What happens mathematically when a wave function collapses

[Christoffelsymbol100]

In Shankars "Principle of Quantum Mechanics" in Chapter 4, page 122, he explains what the "Collapse of the State Vector" means.

I get that upon measurement, the wave function can be written as a linear combination of the eigenvectors belonging to a operator which corresponds to the measurement (because the operators are hermitian and their eigenvectors form a basis).

But then he goes on to say, that the measurement forces the state vector [itex]|\psi>[/itex] into
the eigenstate [itex]|\omega>[/itex].

But HOW does the operator do that? I can't find a single equation which describes the collapse of the wave function.
So my question is, what happens mathematically behind the scenes?

 

[Demystifier]

The collapse is random, and random processes cannot be described mathematically in a way you are used to.

 

[MichPod]

"Operator" does not do that. The operator which corresponds to the measurement is just a mathematical tool to encode eigenvalues and eigenstates.
The collapse of the wave function is a highly controversial topic and one is probably expected in the beginning just to take it as a postulate (Born rule).
It does not follow directly from other postulates of QM and there are no equations in basic QM which describe it.

 

[anorlunda]

But then he goes on to say, that the measurement forces the state vector |ψ>|ψ>|\psi> into
the eigenstate |ω>|ω>|\omega>.

Natural language is our bane. Suppose that it said "once the measurement observes the eigenvalue ##\omega## corresponding to eigenstate ##\omega##, then the state of the system is definitely ##\omega## regardless of what the state was prior to the measurement" That says the same thing, but it avoids "force" or "collapse" or any other verb.

p.s. I just looked up the postulates of quantum mechanics on Wikipedia in search of a quotation to use here. I found the wiki's statements of the postulates so dense and so mathematically correct, that their meaning is obscured.

There is no perfect solution, natural language trips us up all the time, but math can also obscure meaning.

 

[Mentz114]

In Shankars "Principle of Quantum Mechanics" in Chapter 4, page 122, he explains what the "Collapse of the State Vector" means.

I get that upon measurement, the wave function can be written as a linear combination of the eigenvectors belonging to a operator which corresponds to the measurement (because the operators are hermitian and their eigenvectors form a basis).

But then he goes on to say, that the measurement forces the state vector [itex]|\psi>[/itex] into
the eigenstate [itex]|\omega>[/itex].

But HOW does the operator do that? I can't find a single equation which describes the collapse of the wave function.
So my question is, what happens mathematically behind the scenes?

One interpretaqtion is that 'a linear combination of the eigenvectors' is not a physical state, and that the preparation has placed the object in an eigenstate. The measurement then just reports the the (perfect) state it finds. No collapse required.

Now you can ask - how did the preparation apparatus make the choice between between possible state ? IMO that is easier to answer.

 

[Jilang]

No Mentz it cannot be the case. That would correspond to hidden variables. These are ruled out by Bells Theorum.

 

[PeroK]

In Shankars "Principle of Quantum Mechanics" in Chapter 4, page 122, he explains what the "Collapse of the State Vector" means.

I get that upon measurement, the wave function can be written as a linear combination of the eigenvectors belonging to a operator which corresponds to the measurement (because the operators are hermitian and their eigenvectors form a basis).

But then he goes on to say, that the measurement forces the state vector [itex]|\psi>[/itex] into
the eigenstate [itex]|\omega>[/itex].

But HOW does the operator do that? I can't find a single equation which describes the collapse of the wave function.
So my question is, what happens mathematically behind the scenes?

In terms of a continuous process that transforms one state to another, there is nothing in the theory. It's similar to the idea that when an atom absorbs or emits a photon, an electron moves from one energy level to another. This is not a continuous process and there is no time when the electron is halfway between energy levels and the photon is partly created or half absorbed.

 

[Jilang]

In terms of a continuous process that transforms one state to another, there is nothing in the theory. It's similar to the idea that when an atom absorbs or emits a photon, an electron moves from one energy level to another. This is not a continuous process and there is no time when the electron is halfway between energy levels and the photon is partly created or half absorbed.

Not quite, there is an oscillation between the energy levels that lasts a short period of time.

 

[Mentz114]

No Mentz it cannot be the case. That would correspond to hidden variables. These are ruled out by Bells Theorum.

The only way one spin-1/2 particle can have a +/- z-spin at once is if the composite system became a non-local boson with total spin zero.

How likely is that ?

 

[PeroK]

The only way one spin-1/2 particle can have a +/- z-spin at once is if the composite system became a non-local boson with total spin zero.

How likely is that ?

Or, it could have spin-up in the x-direction. Your misunderstanding of superposition which you decided "to live with" on a previous thread is leading you astray.

 

[mike1000]

Or, it could have spin-up in the x-direction. Your misunderstanding of superposition which you decided "to live with" on a previous thread is leading you astray.

That is the proper way to interpret it (spin up in the x-direction), not that the particle was in both the up and down z-spin state at the same time.

 

[Mentz114]

Or, it could have spin-up in the x-direction. Your misunderstanding of superposition which you decided "to live with" on a previous thread is leading you astray.

Hi, thanks for reminding me.

The linear combination cited in the first post belongs to the operator. The result of applying the operator to the state vector is guaranteed to result in an eigenstate - not a linear combination of eigenstates. That is my reading of Ballentine, anyway. So my misunderstanding (apparently) deepens because I still can't see your point.

 

[bhobba]

But then he goes on to say, that the measurement forces the state vector [itex]|\psi>[/itex] into the eigenstate [itex]|\omega>[/itex].

It actually follows from the two principles of QM you will find in Ballentine and continuity.

I could explain the detail, but the fun is in figuring it out for yourself, so after Shankar read Ballentine.

But just as an overview the second axiom of QM is Born's Rule and a little logic explains this 'collapse' idea. Then a little more thought will show collapse doesn't really exist at all - but again best if you figure it out. Hint - in modern times a state and preparation procedure are basically synonymous. So if you subject a system to an observation you are now preparing it differently - so of course, provided the system isn't destroyed by the observation which is what usually happens, all you have done is prepare it differently. Did anything 'collapse'? I won't answer - just leave it for you to nut out. Ballentine gives the gory detail, but if you can figure it out so much the better.

Thanks
Bill

 

[bhobba]

"Operator" does not do that. The operator which corresponds to the measurement is just a mathematical tool to encode eigenvalues and eigenstates. The collapse of the wave function is a highly controversial topic and one is probably expected in the beginning just to take it as a postulate (Born rule). It does not follow directly from other postulates of QM and there are no equations in basic QM which describe it.

Nice answer.

Actually - QM can be deduced from just one axiom:
https://www.physicsforums.com/threads/the-born-rule-in-many-worlds.763139/page-7

See post 137. Note I am being a bit tricky and cheeky in saying its just one axiom - but I will let you spot it. However it doesn't change anything.

Note - the axiom contains nothing about collapse. Its just something some interpretations introduce the formalism says nothing about. Yet we have all these lengthy discussions about it. Its quite simple really.

Thanks
Bill

 

[mikeyork]

An observational apparatus interacts with the system being observed. The "collapse" of the "wavefunction" is the result of that interaction and derives from the re-expression of the combined system of observer and observee in a context defined by the observer.

 

[bhobba]

An observational apparatus interacts with the system being observed. The "collapse" of the "wavefunction" is the result of that interaction and derives from the re-expression of the combined system of observer and observee in a context defined by the observer.

Sort of.

See:
http://www.quantum.umb.edu/Jacobs/QMT/QMT_Chapter1.pdf

If you probe a system to observe it, it actually becomes a POVM rather than the usual Von-Neumann measurement described by an operator.

That's why my link that starts from POVM's in the actual axiom is a better approach and you see immediately collapse has nothing to do with anything, neither does a conscious observer or many of the other things people get confused about.

Thanks
Bill

 

[MichPod]

Actually - QM can be deduced from just one axiom:
https://www.physicsforums.com/threads/the-born-rule-in-many-worlds.763139/page-7

It may take me much time to digest this. I am not well prepared in this topic and it is higher than my level. Meanwhile, let me still ask some questions:

It you have a E_i which is an element of POVM, it may help you to get the probability of the outcome ##i##, but you still need to define what final state Ψ_i of the system corresponds to that outcome ##i##. And after you define that state, you essentially get a collapse i.e. a transition from the original state of the system to the Ψ_i.

Now, I am not claiming that a collapse is a real physical process or just an artifact of the theory, but I claim that you need to have and define such a transition as an additional axiom or part of some axiom (unless I missed that you are able somehow to derive such a transition from the existing axioms).

 

[Jilang]

Sort of.

See:
http://www.quantum.umb.edu/Jacobs/QMT/QMT_Chapter1.pdf

If you probe a system to observe it, it actually becomes a POVM rather than the usual Von-Neumann measurement described by an operator.

That's why my link that starts from POVM's in the actual axiom is a better approach and you see immediately collapse has nothing to do with anything, neither does a conscious observer or many of the other things people get confused about.

Thanks
Bill

Nice link. The wave function as a conditional type probability.. thus being only one half of the story?

 

[mikeyork]

Here's how I see it:

Incoming system + observational system ----->>>> "observed" eigenstate + modified observational system

(It's an S-matrix element and the total system is the same superposition of composite eigenstates both before and after; only the initial and final representations change.)

 

[bhobba]

but you still need to define what final state Ψ_i of the system corresponds to that outcome ##i##. And after you define that state, you essentially get a collapse i.e. a transition from the original state of the system to the Ψ_i.

What you get out of an observation, obviously, is an observational outcome. The state is simply an aid to calculate the probabilities of other observations that might be done. Continuity demands that right after the observation is done its something depending on the outcome. So the change in state follows from other principles. All you can say is what it is after the observation. Did it suddenly collapse, implying instantaneous change, not necessarily - its non committal about that. You have simply prepared it differently. But the real sticker to the collapse idea is we have no collapse interpretations like Many Worlds.

Here is a table of various interpretations and if they have collapse or not:
https://en.wikipedia.org/wiki/Interpretations_of_quantum_mechanics

Thanks
Bill

 

[bhobba]

Nice link. The wave function as a conditional type probability.. thus being only one half of the story?

The wavefunction is simply an aid to calculate probabilities. What it means is interpretation dependent.

I have studied mathematical probability and it is not like conditional probability.

Thanks
Bill