Does a particle really try every possible path?

[bhobba]

If it's at multiple places at the same time

Its been said many many times before but it seems a particularly stubborn idea to try and view quantum systems as doing something when not observed - being in many places at once, taking many parths etc etc. QM is a theory about observations. When not observed the theory is silent. Its not in two places at once, taking multiple paths. As far as we can tell its not doing anything.

The sum over histories approach is simply saying mathematically its like the particle takes many paths simultaneously - it not saying that's what's actually going on. Strictly speaking its a hidden variable interpretation but of a very novel type - the path is the hidden variable - the novelty is it doesn't take one particular path - but all of them.

Thanks
Bill

 

[Ookke]

As far as we can tell its not doing anything.

#30 could be my reply to your post as well. Maybe it's completely irrelevant in physics what is happening between interactions, but it's interesting to think.

 

[phinds]

Ok. Maybe it's then misleading to talk about "paths" at all. However if we let a test particle travel through a room so that it doesn't interact until the back wall, it's tempting to think that something travels through the room, and the mass, energy and momentum are conserved in some form, even though it's not a particle in the classical sense.

Yeah, I think that way too. My logic is that if you move the wall a bit closer, you still get an interaction. Move it a bit closer and you still get an interaction. And so forth. So clearly it's THERE in some sense.

The mistake would be to connect the dots between all those interactions and think that you have found even one path that the particle took on the way to the wall when it was farthest away. There's no path, there's nothing until you get an interaction, but I'm agreeing w/ you that it is in some sense there even if that is undefined and not useful in practice.

 

[bhobba]

#30 could be my reply to your post as well. Maybe it's completely irrelevant in physics what is happening between interactions, but it's interesting to think.

You are looking at it incorrectly. QM is not saying its irrelevant what's happening between observations - its silent about it.

The question is why do you want to add things to a theory that doesn't say anything about it and discuss that with people who have gotten used to this perplexing part of QM? All you are going to get is what I have said - its not part of the theory.

Thanks
Bill

 

[stevendaryl]

No - but its obvious unless <x|x'> is physically resizeable its going to be zero - and since particles, relativistically can't travel FTL or go backward in time that's not possible.

Th caveat here is the Feynman-Stueckelberg interpretation of antiparticles as particles traveling backwards in time. In that view you would include FTL paths - but they aren't really traveling FTL or going backward in time - its simply an elegant way of handling the math. This is seen by the fact a particle can be considered an ani-particle traveling backward in time due to the symmetry of the situation. Its simply an elegant way to interpret it.

Thanks
Bill

A related question on this topic: If you do a path integral using the nonrelativistic lagrangian L = \frac{1}{2} m v^2 - V, you get the nonrelativistic propagator
G(x, t, x, t) = \langle x| e^{\frac{-i}{\hbar} H t} | x&#039;\rangle

If instead you do a path-integral with a relativistic Lagrangian such as: L = - m \sqrt{g_{\mu \nu} U^\mu U^\nu} or L = \frac{1}{2} m U^\mu U^\nu, does that give the Klein-Gordon propagator?

 

[bhobba]

There's no path, there's nothing until you get an interaction, but I'm agreeing w/ you that it is in some sense there even if that is undefined and not useful in practice.

Consistent Histories has an interesting take on this issue. It views QM as a stochastic theory about histories - it doesn't even have observations:
http://quantum.phys.cmu.edu/CHS/histories.html

Thanks
Bill

 

[bhobba]

If instead you do a path-integral with a relativistic Lagrangian such as: L = - m \sqrt{g_{\mu \nu} U^\mu U^\nu} or L = \frac{1}{2} m U^\mu U^\nu, does that give the Klein-Gordon propagator?

Sorry - don't know.

My knowledge of QFT is not as good as I would like it to be.

But it seems conventional to assume all contributions in the propagator and interpret FTL etc as per Feynman-Stueckelberg. Of course they aren't in any sense real like a lot of things we call virtual in QFT.

Thanks
Bill

 

[phinds]

Consistent Histories has an interesting take on this issue. It views QM as a stochastic theory about histories - it doesn't even have observations:
http://quantum.phys.cmu.edu/CHS/histories.html

Thanks
Bill

At a quick look, this makes my head hurt, but it looks interesting so I'll attack it some time when I'm stocked up on aspirin :)

Thanks for the link.

 

[naima]

I come back to the "no ftl path axiom". Suppose we have an atom in a box. its wave function is null outside the box. I open the box at t = 0. the atpm only "explores" the paths with no ftl speed.
The conclusion is that the new wave function is null outside the future cone of the box. No possible gaussian probability is allowed!

 

[bhobba]

I come back to the "no ftl path axiom". Suppose we have an atom in a box. its wave function is null outside the box.
I open the box at t = 0. the atpm only "explores" the paths with no ftl speed. The conclusion is that the new wave function is null outside the future cone of the box. No possible gaussian probability is allowed!

The particle in a box problem is from standard QM which is based on Galilean relativity and hence is not local.

Have zero idea what you comment about Gaussian probability is about.

Thanks
Bill

 

[naima]

It is an example of a wave function which is not null at any place.A consequence of your no ftl path is that at a given time there is a distance L so that all possible wave function are forbidden for the atom from the box

 

[bhobba]

It is an example of a wave function which is not null at any place.A consequence of your no ftl path is that at a given time there is a distance L so that all possible wave function are forbidden for the atom from the box

But no FTL does not apply to a particle in a box - its standard QM to which locality does not apply.

BTW no FTL is not an axiom - its implicit in relativity ie if you assume QFT you assume relativity.

The reason it comes into it is the particles going backward in time trick Feynman used. Particles really aren't going FTL or backwards in time - it simply for mathematical elegance.

Thanks
Bill

 

[stevendaryl]

But no FTL does not apply to a particle in a box - its standard QM to which locality does not apply.

BTW no FTL is not an axiom - its implicit in relativity ie if you assume QFT you assume relativity.

The reason it comes into it is the particles going backward in time trick Feynman used. Particles really aren't going FTL or backwards in time - it simply for mathematical elegance.

Thanks
Bill

Well, what's a little subtle about relativistic quantum mechanics is this: Suppose that at time t_1, you have a state with a single particle, localized near x_1, and later, at time t_2, you find a particle at x_2. Then you might think that no FTL would imply that |x_2 - x_1| &lt; c |t_2 - t_1|. But because relativistic quantum mechanics allows particle creation, you can't conclude that.

 

[naima]

Have zero idea what you comment about Gaussian probability is about.

Read this in wiki
Path integration gives a gaussian. Where can you find that v < c is used?

 

[bhobba]

True - and there is no avoiding it - see section 8.3 - Quantum Field Theory for the Gifted Amateur.

Once you assume relativity you get QFT with its Fock Space.

If you want no FTL you are forced to the full QFT machinery.

Thanks
Bill

 

[bhobba]

Read this in wiki Path integration gives a gaussian. Where can you find that v < c is used?

Yea - they crop up frequently. And you have to use the method of steepest decent to handle it:
http://www.maths.manchester.ac.uk/~gajjar/MATH44011/notes/44011_note4.pdf

This is ordinary QM not QFT - of course it isn't used - it is non local.

For QFT the Lagrangians used are explicitly relativistic - covarience is a symmetry they are required to have.

Added Later:
See Chapter 11, section 11.1 - QFT For The Gifted Amateur where the method is explained in detail.

Thanks
Bill

 

[bhobba]

No possible gaussian probability is allowed!

Before going an further can you explain what Gaussian probability has to do with anything?

Thanks
Bill

 

[sheaf]

A related question on this topic: If you do a path integral using the nonrelativistic lagrangian L = \frac{1}{2} m v^2 - V, you get the nonrelativistic propagator
G(x, t, x, t) = \langle x| e^{\frac{-i}{\hbar} H t} | x&#039;\rangle

If instead you do a path-integral with a relativistic Lagrangian such as: L = - m \sqrt{g_{\mu \nu} U^\mu U^\nu} or L = \frac{1}{2} m U^\mu U^\nu, does that give the Klein-Gordon propagator?

I've only ever seen path integration (over particle trajectories) done in the non relativistic case. In the relativistic case, it's always been classical field configurations i.e. conventional QFT. However I came across this link

The simple physics of a free particle reveals important features of the path-integral formulation of relativistic quantum theories. The exact quantum-mechanical propagator is calculated here for a particle described by the simple relativistic action proportional to its proper time. This propagator is nonvanishing outside the light cone, implying that spacelike trajectories must be included in the path integral. The propagator matches the WKB approximation to the corresponding configuration-space path integral far from the light cone; outside the light cone that approximation consists of the contribution from a single spacelike geodesic. This propagator also has the unusual property that its short-time limit does not coincide with the WKB approximation, making the construction of a concrete skeletonized version of the path integral more complicated than in nonrelativistic theory.

 

[naima]

W

Before going an further can you explain what Gaussian probability has to do with anything?

did you read post 44?

 

[Ookke]

Yeah, I think that way too. My logic is that if you move the wall a bit closer, you still get an interaction. Move it a bit closer and you still get an interaction. And so forth. So clearly it's THERE in some sense.

The mistake would be to connect the dots between all those interactions and think that you have found even one path that the particle took on the way to the wall when it was farthest away. There's no path, there's nothing until you get an interaction, but I'm agreeing w/ you that it is in some sense there even if that is undefined and not useful in practice.

It seems unavoidable that the particle is there all the time in some form, but I don't know if this is correct way to think about it in the end. However, it would be interesting to know if the particle in its interfered state is immune to relatively weak gravity and EM fields.

For example, if we do double-slit experiment with electrons using setup where left wall has positive charge, would the interference pattern have bias toward left? If yes, it would indicate that the particle somehow (weakly) interacts with the EM field even in its interfered state, where it doesn't have exact position until it (strongly) interacts with the back wall by hitting the detector.

 

[bhobba]

\did you read post 44?

Of course.

Again what has that got to do with Gaussian probability? The Gaussian occurs when the wave function is written as the path integral of A exp^iS (S the free particle action) - A exp^iS is a Gaussian function :
http://en.wikipedia.org/wiki/Gaussian_function

But what has it to do Gaussian probability? Indeed in this context what do you mean by 'No possible Gaussian probability is allowed!'. I can make no sense out of it at all.

Thanks
Bill

 

[bhobba]

It seems unavoidable that the particle is there all the time in some form,

Why does a particle have to have the property of 'there' when not observed?

For example, if we do double-slit experiment with electrons using setup where left wall has positive charge, would the interference pattern have bias toward left? If yes, it would indicate that the particle somehow (weakly) interacts with the EM field even in its interfered state, where it doesn't have exact position until it (strongly) interacts with the back wall by hitting the detector.

Ok - look at it via the following analysis
http://arxiv.org/ftp/quant-ph/papers/0703/0703126.pdf

The symmetry behind the screen that leads to equation 9 is broken. You need to analyse the what happens at each slit and take its superposition.

Thanks
Bill

 

[stevendaryl]

But what has it to do Gaussian probability? Indeed in this context what do you mean by 'No possible Gaussian probability is allowed!'. I can make no sense out of it at all.

I took it as a very minor point, and the word "Gaussian" was only used as an example. The point about a Gaussian distribution is that, although it gets small at large distances, it never goes to zero. In contrast, no-FTL would imply that the probability distribution for a particle initially localized at some point would fall to zero outside of the light cone. Very different behaviors.

 

[bhobba]

I took it as a very minor point, and the word "Gaussian" was only used as an example. The point about a Gaussian distribution is that, although it gets small at large distances, it never goes to zero. In contrast, no-FTL would imply that the probability distribution for a particle initially localized at some point would fall to zero outside of the light cone. Very different behaviors.

Ok - fair enough.

The bottom line however is this.

No FTL is only applicable to relativistic theories - not classical ones. The free particle action that appears in the Gaussian form of the path integral is classical - classical physics is inherently non-local.

And as I pointed out if you try to introduce no FTL into the usual concept of a particle in QM it doesn't work - see the reference I gave previously (section 8.3 - Quantum Field Theory for the Gifted Amateur) - one must go to QFT where the concept of a particle is far more subtle.

Watch it though - the calculation is surprisingly complex requiring contour integration and such - I really had to have my thinking cap on when I went through it.

Thanks
Bill

 

[naima]

In his space-time approach Feynman calculates the propagator (x,0) -> (x',t) as an integral of e^{i S(q)/hbar where S(q) is the classical action along the path.
The paths have to be continuous, to start at (x,0) and to end at (x',t) and that is all.
Nothing about speed of light.
The question was can we think that the particle explore all these paths. My answer is that this is a wrong and useless conception. the particle would have to explore a part of them with a ftl speed.

 

[bhobba]

In his space-time approach Feynman calculates the propagator (x,0) -> (x',t) as an integral of e^{i S(q)/hbar where S(q) is the classical action along the path. The paths have to be continuous, to start at (x,0) and to end at (x',t) and that is all. Nothing about speed of light..

What is the symmetry of the Lagrangian p^2/2m? Is it Galilean or relativistic?

Thanks
Bill

 

[gerbilmore]

As a relevant side note, are there any examples of the double slit experiment where the measuring device which establishes which slit the photon goes through is switched OFF and then switched ON to take measurements. I'd expect to see the pattern switch almost immediately from an interference pattern to two lines.

In terms of exploring all paths, does this imply some kind of—and I hate to use the word—'knowledge'. For example, I can arguably explore and rule out all but a few paths from my home to my local shop almost instantaneously in my head. I don't actually physically have to travel them—most are so improbable (via Pluto for example) that they don't even register in my conscious mind. I'm not suggesting a particle has knowledge in this sense, but a particle is part of a larger interconnected whole.

I'm just trying to get a picture of things in my head of what it actually means to travel all paths.

 

[naima]

The question is a mathematical question.
We have an integral to compute along many paths.
Bhobba tells us that the question is to know if we have to integrate a relativistic formula or not.
But he does not accept to write if we have to sum along all the paths.

 

[bhobba]

I'm just trying to get a picture of things in my head of what it actually means to travel all paths.

It's purely a mathematical reformulation - it doesn't mean it literally takes all paths.

When you try and visualise this stuff is when you run into problems.

Thanks
Bill

 

[bhobba]

Bhobba tells us that the question is to know if we have to integrate a relativistic formula or not. But he does not accept to write if we have to sum along all the paths.

And you demonstrate a classical Lagrangian based on Galilean relativity, that right at its foundations is non relativistic, admits any velocity, and offers it as evidence particles travel FTL in the path integral.

If you can't see the logic is circular there is no need to go any further.

This is my last comment on the matter.

Thanks
Bill