup$
    – CR Drost
    Commented Oct 18, 2017 at 22:51

In another manner of seeing the same, the question almost includes the answer. The question states: "If the first measurement of spin is carried out and gives 1/2 for particle one". In that case, the state of the system after the measurement is obtained by collapsing the wave function of the system to a state in which only the 1/2 spin is present. Then, probability of observing spin 1/2 for particle one can be obtained conventionally. To help you in doing this, note that |+,+> is equivalent to |+>|+>, where the first ket applies to the first particle and the second ket to the second particle.


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