Feynman Path Integral

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In a Feynman Path integral, [tex]Z(\phi) = \int \cal D \phi[/tex] [tex] e^{\frac{iS(x)}{\hbar}}[/tex], what does the object [tex]e^{\frac{iS(x)}{\hbar}}[/tex] mean?
 
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fzero

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It's more precise to note that the action [tex]S[/tex] is a functional of the fields [tex]\phi[/tex] and write [tex] S[\phi][/tex] instead of [tex]S(x)[/tex]. The factor [tex] e^{\frac{iS[\phi]}{\hbar}} [/tex] is a phase that provides a weight to each path. In this way paths that tend to minimize the classical action contribute the most to physical amplitudes.
 
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Thanks, for the response. I don't really understand all of the "jargon" since I haven't taken a QM course. I was just curious, Is there any other way to reword it? If not then don't worry about it, I'm going to start reading Roger Penrose's book and I know there is a small section on sum over paths (histories).
 

fzero

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It's tough to give a layman's explanation without drawing lots of pictures, but I can point you to some references. There's a small book by Feynman called "QED: The strange theory of light and matter" which is aimed at a popular audience but still does a great job of keeping the facts intact. There's also a higher-brow discussion at http://www.quantumfieldtheory.info/Path_Integrals_in_Quantum_Theories.htm which is aimed at someone that understands QM but I think the graphic explanation in section 6 there might be ok to follow, since it's based on Feynman's explanation from that book.
 
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901
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I bought QED but haven't started it yet, I'm going to finish Cosmos first, thanks for the insight.
 
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Is it correct to say that [tex]e^{\frac{iS[\phi]}{\hbar}}[/tex] is similar to the traditional formula [tex]e^{ix} = cosx + isinx[/tex] except now x is a functional (action) and there is a factor of [tex]\frac{1}{\hbar}[/tex] do to quantization?
 

fzero

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That formula is mathematically correct, but it's not really used very much in this context. The factor of [tex]1/\hbar[/tex] isn't exactly due to quantization. In the most basic form of the path integral, the paths [tex]\phi[/tex] are all classical solutions [tex]\phi_{cl}[/tex] to the equations of motion that follow from the classical action [tex]S[/tex]. [tex]S[\phi_{cl}][/tex] isn't an observable of the theory and isn't necessarily quantized. In the path-integral formulation, the quantization occurs after the path integral is taken. However the action does have units which are the same as [tex]\hbar[/tex], so it is useful to compute [tex]S/\hbar[/tex] as a dimensionless quantity that we can put into an exponent.

It's a good question, but I'm not sure that my answer will really seem sufficient without more background about what the action means in classical mechanics, not to mention the framework of the path integral itself.
 
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Yea, don't worry about it. Not many people know the actual idea of what a Path Integral is until graduate school so I think your explanation will suffice.
 

A. Neumaier

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Is it correct to say that [tex]e^{\frac{iS[\phi]}{\hbar}}[/tex] is similar to the traditional formula [tex]e^{ix} = cosx + isinx[/tex] except now x is a functional (action) and there is a factor of [tex]\frac{1}{\hbar}[/tex] do to quantization?
The factor is there to make the exponent dimensionless, since S and hbar both have the dimension of action, and the exponential of a dimensional quantity doesn't make sense.

The path integral itself is an infinite-dimensional generalization of integrals over the real line such as those occurring in the definition of a Gaussian distribution, except that the fact that the exponent is purely imaginary makes the integral somewhat difficult to define.

The integral of e^{ix^2} over R is neither Riemann integrable nor Lebesgue integrable,
and Henstock integrability, which fits here, is not common knowledge. In infinite dimensions, the difficulties are much worse....

Therefore one often (e..g., in lattice simulations of QFT) uses so-called Euclidean path integrals, where i is replaced by -1 to make it mathematically tractable. This can be justified to some extent by analytic continuation.
 
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The integral of e^{ix^2} over R is neither Riemann integrable nor Lebesgue integrable,
and Henstock integrability, which fits here, is not common knowledge. In infinite dimensions, the difficulties are much worse....

Therefore one often (e..g., in lattice simulations of QFT) uses so-called Euclidean path integrals, where i is replaced by -1 to make it mathematically tractable. This can be justified to some extent by analytic continuation.
When you replace i by -1, paths with high action no longer vary wildly in phase and hence cancel, but instead become suppressed.

This to me is radically different. In the former case, there is a collective canceling of high action paths. In the latter case, there is individual suppression of each individual high action path.

Is this the right interpretation of the difference between working in Minkowski space and Euclidean space?

For example, take a two slit experiment. Why do you only have to sum over those two slits, and not consider paths that go through the wall? In the Minkowski interpretation, you do add all the paths that go through the wall, but since the action is high there due to a high potential, they all cancel. In the Euclidean intepretation, each path that goes through the wall is suprresed by the high potential energy (in Euclidean space the Lagrangian pretty much turns into the Hamiltonian, so [tex]e^{-S_E}=e^{-\int L_E d\tau}=e^{-\int H d\tau} [/tex]).
 

A. Neumaier

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When you replace i by -1, paths with high action no longer vary wildly in phase and hence cancel, but instead become suppressed.

This to me is radically different. In the former case, there is a collective canceling of high action paths. In the latter case, there is individual suppression of each individual high action path.

Is this the right interpretation of the difference between working in Minkowski space and Euclidean space?
Yes. This means that the latter is well-behaved and can be studied by much more powerful techniques than the former, where one mainly relies on purely formal manipulations that are known to be sometimes wrong in much simpler situations.

if you replace the i by a variable j then you can interpolate between the regimes. For j=-1 one has the Euclidean integral and as one shifts more and more of the real part to the imaginary part one tends to the Minkowski integral. In the limit j-> i one gets the latter.
This is called analytic continuation, and in all cases where it can be proved to be valid, things are fine. The precise conditions are determined by the Osterwalder-Schrader theorem.
 
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Yes. This means that the latter is well-behaved and can be studied by much more powerful techniques than the former, where one mainly relies on purely formal manipulations that are known to be sometimes wrong in much simpler situations.
It seems to me that [tex]e^{\int -\frac{\mathcal H}{\hbar}d\tau}[/tex] is much easier to understand than [tex]e^{\int i\frac{\mathcal L}{\hbar}dt}=e^{iS} [/tex] when you take the limit [tex] \hbar[/tex] goes to zero (i.e., when the action is really big compared to [tex]\hbar [/tex] ), yet most popular books insist on the latter and talk about phase cancellations at large actions (Minkowski) instead of supression because of too much energy on the path (Euclidean).

if you replace the i by a variable j then you can interpolate between the regimes. For j=-1 one has the Euclidean integral and as one shifts more and more of the real part to the imaginary part one tends to the Minkowski integral. In the limit j-> i one gets the latter.
This is called analytic continuation, and in all cases where it can be proved to be valid, things are fine. The precise conditions are determined by the Osterwalder-Schrader theorem.
Actually if you take the Hamiltonian H in the path integral and replace it by [tex]H(1-i\epsilon)[/tex] for infinitismally small positive [tex]\epsilon [/tex], then you get the [tex]i\epsilon [/tex] prescription which can take the place of a Wick rotation. Is this the same thing as your variable complex variable j?
 

A. Neumaier

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It seems to me that [tex]e^{\int -\frac{\mathcal H}{\hbar}d\tau}[/tex] is much easier to understand than [tex]e^{\int i\frac{\mathcal L}{\hbar}dt}=e^{iS} [/tex] when you take the limit [tex] \hbar[/tex] goes to zero (i.e., when the action is really big compared to [tex]\hbar [/tex] ), yet most popular books insist on the latter and talk about phase cancellations at large actions (Minkowski) instead of supression because of too much energy on the path (Euclidean).
This is actually done in lattice QFT.

But of course it alters the whole theory - the symmetry group is then SO(4) in place of the Lorentz group SO(1,3)! Thus all results must be reinterpreted at the end via analytic continuation (hoping that it is valid). That's why one _must_ start with the Minkowski version.


Actually if you take the Hamiltonian H in the path integral and replace it by [tex]H(1-i\epsilon)[/tex] for infinitismally small positive [tex]\epsilon [/tex], then you get the [tex]i\epsilon [/tex] prescription which can take the place of a Wick rotation. Is this the same thing as your variable complex variable j?
j=1-i eps (roughly - but it must be applied to the action, not to H). For tiny eps (and the prescription requires the limit eps-->0), this makes the integral only marginally better behaved than j=1. Only continuation to real j removes all oscillations.
 

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