Is the Path Integral Formulation of Quantum Mechanics Violating the Speed Limit?

[ahmedcoe]

As i understand as a solution to the double slit experiment is the path integral formulation.

Since a particle fired at one slit will interfere with its all other trajectories and will formulate that pattern we all know, doesn't this imply that information is exchanged between it and all those other possible paths which some of them can be on the other side of the galaxy. Doesn't this mean that the speed limit is violated in some manner.

 

[sheaf]

I asked a similar question a long time ago (although the thread title wasn't particularly good). The discussion is https://www.physicsforums.com/showthread.php?t=152390", maybe some of it is relevant.

 

[K^2]

Particle doesn't decide based on path integration. Particle actually takes all possible trajectories, the same way a wave does. Away from physical path, the action is not extreme, and so a small change in path causes an almost random change in phase. That means phases from all path contributions at these points interfere, and you end up with probability of detecting particle there being zero. Along the path of extreme action, small variations in path do not change the phases. That means that all paths interfere constructively. That produces a finite probability amplitude, allowing detection of the particle. There is no interaction between possible paths except at a point, so no information is actually being transmitted, and light cone never enters into consideration.

 

[Demystifier]

Doesn't this mean that the speed limit is violated in some manner.

Yes it does. However, one should have two facts in mind:

1. No real particles actually move along these trajectories appearing in the path integral. The path integral is merely a mathematical tool useful in many linear systems with first derivatives. This includes quantum mechanics, classical statistical mechanics, and even finance.

2. The trajectories that violate the relativistic speed limit are not solutions of the classical equations of motion. Thus, they are not in conflict with the theory of relativity which (assuming that mass squared cannot be negative) says that no solution of classical equations of motion can exceed the velocity of light.

 

[K^2]

Yes it does.

No, it doesn't. The basic propagator is contained within a light cone, so all of the paths are valid world lines.

 

[Demystifier]

No, it doesn't. The basic propagator is contained within a light cone, so all of the paths are valid world lines.

The propagator is contained within a light cone, but the path integral does contain trajectories the pieces of which exceed the velocity of light. You seem to be missing the fact 1. in my previous post.

 

[K^2]

If the propagator is contained within the light cone, there is no path that exceeds speed of light locally that enters the path integral. If you really need me to explain this in detail, I will, but it might be simpler if you just sit down and think about it. Try using Dirac equation as an example.

 

[Demystifier]

If the propagator is contained within the light cone, there is no path that exceeds speed of light locally that enters the path integral. If you really need me to explain this in detail, I will, but it might be simpler if you just sit down and think about it. Try using Dirac equation as an example.

Please, explain it to me in detail! Thanks!

 

[K^2]

Suppose you have a simple propagator S(x-y), such that S(y-x)=0 whenever y-x is space-like. Any path taken can be written as a sequence of points from origin to destination. Amplitude/phase shift for this path is S(x₁-x₀)*S(x₂-x₁)*...*S(x<sub>n-xn-1). In general, the number of points is uncountable, so pardon my notation. Do I need to explain why in order for speed to be faster than speed of light at some point along the path, xi-xi-1 has to be space-like for some i? If you accept that, corresponding S(xi-xi-1)=0, and amplitude of the path is zero. Id est, it does not contribute.

Is that clear, or do I need to get into more detail?</sub>

 

[Demystifier]

K^2, your argument makes sense, but I am still confused.
In particular, you say

... amplitude of the path is zero.

But amplitude of the path q(t) is
exp{iA[q(t)]}
where A[q(t)] is the classical action of the path q(t). How THAT amplitude can be zero (for faster than light trajectory q(t))?

Let me try to solve this puzzle by myself:
- If we consider the Dirac particle, then there is no such thing as classical action for the particle, because there is no spin in the classical theory. Therefore, my argument above cannot be applied.
- If we consider a spinless particle, then this particle is described by the Klein-Gordon equation. But this is an equation with the second-time derivative, so it is questionable if it can be formulated in the path-integral form in the same sense in which Schrödinger-like equations can.
- If we forget about first quantization and go to quantum field theory, then the functional-integral formulation is well understood. However, this is an integral over field configurations, so particle paths do not occur at all.

In any case, to derive the path integral one needs a resolution of the unit operator, something like
1=sum_q |q><q|
However, in the relativistic case it is not clear what the complete basis {|q>} is.

Perhaps the most sensible possibility is to bypass the resolution of identity and to DEFINE the theory through a path integral from the very beginning. Indeed, this method of quantization is widely used in string theory and can be modified for relativistic particles too. (See the attachment!) But in THIS approach your argument cannot be applied and my argument at the top of this post is correct; even faster-than-light paths contribute.

What are your thoughts?

 

[K^2]

Good points. Any chance that A[q(t)] is imaginary for illegal paths when relativity is taken into account? It wouldn't be zero, but it can go to zero. Which would also work with some limited non-locality of propagators. Tunneling, after all is non-local.

I'm going to go ahead and try it with a Lagrangian that actually makes sense for some real field and see if I can come up with a better answer.

 

[Demystifier]

Any chance that A[q(t)] is imaginary for illegal paths when relativity is taken into account?

Yes, if you use action with a square root.

It wouldn't be zero, but it can go to zero. Which would also work with some limited non-locality of propagators.

For some faster-than-light trajectories it will be far from zero, so it doesn't help much.
But of course, it is well known that classical trajectories (which, of course, are not faster than light) have the largest contribution to the path integral. The contributions of other trajectories partially cancel out. The cancellation is exact in the limit h->0.

 

[K^2]

That still makes no sense with propagator formulation. I should be able to get the same result regardless of whether I'm getting phase by integrating over action or if I'm getting phases by multiplying propagators together.

 

[Demystifier]

See the pages 4-5 in the attached paper above. There the expected relativistic propagator is derived from the path integral. However, the contour of integration around poles at p^2=m^2 is not specified. As is well known, certain contours lead to propagators that are small BUT NOT VANISHING at spacelike separations. This suggests to me that perhaps your assumption that the propagator should strictly vanish at spacelike separations should be relaxed. What do you think?

 

[K^2]

If the propagator does not vanish outside the light cone, yes, the particle obviously takes paths that violate locality. But then the overall signal can violate causality.

Honestly, I'm just being lazy. I should sit down with Dirac's equation and just derive all that. I'll try to do that over the weekend, see what I come up with.

 

[Demystifier]

If the propagator does not vanish outside the light cone, yes, the particle obviously takes paths that violate locality. But then the overall signal can violate causality.

I think one should distinguish a more general concept of a propagator from a less general notion of a Green function. Here by Green function I mean a function used to solve the wave equation through the Green-function method. Clearly, the Green function in this sense MUST be causal, because the wave equation is. On the other hand, the propagator can be defined through ANY contour around the poles, and some choices of the contour may lead to acausal propagators.

Having this in mind, perhaps in relativistic QM the path integral gives a propagator, but not a Green function.

 

[K^2]

The ambiguity is only in which poles you include/exclude. Dirac's equation only has two. They give you the particle and anti-particle propagators. That's it. And both of these also work as Green's functions for the wave equation.

 

[Demystifier]

The ambiguity is only in which poles you include/exclude. Dirac's equation only has two. They give you the particle and anti-particle propagators. That's it. And both of these also work as Green's functions for the wave equation.

You have not counted all the possibilities. You may also include them both. Or exclude them both. Finally, you may take superpositions of these 4 propagators.

See also
http://en.wikipedia.org/wiki/Propagator#Relativistic_propagators

 

[The_Duck]

I haven't gone through this thread carefully, but I am a bit confused about what is being discussed. As Demystifer said in post #10, isn't the path integral in QFT over field configurations and not particle paths? If you are doing a path integral over particle trajectories aren't you doing regular quantum mechanics and not QFT? And doesn't this have the well-known problem that particles can propagate faster-than-light, so that you don't get a consistent relativistic theory? E.g. it seems common in introductions to QFT to talk about the attempt to get a relativistic quantum theory from the Hamiltonian

$$\hat{H} = \sqrt{\hat{p}^2 + m^2}$$

But if you take this as your Hamiltonian in single-particle QM your particle can propagate faster than light. If you are doing a path integral over particle trajectories with the paths weighted by the classical Lagrangian for a free relativistic point particle, which is the same as the above Hamiltonian, your paths will not cancel out at spacelike separations.