Feynman Path Integral's Meaning

jaketodd

Gold Member
Does the math of the Feynman path integral dictate particles taking many paths or is that just an english language gloss-over?

Thanks,

Jake

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ansgar

all paths contribute, we dont know and can not know which path the particle took

jaketodd

Gold Member
all paths contribute, we dont know and can not know which path the particle took
But some paths have a greater probability than others, correct?

ansgar

But some paths have a greater probability than others, correct?
yes, each path is weighted with exp(-i/hbar * S ) where S is the classical action

but since we can't tell which path the particle took, we must add these probabilites coherently

LukeD

Consistent Histories provides a way of assigning probabilities to non-interfering sets of paths though. So we can assign probabilities, but not to individual paths.

IttyBittyBit

Why don't you get the answer from the father of the path integral himself?

In QED: The Strange Theory of Light and Matter, Feynman explains, in a highly accessible way, what path integrals are and how they relate to QED. You can find that book in just about any library.

If you want nuts-and-bolts explanations there is no easy way other than learning about quantum field theory.

jaketodd

Gold Member
Why don't you get the answer from the father of the path integral himself?

In QED: The Strange Theory of Light and Matter, Feynman explains, in a highly accessible way, what path integrals are and how they relate to QED. You can find that book in just about any library.

If you want nuts-and-bolts explanations there is no easy way other than learning about quantum field theory.
Why not? Because I think even him dismissed the interpretation that the particle takes infinitely many paths and those paths cancel to probabilities for paths, yet, to my humble knowledge, that is what the math of the path integral states happens.

jrosen13

In physics, when one has a crazy idea that works, it may be necessary to state that it is not the only way to look at it so people dont simply dismiss it because it seems crazy.

The essence is that the classical path is a stationary point of the action, i.e. a maximum/minimum/ or saddle point. However in QM, we know that uncertainty is built in and for short times we can have large energy fluctuations and short distances large momentum fluctuations. The leading quantum corrections to the classical physics comes from taking into account these quantum fluctuations on the classical path, hence many paths are now accessible and "feeling" these paths leads to nontrivial quantum physics.

Frame Dragger

Does the original "Sum Over All Histories" for a particle make the notion of a PI any clearer?

@jaketodd: He did, but that doesn't mean it isn't useful or worth learning.

FireBones

My understanding is that Feynman did not care much for interpretation anyway. "Shut up and calculate!"

yenchin

Incidentally, this is a nice video:

Last edited by a moderator:

IttyBittyBit

My understanding is that Feynman did not care much for interpretation anyway. "Shut up and calculate!"
Yup.

My understanding is that Feynman did not care much for interpretation anyway. "Shut up and calculate!"
As first said by Mermin :)

jaketodd

Gold Member
@jaketodd: He did, but that doesn't mean it isn't useful or worth learning.
I didn't say learning about the path integral and related things isn't useful or worth learning. I was saying that I think Feynman didn't publicly support the implications of the math that his path integral stated, and therefore he is not the person to get the proper interpretation from.

Doesn't it all come down to the math in the end? If the math says a particle takes infinitely many paths, then that's what is happening if it's correct regardless of the shrink wrap you put around it. However, a good question would be: "Is there a different way to mathematically reproduce the same results (and therefore the same accuracy) without the infinitely many paths being built into the math?" Or, am I wrong and that interpretation is not built into the math?

Thanks,

Jake

Frame Dragger

I didn't say learning about the path integral and related things isn't useful or worth learning. I was saying that I think Feynman didn't publicly support the implications of the math that his path integral stated, and therefore he is not the person to get the proper interpretation from.
I would say that makes him uniquely qualified, being able to use a tool without believing that it describes a complete physical reality.

jaketodd said:
Doesn't it all come down to the math in the end? If the math says a particle takes infinitely many paths, then that's what is happening if it's correct regardless of the shrink wrap you put around it. However, a good question would be: "Is there a different way to mathematically reproduce the same results (and therefore the same accuracy) without the infinitely many paths being built into the math?" Or, am I wrong and that interpretation is not built into the math?
You're right. You're wrong. What you just said is why it's a useful mathematical tool, but NOT a physical description. In other words, the math is just fine, but if it has no physical reality it is still an approximation, or tool. That's not a bad thing, but it doesn't mean that until we can determine what occurs at that scale, our description reflects reality.

jaketodd

Gold Member
Isn't the path integral the best we have in terms of what happens on that scale? Could it be the best we will ever have given the inherent uncertainty in quantum theory? If not, is it not reasonable to think what would be an improvement on it would contain the ideas of the math of the path integral and then the something else?

Thank you,

Jake

Frame Dragger

Isn't the path integral the best we have in terms of what happens on that scale? Could it be the best we will ever have given the inherent uncertainty in quantum theory? If not, is it not reasonable to think what would be an improvement on it would contain the ideas of the math of the path integral and then the something else?

Thank you,

Jake
The first, I would agree with, but the second... how can we tell? I would say an understanding of the Planck (and sub-Planck) scale would help to eliminate these approximations.

RUTA

My understanding is that Feynman did not care much for interpretation anyway. "Shut up and calculate!"
“There was a time when the newspapers said that only twelve men understood the theory of relativity. I do not believe there ever was such a time. There might have been a time when only one man did, because he was the only guy who caught on, before he wrote his paper. But after people read the paper a lot of people understood the theory of relativity in some way or other, certainly more than twelve. On the other hand, I think I can safely say that nobody understands quantum mechanics. … Do not keep saying to yourself, ‘But how can it be like that?’ because you will get ‘down the drain,’ into a blind alley from which nobody has yet escaped. Nobody knows how it can be like that.” Richard Feynman, The Age of Entanglement, Lousia Gilder, Vintage Books, New York (2008), pp 228-229. She cites “November 1964…Lectures”: Feynman, Character of Physical Law, chapter 6.

RUTA

yes, each path is weighted with exp(-i/hbar * S ) where S is the classical action but since we can't tell which path the particle took, we must add these probabilites coherently
The phasor exp(-iS/hbar) is unit length for any S. What S changes for the phasor of each path is its angle, not its length, so the phasors add more coherently about the extremum of S. Of course, the extremum of S gives the classical equations of motion via the least action principle.

RUTA

Does the math of the Feynman path integral dictate particles taking many paths or is that just an english language gloss-over?
It does not even require the existence of particles, a fortiori paths. Thus, the answer to the first part of your question is "no."

edpell

The idea that a particle diffuses from point A to point B via all the paths available and favors those with least "resistance" seem beautiful and simple. Why the strong objection?

Frame Dragger

The idea that a particle diffuses from point A to point B via all the paths available and favors those with least "resistance" seem beautiful and simple. Why the strong objection?
Why the objection... hmmm... well it was invented as a mathematical artifact for one. Another would be... how does a particle "diffuse" and encounter "resistance" to find a single path from A to B, and then apparantly coalesce at B? If a particle is capable of such diffusion, why the need for a path from A to B? It also doesn't do much to explain DCQE experiments, or a number of other glaring issues.

I think the question is, how do you feel that is simple or beautiful? It's the equivalent of brute-force hacking a password (not efficient) and applying that concept to nature as a whole because it works on paper.

Lots of things seem simple and lovely until you really consider the ramifications of them (the Geocentric universe for instance).

zenith8

Doesn't it all come down to the math in the end? If the math says a particle takes infinitely many paths, then that's what is happening if it's correct regardless of the shrink wrap you put around it. However, a good question would be: "Is there a different way to mathematically reproduce the same results (and therefore the same accuracy) without the infinitely many paths being built into the math?" Or, am I wrong and that interpretation is not built into the math?
No - this is not true. Let me show how you to achieve exactly the same result using only one path, rather than an infinite number of them (this is an edited version of a post I made in a thread ages ago):

The main point about Feynman's theory is to calculate the propagator (essentially, the mathematical object that enables you to calculate the wave function at some space point and time in the future, given the wave function now).

Now if you subscribe to the view that electrons have trajectories (i.e. the de Broglie-Bohm view) and you use the obvious trajectory implied by the quantum formalism, then you can compute the propagator using only that single 'quantum' path rather than Feynman's infinite number of trajectories. Look at the interesting similarity between the following formulae for the propagators:

DE BROGLIE-BOHM

$$K^Q({\bf x}_1,t_1;{\bf x}_0,t_0) = \frac{1}{J(t)^ {\frac{1}{2}} } \exp\left[{\frac{i}{\hbar}}}\int_{t_0}^{t_1}L(t)\;dt\right]$$

FEYNMAN

$$K^F({\bf x}_1,t_1;{\bf x}_0,t_0) = N \sum_{all paths} \exp\left[{\frac{i}{\hbar}}\int_{t_0}^{t_1}L_{cl}(t)\;dt \right]$$

In the Feynman case the propagator linking two spacetime points is calculated by linearly superposing amplitudes $$e^{iS/\hbar}$$ obtained by integrating the classical Lagrangian $$L_{cl}(t)={\frac{1}{2}}mv^2-V$$ associated with the infinite number of all possible paths connecting the points.

In the de Broglie-Bohm approach, you achieve the same effect by integrating the 'quantum' Lagrangian $$L(t)={\frac{1}{2}}mv^2-(V+Q)$$ along precisely one path (the one the electron actually follows). Here Q is the potential associated with the 'quantum force' (the particle is being pushed around or 'guided' by a physical field represented mathematically by the wave function).

So it's all a question of knowing the correct path/trajectory. Not a lot of people know this.. Note that this elevates the de Broglie-Bohm theory from being an 'interpretation' to a mathematical reformulation of quantum mechanics equivalent in status to Feynman's.

A couple of additional technical points:

(1) In the Feynman path integral method, computing the propagator by summing over all possible paths is only half of it. The Feynman propagator $$K^F$$ is a many-to-many mapping i.e. all points are linked by all possible paths. So the full $$\Psi({\bf x}_1,t_1)$$ is found from Huygen's principle by summing the contributions coming from all possible start points - you multiply the amplitude at $${\bf x}_0,t_0$$ by the transition amplitude $$K^F$$ for 'hopping' to $${\bf x}_1,t_1$$. Then you sum (integrate) over all $$x_0$$:

$$\Psi({\bf x}_1,t_1)=\int K^F({\bf x}_1,t_1; {\bf x}_0,t_0) \Psi({\bf x}_0,t_0)\;d x_0$$

In the deBB approach you achieve the same end as in the path integral method - the computation of $$\Psi$$ given the initial value - in a quite different and conceptually simpler manner with two spacetime points connected by at most a single path. The two steps in Feynman's approach (propagator then Huygens) are thus condensed into one (propagator). $$\Psi$$ is generated from its initial form by a single-valued continuum of trajectories.

(2) You might think you need the wave function over all space to compute the propagator in the deBB case but this is not true. You just need the second derivative of the wave function - or more accurately the second derivative of its amplitude (for the quantum potential Q) and of its phase (for the $$\nabla\cdot{\bf v}$$ in the Jacobian $$J$$) at the points along the track. In a practical numerical method, these can be calculated by sending a particle down the trajectory $${\bf v} = \nabla S$$ (i.e. following the streamlines of the quantum probability current) and then evaluating the required derivatives numerically using finite differencing or whatever. There is a whole community of physical chemists (believe it or not) who do precisely this to solve chemistry problems.

So to conclude, Feynman's paths are mathematical tools for computing the evolution of $$\Psi$$, while (if it is the case that particles actually exist) one among the de Broglie-Bohm paths is the actual motion of the particle as deduced from the equations of QM, which exists in addition to the wave field $$\Psi$$. Keep finally in mind that path integrals are not exclusive to QM; one can write any linear field equation (e.g. Maxwell) in terms of path integrals.

OK - the last time I posted this observation it apparently caused one of the homework helpers to have a complete mental break down. Many apologies if this happens again!

edpell

and then apparently coalesce at B
I think we need to stop thinking in terms of points. QM wise it is not at a point A it is more a blob around A and it does not go to point B it is more a blob around B. If it starts and stops as a blob why would we think it would take one infinity thin path from blob volume A to blob volume B?

Frame Dragger

I think we need to stop thinking in terms of points. QM wise it is not at a point A it is more a blob around A and it does not go to point B it is more a blob around B. If it starts and stops as a blob why would we think it would take one infinity thin path from blob volume A to blob volume B?
I have no idea, but that doesn't make the Path Integral formulation any less of a mathematical artifact. Zenith8 gave you the Bohmian interpeation (and more good info), which is really the only alternative to SQM + Some Interpreation.

I'm not saying that the HUP doesn't have a physical reality, and if that is the case, then maybe particles DO exactly what you suggest. Then again, someone with the MWI would just say the particlee takes ALL possible paths, but we observe only one. None of that is testable right now, so I don't accept it as anything more than a great approximation.

Stephen Hawking for one, makes a very convincing argument for a "Sum Over All Histories" aka Path Integral for particles, but without unification... who knows?

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