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I gather that Bohm denies the notion that the act of measurement decides whether a photon will be a wave or a particle. Bohm's idea seems to be that the photon is always a particle with a real trajectory that always passes through one OR the other slit, NOT bizarrely through both slits simultaneously, in the double-slit experiment. The photon, Bohm says, also has a "pilot wave" that causes the photon to land on a photographic detecting plate according to an interference pattern (which pattern shows up on the plate after the physicists fire off, one at a time, a sufficient number of photons).

My question: if the photon always has Bohm's pilot wave, and if both the photon itself and its pilot wave objectively exist prior to and independent of measurement and detection, then how does Bohm explain the lack of any interference pattern when the double-slit experiment is set up to detect not waves but particles? I gather that when the physicists set up the double-slit experiment to detect a particle, and the physicists fire off one photon at a time toward the double slit barrier, then after enough photons have been fired off, the result on the detecting photographic plate is NOT an interference pattern, but only two bands of light corresponding to the two slits, just as one would expect if light were corpuscular. In other words, no indication of wave behavior, which means no indication of Bohm's pilot wave. If, as Bohm says, the photon is always a real particle that has a pilot wave that really exists objectively, prior to and independent of measurement or detection, shouldn't an interference pattern ALWAYS show up with the two slit experiment (after enough photons have been fired off), even when physicists set up the experiment to detect particles, not waves?

If possible, explain your answer without assuming I have any knowledge of physics. Thanks very much.

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  • $\begingroup$ "...and the physicists fire off one photon at a time toward the double slit barrier, then after enough photons have been fired off, the result on the detecting photographic plate is NOT an interference pattern, but only two bands of light corresponding to the two slits, just as one would expect if light were corpuscular. In other words, no indication of wave behavior..." This is not true. The interference pattern builds up gradually, one spot at a time. See the diagram in en.wikipedia.org/wiki/Double-slit_experiment $\endgroup$ – D. Halsey Oct 20 '18 at 20:38
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If you have a detector at one of the slits to detect which slit the particle goes through, the interference pattern disappears. In the de Broglie-Bohm pilot wave theory, there is still a wavefunction which is defined at both slits. However, each particle's trajectory is well-defined, and carries it through only one slit.

The full answer to this question is rather complicated due to the fancy "conditional wavefunction" used in de Broglie-Bohm theory. In dBB theory, there is a universal wavefunction whose evolution is governed by the Schroedinger equation. However, we can describe smaller systems using this so-called conditional wavefunction. An interesting aspect of these conditional wavefunctions is that they do not necessarily evolve based on the Schroedinger equation, although they often do.

In pilot wave theory, the future positions of a particle are fully knowable given the initial conditions. The double slit experiment in this theory takes as a presumption that the initial states of the particles shot at the slit are distributed randomly, and that these positions are individually unknowable. Because of this, we observe an interference pattern. If we measure the slit through which a particle travels, though, then we necessarily know its state and so can know with certainty its future trajectory. Essentially, when we register a detection, i.e., make a measurement, we necessarily have to interact with the pilot wave (that is what it means to take a measurement, after all), which causes the collapse of the wavepacket traveling through the other slit, removing anything with which to interfere.

Any more "why" than this requires getting into the math, which without a physics background is relatively complicated.

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  • $\begingroup$ "complex", not "complicated" $\endgroup$ – Gimme the 411 Oct 20 '18 at 16:36

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