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$\sigma$-additivity - probability of a sum of countable number of pairly disjoint events equals a sum of probabilities of these events. (3. Axiom of Probability)

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For pairly disjoint sets $A_k$

Since there are different methods of summing amplitudes in QM depending on whether outcomes of measurement are distinguishable or not (i.e. in case of identical particles and non-identical particles [Feynman vol. 3]), does it mean that "probability" in QM does not satisfy the $\sigma$-additivity which is crucial property in classical probability?

If there are several mutually exclusive, indistinguishable alternatives in which an event might occur, the probability amplitude of all these possibilities add to give the probability amplitude for that event: enter image description here

When an experiment is performed to decide between the several alternatives, the same laws hold true for the corresponding probabilities and not the probability amplitudes:

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  • In quantum mechanics it's not the probabilities that are additive, but the states that make up a wave function. Even the wave functions of different particles don't live in a direct sum but in a product space. Is that what you mean? – CuriousOne Oct 07 '14 at 17:06
  • I don't quite understand. Given a state and a positive operator valued measure, there is a probability measure associated to this measurement. As a probability measure, this is sigma-additive. – Martin Oct 07 '14 at 17:07
  • Quantum mechanical probabilities do not describe the dynamics of the system, they only describe the outcomes of measurements. The outcomes of measurements are NOT additive. – CuriousOne Oct 07 '14 at 17:08
  • Edited to make in clearer. – Ghostwriter Oct 07 '14 at 17:18
  • In quantum mechanics all "mutually exclusive indistinguishable" alternatives interfere with each other trough a complex phase. See Feynman's path integrals for a correct description of a QM system based on "alternative paths" trough phase space. It's not even correct to say that the system has taken one of these paths (as is falsely suggested by classical interpretations of the double slit experiment), because the outcome is only predicted correctly by assuming that it has taken ALL possible paths simultaneously. – CuriousOne Oct 07 '14 at 17:29
  • see my answer: http://physics.stackexchange.com/questions/116595/why-is-the-application-of-probability-in-qm-fundamentally-different-from-applica/116609#116609 – Valter Moretti Oct 07 '14 at 18:11
  • Thank you Valter Moretti. So have I understood it correctly, that $\sigma$-algebra (and $\sigma$-additivity) can be taken from classical probability to the quantum world only if we define AND and OR operators using approach with commutation? – Ghostwriter Oct 07 '14 at 18:50
  • As you have seen, in qm it is used a generalized notion of (say non commutative) sigma algebra and sigma additive probability, whose properties give rise to the more familiar composition of complex amplitudes. However as soon as one specialzes this probability to a maximal set of commuting elementary properties of the system, he/she recovers a standard sigma algebra and a standard sigma. additive probability – Valter Moretti Oct 07 '14 at 19:57

1 Answers1

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QM probabilities do obey $\sigma$-additivity, however your sample space and probability measure change depending on what measurements you are making.

To use the classic example, consider a double slit experiment in which a particle can pass through two slits, $A$ and $B$, and hits a screen at point $x$. If we don't measure which slit the particle went through we can ask for, say, the probability that it hits the screen in the interval $[x_0,x_1]$ or the in the interval $[x_2,x_3]$ and we find that \begin{align} P(x_0\le x\le x_1) = \int_{x_0}^{x_1}\mathrm{d}x\,|\psi_A(x)+\psi_B(x)|^2 \end{align} and similarly for $P(x_2\le x \le x_3)$. If the two intervals are disjoint it is clear that \begin{equation} P(x_0 \le x\le x_1 \;\mathrm{or}\;x_2\le x \le x_3) = P(x_0\le x\le x_1) + P(x_2\le x \le x_3) \end{equation} so we have $\sigma$-additivity. However notice that our sample space is the space of possible $x$ values and makes no reference to which slit the particle went through. If we ask which slit the particle went through, we find that we no longer have a well defined probability, which is why we can't even sensibly talk about the question.

If we measure which slit the particle went through then we can ask for, say, the probability that it went through slit $A$ and hit the screen in the interval $[x_0,x_1]$. If we call the slit the particle went through $S$ we get an expression \begin{equation} P(S = A,\, x_0 \le x \le x_1) = \int _{x_0}^{x_1}\mathrm{d}x\,|\psi_A(x)|^2\end{equation} and we can come up with similar expressions for things like $P(S = B,\, x_0 \le x \le x_1)$ or $P(S = A,\, x_1 \le x \le x_2)$. If we now look at the probability for the particle hitting a point on the screen, regardless of which slit it went through, we do not get an interference pattern. i.e.\begin{align} P(S = A\;\mathrm{or}\;B,\, x_1 \le x \le x_2) &= \int _{x_0}^{x_1} \mathrm{d}x\, |\psi_A(x)|^2 + |\psi_B(x)|^2\\ &= P(S = A,\, x_0 \le x \le x_1)+ P(S = B,\, x_0 \le x \le x_1) \end{align} so, again, we have $\sigma$-additivity, but we also have a different sample space, which now includes which slit the particle went through, and even if we some over this we have a different probability measure on the values of $x$.

In short quantum mechanical probabilities are always well-defined they are just not necessarily as well-behaved as you might classically expect, so we don't tend to think about them in those terms most of the time.

Drarp
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