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I am having a tough tough stressful week trying to write an article about quantum mechanics. I know there was a question asked the same thing before, but I didn't understand nor I did not want to wake up an old thread.

The double slit experiment we all know, if we put an apparatus at the slit to detect which slit the electron has gone through but not look at the data it spits out, what happens to the screen? (Looking at the screen doesn't do any change right?)

1) If an active observer is not necessary for the wave function to collapse, the existence of a sensor will destroy the interference pattern because the sensor has interfered with the system.

2) If there needs to be an observer to read off the data from the sensor and acknowledge that the electron has acted like a particle and gone through either of the slit, the interference pattern will not be destroyed when the sensor is just there but no one has read the data.

Which one is it?

Jasper
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Seung
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  • For reference, I think the nicest previous SE answer to this question is https://physics.stackexchange.com/questions/59778/interference-and-which-path-information/62042#62042 – isometry Jan 07 '19 at 16:32
  • @BruceGreetham That is a lot to digest! Thank you for stopping by with a great link. – Seung Jan 07 '19 at 18:17
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    OK in a nutshell it is 1). If there is a sensor interacting with the system then that is what physicists mean by measurement. The human observer is not an essential part of the measurement process. – isometry Jan 07 '19 at 18:54
  • @BruceGreetham Do you know any source that has done this experiment? The experiment itself is trivial and someone must have done it which I cannot find. Why are there still people who believe in "consciousness collapses wavefunction" or any other interpretations that require an observation? If this is indeed 1), that directly denies that particular interpretation. – Seung Jan 07 '19 at 19:49
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    As to your first question: look at https://www.youtube.com/watch?v=cdLF6DEqtjQ minute 2:00-4:00. Here it is just a gas doing the "observing". As you turn the gas pressure up the interference gradually disappears. – isometry Jan 07 '19 at 20:20
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    As to your second question: I can't control what people choose to believe, but you'll find that most of the serious contributors to physics SE do not believe that. – isometry Jan 07 '19 at 20:24
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    @BruceGreetham That youtube video was so hard to interpret in terms of my thought experiment. At first I didn't know what it got to do with my experiment. I do now. That is deep (or I could be stupid) Much much appreciated for your help! – Seung Jan 07 '19 at 20:55

2 Answers2

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The interference pattern will be destroyed even before the wavefunction-collapse. Let's say your particle is described by the wavefunction $|\text{p}>$. When passing the double slit (I call those the up and down slits), it becomes entangled with the sensor, and your wavefunction becomes : $|\psi> \equiv|\text{sensor}_{up}>\otimes |\text{p}_{up}>+|\text{sensor}_{down}>\otimes |\text{p}_{down}>$ When you want the probability to find the particle at a certain position on the screen, you will calculate $<\psi|\psi>$ but since $<\text{sensor}_{up}|\text{sensor}_{down}>=0$, you won't get the term $<\text{p}_{up}||\text{p}_{down}>$ which is the one you need to have an interference pattern. Simply adding the sensor destroyed the interference, whether you look at it or not.

Ewen Bellec
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  • That was a lot technical but I get the point where the existence of sensor gets entangled with the particle and destroys the interference pattern regardless of the one looking at the data. Thanks! – Seung Jan 07 '19 at 14:52
  • Sorry about that. The tensor product $\otimes$ is a nice way to avoid the question of "when does the wavefunction collapse?". Instead, at each interaction with a new sensor, particle... you add something to $|\psi>$. And you "collapse" $|\psi>$ at the end. – Ewen Bellec Jan 07 '19 at 15:52
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You are describing a situation equivalent to Schroedinger's Cat. If the recording apparatus is isolated from the rest of the universe, then passage of an electron wavefunction through the double slit puts the recording apparatus and its records into a mixed state. When you then "measure" the record of the experiment by reading it, you "collapse" the wavefunction of the apparatus, the record, and the electron.

It is this kind of thought experiment that leads to the Many Worlds view of quantum mechanics: provide an observer to read the results of the experiment, but isolate him/her along with the apparatus and record, and the experiment puts the observer into a mixed state.

S. McGrew
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  • So classical things can exist in a superposition of states? It would also be great if you can answer 1 or 2 in the question. It wasn't quite Shchroedinger's Cat when I thought about this because I can look at the cat. – Seung Jan 07 '19 at 14:45
  • Yes, classical things can exist in a superposition of states. This has been demonstrated for molecules and even for very small mechanical resonators. The difficulty of doing such demonstrations gets huge for big systems because of the difficulty of isolating big systems well enough. However, the principle has been proven experimentally. If you remove all metaphysical meanings for "observer" by substituting a robot for each observer, it's much easier to make sense of the physics. – S. McGrew Jan 07 '19 at 14:53
  • See for example [https://en.wikipedia.org/wiki/Quantum_machine]. – S. McGrew Jan 07 '19 at 14:55
  • The more I learn the urge to kill myself spikes :( What a lovely world we are living in thanks to not classic things. Thanks anyway! – Seung Jan 07 '19 at 14:58
  • The puzzles posed by QM are invigorating, IMHO. The Many Worlds idea is much more fun that the frozen spacetime block described by classical physics, connecting a definite set of initial conditions to an inevitable, fully determined future course of events. At least in the Many Worlds view, all physically possible futures can (and do) occur. – S. McGrew Jan 07 '19 at 15:09