schrodinger wave equation

nouveau_riche

I am new to quantum mechanics and trying to combine the pieces.

If I am looking into the quartum world, first I prepare a mechanism with which i can bring the properties and behavior of the particles i.e. an experiment to study them, but the information i emphasize to look on in the experiment will alter the results. say for instance that i am interested in knowing the position of the particle.

now the device that interacts with the particle and give me the information abouts its presence cannot be localized with certainity as the particle interaction with my device doesn't allow so, therefore the information that i have in hand, is about the region where it is likely to be found.

according to quantum mechanics: if i look for the particle in some region in time, the wavefunction will collapse to peak the probability amplitude there.

now in whatever region i look say at point P, the wavefunction will collapse to peak its probability amplitude, and it is likely that i can localize the particle with some uncertainity.

therfore i can get the sense for particle position through its collapse.

what i did not understand is that where i use the schrodinger wave equation and what is its use once i have it. what is the point to know the probability of localizing a particle say at point P when i know that the moment i tuned my experiment, the particle will be found there as a result of collapse?

please help

PhysicsGente

You use the SE to find a wave function that describes your system right? Besides, what happens if you repeat the experiment over and over again? I guess the goal is to be able to predict what happens at a later time.

nouveau_riche

my point is that when schrodinger itself would have looked at the quantum particle, he must have seen the collapsed wavefunction but not the schrodinger wavefunction.
In the experimental setup, how could he come to probabilistic evolution of particle knowing that it is the collapsed, unity going, probability amplitude observed by him.

I read in griffiths that if you do not measure the particle , its behavior is governed by the schrodinger equation, i do not understant the point that how do we know that the particle is described by a probability wave when the only way i can verify it is by interacting which will alter the results?

bhobba

Because if you prepare a system in exactly the same way so it is in exactly the same state then we find, if the experiment is carried out enough times, the result follows the probability law predicted by QM. You may retort that is because we have missed something. Trouble is no one has ever been able to find that something. In fact theorems exist that show that something would have some some pretty weird properties such as not being local (Bells Theorem).

Thanks
Bill

nouveau_riche

then what is the role of collapsing wave function, because i don't see it in your answer

bhobba

Its like when you toss a coin - prior to the toss its 50-50. After the toss its a dead cert. The probabilities 'collapsed' into something else.

Remember a wave function is waves of probability - not something 'physical'.

Thanks
Bill

vanhees71

The collapse of the wave function is only specific to the Copenhagen and related interpretations. It is not a necessary ingredient of the theory and in my opinion causes more inconsistencies than it helps to understand nature with help of quantum theory, which is the most comprehensive and successful description of nature we have so far.

Bhobba, gave already the correct answer. What you can do to experimentally check quantum theory is to prepare the system under consideration (say a particle with a quite well definined momentum) for many times as good as you can and then measure some quantity (e.g., the position of the particle when hitting a photo plate after running through a double slit). The quantum theoretical formalism will predict a certain probability to find the particle at a certain place at the photo plate, and this you can check by comparing the pattern formed when the experiment is performed at an ensemble of equally independently prepared systems.

Nowhere in this socalled "Minimal Statistical Interpretation" (MSI) I had to invoke somthing as a collapse of the wave function (or quantum state), and with this MSI there are no troubles with the interpretation whatsoever (particularly EPR like inconsistencies do not appear).

bhobba

Indeed. Although I will occasionally use terms like 'collapse' it is a very unfortunate choice of words, not really required, and best avoided.

The best book to understand this interpretation and whats really going on is the book by Ballentine mentioned previously in the thread. I cant even recall seeing the word collapse in that book - its certainly not in the index.

Added later:

I should add the ensemble interpretation of Ballentine is not strictly speaking the same as MSI - it is very close to it.

Thanks
Bill

markusgo

Hello . Today we were talking about the SE in chemistry class and we applied it 2 the "simple" hydrogen case.

My question is "how can we apply the SE to big molecules like organic molecules,DNA,...?"

mikeph

Answer: "with great difficulty"

In the case of the hydrogen atom there are two particles mutually interacting, and this can be simplified to a single electron moving in a Coulomb potential. For a water molecule, you have 10 electrons and 10 protons, with a separate Coulomb force between every possible pairing, and these are all gathered into one huge "Hamiltonian" term which depends on the spatial coordinates of each of the 20 charges. So right away your Schrodinger equation is a differential equation in at least 60 variables, and that's only for water.

It's simply impossible to analytically calculate energy levels. There is a lot of work going on right now using computers to do this, and it takes a long time.

nouveau_riche

the 50-50 probability is also a result of repeated experimentation.The probability of an outcome taken to be as the ratio of favorable outcome to total outcome.

If you haven't played with dice enough, you can't say about the probabilistic behavior of a particular number on it. In the similar way you have to play and interact with quantum particle and by doing so you collapses it.

bhobba

I am not sure I see your point. My point, using your dice analogy, is if you make a dice exactly the same as another dice you know from previous experimentation what the probabilities will be without further ado.

Thanks
Bill

nouveau_riche

okay i will explain it to you.
consider i am new to probability, i started to play with dice in order to predict how likely do i get 2 if i repeat my experiment.
i performed it many times and see that the ratio of favorable outcome to total comes 1/6 but i would still not be sure if i get 2 in my next 6 throws.

now lets go to the quantum world, how will you evaluate the probability for localizing the particle? you will look for it, perform the experiments many times and then qualitatively use it for future experiments prepared in the same way. The question is that, if collapse really occurs, then as soon as you inteact with the system,instead of the showing its schrodinger probabilistic behavior the localized particle hits you with its peak probability amplitude i.e. it is there when you saw it?

how should i resolve for verification of schrodinger wave equation in the picture i described for quantum world relative to what i did with dice thing.

bhobba

It is true you cannot say, without further elaboration, what property it has prior to observation like you can with a dice - that's Kochen-Sprecker. However you can say the property the observational apparatus and system taken together has as a result of the observation which is why you view it as the outcome of the system and observational apparatus combined and not ascribe any property to the system aside from what the apparatus registers at the time of observation. This is the issue that troubled Ballentine with his older Ensemble interpretation before the full import of Kochen-Sprecker was taken into account, and led him to think its best to view it as the result of some sub quantum process that evades Kochen-Sprecker. Since then though he seems to have taken to viewing it as an ensemble of observational apparatus and observed system so the issue does not arise. The minimalist interpretation views it that way from the start and doesn't ascribe any view of probability to the situation such as considering it as an ensemble (which is one view of probability - but not the only one eg you can use the Kolmogorov axioms). This is why it's not exactly the same as the Ensemble interpretation - very close - but not the same.

However the kicker is if you take decoherence into account (thats the further elaboration) a superposition is transformed into a mixed state by the observation and you can view it as having the property prior to observation in exact analogy with a dice. This is why when Wigner first heard about decoherence from some of the early work of Zurek he recognised immediately stuff like Schodengers Cat, Wigners Friend etc had been completely resolved and abandoned the consciousness causes collapse views he previously held.

However we are now getting way way off topic onto issues that have been thrashed out many times in many different threads (and not resolved to everyone's satisfaction - which is why these discussion can go on and on and get nowhere) and really it needs to be taken up elsewhere perhaps in a new thread under foundational issues. Basically its the old Einstein-Bohr debates rehashed over and over - each side at the end of the day retreats to their impregnable positions and it gets nowhere in the sense of resolving anything. You understand the issues better but no definite answer results. My suggestion is instead of posting here is to read about the fundamental issues involved and form your own view eg:
http://www.amazon.com/Quantum-Einste.../dp/0393339882

I did that many years ago and its only recently I resolved it in my mind to my satisfaction - but having discussed it with others many times what I find 'satisfying' is not neceassarily what others do.

Thanks
Bill