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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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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.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?
When you replace i by -1, paths with high action no longer vary wildly in phase and hence cancel, but instead become suppressed.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.
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.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?
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).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.
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?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.
This is actually done in lattice QFT.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).
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.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?