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From what I gather, quantum "particles" travel in a way that's wave-like, but as soon as they interact with another quantum particle, the wave collapses into a

by hasenj 15y ago

From what I gather, quantum "particles" travel in a way that's wave-like, but as soon as they interact with another quantum particle, the wave collapses into a single point, and then starts traveling as a wave again starting from the point of collapse. (or something along these lines). This wave-like form is basically the "wave function"; it describes how the quantum particle moves, and where we might find it. If this is right, then I don't see anything "mysterious" in quantum entanglement. If anything, quantum entanglement seems to be a proof that the quantum-particle (aka wave function) has an internal state that can be known without necessarily having to interact with it directly. Somebody please throw some sense into me if I'm spewing non-sense.

SoftwareMaven 15y ago
The problem is that the "collapse" (from wave to particle) occurs as a result of somebody measuring. Google the double-slit experiment: when not measuring individual photons, you get wave-like interference. When measuring individual photons, you get particle behavior. The conundrum is what is it about measuring that collapses the waveform? It is quite a mystery, and, honestly, very exciting. (btw, IANAP, but I love reading about it)
- ivan_ah 15y ago
  > When measuring individual photons, you get particle behavior. Actually, you get wave properties (interference) even with a single photon at a time. This is why it is so fucked up... http://en.wikipedia.org/wiki/Double-slit_experiment#Interference_of_individual_particles http://en.wikipedia.org/wiki/Double-slit_experiment#Interfer...
  - Cushman 15y ago
    Not exactly. The collection of particles, even if you detect them one at a time, displays an interference pattern. However, if you make an effort to detect which slit each particle goes through, you see that each one behaves as a particle, like you would expect, and the interference pattern vanishes. And this is true whether you detect which slit it passes through before or after its point of impact has been detected. That's what's weird. A particle can be particle-like or wave-like, but not both, and the thing that determines which it is is whether or not you will look.
    - gnaritas 15y ago
      Looking however means shooting particles at something to see what bounces off, so it must necessarily affect the system.
      - StacyC 15y ago
        This is what I was wondering about: how does a particle ‘know’ it is being observed? Something has to interact with it physically, and once affected its physical properties change. Is that right? I’m a total layman here but so fascinated with all this. Trying to get a better understanding.
        gnaritas 15y ago
        Particles don't know anything. The way we see anything, even with our own eyes, is to analyze reflected photons. You can't see something without hitting it with something first and analyzing what bounces back. Obviously, if you're bouncing photons off something, you're going to affect its state. Usually this effect is negligible, but in the case of trying to watch individual particles, well, it's no longer negligible, it affects the experiment. The problem is people tend to assume observation is passive, they think seeing something does't affect it; this simply isn't true at all. Eyes only work because the sun is continually spraying and bouncing photons off everything. Sight isn't passive at all, you need a photon source spraying photons everywhere, like a flash light. What's strange about the doulbe slit experiment isn't that watching forces the particles to act like particles, that's the expected behavior all the time, the same as a single slit. What's strange is that when you don't watch it, you get the interference pattern indicating a wave that isn't acting like a particle. That's not at all expected and indicates the particle is going through both slits and interfering with itself. I'm a total layman as well, but that's my take on it.
        Anderkent 15y ago
        That is very wrong. A particle doesn't 'know' it is being observed. However, when it interacts with other particles, the new state of both particles depends on the old states of both particles. For a better explanation than I could possibly write here, see http://lesswrong.com/lw/r5/the_quantum_physics_sequence/ http://lesswrong.com/lw/r5/the_quantum_physics_sequence/.
- hasenj 15y ago
  > as a result of somebody measuring > The conundrum is what is it about measuring that collapses the waveform? It is quite a mystery, and, honestly, very exciting. Isn't it obviously the fact that it's interacting with another object? - wave-function interacts with your wall/measuring-device -> wave-function collapses into a point (or: disappears and spawns a new wave-function at said point (which we end up interpreting as the "position" of the particle)) - wave-function doesn't interact with anything -> happily keeps propagating as a wave
  - othermaciej 15y ago
    Two quantum particles can interact with each other without either's wavefunction collapsing. That's pretty much what makes the double slit experiment work. It seems that generally, interacting with a "macroscopic" object tends to collapse the wave function. You can find an interesting discussion of one way scientists try to explain this phenomenon here: http://en.wikipedia.org/wiki/Quantum_decoherence http://en.wikipedia.org/wiki/Quantum_decoherence
- lisper 15y ago
  > It is quite a mystery It is not a mystery at all. http://flownet.com/ron/QM.pdf http://flownet.com/ron/QM.pdf http://www.youtube.com/watch?v=dEaecUuEqfc http://www.youtube.com/watch?v=dEaecUuEqfc
bitcracker 15y ago
> as soon as they interact with another quantum particle, the wave collapses into a single point, and then starts traveling as a wave again starting from the point of collapse. What does "collapse" mean? Is it the overlapping or "sum" of different wave functions? Isn't the measuring instrument itself a wave function?
- archgoon 15y ago
  Collapse means that any future measurements/interactions of the wave function can be assumed to have evolved from the measured state. Most measurements themselves normally only tell you with high probability information about state, but that's fine. If you have a wave function which can be in two different states X1 and X2, then it's full state is going to be some linear combination of the states X= a X1+b X2. It can, over time, evolve deterministically, with a and b potentially becoming larger and smaller over time. When a measurement occurs, such as the detection of a photon of a certain frequency which indicates that the particle made a state transition, you then assign all but one of the coefficients to zero (or approximately zero depending on the measurement). If the particle is in a certain compatible potential (one in which the measured state has a specific energy), the coefficients will stay at zero. To answer your question: " Isn't the measuring instrument itself a wave function?" Yes it is. Unfortunately, if you have two particles, to describe the system with both particles in it, as opposed to just one, you don't just add the particle wave functions together, you have to take the cross (well tensor, but it's similar) product of the two vector spaces of the particles (remember, function spaces are vector spaces). This is true _even if they do not interact with each other_. Yudkowsky has a pretty good description of this http://lesswrong.com/lw/pp/decoherence/ http://lesswrong.com/lw/pp/decoherence/
  - bitcracker 15y ago
    Thanks for your good answer! Quantum theory is really strange and amazing.
aufreak3 15y ago
Wave function collapse is not tenable as part of quantum theory. The "decoherence" process holds ground currently as a quantum mechanical description of the measurement process and it doesn't make absurd distinctions like "observer versus observed". Less wrong has a layman's description of decoherence here - http://lesswrong.com/lw/pp/decoherence/ http://lesswrong.com/lw/pp/decoherence/

by hasenj 15y ago

SoftwareMaven 15y ago

The problem is that the "collapse" (from wave to particle) occurs as a result of somebody measuring. Google the double-slit experiment: when not measuring individual photons, you get wave-like interference. When measuring individual photons, you get particle behavior. The conundrum is what is it about measuring that collapses the waveform? It is quite a mystery, and, honestly, very exciting. (btw, IANAP, but I love reading about it)

ivan_ah 15y ago
> When measuring individual photons, you get particle behavior. Actually, you get wave properties (interference) even with a single photon at a time. This is why it is so fucked up... http://en.wikipedia.org/wiki/Double-slit_experiment#Interference_of_individual_particles http://en.wikipedia.org/wiki/Double-slit_experiment#Interfer...
- Cushman 15y ago
  Not exactly. The collection of particles, even if you detect them one at a time, displays an interference pattern. However, if you make an effort to detect which slit each particle goes through, you see that each one behaves as a particle, like you would expect, and the interference pattern vanishes. And this is true whether you detect which slit it passes through before or after its point of impact has been detected. That's what's weird. A particle can be particle-like or wave-like, but not both, and the thing that determines which it is is whether or not you will look.
  - gnaritas 15y ago
    Looking however means shooting particles at something to see what bounces off, so it must necessarily affect the system.
    - StacyC 15y ago
      This is what I was wondering about: how does a particle ‘know’ it is being observed? Something has to interact with it physically, and once affected its physical properties change. Is that right? I’m a total layman here but so fascinated with all this. Trying to get a better understanding.
      - gnaritas 15y ago
        Particles don't know anything. The way we see anything, even with our own eyes, is to analyze reflected photons. You can't see something without hitting it with something first and analyzing what bounces back. Obviously, if you're bouncing photons off something, you're going to affect its state. Usually this effect is negligible, but in the case of trying to watch individual particles, well, it's no longer negligible, it affects the experiment. The problem is people tend to assume observation is passive, they think seeing something does't affect it; this simply isn't true at all. Eyes only work because the sun is continually spraying and bouncing photons off everything. Sight isn't passive at all, you need a photon source spraying photons everywhere, like a flash light. What's strange about the doulbe slit experiment isn't that watching forces the particles to act like particles, that's the expected behavior all the time, the same as a single slit. What's strange is that when you don't watch it, you get the interference pattern indicating a wave that isn't acting like a particle. That's not at all expected and indicates the particle is going through both slits and interfering with itself. I'm a total layman as well, but that's my take on it.
      - Anderkent 15y ago
        That is very wrong. A particle doesn't 'know' it is being observed. However, when it interacts with other particles, the new state of both particles depends on the old states of both particles. For a better explanation than I could possibly write here, see http://lesswrong.com/lw/r5/the_quantum_physics_sequence/ http://lesswrong.com/lw/r5/the_quantum_physics_sequence/.
hasenj 15y ago
> as a result of somebody measuring > The conundrum is what is it about measuring that collapses the waveform? It is quite a mystery, and, honestly, very exciting. Isn't it obviously the fact that it's interacting with another object? - wave-function interacts with your wall/measuring-device -> wave-function collapses into a point (or: disappears and spawns a new wave-function at said point (which we end up interpreting as the "position" of the particle)) - wave-function doesn't interact with anything -> happily keeps propagating as a wave
- othermaciej 15y ago
  Two quantum particles can interact with each other without either's wavefunction collapsing. That's pretty much what makes the double slit experiment work. It seems that generally, interacting with a "macroscopic" object tends to collapse the wave function. You can find an interesting discussion of one way scientists try to explain this phenomenon here: http://en.wikipedia.org/wiki/Quantum_decoherence http://en.wikipedia.org/wiki/Quantum_decoherence
lisper 15y ago
> It is quite a mystery It is not a mystery at all. http://flownet.com/ron/QM.pdf http://flownet.com/ron/QM.pdf http://www.youtube.com/watch?v=dEaecUuEqfc http://www.youtube.com/watch?v=dEaecUuEqfc

bitcracker 15y ago

> as soon as they interact with another quantum particle, the wave collapses into a single point, and then starts traveling as a wave again starting from the point of collapse. What does "collapse" mean? Is it the overlapping or "sum" of different wave functions? Isn't the measuring instrument itself a wave function?

archgoon 15y ago
Collapse means that any future measurements/interactions of the wave function can be assumed to have evolved from the measured state. Most measurements themselves normally only tell you with high probability information about state, but that's fine. If you have a wave function which can be in two different states X1 and X2, then it's full state is going to be some linear combination of the states X= a X1+b X2. It can, over time, evolve deterministically, with a and b potentially becoming larger and smaller over time. When a measurement occurs, such as the detection of a photon of a certain frequency which indicates that the particle made a state transition, you then assign all but one of the coefficients to zero (or approximately zero depending on the measurement). If the particle is in a certain compatible potential (one in which the measured state has a specific energy), the coefficients will stay at zero. To answer your question: " Isn't the measuring instrument itself a wave function?" Yes it is. Unfortunately, if you have two particles, to describe the system with both particles in it, as opposed to just one, you don't just add the particle wave functions together, you have to take the cross (well tensor, but it's similar) product of the two vector spaces of the particles (remember, function spaces are vector spaces). This is true _even if they do not interact with each other_. Yudkowsky has a pretty good description of this http://lesswrong.com/lw/pp/decoherence/ http://lesswrong.com/lw/pp/decoherence/
- bitcracker 15y ago
  Thanks for your good answer! Quantum theory is really strange and amazing.

aufreak3 15y ago

Wave function collapse is not tenable as part of quantum theory. The "decoherence" process holds ground currently as a quantum mechanical description of the measurement process and it doesn't make absurd distinctions like "observer versus observed". Less wrong has a layman's description of decoherence here - http://lesswrong.com/lw/pp/decoherence/ http://lesswrong.com/lw/pp/decoherence/