So let's say you are doing a double slit experiment. Also, let's use electrons.

My question is, won't the gravity of the electron affect the earth, thereby causing it decoherence and its wave function to collapse (or for MWI, entanglement and loss of information to the environment, preventing interference)?

The reason why I think this would happen is because you could tell which path the electron took based off its tidal effects on the earth: everything is a detector.

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    $\begingroup$ Why would Earth's gravity cause decoherence? It's a simple effective potential. You can get yourself a neutron fountain and use it to measure the energy eigenstates of quantum mechanical objects trapped by Earth's gravity, if you like... will that cause decoherence? No. As for slit experiments... the correction from gravity is too small to worry about it. In general slit experiments are too trivial to worry about them, at all. $\endgroup$
    – CuriousOne
    Jan 11, 2016 at 3:55
  • $\begingroup$ Experiments say otherwise... but if you want to believe it that. Yes, there is probably an interaction term that will cause decoherence in ultra strong gravitational fields... but you sure won't be able to measure that on Earth. Maybe inside a black hole, very close to the singularity, where the field will cause pair production etc.. $\endgroup$
    – CuriousOne
    Jan 11, 2016 at 3:59
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    $\begingroup$ @CuriousOne Well, yes the experiments say otherwise. That is the point of my question. $\endgroup$ Jan 11, 2016 at 3:59
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    $\begingroup$ QED effects, like soft photon emission, interfere with entanglement much more than Earth's gravity, physics.stackexchange.com/questions/194458/… But even for them the deviations from QM probabilities are too small to be detected at present. Yes, if you couple classical gravity with QM you will be theoretically able to track electrons, break uncertainty and conservation of energy, etc. That's because such coupling is mathematically inconsistent, and we don't have quantum gravity to tell us how to fix it. $\endgroup$
    – Conifold
    Jan 11, 2016 at 5:36
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    $\begingroup$ @Conifold possibly arxiv.org/abs/gr-qc/0311082 is all you would need. But as you say, the charge of the electron will affect its environment much much more than its mass. Anyway, the real answer to this question would involve density matrices, and interactions that aren't macroscopically amplified. $\endgroup$ Jan 11, 2016 at 10:41

2 Answers 2


Yes, everything is a detector, but you need to quantify which things your system interacts with (and how strongly). Gravity is in some sense a poor example, because the quantum details of gravity are still an unsettled question (and gravity is a weak force regardless), so let's bypass that red-herring by replacing gravity with the electromagnetic field:

As your charged electron accelerates one way or another in a Stern-Gerlack apparatus or double-slit experiment, in theory it should radiate electromagnetic waves. Moreover, you would expect to be able to determine its position, by measuring differences of how particles in the environment are affected by the electron's EM field, right?

Basically, the reason you still observe interference fringes is because the coupling with the environment is weak. (Whereas if you gradually adjust the experimental parameters to increase the strength of coupling to the environment, then the fringes gradually fade.) Weak means that if you do the math, it isn't possible even in principle to infer sufficient information from the environment.

You might enjoy some of Zeilinger's journal-papers such as experimental demonstration of slit interference fringes with buckyballs (which are over a million times more massive than electrons), including demonstration of gradual decoherence (controlling the strength of interaction with the environment). You could also look at QM papers on weak measurement, or decoherence theory.

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    $\begingroup$ "it isn't possible even in principle to infer sufficient information from the environment" is this related to the uncertainty principle? (In general, does the uncertainty principle say strength of coupling is proportional to information, in some sense?) $\endgroup$ Jan 11, 2016 at 12:53
  • $\begingroup$ Is there a hard theoretically cut-off for massive something has to be for its gravitational force to cause any decoherence at all? $\endgroup$
    – Qubei
    Nov 11, 2020 at 10:15

"everything is a detector"

This can't be true, or else there would be no such thing as persistent entanglement.

As @Conifold points out, the electron's charge should be a far more potent source of environmental disturbance, anyway. Why doesn't the charge of the electron leave a trace as it passes through the slits - some persistent disturbance of the charged particles that make up the atoms that make up the filter?

The answer must be that the coupling in both cases (electromagnetic, gravitational) genuinely does not cause decoherence. In the case of gravitation, I would think it's just the extreme weakness of the interaction. In the case of electromagnetism, I'm not so sure.

This answer is a placeholder, written in a rush. I will return to it and improve it a few days from now, if no-one has written a better one.

  • $\begingroup$ One of the thoughts is that all systems are obviously quantum systems as they are all made up of quantum particles. The idea is that wave collapse is a frame of reference of when the detector's system becomes entangled with the other system. This would mean that a particle in a super position to us, we are in a super position from the particle's frame of reference. With this, everything is a detector as "detecting" is just the event of becoming entangled. $\endgroup$
    – Bengie
    Nov 1, 2019 at 19:45

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