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It seems like some experiments on quantum systems, like the electron $g-2$ measurement, do not rely directly on the Born rule, since they are more so measuring inherent characteristics of the evolution of the wavefunction.

Whereas, an experiment that obviously does rely on the Born rule is the single-electron double-slit experiment, since in principle you could tell whether the distribution of electrons looks more like $|\psi|^2$ versus, let's say, just $|\psi|$. However, I've watched a couple videos of that experiment, and to me it doesn't look precise enough to distinguish between $|\psi|^2$ and $|\psi|$. (I mean, it's a mammoth achievement just to get that experiment working at all)

So maybe it's just my utter lack of experimental knowledge, but I'm wondering, can we really be sure the probability goes as $|\psi|^2$ and not $|\psi|$?

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    $\begingroup$ some people will rush to reply that "but only the squared version preserves probability", but this is simply fixed by diving by the evolving norm. Also not a strong theoretical argument as particle conservation is a low energy illusion. True quantum mechanics must predict rates of events, not specific probabilities. Good question +1 $\endgroup$
    – lurscher
    Sep 12, 2022 at 19:21
  • $\begingroup$ The TDSE implies a continuity equation, equivalent to conservation of total probability, in which the probability density is $|\psi|^2$. So empirical tests of the TDSE fit the bill. $\endgroup$
    – J.G.
    Sep 12, 2022 at 22:15
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    $\begingroup$ "can we really be sure the probability goes as $|\psi|^2$ and not $|\psi|$ ?" Why not $|\psi|^3$? Why not $|\psi|^{1.42}$? Why not $|\psi|^{2.001}$? $\endgroup$
    – hft
    Sep 12, 2022 at 22:32

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I might be missing the depth somewhere, but I would suggest the electron double-slit experiment first: $|\sin(kx)|$ clearly looks different as compared to $\sin^2(kx)$.

The second example is the Malus' law in optics. Polarizers are performing the quantum measurement on photons' polarization state. The fact that the measured intensity goes as $\cos^2(\theta)$ seems like a proof enough to me of the "squared version" of the Born rule.

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    $\begingroup$ This is the correct answer $\endgroup$
    – lurscher
    Sep 12, 2022 at 23:07
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Any photon-counting diffractive spectroscope. We see a rate of photon arrival proportional to the intensity of the diffracted wave, that is the square of the amplitude.

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So maybe it's just my utter lack of experimental knowledge, but...

It's effectively unrelated to your knowledge (or lack thereof) of experiments.

...I'm wondering, can we really be sure the probability goes as $|\psi|^2$ and not $|\psi|$?

By definition, given a wavefunction solution to the Schrodinger equation $\psi(x_1, x_2, \ldots)$, the probability density is (see, e.g., Messiah, Quantum Mechanics, Volume 1, Chapter IV, Section 2): $$ |\psi(x_1, x_2, \ldots)|^2 $$

You could re-write a bunch of assumptions and conventions and thereby replace $|\psi|^2$ with some other monotonic function of $|\psi|^2$ and call that a probability density instead.

But, all the the conventions we use in practice are already directed at using $|\psi|^2$ as the probability density, not some other function.

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    $\begingroup$ That waves diffract according to amplitude but deposit energy according to its square was understood a century before Born. Adam Herbst's question is "why not linear in amplitude?". You're taking the "standard/typical meaning of $|\psi|^2$" as an axiom. But that emerged from the requirement to match long-held experimental knowledge. Physics doesn't come from definitions and axioms. $\endgroup$
    – John Doty
    Sep 12, 2022 at 21:21
  • $\begingroup$ @JohnDoty I am taking it as an axiom. And I really don't care where it came from. It has been a standard axiom for almost a hundred years, and I don't feel a great need to justify it by saying that ultimately it must be checked against experiment. $\endgroup$
    – hft
    Sep 12, 2022 at 21:42

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