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Comment on It's not supposed to make sense...

kayzeekayzee@lemmy.blahaj.zone ⁨3⁩ ⁨days⁩ ago

Not exactly. Quantum physics applies no matter how you measure it. The double-slit experiment is an example of this: Photons moving through two slits will form a wave interference pattern on a detector plate, even though the detector doesn’t affect the position of the photons beforehand.

It’s more like: when you become aware of the results of a quantum measurement, you yourself become a part of the quantum system, and being a part of the system requires measurements to have real values. Whether you should interpret this as a wave-function collapse or branching into multiple parallel universes is up for debate though.

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Trail@lemmy.world ⁨3⁩ ⁨days⁩ ago
When you perform the measurement on which slit the particle passes through, then the measuring device is also part of the system and it affects it. The measurement reduces the degrees of freedom in the system so there are no longer two equivalent ways for the particle to pass through the slits (either A or B), but rather you now have a measured slit and an unmeasured slit. Since there are no longer multiple ways to achieve the same result, the is no longer interference due to equivalent probabilities.

Matt Stassler has a nice series of blog posts on this.

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- Serinus@lemmy.world ⁨3⁩ ⁨days⁩ ago
  Yes, but that’s semantics. Clearly the observation has some effect, but it’s not from any force we recognize.
  
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Sas@beehaw.org ⁨3⁩ ⁨days⁩ ago
It gets even more interesting: to interfere in the double slit experiment, the light has to take a longer path for some points and light is really good at finding the shortest path. And, since you can extend the double slit experiment to infinite slits with infinitely thin blockers between the slits, you can leave away the slits entirely and still have a valid version of that experiment and get interference. It’s just, that most interference is destructive.

Veritasium had a very interesting video about that recently and my extrapolation of this is that there is neither a collapse of wave functions nor multiple parallel universes.

My intuition says that the wave function is there after being “observed”. There is no multiple possible outcomes, just very visible ones and a lot of destructive interfered ones.

However what i just wrote is not science but me extrapolating from science so don’t take it for anything more than that. It somehow causes quantum physics to make intuitive sense for me so i like it. Nothing more than that.

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Brainsploosh@lemmy.world ⁨3⁩ ⁨days⁩ ago
Honest question: what happens afterwards? When we’ve stopped observing, does it reassemble into it’s superpositive form? Are we depleting quantum states somehow?

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- kayzeekayzee@lemmy.blahaj.zone ⁨3⁩ ⁨days⁩ ago
  Sorta! According to the Heisenberg Uncertainty Principle, there’s an upper limit to how much we can “know” about the given state of a particle. This isn’t an issue with our measurements, but a fundamental property of the universe itself. By measuring one aspect of a quantum system (for example, the momentum of a particle), we become less certain about other aspects of the system, even if we had already measured them before (such as the position of the same particle).
  
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  - Brainsploosh@lemmy.world ⁨3⁩ ⁨days⁩ ago
    Thank you for your answer!
    
    Maybe I’m too dense, but what happens with other quantum states that aren’t position/velocity based? I’m thinking things like when we collapse spin, e.g. in entangled particles.
    
    I’ve heard that entangled particles are “one use”, I’d assume they can be restored and possibly re-entangled, but how?
    
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    kayzeekayzee@lemmy.blahaj.zone ⁨3⁩ ⁨days⁩ ago
    Good question! You are certainly not dense!
    
    The position-momentum uncertainty relationship is just a specific case of a more general relationship. There are other uncertainty relationships, such as between time and energy or between two (separate/orthogonal) components of angular velocity. The relationships basically state that whenever you measure one of the two values, you are required to add uncertainty to the other.
    
    Unfortunately, this is kinda where my knowledge on the subject starts to hit its limits. As for spin, it has a lot of effects on the energy of the system it’s involved with, so I believe the energy-time or angular momentum exclusion principles would apply there.
    
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    Krudler@lemmy.world ⁨3⁩ ⁨days⁩ ago
    Entanglement is a fucked up concept that I’m just starting to grasp.
    
    Parametric down sampling, say, causes particles to split, and then certain states of the resultant particles must respect conservation.
    
    So if the source particle had no spin, but the resultant particles must have spin, one will have minus half spin and the other will have plus half spin. Equals zero, which was the total spin of the source particle. Same deal with charges.
    
    But they are not resolved until this wishy-washy “measurement” thing comes into the picture
    
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- pcalau12i@lemmygrad.ml ⁨3⁩ ⁨days⁩ ago
  The double-slit experiment doesn’t even require quantum mechanics. It can be explained classically and intuitively.
  
  It is helpful to think of a simpler case, the Mach-Zehnder interferometer, since it demonstrates the same effect but where where space is discretized to just two possible paths the particle can take and end up in, and so the path/position is typically described with just with a single qubit of information: |0⟩ and |1⟩.
  
  You can explain this entirely classical if you stop thinking of photons really as independent objects but just specific values propagating in a field, what are sometimes called modes. If you go to measure a photon and your measuring device registers a |1⟩, this is often interpreted as having detected the photon, but if it measures a |0⟩, this is often interpreted as not detecting a photon, but if the photons are just modes in a field, then |0⟩ does not mean you registered nothing, it means that you indeed measured the field but the field just so happens to have a value of |0⟩ at that location.
  
  Since fields are all-permeating, then describing two possible positions with |0⟩ and |1⟩ is misleading because there would be two modes in both possible positions, and each independently could have a value of |0⟩ or |1⟩, so it would be more accurate to describe the setup with two qubits worth of information, |00⟩, |01⟩, |10⟩, and |11⟩, which would represent a photon being on neither path, one path, the other path, or both paths (which indeed is physically possible in the real-world experiment).
  
  When systems are described with |0⟩ or |1⟩, that is to say, 1 qubit worth of information, that doesn’t mean they contain 1 bit of information. They actually contain as much as 3 as there are other bit values on orthogonal axes. You then find that the physical interaction between your measuring device and the mode perturbs one of the values on the orthogonal axis as information is propagating through the system, and this alters the outcome of the experiment.
  
  You can interpret the double-slit experiment in the exact same way, but the math gets a bit more hairy because it deals with continuous position, but the ultimate concept is the same.
  
  A measurement is a kind of physical interaction, and all physical interactions have to be specified by an operator, and not all operators are physically valid. Quantum theory simply doesn’t allow you to construct a physically valid operator whereby one system could interact with another to record its properties in a non-perturbing fashion. Any operator you construct to record one of its properties without perturbing it must necessarily perturb its other properties. Specifically, it perturbs any other property within the same noncommuting group.
  
  When the modes propagate from the two slits, your measurement of its position disturbs its momentum, and this random perturbation causes the momenta of the modes that were in phase with each other to longer be in phase. You can imagine two random strings which you don’t know what they are but you know they’re correlated with each other, so whatever is the values of the first one, whatever they are, they’d be correlated with the second. But then you randomly perturb one of them to randomly distribute its variables, and now they’re no longer correlated, and so when they come together and interact, they interact with each other differently.
  
  There’s a paper on this here and also a lecture on this here. You don’t have to go beyond the visualization or even mathematics of classical fields to understand the double-slit experiment.
  
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AppleTea@lemmy.zip ⁨3⁩ ⁨days⁩ ago
My example is more in regards to wave/particle duality as it shows up in variations of the double slit experiment. Putting a detector at one of the slits is an active interaction, giving you the particle-like behavior rather than the interference pattern.

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- kayzeekayzee@lemmy.blahaj.zone ⁨3⁩ ⁨days⁩ ago
  What I mean to say is that the detector is not what’s changing the particle; It’s the process of learning about an aspect of the quantum system that forces it into one state or another (at least from our own personal perspectives).
  
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  - AppleTea@lemmy.zip ⁨3⁩ ⁨days⁩ ago
    uh, I’m a total quantum layman, but I’m pretty sure its the detector.
    
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pcalau12i@lemmygrad.ml ⁨3⁩ ⁨days⁩ ago
Why interpret it as either? The double-slit experiment can be given an entirely classical explanation. Such extravagances are not necessary.

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- barsoap@lemm.ee ⁨3⁩ ⁨days⁩ ago
  
  As the old saying goes “extraordinary claims require extraordinary evidence.”
  
  It may be convenient to look at classical interpretations but “The intuition we evolved to interact with macro systems is also applicable to the micro level” is in itself an extraordinary claim.
  
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