A silly question I'm sure about Feynman's many paths

1. Jul 13, 2009

jaketodd

A silly question I'm sure about Feynman's "many paths"

I've been reading about Feynman's many-paths idea. And I've read that according to his idea, an electron, for example, takes every possible path in the universe and they cancel out (the arrows pointing different directions) to a definite path. But if the electron took every possible path, then wouldn't they all cancel and the electron would go nowhere since the arrows point in every direction and would cancel completely to no path? Don't be too hard on me.

2. Jul 13, 2009

malawi_glenn

Re: A silly question I'm sure about Feynman's "many paths"

This was a very oversimplification view on Feynman path integrals, I assume this was not mentioned in a physics textbook?

The thing you mention is the classical regime of the path integrals, that only ONE path contribute. In Quantum mechanics ALL paths contributes, and it is meaningless to ask which path it took.

Each path is weighted by exp(iS/hbar), where S is the classical action. You are thinking of the paths as vectors, which is not true. In the classical regime, the action S is much larger than hbar - the typical QM scale - and hence the path integral will receive a fast oscillating phase making that paths will interfere and cancel out and only the classical path (where S is extremum) will contribute.

http://en.wikipedia.org/wiki/Path_integral_formulation

http://www.quantumfieldtheory.info/Path_Integrals_in_Quantum_Theories.htm [Broken]

Last edited by a moderator: May 4, 2017
3. Jul 13, 2009

zenith8

Re: A silly question I'm sure about Feynman's "many paths"

Here we go again with your 'meaningless' stuff.

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

It's very interesting to note that if you subscribe to the view that electrons have trajectories (i.e. the de Broglie-Bohm interpretation) and you use the obvious trajectory implied by the quantum formalism, then you can compute the propagator using only that single 'quantum' track rather than Feynman's infinite number of trajectories. Perhaps the OP won't be able to follow the meaning of the equations, but he can certainly appreciate the similarity between the following formulae for the propagators:

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 pilot-wave 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 being pushed by the wave function).

It's all a question of knowing the correct path/trajectory. Not a lot of people know this..

Note finally that knowing this elevates the de Broglie-Bohm theory from being an 'interpretation' to a mathematical reformulation of quantum mechanics equivalent in status to Feynman's.

4. Jul 13, 2009

malawi_glenn

Re: A silly question I'm sure about Feynman's "many paths"

I will, as I have stated many times, go for the default, Copenhagen interpretation/formulation is default and implicit when one says QM.

He specifically asked also for Feynman approach, not Bohm..

And it is not "my" meaningless stuff, it is the meaningless stuff inherited in the Copenhagen formulation of QM used by the majority of physics community. When will you stop with these personal assaults?

5. Jul 13, 2009

Count Iblis

Re: A silly question I'm sure about Feynman's "many paths"

Compute the amplitude:

<zenith8 stops assults(t)|zenith8 is assulting(t=0)>

using the path integral formalism.

6. Jul 13, 2009

Demystifier

Re: A silly question I'm sure about Feynman's "many paths"

This is certainly not true. Electron either takes exactly one path (as in the Bohmian interpretation) or does not take any path at all (as in the Copenhagen interpretation). What is true in both interpretations is that the wave function can be calculated such that certain quantity is mathematically calculated over all paths and that results obtained from different paths are summed up. However, this mathematical method for calculating the wave function (or more precisely, the propagator) has nothing to do with actual particle paths. In fact, this mathematical method can be used for solving any first-order linear differential equation, which, in general, may have nothing to do with quantum mechanics and particles.

Last edited: Jul 13, 2009
7. Jul 13, 2009

Demystifier

Re: A silly question I'm sure about Feynman's "many paths"

Zenith8, even though, as you probably know, I am also a supporter of the Bohmian interpretation, I must criticize the assertion that in the Bohmian approach one can calculate the propagator with one path only. Namely, to do that one must first know the quantum potential. But to calculate the quantum potential, one must first know the wave function. But to know the wave function, one must first calculate the wave function by a more conventional method, e.g., by the path-integral method that involves a sum over ALL paths with classical action. In this sense, the Bohmian approach with one path only does not simplify the calculation of the propagator.

8. Jul 13, 2009

Tac-Tics

Re: A silly question I'm sure about Feynman's "many paths"

Forum drama aside....

OP: When Feynman says "cancel out" he doesn't mean the paths themselves. He is talking about the amplitudes.

For each possible path an electron could take, there is an amplitude assigned to it.

An amplitude is a quantum probability, given as a complex number. Instead of saying something has a 50% chance of happening, you say it has an amplitude of $$\frac{1}{\sqrt{2}} e^{i \frac{\pi}{4}}$$. We use complex numbers because they are convenient, but intuitively, you should think of it as a "probability with a direction". When you square the length of an amplitude (its modulous or "absolute value"), you get the classical probability for it.

The key difference between a probability and an amplitude is how they add up. When you add classical probabilities, your chances always increase. If I have a 1% chance to win the lottery (and those are fantastic odds, btw!), I always increase my chances by buying more tickets. In QM, though, because amplitudes have a direction associated with them, adding them together doesn't always increase their length. In other words, when you play the quantum lotto, buying more tickets might increase you chances, but it could also lower them as well.

So back to physics. An electron moves. Each possible path has an amplitude. When you sum together the amplitudes for each path, you find that most of them end up canceling out. The paths that don't dictate the probability distribution of finding the particle in a given place. The details are encoded into the formulas given by the above posters, but I thought I'd put the layman's explanation out there, since it sounds like you aren't quite ready for the ugly details yet.

9. Jul 13, 2009

Demystifier

Re: A silly question I'm sure about Feynman's "many paths"

Such a claim is at best misleading. First, such a claim is meaningfull only within one (among many) formulations of QM. Second, many phenomena in classical physics (such as classical Brownian motion) can also be calculated with the path-integral method, so would you say that in classical physics all paths contribute as well? (Of course you wouldn't.)

Generally, it is not meaningless to ask this question, unless you assume that you know what is the correct interpretation of QM. But nobody knows that yet (even though many prefer some interpretations over the others), so this question is still meaningless and legitimate. The correct answer may be - neither (we do not know yet), but even if we knew that it was the right answer, the question would still be meaningfull.

10. Jul 13, 2009

haushofer

Re: A silly question I'm sure about Feynman's "many paths"

I also had a question about this here a time ago; maybe it's best to think about these paths in configuration space and think in terms of waves, not in terms of pinpointed particles.

11. Jul 13, 2009

malawi_glenn

Re: A silly question I'm sure about Feynman's "many paths"

Given the status of the OP, and that I always choose Copenhagen formulation, all paths will contribute weighted by the exponent of i times the action divided by hbar. Saying so does not rule out that Path integral formalism can be used for classical phenomenon, such as Brownian motion. I only tried to give an answer from the pragmatic point of view which is often the first view one starts with.

Same with the "meaningless", I always answer in Copenhagen, that is default and is what is taught in first classes in QM, which I hope and persume that the OP will do one one day.

I wonder what has made the OP really confused? This whole discussion of the different interpretations and formulations of QM perhaps?

12. Jul 13, 2009

Demystifier

Re: A silly question I'm sure about Feynman's "many paths"

Malawi_glenn, it is fine to be pragmatic and to assume the Copenhagen approach when one asks a practical question.
However, jaketodd clearly does not ask a practical question, but a conceptual one. A conceptual question requires a different type of answer. If you are not interested in conceptual questions (because you find them irrelevant or whatever) then you should leave answering such questions to others. Otherwise, you confuse non-expert readers who cannot easily distinguish between practical and conceptual questions and answers.

13. Jul 13, 2009

Hans de Vries

Re: A silly question I'm sure about Feynman's "many paths"

Indeed.

Technically it's not about "the paths which the particle takes" but about how
the wavefunction propagates. All the paths arise because each point of the
wavefunction acts as a new source for propagation.

In this regard the path integral does not depend on the interpretation of QM.

Regards, Hans

Last edited: Jul 13, 2009
14. Jul 13, 2009

Demystifier

Re: A silly question I'm sure about Feynman's "many paths"

I agree. The path integrals tell us about the wave aspects of quantum phenomena and do not much help to understand their particle aspects.

15. Jul 13, 2009

malawi_glenn

Re: A silly question I'm sure about Feynman's "many paths"

OP clearly asked about FEYNMANS "many-path idea", and I gave him the answer which can be found by reading about Feynmans own explanations of his "idea". Feynmans approach was "pragmatically conceptual" one could say. If someone ask about Feynmans way, why should I give him/her Bohms way?

Now to do this fair, one COULD mention that there are other ways to see this than Feynmans way of explaning this, like you guys did - introducing Bohmian way - but can we do this without personal assaults?

16. Jul 13, 2009

Demystifier

Re: A silly question I'm sure about Feynman's "many paths"

I'm sure we can.

17. Jul 13, 2009

malawi_glenn

Re: A silly question I'm sure about Feynman's "many paths"

So instead of calling my answer "confusing" one can say that the answer is not complete, since there are a lot more ways both to interpret QM and formulate it. And that the answer is accordance to the mainstream, and there included Feynmans original, interpretation and formulation.

But what we all agree on is that one should not think the paths as paths in space, like vectors, but rather something like "probability with direction" as Tac-Tics formulated it, which actually was the main question of the OP, how this "path-assignment" works in detail and conceptually.

18. Jul 13, 2009

Demystifier

Re: A silly question I'm sure about Feynman's "many paths"

In fact, I think Feynman originally believed that, in some weird sense, particles really DO take all these paths at once. But later he gave up of such a weird interpretation and accepted the path-integral method, as most physicists today do, merely as a calculation tool.

19. Jul 13, 2009

Demystifier

Re: A silly question I'm sure about Feynman's "many paths"

Well, it is certainly not my intention to insult you. Still, the claim that paths should be understood as "probabilities with direction" is still confusing to me. It's nothing personal against you, but such an explanation is confusing to me.

20. Jul 13, 2009

malawi_glenn

Re: A silly question I'm sure about Feynman's "many paths"

Well by assault I was mainly considering Zenith, who thinks that I am the only person in the world that uses Copenhagen as default.

Yeah, but that is Toc-Tics explanation, I found it good - but is perhatps that I am so used with this probabilistic chat around QM. What "is happening" (on the mathematical level) is that there is destructive interference..

Maybe one should wait til/if the OP returns, QM tends to become very mathematically

21. Jul 13, 2009

friend

Re: A silly question I'm sure about Feynman's "many paths"

Yes, I've also been told that every possible path in the Path Integral includes paths that are faster than light, and paths backwards in time. Is this true? Thanks.

22. Jul 13, 2009

malawi_glenn

Re: A silly question I'm sure about Feynman's "many paths"

23. Jul 13, 2009

jambaugh

Re: A silly question I'm sure about Feynman's "many paths"

But of course if you physically block some of the other paths you do not get the same dynamics. OK so you (re)interpret this with pilot waves. But then you can't really phenomenologically separate the electron from this pilot wave. So how can you then say "the electron took one path" when in fact the electron system (particle plus pilot wave) does not?

What's more there is no unique path for situations such as a symmetric double-slit trajectory. You rather get equal contributions form two paths. Does the electron split in half?

Other than satisfying some emotional need to recast quantum physics in terms of a classical object world picture what does tacking on pilot waves add to the physics?

Remember CI doesn't so much assert the absence of e.g. pilot waves or even multiple worlds. It rather insists on agnosticism about these theological speculations. It is the same as SR's agnosticism about the unobservable luminiferous aether, or science's general agnosticism about God & friends. Assertions about the nature of reality beyond the observable is a departure from the domain of science.

Finally your reasoning about the "proper path" is not much different from say my arguing that since the area under a smooth curve can be calculated by picking a specific point on that curve and calculating the rectangular area under a constant function with that value means that "we could say that the function really is a constant defined at this point". (Note my analogy addresses the form of reasoning not the relationship between the actual systems.)

Note as well that the Feynman many paths is understood to be a method of calculation and not an ontological assertion. You may hear it said that "the particle really takes all possible paths" but this is usually either a naive misreading of Feynman or more likely they're being loose with the language in the same sense as someone speaking about the "value of a function at infinity" when it is understood that they mean the limit of its value as one approaches infinity.

24. Jul 13, 2009

malawi_glenn

Re: A silly question I'm sure about Feynman's "many paths"

Maybe we should start a new thread where we can throw in this "interpretation" "formulation" issue again? I am definitely open for such a "battle" again, but not at the cost of this thread, I think it will be come to blurry overwhelming for the OP :-)

25. Jul 13, 2009

zenith8

Re: A silly question I'm sure about Feynman's "many paths"

Hi Demystifier,

I think you're missing the point actually, for several reasons (notation from my earlier post #3):

(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 $${\bf x}_0$$:

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

In the pilot-wave method 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 don't need to know the wave function over all space to compute the propagator. You just need the second derivative of the wave function - or more accurately the second derivative of its amplitude (for the quantum potential) 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.

(3) The point of my original post was anyway not to claim the superiority of the de Broglie-Bohm method over Feynman's method. I was just trying to make a conceptual point i.e. isn't it interesting that you can achieve exactly the same end result (calculating the wave function at some point in the future) by either Feynman's summing over the integrals along the infinite number of all possible trajectories in the entire universe to calculate the propagator then summing over all possible propagators so calculated for the infinite number of all possible different starting points or by er.. de Broglie-Bohm's integrating along one single track. And the latter track must be the actual path of the particle (for the quantum probability current to make sense). From this point of view it doesn't matter whose method is more numerically tractable. It's just highly interesting.

In actual fact Feynman's paths are mathematical tools for computing the evolution of $$\Psi$$, while (if 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 $$\Psi$$. Keep 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.

Note also that in one philosophical interpretation of quantum mechanics, the 'sum over histories' interpretation, the path integral is taken to be fundamental and reality is viewed as a single indistinguishable 'class' of paths which all share the same events. Given the above, you have to say Hmm to that..