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.
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).
Brainsploosh@lemmy.world 2 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?
kayzeekayzee@lemmy.blahaj.zone 2 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.
Krudler@lemmy.world 2 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