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Design and Development By. My Cart - 0 item. Add to Cart. Description Tags Reviews 0 Additional Table of contents -. Product Tags Add Your Tags:. About Market. The same is valid for a path-integral formulation. It is difficult to present a general weight of these or those paths.
It depends. Sign up to join this community. The best answers are voted up and rise to the top. Home Questions Tags Users Unanswered. What's the role of classically forbidden paths in path integral?
Ask Question. Asked 6 years, 8 months ago. Active 6 years, 7 months ago. Viewed times. Now it doesn't seem appropriate any more so let me reformulate the question: First of all let me apologize for the bad terminology.
Jia Yiyang. Jia Yiyang Jia Yiyang 2, 1 1 gold badge 16 16 silver badges 39 39 bronze badges. In quantum mechanics, for example, the set of classical paths is a set of measure zero for the path integral. In fact, even the set of differentiable paths has measure zero. Not sure why someone downvoted it. Have an upvote.
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My comment isn't technically an answer: I asked Jia to clarify the question because I wasn't completely sure if we're discussing support of path integral measures or something like tunnelling. I suggest to roll back the question formulation to version 2. The notion of classically forbidden traditionally refers to quantum tunneling, but here I would if I had no prior information assume it refers to situations where there are no classical paths.
After a roll-back, you could post your remaining question about non-smoothness in a new separate post using correct terminology. These just all seem too fuzzy to me, I can't help wondering if there's some subtlety. If we didn't ignore the differentiable paths, we'd find that the commutation relations for position and momentum were altered. Or else the commutation relation will be wrong? The commutation relations can be derived from the path integral.
In the Fourier transform, this means shifting the pole in p 0 slightly, so that the inverse Fourier transform will pick up a small decay factor in one of the time directions:. And yes, it's true that it should be more appropriately named principle of stationary action, but I think for many cases the action functional also happens to be the least. Hot Network Questions. The constraint 12 was motivated by 't Hooft by information loss . But they would't contribute. Our Awards Booktopia's Charities. The price of a path integral representation is that the unitarity of a theory is no longer self-evident, but it can be proven by changing variables to some canonical representation.
If the path integral gives non-zero weight to the differentiable paths, the commutation relations will be altered. The portion of the measure supported on differentiable paths predicts that position and momentum commute. Non-differentiability of paths is essential for deriving the commutation relations.
However, does the fact that differentiable paths are of measure 0 necessarily implies these paths contribute nothing to the physics? If you're asking about whether there will be solutions to the EL eqn, I think that's given by some regularity conditions of the Lagrangian. And, more importantly, even when CM can be recovered, it doesn't mean that the classical path contributes to the path integral measurably.
And I agree with "[…] the path integral formalism still holds.
Jan 18 '13 at I did say it's not very rigorous, but I was trying to convey the idea of the path integral in the spirit that Feynman might have. And yes, it's true that it should be more appropriately named principle of stationary action, but I think for many cases the action functional also happens to be the least. With regard to recovering CM, the path integral should, or rather, must reduce to CM in the cases when the Lagrangian is 'derived' from a classical system.
Added after the edit to the question: I gather this picture, also taken from Feynman's QED , should clear up the "last but not least" part. Keep these mind Keep these mind 5, 2 2 gold badges 24 24 silver badges 51 51 bronze badges. As you start to move towards the path integral the number of paths and therefore little arrows will increase, however combined they will keep roughly forming the same "S" shape. Therefore, more arrows means that they each become smaller, that is, their individual contribution to the path sum will decrease. In the limit, the amplitude of that arrow will have zero contribution, but we still have the "S" shape.
I hope that helps. And keep in mind that sometimes no classical path even exists. Vladimir Kalitvianski Vladimir Kalitvianski Sign up or log in Sign up using Google. Sign up using Facebook. Sign up using Email and Password. Post as a guest Name.
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