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I'm interested in how and how much classically-forbidden paths contribute to a path integral? Is there any good reference on the issue? Any discussion in QM or QFT context would be appreciated.

EDIT: I have a more specific question than the above but initially it was a reference request so I decided to make it more general. Now it doesn't seem appropriate any more so let me reformulate the question:

First of all let me apologize for the bad terminology. By "classically forbidden" I actually meant "differentiable"(i.e. for QM differentiable in time direction, for QFT in both space and time direction) instead of "forbidden by classical dynamics".

My motivation comes from path integral of QED, if we only integrate the fermionic degrees of freedom under some smooth gauge field, we will get a quantized theory of many electrons with a classical gauge background, and the fully quantized theory will emerge after we also integrate over gauge fields. This seems to be a reasonable way of thinking, but some of my subsequent derivations seem to suggest some quantum effects will disappear such as photon-photon scattering, but something is still preserved like the many-body feature of QFT(I'm not very sure about calculation yet so I'd rather not show it here). It occured to me it might be because I'm only including smooth backgrounds.

This motivates me to ask, what exactly is the role of smooth and non-smooth paths in path integral? Do they result in different and isolated features of QFT so that it's ok to consider them separately, or do their effects just mix with each other so that we always have to consider them as a whole?

Last but not least, the comments and answers below remind me of another question, if the classical path(this time I mean path predicted by classical dynamics) always contributes 0 to the path integral for any value of $\hbar$, then what do we mean by saying the classical path will dominate in $\hbar\to0$ limit? After all a simple fact of math is that a sequence of 0's cannot give you a limit of 1.

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What do you mean by classically forbidden? 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. –  user1504 Jan 17 '13 at 21:59
    
@Gugg: I think your answer looks correct, but the question seems a little vague. (Not sure why someone downvoted it. Have an upvote.) 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. –  user1504 Jan 17 '13 at 22:57
    
Dear @Jia Yiyang. 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. In any case, the 4 answers so far have paid no attention to your later revisions $\geq 3$, and only few readers would associate classically forbidden with non-smoothness. After a roll-back, you could post your remaining question about non-smoothness in a new separate post using correct terminology. –  Qmechanic Feb 8 '13 at 22:46
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4 Answers 4

Other people have already addressed quantum mechanics, so let me comment on the field theory case.

In all of the QFTs which have been rigorously constructed, in spacetime dimension 2 & 3, the Euclidean path integral is supported on a space of distributions. The set of continuous classical fields sits inside this space of distributions, but it has measure zero with respect to the path integral. I see no reason to expect the QFTs that describe real world physics to be any better behaved. The path integral measure has to be supported on distributions to give an OPE with short distance singularities.

So yes, just summing over classical fields will probably not give you a good approximation to the path integral.

The only reference I know on this stuff is Glimm & Jaffe. (There may be more accessible references somewhere in the literature. I just don't know them.)

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So if we ignore all the differentiable paths, the physics remains the same? These just all seem too fuzzy to me, I can't help wondering if there's some subtlety. –  Jia Yiyang Jan 21 '13 at 16:43
    
It's really a physical thing. If we didn't ignore the differentiable paths, we'd find that the commutation relations for position and momentum were altered. –  user1504 Jan 21 '13 at 17:22
    
I'm more confused, are you saying not just that differentiable paths can be ignored, but they have to be ignored? Or else the commutation relation will be wrong? –  Jia Yiyang Jan 22 '13 at 6:00
    
Yes. The commutation relations can be derived from the path integral. 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.) –  user1504 Jan 22 '13 at 14:36
    
I see what you mean, and I agree with the commutation relation part. However, does the fact that differentiable paths are of measure 0 necessarily implies these paths contribute nothing to the physics? One thing for example, how does the classical path (which is differentiable) emerge from the theory in $\hbar\to0$ limit? –  Jia Yiyang Jan 22 '13 at 19:06
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Unfortunately, after the edit to the question, this answer only refers to the final "last but not least" part of it.

In certain experiments there is no classical path that leads to the outcome. In such cases classically-forbidden paths contribute 100% to the path integral. See Feynman's QED, fig. 27. The picture is on the internet.

The reverse "might" be said of (most) macroscopic objects (in particular not QM-experimental set-ups). There the classical path would in itself describe the total path integral. (But they would't contribute. See below.)

But I take it that whenever there are non-classical paths, you could always take out the exact classical paths and still end up with the same path integral, because the classical paths would have measure $0$.

So that would mean that, if there are (a measurable amount of) non-classical paths, they contribute 100%.

But there are always (a measurable amount of) non-classical paths, therefore:

Classically forbidden paths contribute 100% to a path integral.

(Plenty of handwaving here, e.g., you might want to show that all non-classical paths together don't destructively interfere. But they don't, because then there wouldn't be any amplitude left.)

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. (And in the limit, amplitudes are amplitude densities.) To me, "dominate" appears to be the wrong word. I think it is the other way around: The nearby paths determine the classical path (in situations where it is OK to use CM).

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I don't know if I understand the picture correctly, it seems to suggest all the paths sum up to the classical path, and it tells us nothing about the how the transition from quantum to classical happens as $\hbar\to 0$ –  Jia Yiyang Jan 21 '13 at 16:49
    
In Fig. 32, the path sum (of amplitudes) consists of 7 paths represented by the 7 small arrow sticked together in the picture on the right. The little arrow $D$ represents the classical path. 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. [nxt cmt] –  Glen The Udderboat Jan 21 '13 at 22:27
    
In particular, the arrow that represents the classical path will decrease in amplitude (and it will stay roughly where it is). In the limit, the amplitude of that arrow will have zero contribution, but we still have the "S" shape. So, we still have full amplitude for moving from $S$ to $P$, which means that the "illusion" of CM seems retrieved, but it remains an illusion since the contribution of the classical path $D$ has measure zero... I hope that helps. –  Glen The Udderboat Jan 21 '13 at 22:33
    
The point being: Once your in the limit, you can remove the arrow associated with the classical path from the "S" shape, and it wouldn't change the "S" shape. (And keep in mind that sometimes no classical path even exists.) –  Glen The Udderboat Jan 21 '13 at 22:39
    
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I think I know what you are thinking. Here's how it goes: in classical mechanics, we have a Lagrangian describing the system. Our principle of least action says that the system will follow a path that extremizes $S = \int L dt$. This amounts to taking the E-L equations on it. The resulting path is called the classical path and we say that it is the only path that the system follows.

In QM, the Feynman path integral says, let us talk about the amplitude for a particle to go from $a$ to $b$. Let us call this $K(b,a)$. Heuristically this is given by $K(b,a) = \sum_{all paths} \phi[x(t)]$, where the contribution of each path that goes from $a$ to $b$ is $\phi[x(t)] = \text{const} e^{i/\hbar S[x(t)]}$, where $S[x(t)]$ is the classical action of that path.

So to answer your question, how much do classically forbidden paths contribute (and I'll add, to the transition amplitude)? It's just $e^{i/\hbar S[x(t)]}$.

Now how do we recover classical mechanics? Send $\hbar \to 0$, and we see that the most contribution to the sum comes from the smallest value of $S$, as $e^{i/\hbar S[x(t)]}$ oscillates the least. (This is, or is like, the saddle point approximation). The path that gives the most contribution is the classical path, and it is that path that the particle takes classically.

Of course that wasn't very rigorous, but I would recommend 'Quantum mechanics and path integrals' by the guy who came up with this himself, Richard P. Feynman.

Cheers.

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Doesn't this assume that there is a classical path to the outcome? –  Glen The Udderboat Jan 17 '13 at 22:38
    
No? since when? 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. there are systems where we start with some Lagrangian which doesn't have a classical analogue, but the path integral formalism still holds. –  nervxxx Jan 17 '13 at 22:57
    
Since "Now how do we recover classical mechanics?" seems to assume that CM always holds. 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.") –  Glen The Udderboat Jan 17 '13 at 23:21
    
It doesn't have to be the "smallest value of S", classical paths must only be extrema of the action functional –  Raphael R. Jan 18 '13 at 4:08
    
lol you guys are reading too much into this. 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. i was just showing heuristically how it does so. –  nervxxx Jan 18 '13 at 4:23
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QM deals with waves, and for waves every point of space takes part in creating the resulting wave. The same is valid for a path-integral formulation. It is difficult to present a general weight of these or those paths. It depends.

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