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How did you conclude that there's no need for thinking that there's a collapse of the wavefunction? How does wavefunction collapse arise from the Schrodinger eq
by millstone 7y ago
How did you conclude that there's no need for thinking that there's a collapse of the wavefunction? How does wavefunction collapse arise from the Schrodinger equation?
- scotty79 7y agohttps://en.m.wikipedia.org/wiki/Ehrenfest_theorem https://en.m.wikipedia.org/wiki/Ehrenfest_theorem Basically classical mechanics is just Schrodinger equation for sharp, small wave functions. So you don't need to think that due to measurement classical particle pops up out of fuzzy thing that the wave function is. Instead you can think instead that due to measurement wavefunction itself reshapes to be really narrow and sharp but it still obeys same shrodinger equation. Just now thanks to sharpness you can simplify the equation a lot so it becomes classical motion equations.
- username90 7y agoAn integral part of quantum mechanics is the actual discrete states, ie the eigen values which it probabilistically chooses during a wave function collapse. Hence it doesn't behave like a wave, also it doesn't behave like a particle, it behaves like a mix of both. That is quantum mechanics and we don't fully understand the maths of it yet, ie when do these waves behave like particles and when do they behave like waves ie when do the wave function collapse?
- scotty79 7y agoWhat I'm saying is you don't have to see eigen values as something of different nature than the wave function. They represent very condensed wave function. Instead of getting rid of all probability by calculating eigenstates you could get rid of just a part of probability (sort of like by using fuzzy numbers in your calculation instead of real numbers) and get something that is close to eigenstates but still is a wave function. If you track such object in time and space you'll see it still obeys Schrodinger equation. Nobody does that because that won't help you with predicting results of your experiments and it's way more work. But it gives you insight some people crave while others feel it's completely unnecessary.
- username90 7y ago> What I'm saying is you don't have to see eigen values as something of different nature than the wave function. They represent very condensed wave function. They do not represent very condensed wave functions, have you taken even a single course in quantum mechanics? You should have learned about the Double slit experiment in high school at least, how do you explain that using only the schrödiner equation? https://en.wikipedia.org/wiki/Double-slit_experiment https://en.wikipedia.org/wiki/Double-slit_experiment
- kgwgk 7y agoWhy do you think the result of the double slit experiment cannot be explained using Schroedinger's equation? If there is a problem at the time of the measurement this is not something specific to the double slit, the same problem exists with a single slit and in any detection of any particle ever.
- scotty79 7y agoThe wave function are reshaped by the slits. When it interacts with the barrier with two slits it gets reshaped to have for example very low value directly after the barier far away from the slits. And then final reshaping into something extremely sharp occurs at the screen. The final reshaping is so precise that given photon wave function gets condensed precisely to a single atom of a particle of the photographic dye. Anything except for those reshapings is governed by Schrodinger equation. > [Eigenvalues] do not represent very condensed wave functions User kgwgk said: "Note that eigenvalues are wavefunctions just like any other state" Even if mathematically they don't. Same way as limit of a function at some point outside the domain of a fuction is not a value of that function. But if you consider something close to that limit it is still value of that function. Limits are useful for calculations but if the function represents someting real, limits unlike the function don't actually exist. They are one step more removed from reality than the function is. What I'm saying is that it might be useful for developing intuition for quantum mechanics to consider that eigenvalues don't exist. That they are just mathematical constructs that are useful to represent wavefunctions that are close in shape to them. The genral tendency is exactly opposite. To consider wave function to be mathematical construct (becouse complex numbers and no macroscopic analog) and eigenvectors and states they describe as reality (just because we measured something close to their precisely calculated values).
- millstone 7y agoBut how does the wavefunction become narrow and sharp "due to measurement?" That's the crux; that's what the measurement problem is. Making a wavefunction "narrow and sharp" means projecting it onto an eigenvector of your measurement basis (e.g. "spin up.") Projection is non-unitary and irreversible. Whether and how this can be achieved via the Schrodinger equation is what all the fuss is about.
- scotty79 7y agoSchrodinger equation describes evolution, but not reshaping. Reshaping is caused by interaction. Interaction occurs with probability defined by wave functions. The thing is, to predict results of experiments we don't really need to know how exactly wave function reshapes. It's enough to know how it "reshapes" to a specific point, that we from that moment on call a particle that came out of wavefunction collapse. You can consider as many points in as you need to calculate everything you want to know. Hence current state of quantum mechanics where we can calculate everythig but don't feel as we have a full picture of what happens.
- username90 7y agoSaid reshaping is the hard part, basically everybody who have studied a bit of physics understands the schrödinger equation and the wave part, the hard part is the particle like behavior which causes the wave to break down. Meaning it isn't a wave, waves doesn't break down like that, waves doesn't just gather together into some coherent state randomly. That is the hard part, if you can't explain that then don't say that QM is easy.
- kgwgk 7y agoMany interactions are perfectly described by Schroedinger's equation. The interactions inside an atom, for example. Or the operation of a quantum computer. What "occurs" (that's the part we don't really understand) with probability defined by wave function is "measurement" but not every interaction is a measurement. A measurement is the macroscopic effect of interactions (which may or may not be fully described by Schroedinger's equation).
- username90 7y agoNote that you are wrong, it doesn't become classical mechanics for small sharp wave functions. It has similar movement, but the wave function would also thin out until it is smoothed out over the entire universe. We all know particles doesn't do that, hence schrödinger equation alone isn't enough to explain observations about classical mechanics.
- scotty79 7y agoCould you elaborate on that? How do we know particles don't do that? Being high near one spot and smooth, near zero over entire rest of universe looks exactly like a particle to me.
- username90 7y agoNo, I mean it will be evenly spread out in all directions and not have any location where it is dense.
- scotty79 7y agoI don't see what you are saying in the description of Erhenfest theorem. They are saying: "For general systems, if the wave function is highly concentrated around a point [... the thing different between classical and quantum ...] will be almost the same, since both will be approximately equal to [...]. In that case, the expected position and expected momentum will approximately follow the classical trajectories, at least for as long as the wave function remains localized in position."