This question relates to the paper commented in this 2010 article.

The paper itself is Ruling Out Multi-Order Interference in Quantum Mechanics; it is the discussion of a triple-slit interference experiment testing the validity of Born rule.

The abstract asserts that

Born’s rule predicts that quantum interference, as shown by a double slit diffraction experiment, occurs from pairs of paths.

I do not understand this statement.

The theory behind it originates from Quantum Mechanics as Quantum Measure Theory (Sorkin, 1994) where is is said

Why are probabilities squares of amplitudes; why are they expressed most naturally in terms of pairs of paths rather than individual paths?

Why indeed? If we refer to the path integral description all possible paths are taken into account, including the very weird ones, and there is an infinity of them. What is the point in decomposing the interference pattern of a three-slit diffraction in terms of pairs of paths ?

When the first article says

Born's rule is one of the key laws in quantum mechanics and it proposes that interference occurs in pairs of possibilities. Interferences of higher order are ruled out.

what does it means exactly? Is there something here that gives any real insight about Born rule, or is this way of picturing interferences just the effect of a specific mathematical treatment, while being actually equivalent to the path integral formalism?

More generally, what is the significance of this experiment?

  • $\begingroup$ You are looking at another naively designed quantum mystic experiment that has absolutely no consequence for physics proper. For one thing the ontology employed by these folks is completely false (there are no paths in quantum mechanics), for another, the Born rule is an experiment dependent approximation that relies on interactions in the detectors destroying off-diagonal elements in the density matrix, rather than the time evolution of the wavefuntion. In other words... these guys are basically testing trough a Rube-Goldberg machine how "black" their detector is. The significance is none. $\endgroup$
    – CuriousOne
    Commented May 24, 2016 at 20:29

2 Answers 2


I'm just guessing here, but one way to interpret it goes like this.

If you have the amplitude $\Phi$ for an outcome that can be achieved in multiple indistinguishable ways each described by an amplitude $\phi_i$ with $i \in 1, 2, \dots , n$ we write $$\Phi = \sum_{i=1}^n \phi_i \,.$$

That makes the probability for the outcome \begin{align*} P &= \Phi^* \Phi\\ &= \left( \sum_{i=1}^n \phi_i \right)^* \left( \sum_{i=1}^n \phi_i \right) \\ &= \left( \sum_{i=1}^n \phi_i^* \phi_i \right) + \left[ \sum_{i=1}^{n-1} \sum_{j=i+1}^n \left( \phi_i^* \phi_j + \phi_j^* \phi_i\right)\right] \,. \end{align*} The single sum is over the squares of the individual amplitudes each of which is non-negative. All the interference comes from the cross-terms each of which involves a pairing of terms from exactly two possible ways for the outcome to be achieved.

So, no single term involving more than two ways of achieving the outcome appears in the probability of that outcome, and I suppose that this is what is referred to. But I don't understand what significance the authors ascribe to the fact.


Basically, interference of higher order means an interference term that involves three or more slits.

In the double slit experiment, according to Born's rule, $$\lvert A+B\rvert^2 = \lvert A\rvert^2 +\lvert B\rvert^2 + A^*B + B^*A$$

$$\lvert A+B\rvert^2 = \lvert A\rvert^2 +\lvert B\rvert^2 + I_{AB}$$

Notice that the interference term $I_{AB}$ involves the two slits $A$ and $B$.

For three slits, according to Born's rule, $$\lvert A+B+C\rvert^2 = \lvert A\rvert^2 +\lvert B\rvert^2 + \lvert C\rvert^2+ I_{AB} + I_{AC} + I_{BC}$$

Notice that there is no interference term that involves A, B and C all at once, i.e. there is no $I_{ABC}$.

In this 2010 article, they used the fact that the probability for three slits $P_{ABC}$, could also be expressed as: $$\lvert A+B+C\rvert^2 = P_{ABC} = P_{AB} + P_{AC} + P_{BC} - (P_{A} + P_{B} +P_{c})$$

$$I_{ABC} = P_{ABC} - (P_{AB} + P_{AC} + P_{BC}) + P_{A} + P_{B} +P_{c}$$

They measured all seven quantities in the right side and concluded that there was no Born rule violation. In the paper, they provided an upper bound for $I_{ABC}$ which is $10^{-2}$ of the expected pairwise interference.

The significance of the experiment is that it provided an upper bound for the possible deviations from Born rule. In other words they concluded that Born rule is true within their bound of experimental errors.

  • $\begingroup$ Well according to the path integral approach there is always interference of all possible paths. This means that even for a single slit you get an infinity of paths interfering, so changing the number of slits will not add any higher-order effect. As the accepted answer shows, pairs simply appear as cross-terms in the (mathematical) expansion of an outcome probability as per Born rule. There does not seem to be anything special going on here, physics-wise. $\endgroup$ Commented Sep 19, 2017 at 14:24
  • $\begingroup$ @StéphaneRollandin This paper was testing the validity of Born rule. In other words, what if the probability is not strictly modulus squared? If Born rule is just an approximation then they would have detected contributions that are not accounted by Born rule. $\endgroup$
    – Anon
    Commented Sep 19, 2017 at 14:30
  • $\begingroup$ @StéphaneRollandin If the accuracy of their experiment was much higher and if they detected a violation of the Born rule then Schrodinger's equation must be modified. $\endgroup$
    – Anon
    Commented Sep 19, 2017 at 14:35
  • $\begingroup$ It is certainly worthwhile to experimentally test the validity of Born rule, no problem with that. Only, I can still see no legitimation to the notion that "quantum interference occurs from pairs of paths". This is what my question was about. $\endgroup$ Commented Sep 19, 2017 at 14:37
  • 1
    $\begingroup$ @StéphaneRollandin This answer adds something that my answer explicitly disclaims: an interpretation of the significance: working from just your quotes I didn't follow that the report on an experimental limit. This kind of quantitative 'we found exactly what the theory says' paper is important, but it helps to rule out alternate theories that make almost but not quite identical predictions. $\endgroup$ Commented Sep 19, 2017 at 15:30

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