darkenlighten Posted March 22, 2011 Share Posted March 22, 2011 (edited) Okay I am here to help. It seems like a lot of people misunderstand what exactly the wave function is, so I want to give my educated understanding of this. The wave function itself is not physical, but it is mathematical, hence a function. The wave function is a description of the state, the QM equivalent of "particle is at x and has a momentum p". It is the Schroedinger equation that determines the time evolution of this state (and thus the time evolution of the wave function). So when someone says, the wave function collapsed, it does not mean something physically collapsed, but that the mathematical model collapsed. To say further, the wave function describes all possible states in which the system can be in. For example the photon traveling towards a double slit; now while the photon has a particle-wave duality, this is separate from its wave function, though the wave function might describe the physical property. The wave function mathematically describes what the possible states of the photon are, given the initial conditions. So when we observe the photon at say slit 1, we have collapsed the wave function, because now there is no longer a superposition (addition) of all possible states, but now just a single state, the state that we observed. Thus setting up another set of initial conditions to our wave function, changing the possible states from there on out. Edited March 22, 2011 by darkenlighten Link to comment Share on other sites More sharing options...

ecoli Posted March 22, 2011 Share Posted March 22, 2011 What does it mean when a mathematical model collapses? Link to comment Share on other sites More sharing options...

darkenlighten Posted March 22, 2011 Author Share Posted March 22, 2011 In this case it means it is no longer a superposition of multiple possible outcomes, but now just the one known state. Link to comment Share on other sites More sharing options...

timo Posted March 22, 2011 Share Posted March 22, 2011 I appreciate your intent to helping interested people understanding basic Quantum Mechanics. But I think you have a different notion of "educated understanding" than me or -more important- most other people. Trying to explain something when no one actively asked for it should raise the minimum quality of an explanation: The explanation now must be able to compete with standard sources like textbooks or Wikipedia because otherwise it is just noise. "The wave function is the time evolution of a system derived from Schroedinger's equation given the Hamiltonian (the Hamiltonian describes the energy of the system) of a system" is essentially wrong. The wave function is a description of the state, the QM equivalent of "particle is at x and has a momentum p". It is the Schroedinger equation that determines the time evolution of this state (and thus the time evolution of the wave function). 1 Link to comment Share on other sites More sharing options...

swansont Posted March 22, 2011 Share Posted March 22, 2011 "The wave function is the time evolution of a system derived from Schroedinger's equation given the Hamiltonian (the Hamiltonian describes the energy of the system) of a system" is essentially wrong. The wave function is a description of the state, the QM equivalent of "particle is at x and has a momentum p". It is the Schroedinger equation that determines the time evolution of this state (and thus the time evolution of the wave function). And to continue on this, operating on the wave function with the Hamiltonian will tell you the energy, and similarly operating on the wave function with another operator will yield some other information. Link to comment Share on other sites More sharing options...

darkenlighten Posted March 22, 2011 Author Share Posted March 22, 2011 Yea I messed that up didn't I haha. I'll admit its been a little bit since I studied quantum mechanics. My main motivation was the several threads previous to this one about the wave function. I made a mistake, but that is why I have guys. Nonetheless, the following of that is correct. Link to comment Share on other sites More sharing options...

michel123456 Posted March 22, 2011 Share Posted March 22, 2011 (...) now while the photon has a particle-wave duality, this is separate from its wave function, though the wave function might describe the physical property. The wave function mathematically describes what the possible states of the photon are, given the initial conditions. So when we observe the photon at say slit 1, we have collapsed the wave function, because (...) (emphasis mine) IMHO: The photon has no wave function. We are the ones who are in search of explanations and "give" to the photon a wave function. As you said, the wave function is nothing physical. The photon has no idea about how humans have decided to mathematize, the photon don't "care" about wave function, scientists do. Reversely when you say "we have collapsed the wave function" it looks like humans have stolen or destroyed something from the photon. But nothing like that happens. The photon never "had" a wave function and we never took the wave function away. Simply, mathematization is no more necessary to calculate position & momentum. I hope that is what you wanted to explain. Link to comment Share on other sites More sharing options...

darkenlighten Posted March 22, 2011 Author Share Posted March 22, 2011 Indeed michel. Link to comment Share on other sites More sharing options...

ajb Posted March 23, 2011 Share Posted March 23, 2011 Very wrongly I tend to think of the wave function of particle as a classical scalar field over space. That is to every point in a space we assign a complex number. So, you can think in terms of sections of complex lines bundles if you wish. The dynamics of this scalar field is given by the Schrödinger equation. The physical interpretation is that the absolute value of this complex field is as a probability density for the position. Link to comment Share on other sites More sharing options...

swansont Posted March 23, 2011 Share Posted March 23, 2011 Reversely when you say "we have collapsed the wave function" it looks like humans have stolen or destroyed something from the photon. But nothing like that happens. The photon never "had" a wave function and we never took the wave function away. Since that's not what collapsing the wavefunction means, this is moot. Link to comment Share on other sites More sharing options...

michel123456 Posted March 23, 2011 Share Posted March 23, 2011 Since that's not what collapsing the wavefunction means, this is moot. I don't understand your comment. Link to comment Share on other sites More sharing options...

swansont Posted March 23, 2011 Share Posted March 23, 2011 A wavefunction collapse is not referring to taking anything away from the photon. Link to comment Share on other sites More sharing options...

sysD Posted March 23, 2011 Share Posted March 23, 2011 So a wave function is a function of probability and position for a specific point in time (eg. t=0.5 seconds)? Can someone post a picture of a wave function with explanations of its properties? Link to comment Share on other sites More sharing options...

mississippichem Posted March 24, 2011 Share Posted March 24, 2011 Very wrongly I tend to think of the wave function of particle as a classical scalar field over space. That is to every point in a space we assign a complex number. So, you can think in terms of sections of complex lines bundles if you wish. The dynamics of this scalar field is given by the Schrödinger equation. The physical interpretation is that the absolute value of this complex field is as a probability density for the position. Interesting that you say that, I've found myself guilty of a similar morbid analogy myself. Though technically incorrect it seems to be a pretty good qualitative analogy, something that is hard to come by when talking about such things. Link to comment Share on other sites More sharing options...

ajb Posted March 24, 2011 Share Posted March 24, 2011 I meant to say roughly, you can quite correctly think in terms of complex line bundles. My brain and fingers don't always cooperate when writing! Sorry Link to comment Share on other sites More sharing options...

swansont Posted March 24, 2011 Share Posted March 24, 2011 So a wave function is a function of probability and position for a specific point in time (eg. t=0.5 seconds)? Can someone post a picture of a wave function with explanations of its properties? http://simple.wikipedia.org/wiki/File:Particle_in_a_box_wavefunctions_2.svg This is a particle in a box solutions for different energies. For probabilities you would square the functions. The n=1 solution shows that the particle is most likely to be found in the center of the box. For n=2, the probability of being in the center is zero, and there are more nodes as you go to higher energy states. Link to comment Share on other sites More sharing options...

sysD Posted March 26, 2011 Share Posted March 26, 2011 For n=2, wouldn't the lowest probability be where the wavelength is lowest? Link to comment Share on other sites More sharing options...

swansont Posted March 26, 2011 Share Posted March 26, 2011 For n=2, wouldn't the lowest probability be where the wavelength is lowest? The wavelength is a constant — it's a sine function. I think you mean amplitude. After you square it, that is zero at x=0, L/2 and L for n=2 Link to comment Share on other sites More sharing options...

michel123456 Posted March 26, 2011 Share Posted March 26, 2011 (edited) The wavelength is a constant — it's a sine function. I think you mean amplitude. After you square it, that is zero at x=0, L/2 and L for n=2 I suppose he (Swansont) means it is zero where the sinusoidal intersects the horizontal axis. IOW at both ends for all cases, plus in the middle for N=2. It should be clear that the name "particle in a box" may mislead. This is a case of a "particle on a rope", or One-dimensional solution of the particle-in-a-box example.As explained in the wiki article (see previous link): The simplest form of the particle in a box model considers a one-dimensional system. Here, the particle may only move backwards and forwards along a straight line with impenetrable barriers at either end.[1] The walls of a one-dimensional box may be visualised as regions of space with an infinitely large potential. Conversely, the interior of the box has a constant, zero potential.[2] This means that no forces act upon the particle inside the box and it can move freely in that region. However, infinitely large forces repel the particle if it touches the walls of the box, preventing it from escaping. What I don't understand is how one can conciliate the 2 statements: 1_that no forces act upon the particle inside the box 2_infinitely large forces repel the particle if it touches the walls of the box How do "forces repel the particle" if the probability to find the particle next to the wall is zero? If it was a classical system, the particle should hurt the wall in order to "feel the force" in the first place. Like a ball bouncing. In this case the probability to find the particle next to the wall would be non-zero. The only other solution should be to admit that repelling forces acting at a distance do exist inside the box. I don't get it. Edited March 26, 2011 by michel123456 Link to comment Share on other sites More sharing options...

swansont Posted March 26, 2011 Share Posted March 26, 2011 I suppose he (Swansont) means it is zero where the sinusoidal intersects the horizontal axis. IOW at both ends for all cases, plus in the middle for N=2. It should be clear that the name "particle in a box" may mislead. This is a case of a "particle on a rope", or One-dimensional solution of the particle-in-a-box example.As explained in the wiki article (see previous link): No, it's in a box. The only thing affecting it are the walls. This is one dimension of the solution, but the solutions are orthogonal, so that doesn't really matter. What I don't understand is how one can conciliate the 2 statements: 1_that no forces act upon the particle inside the box 2_infinitely large forces repel the particle if it touches the walls of the box No forces other than the walls are present. The walls are not inside the box. Hence there are no forces inside the box acting on the particle. How do "forces repel the particle" if the probability to find the particle next to the wall is zero? If it was a classical system, the particle should hurt the wall in order to "feel the force" in the first place. Like a ball bouncing. In this case the probability to find the particle next to the wall would be non-zero. The only other solution should be to admit that repelling forces acting at a distance do exist inside the box. I don't get it. Your complaint seems to be that quantum-mechanical systems do not behave classically. The answer to that is yes, that's why QM had to be developed. Link to comment Share on other sites More sharing options...

michel123456 Posted March 26, 2011 Share Posted March 26, 2011 That was the bad answer I expected. -1 Link to comment Share on other sites More sharing options...

sysD Posted March 26, 2011 Share Posted March 26, 2011 The wavelength is a constant — it's a sine function. I think you mean amplitude. After you square it, that is zero at x=0, L/2 and L for n=2 Yeah, sorry, lol. Okay, so for that example, if we assign some "x" values for explanation purposes... for example, in n=1, the node in the middle can be x=2. for n=2, the peak of of the wave is at x=1 and the trough is at x=3 in n=1, its obvious from the peak amplitude that the probability of finding the particle is highest in the centre of the box (x=2) but in n=2, wouldn't the probability be lowest in the trough (x=3) and highest at the peak (x=1)? Link to comment Share on other sites More sharing options...

timo Posted March 26, 2011 Share Posted March 26, 2011 sysD, the images are crappy in this regard because they lack the y=0 line. The trough at x=3 is exactly the negative of the peak at x=1. So after squaring the function, they look exactly the same. Note also Swansont's previous statement that This [set of wave-functions you are talking about] is a particle in a box solutions for different energies. For probabilities you would square the functions. Link to comment Share on other sites More sharing options...

swansont Posted March 26, 2011 Share Posted March 26, 2011 Yeah, sorry, lol. Okay, so for that example, if we assign some "x" values for explanation purposes... for example, in n=1, the node in the middle can be x=2. for n=2, the peak of of the wave is at x=1 and the trough is at x=3 in n=1, its obvious from the peak amplitude that the probability of finding the particle is highest in the centre of the box (x=2) but in n=2, wouldn't the probability be lowest in the trough (x=3) and highest at the peak (x=1)? No, because you square the wave function to find the probability. The trough of the sin function becomes a peak when you do that. Link to comment Share on other sites More sharing options...

sysD Posted March 28, 2011 Share Posted March 28, 2011 ah forgot about the squaring. gotcha. Link to comment Share on other sites More sharing options...

## Recommended Posts

## Create an account or sign in to comment

You need to be a member in order to leave a comment

## Create an account

Sign up for a new account in our community. It's easy!

Register a new account## Sign in

Already have an account? Sign in here.

Sign In Now