Combining Concepts from Ordinary and Quantum Probability

Question

In calculating ordinary (non-quantum) probabilities, we often use $$P(A \lor B)=P(A)+P(B)-P(A \land B)$$ Where $P(A \land B)$ is the overlap term.

In the the quantum we use $$P=\psi^ \star \psi$$ For problems like the double slit experiment we use $$\psi(A \lor B)=\psi(A)+\psi(B) $$ Where A represents one possible way an event can happen and B represents a different possible way it can happen. For example, in the double slit experiment A is the possibility that the particle went through one slit, while B is the possibility that the particle went through the other slit.

But this applies to a variety of quantum experiments where two different situations are possible and so must be considered.

$$P(A \lor B)= \psi(A \lor B)^\star \psi(A \lor B) $$ $$P(A \lor B)= (\psi(A)+\psi(B))^\star (\psi(A)+\psi(B))$$ $$P(A \lor B)= \psi(A)^\star \psi(A) + \psi(A)^\star\psi(B)+\psi(B)^\star \psi(A)+\psi(B)^\star \psi(B)$$ $$P(A \lor B)=P(A)+P(B)+\psi(A)^\star\psi(B)+\psi(B)^\star \psi(A) $$ So how crazy would it be to connect those two ideas and then have, $$P(A \land B)=-(\psi(A)^\star\psi(B)+\psi(B)^\star \psi(A)) $$ If so, what would this even mean?

(I know that it doesn't mean that if we can tell which of the two slits the particle goes through that we could catch it in the act of being split in two.)

For reference, see Feynman's excellent discussion on Probability and Uncertainty: The Quantum Mechanical View of Nature.

Answer 1

In a quantum theory of probability (see also this answer by Valter Moretti), the space of meaningful propositions does not really have an "and" operation like we are used to - given two propositions about measurement results $A$ and $B$, the expression $A\land B$ does not necessarily make sense (Moretti calls such events "quantistically incompatible" in the linked answer).

For instance, say $A$ and $B$ each correspond to measurements of non-commuting operators $O_A,O_B$. Then we can't really say what $A\land B$ means - we cannot measure both simultaneously, and measuring one before the other changes the state. One can define a notion of "and/or" for quantum propositions, but their physical interpretation is not clear at all for incompatible propositions, see again the linked answer.

Answer 2

For problems like the double slit experiment we use $$\psi(A \lor B)=\psi(A)+\psi(B) $$

No, that's not how wavefunctions work. $\psi$ is a function of position, not a function of an event, so it's meaningless to write $\psi(A)$. Plus, even $\lvert\psi\rvert^2$ represents a probability density. The actual probability comes from multiplying by a volume - for instance the probability associated with some region $\mathcal{R}$ is $$P(\text{particle in }\mathcal{R}) \equiv P(\mathcal{R}) = \int_{\mathcal{R}} \psi^*(\mathbf{x})\psi(\mathbf{x})\,\mathrm{d}\mathbf{x}$$ If you're going to apply the rules of probability, like $P(A\lor B) = P(A) + P(B) - P(A\land B)$, to anything in quantum mechanics, you need to apply it to this, the actual probability. If you do this, you wind up finding that the expression winds up translating into a statement about regions: $$\begin{align} P(\mathcal{R}_A\lor \mathcal{R}_B) &= P(\mathcal{R}_A) + P(\mathcal{R}_B) - P(\mathcal{R}_A\land \mathcal{R}_B) \\ \int_{\mathcal{R}_A \cup \mathcal{R}_B} \psi^*(\mathbf{x})\psi(\mathbf{x})\,\mathrm{d}\mathbf{x} &= \int_{\mathcal{R}_A} \psi^*(\mathbf{x})\psi(\mathbf{x})\,\mathrm{d}\mathbf{x} + \int_{\mathcal{R}_B} \psi^*(\mathbf{x})\psi(\mathbf{x})\,\mathrm{d}\mathbf{x} - \int_{\mathcal{R}_A \cap \mathcal{R}_B} \psi^*(\mathbf{x})\psi(\mathbf{x})\,\mathrm{d}\mathbf{x} \end{align}$$ which is a fairly straightforward statement about integrals.