Jump to content

What happens when the waveform collapses?


geordief

Recommended Posts

An object is apparently in some sort of "suspended animation" as it crosses "empty space" and then when an "observation " is made   the "waveform collapses" and its probable location is "set in stone"

 

(this is  exemplified ,it seems to me in the double slit experiment)

 

Is it possible to say  with more detail  what "collapsing the waveform" actually means? Or do we just have to say "well this is what happens and   that is how we describe the phenomenon  with words"  ?

 As a side thought ,is there any mileage in the idea that when any "observation" is made , we are looking at two  probability waveforms meeting/interacting/superimposing  (that of the object and also that of the detector) ? 

Link to comment
Share on other sites

I've only studied this stuff to a superficial degree but here's my understanding, hope it makes sense.

Before you measure a wavefunction it exists as a linear superposition of a number of composite wavefunctions. That means the wavefunction is made out of lots of other wavefunctions all added together. The contribution of some of those other wavefunctions is greater than others. When we make an observation the wavefunction collapses to just one of those composite wavefunctions, with a probability proportional to its contribution to the overall wavefunction. I believe the collapse of the wavefunction refers to just this: what was once described by a sum a wavefunctions is now observed to be just one of those wavefunctions.

Link to comment
Share on other sites

Thanks . I wonder what determines the make up of the composite  waveform. Are there a finite amount of these component  parts? (is everything smooth? )

What about my idea that the detecting object also has its own set of waveforms which also collapse when the detection event occurs ?

Is this actually part of the model  (just that I have not come across it explicitly stated ) ?

Edited by geordief
Link to comment
Share on other sites

3 hours ago, geordief said:

Thanks . I wonder what determines the make up of the composite  waveform. Are there a finite amount of these component  parts? (is everything smooth? )

What about my idea that the detecting object also has its own set of waveforms which also collapse when the detection event occurs ?

Is this actually part of the model  (just that I have not come across it explicitly stated ) ?

You need to understand the mathematics of Hilbert spaces to understand this issue.

The linear combination may be finite or infinite, there is not restriction.

The observables correspond one-to-one to the self adjoint operators in a separable Hilbert space of infinite dimension, H. The pure states correspond one-to-one to the one dimensional subspaces of H. Every state is a (possibly infinite) convex combination of pure states. (This is a mathematical statement of waveform generation and collapse)

'Collapse of the waveform' is linked to the so called 'measurement problem' or measurement paradox and is very difficult to discuss without some higher maths.

Try reading here

http://www.informationphilosopher.com/problems/measurement/

Essentially the measurement problem, waveform collapse and 'paradoxes' like Schroedinger's cat are about the crossover point from deterministic classical mechanics to probabilistic quantum mechanics and how this is handled.

Edited by studiot
Link to comment
Share on other sites

thanks, i will have a look.

 

btw. can you remind me again  of that Bensen (?) book you recommended  to me  a few months ago (it was a historical account of the leadings up to modern Relativity  )  i can't find it  now without fine tooth combing my posts and I want to order it on ebay or Amazon.

 

ps no chance of an (understandable) example of a  waveform with just 2 components, I suppose.(since you imply they can be finite in number)

Edited by geordief
Link to comment
Share on other sites

Fields of force

William Berkson

Routledge Kegan & Paul

1974

 

Here is the beginning of the axiomatic mathematical process of deriving QM.
Another dozen pages of development will arrive at axiom VIII which is the part of my previous post.

 

mathqm1.jpg

Link to comment
Share on other sites

I have question ....when a photon waveform collapses, does the entire energy that is contained in the waveform get "collapsed" to a localized point, conserving the energy, or is some inefficiency observed in the transformation process?

Link to comment
Share on other sites

2 hours ago, hoola said:

I have question ....when a photon waveform collapses, does the entire energy that is contained in the waveform get "collapsed" to a localized point, conserving the energy, or is some inefficiency observed in the transformation process?

It doesn't work like that.

Energy is never all at one point in QM

The total energy doesn't change because of the collapse, just the distribution.

edit.

Note in my extract, halfway down the left hand page "A point in phase space has no physical meaning"

In classical science we attribute properties to things.

So, for instance, the mass of an oxygen atom is 16 units whether it is buzzing about in an oxygen atom in the atmosphere , or combined with hydrogen atoms to form ocean water or locked into the rocks in the form of silica or iron oxide.

But in QM we cannot attribute properties to a particle alone.

The properties depend upon the environment.

So the mass of an electron appears different depending upon whether it is within an ionic lattice, travelling freely in empty space or part of a free atom. Our method of measurement also affects the answer.

This is quite separate from relativistic effects. I have not included those as they would only complicate matters and obscure the QM aspect.

So when we talk about the properties of a particle in QM we always have to do this in relation to the environment.

Edited by studiot
Link to comment
Share on other sites

15 hours ago, hoola said:

I have question ....when a photon waveform collapses, does the entire energy that is contained in the waveform get "collapsed" to a localized point, conserving the energy, or is some inefficiency observed in the transformation process?

Which wave are you discussing? The wave function, or the deBroglie wave?

I'm guessing the former, since you ask about collapse. It doesn't have any energy. It's a description; when you square it it tells you the probability of being in a location or having a particular momentum. When you operate on it with the hamiltonian it tells you the energy of the system. It's not something that physically exists

Link to comment
Share on other sites

55 minutes ago, swansont said:

Which wave are you discussing? The wave function, or the deBroglie wave?

I'm guessing the former, since you ask about collapse. It doesn't have any energy. It's a description; when you square it it tells you the probability of being in a location or having a particular momentum. When you operate on it with the hamiltonian it tells you the energy of the system. It's not something that physically exists

Can that "energy of the system" principle be extrapolated to encompass all connected systems? (there are  no isolated systems ,are there?)

Is it possible that the energy of the entire  universe (to include the (unobservable part) is  theoretically zero at all times?

 

 

 

 

Link to comment
Share on other sites

58 minutes ago, geordief said:

Can that "energy of the system" principle be extrapolated to encompass all connected systems? (there are  no isolated systems ,are there?)

People have argued that you can (in principle) write a wave function of the universe. But that's in part because you can make a composite out of independent wave functions.  There are systems where treating them as isolated is a reasonable assumption.

58 minutes ago, geordief said:

Is it possible that the energy of the entire  universe (to include the (unobservable part) is  theoretically zero at all times?

Possible, and one scenario that's considered.

Link to comment
Share on other sites

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
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.