Does a particle really try every possible path?

[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 $|x_2 - x_1| < c |t_2 - t_1|$. 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
$G(x, t, x, t) = \langle x| e^{\frac{-i}{\hbar} H t} | x'\rangle$

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

 

[ChrisVer]

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

Try visualizing it you get problems: Well that depends on the person I guess... I find it an extreme beautiful idea and easy to "visualize"..
And in general many things are just mathematical. I mean, you have an unknown region, where you don't know what is happening because you can't observe it. It's amazing that you can say "everything happens" but also "everything ends up in what I measure".

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.

What exactly are you trying to say here?

 

[naima]

I wanted to say that path integral is the integral along ALL paths.

 

[Atomic squire]

Would you mind explaining what you mean? Why a misstatement? I could, in theory, travel at the speed of light, couldn't I?

Well, no matter can travel at or above the speed of light without disintegrating, even in theory. The particles that create the field that holds together the quarks that make up protons and neutrons move only at the speed of light, so traveling at that speed would basically cancel out the movement and destroy the field. Thus, sir atom turns to quark mush.

 

[PeterDonis]

The particles that create the field that holds together the quarks that make up protons and neutrons move only at the speed of light

No, they don't. The particles that create the field are virtual particles (more precisely, in the appropriate approximation, the field can be viewed as being mediated by virtual particles--but there are field phenomena that cannot be modeled in this approximation), and virtual particles have a nonzero amplitude to travel faster than light. (They also have a nonzero amplitude to move slower than light even if they are massless--for example, virtual photons have a nonzero amplitude to move slower than light. Virtual particles that move in a way that violates the usual energy-momentum relation for their particle type are called "off-shell", and they must be included to get the right answers out of the path integral.)

 

[PeterDonis]

since ftl paths can't exist, neither can the paths in the Feynman integral.

I'm not sure this is correct. As I understand it, in the general form of the path integral you cannot restrict the paths to only non-FTL ones. Feynman, IIRC, explained it as being due to all the energies being positive; restricting to non-FTL paths amounts to restricting the 4-momentum to a finite interval, and all the energies being positive means only positive frequencies, and known properties of the Fourier transform say that a function with a Fourier transform with only positive frequencies that vanishes outside a finite interval must be identically zero. So to have a non-vanishing path integral at all with only positive energies, you need to include FTL paths.

I understand that the above does not require that the Feynman-Stueckelberg interpretation of virtual particles moving outside the light cone as moving "backwards in time" is correct. But AFAIK the need to include FTL paths in the path integral does not depend on a particular interpretation.

 

[bhobba]

Well, no matter can travel at or above the speed of light without disintegrating, even in theory

I agree with Peter.

There are some parts of your post I can't really make sense of, but the above is incorrect.

The reason particles can't travel above the speed of light (they can travel at the speed of light if massless) has to do with space time geometry and is a consequence of the fact regardless of how fast you are traveling the speed of light is always the same. As Wheeler says forward is always forward and since a beam of light always moves away from you at the speed of light you can't catch up to it, hence an object traveling slower than the speed of light can't reach that speed.

Of relevant to this thread however is that the path integral formalism is formulated for ordinary QM, and although it not usually emphasised, classical mechanics is not local and based on the Galilean transformations (see page 8 - Landau - Mechanics) so any path or speed is allowed. In QFT things are different and we have virtual particles (they don't really exist - they are simply an artefact of the perturbation formalism used) that do all sorts of crazy things like going FTL etc - bit since they aren't real who cares.

Thanks
Bill

 

[bhobba]

I'm not sure this is correct. As I understand it, in the general form of the path integral you cannot restrict the paths to only non-FTL ones.

Sure - as far as I understand it one must consider all contributions of the propagator - even FTL ones - but arent they called off-shell and not real?

Thanks
Bill

 

[Joel A. Levitt]

How do you define PARTICLE?
How do you understand the FTL transmission of information observed in recent entanglement experiments?

 

[PeterDonis]

the path integral formalism is formulated for ordinary QM

It may have originally been formulated for ordinary QM, but it certainly works for quantum field theory (with appropriate changes). Zee's Quantum Field Theory in a Nutshell has a good review.

as far as I understand it one must consider all contributions of the propagator - even FTL ones - but arent they called off-shell and not real?

"Off shell" means "not obeying the appropriate energy-momentum relation". Plenty of non-FTL paths are also off-shell.

As for "not real", either a path is in the integral or it isn't. What does "real" have to do with it?

 

[PeterDonis]

How do you define PARTICLE?

A particular type of excitation of a quantum field.

How do you understand the FTL transmission of information observed in recent entanglement experiments?

Reference, please?