I'm trying to wrap my head around negative-result, or "interaction-free" measurements that seem to provide very real, actionable information about a system.

Such measurements are described in the Wiki article on the Elitzur-Vaidman setup https://en.wikipedia.org/wiki/Elitzur%E2%80%93Vaidman_bomb_tester and similar experiments that have followed, however to my read of it there doesn't really seem to be anything new conceptually here that wasn't covered by Renninger in 1953 with his "Negative Result" thought experiment.

Renninger, and now more recently Vaidman and co have pretty effectively shown that a "negative" or "interaction-free" measurement of a quantum system is just as effective as a traditional measurement in collapsing the system (if one wants to speak of wavefunction collapse).

So my question is really in 2 parts:

1. What, if anything, do interaction-free measurements imply about the nature of the wavefunction? It's easy to conceptualize direct measurement/interaction affecting a system, but it seems incredibly strange that the absence of interaction can be equally effective at steering a quantum system.

2. If a negative result is as effective as a positive result in steering a quantum system, why are Quantum Zeno effects not induced by the mere presence of a detector?

For example:

I surround my radioactive isotope with a fully spherically symmetrical detector that provides continuous monitoring of whether a particle has released decay products. For every instant that goes by and my detector does not click, this is just a "negative" measurement on the system, and so I can update the state of the system as having not evolved beyond its prepared state. Every further instant that passes is the same, and so it seems I should expect to observe a quantum zeno effect where my sample doesn't decay in the time expected.

My issue is this:

If both positive and negative measurements are both capable of disturbing the wavefunction, it seems anything at all should qualify as a measurement. But as that doesn't seem to be the case, I'd like some clarification from the community on what the consensus interpretation is here.

  1. What, if anything, do interaction-free measurements imply about the nature of the wavefunction?

Writing $|a\rangle$ and $|b\rangle$ for two particle states, $|O\rangle$ for the original state of a detector, and $|A\rangle$ and $|B\rangle$ for the detector's possible states after interacting with the particle, in an ordinary measurement we start with $$(α|a\rangle + β|b\rangle) \, |O\rangle = α|a\rangle|O\rangle + β|b\rangle|O\rangle$$ then that evolves to $$α|a\rangle|A\rangle + β|b\rangle|B\rangle$$ and then there's a mysterious wavefunction collapse process that selects one of those outcomes (B, say) and deletes the other, and we ultimately end up with $$|b\rangle|B\rangle.$$

Interaction-free measurement is the same, except that the state $|B\rangle$ is more or less just the time evolution operator applied to $|O\rangle$, instead of being the result of an interaction between the particle and the detector.

This makes no difference as far as the formalism is concerned. In principle, one could probably come up with a model of wavefunction collapse in which it does matter, but that model would be ruled out by experiment. What seems to actually trigger wavefunction collapse is the difference between the states $|A\rangle$ and $|B\rangle$, not how much either of them resembles $|O\rangle$. And that difference still exists (and is still large) because $|A\rangle$ did interact with the particle.

In other words, in an "interaction-free" measurement, there is an interaction, and it triggers the collapse process, which then impertinently deletes it from existence.

  1. If a negative result is as effective as a positive result in steering a quantum system, why are Quantum Zeno effects not induced by the mere presence of a detector?

It's clear from the formalism that the mere presence of a detector does cause a watched-pot effect. To the extent that this isn't noticeable, it's probably because nondetections are vaguer than detections in realistic experimental setups, and so the state that you end up in after each collapse will never be as sharp as it is in this toy example.

Also, lack of observation can't cause a collapse at every instant; it must be discrete. I don't really understand this, but the detector has to continuously evolve (in one branch) from an undetected state to a detected state to trigger collapse, and in the early stages of this process the deviation from the undetected state is small enough that quantum interference is still in play, so I suppose there will be quantum interference between detections starting at slightly different times, leading to a quantization of the detection time.

There was a recent question that covered similar territory (but I see that you already know about it since you commented on it).

  • $\begingroup$ Thanks for the reply - I think I'm following you but not tracking how |𝐴⟩ can be said to interact with the particle in your example. $\endgroup$ – JPattarini Nov 4 '19 at 18:25
  • $\begingroup$ @JPattarini The interaction between $|a\rangle$ and $|O\rangle$ producing $|A\rangle$ is just ordinary interaction, e.g. a charged particle ionizing molecules it passes in a cloud chamber. The main argument I'm making is that this interaction has to happen in the wave function, in superposition with other potential outcomes, even if it's erased later by the collapse. You can't think of measurement devices as purely classical objects that just trigger the collapse rule. They have to be quantum at a low level because they interact with the particle in this fundamentally quantum way. $\endgroup$ – benrg Nov 6 '19 at 6:08
  • $\begingroup$ I think I follow, but a track in a cloud chamber is a result of actual (positive) interaction. A negative result measurement like the Vaidman setup by definition doesn’t interact with the target, so I’m struggling with how it can result in real collapse of the wavefunction. $\endgroup$ – JPattarini Nov 8 '19 at 1:50
  • $\begingroup$ thanks for the effort, if you could take a look at my prior comment as time allows I’d appreciate the help with how to conceptualize $\endgroup$ – JPattarini Nov 9 '19 at 15:42

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