0
$\begingroup$

Suppose we have double-slit experiment. Before the right slit, we add a particle detector that allows the particle to go through. It is well-known that if we carry this out, then the wave interference effect goes away, and the distribution of the particles on the opposite wall looks like the distribution of bullets.

Now suppose we put this whole apparatus in a black box and make sure that once the box is opened, the data from the right-slit detector is destroyed. Say it is just never hooked up to any storage device in the first place. All we can observe is the collision pattern on the back wall. (I don’t know if it’s really possible to “destroy” this information; so this is probably a good place to poke holes in the setup.)

So if the collapse of the wave function depends on observers and consciousness, the result should be as if there were no right-slit detector, and we should see a wave interference pattern. But if the measuring device objectively causes the wave to collapse, then the result will be like bullets, as if we had been watching the whole time.

Which will happen?

$\endgroup$
2
  • $\begingroup$ Does your proposed experiment really require the information detected by the detector to be destroyed? Does it not suffice that an observer is not accessing that information? $\endgroup$
    – user87745
    Apr 1, 2021 at 19:26
  • $\begingroup$ @dvij Yes we could just say no one ever sees it. $\endgroup$ Apr 1, 2021 at 20:16

2 Answers 2

0
$\begingroup$

The result on the screen behind the double slit would be the same in both situations: there would be no interference pattern. It doesn't matter if a conscious observer inspects the detector result or not, and in that sense, the "collapse" is objective.

What counts is the presence of the detector itself: by definition, a detector changes its state in a specific way if and only if a particle goes through the associated slit. This requires the detector to interact with the particles in a specific way. This interaction also modifies the wave function of the particles in a way that produces the pattern without interference.

As you guessed, there is a problem with your suggested memory-less setup of the detector: a detector which does not record the measurement result in one way or the other is not a detector at all (a contradiction in terms). But it doesn't matter for the pattern on the screen if a conscious being looks at this record after the experiment or not. It doesn't matter either if the detector is in fact a conscious being or not.

$\endgroup$
5
  • $\begingroup$ Can you give a citation to support this claim? $\endgroup$ Apr 1, 2021 at 19:04
  • $\begingroup$ See, e.g., Chapter 9 of Leslie E. Ballentine's Quantum Mechanics (1998) which contains a detailed discussion of what it means to measure a quantum state by means of a detector, and how a measurement affects the states of the measured system and of the measurement device. $\endgroup$
    – Figaro
    Apr 1, 2021 at 19:37
  • $\begingroup$ Ok I will take a look. My impression was that the measurement problem of what exactly a measurement is, is still a big open question. $\endgroup$ Apr 1, 2021 at 20:17
  • $\begingroup$ You are right, there's still a lot of debate on the measurement problem, and I didn't want to get into this topic. Whatever position you take in this debate, my point is that the result in your example does not depend on there being a conscious "subjective" observer or not. At least if we consider a conscious being just a special kind of detector. Of course, this by itself doesn't tell us how to interpret the "collapse" of the wave function. Personally, I find the ensemble interpretation (as formulated by Ballentine, e.g.) quite compelling. $\endgroup$
    – Figaro
    Apr 1, 2021 at 20:53
  • $\begingroup$ My understanding of the “Wigner’s Friend Paradox” is that the notion of when a measurement has occurred can be relative. At least, that’s a respectable viewpoint that’s been advanced. $\endgroup$ Apr 3, 2021 at 11:56
0
$\begingroup$

What counts is the wave function. The presence of the detector alters the wave function in such a way that the interference is destroyed. It does not even matter if actual particles are sent through the slits, let alone whether they are observed or not.

In the simplest representation the detector has to orthogonal states, 0 or 1 particle seen, say in the left slit. These to states are entangled with the two nonorthogonal states of the particle slit system, left and right. The resulting two states are $|0R\rangle$ and $|1L\rangle$, and these are orthogonal so no interference occurs.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.