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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.

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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.

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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.

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  • $\begingroup$ Are you certain about that? Check out the Feynman video I put a link to in my last edit. $\endgroup$
    – David Elm
    May 19, 2017 at 23:12
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    $\begingroup$ David Z - I disagree with your negative statements. It is meaningful to write $\psi(A)$ in the sense of $|\psi_A\rangle$ representing an event, or a result of a measurement, and "or" may be encoded by the sum as the OP indicated for two orthogonal i.e. mutually exclusive terms. The statement that "the wave function is a function of all possible events" is much more accurate than "it is a function of positions". The latter creates the wrong feeling that it is a classical field in space but it's not. Instead, it's the encapsulation of knowledge about everything in the world. $\endgroup$ May 20, 2017 at 5:43
  • $\begingroup$ @LubošMotl Of course one can have quantum states associated with events, but within basic quantum mechanics at least, the wavefunction is a function of position associated with the quantum state, not the state itself, i.e. $\psi(\mathbf{x}) = \langle \mathbf{x}\vert\psi\rangle$. If I understand your comment correctly, $A$ is a label identifying one of a class of related wavefunctions, not a parameter to the wavefunction as $\mathbf{x}$ is. $A$ might represent, for instance, "particle in region A". $\endgroup$
    – David Z
    May 20, 2017 at 5:55
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    $\begingroup$ David Z: the question doesn't have any specific attachment to what you call "basic quantum mechanics" in the sense of non-relativistic mechanics of 1 particle's position. It is a general question about the universal quantum mechanical interpretation of AND and OR prepositions in terms of products and interference terms. $\endgroup$ May 20, 2017 at 6:55
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    $\begingroup$ The wave function and the quantum state are exactly synonymous. So a question saying "one but not the other" is logically inconsistent. Quantum states are always represented as collections of complex amplitudes. The index labeling "which amplitude" is the same thing as an "argument of the wave function" and they're in one-to-one correspondence with basis vectors of a basis of the Hilbert space. And every basis vector may always be identified with an event, for example with the event that the physical system was observed in the given state. $\endgroup$ May 21, 2017 at 5:39

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