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friend
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Where can I find an On-line exposition of the undefined nature of the measure in the Feynman path integral? Thanks.
AEM said:Just curious, what do you mean by the phrase "the undefined nature of the measure" ?
Wiki has a bit more:friend said:I hear that the measure of the path integral, which seems to be a measure on the space of paths is not well defined. Or that maybe the infinite dimensional measure, dx1*dx2*dx3*...*dxn is not well defined. But that's all the explanation I seem to be able to get.
(The last bit is important in the case of the path integral over a nonseparable space of paths.)Let (X, || ||) be an infinite-dimensional, separable Banach space. Then the only locally finite and translation-invariant Borel measure μ on X is the trivial measure, with μ(A) = 0 for every measurable set A. Equivalently, every translation-invariant measure that is not identically zero assigns infinite measure to all open subsets of X.
(Many authors assume that X is separable. This assumption simplifies the proof considerably, since it provides a countable basis for X, and if X is a Hilbert space then the basis can even be chosen to be orthonormal. However, if X is not separable, one is still left with the undesirable property that some open sets have zero measure, so μ is not strictly positive even if it is not the trivial measure.)
strangerep said:Wiki has a bit more:
http://en.wikipedia.org/wiki/There_is_no_infinite-dimensional_Lebesgue_measure
(The last bit is important in the case of the path integral over a nonseparable space of paths.)
friend said:I'm not see the nonseparable nature of the space of paths. It sounds like you may have more information than you are telling us here. This may be a nice start. But at least for me it needs to be fleshed out a bit more. Thank you.
friend said:I'm not see the nonseparable nature of the space of paths. [...]
No, it's ill-defined if n is infinite. That's the main point of the extract fromfriend said:So what about the dx1dx2...dxn. Is this a well defined measure, even if n is infinite?
No. A path is just a particular kind of function. If we have a space of such functions whichAre we talking about an ill-defined measure only when we interpret this as a space of paths?
Fra said:What they attempt is to constrain the integration space to what they think are the set of PHYSICAL possibilities, so that all so called non-physical paths have a zero measure, so these possibilities need not even be included in the integral. This as a general idea is IMHO PHYSICALLY sound, but I think they fail to produce are argument to show the uniqueness and observer independence of this choice.
/Fredrik
strangerep said:No, it's ill-defined if n is infinite. That's the main point of the extract from
http://en.wikipedia.org/wiki/There_is_no_infinite-dimensional_Lebesgue_measure
which I quoted earlier.
friend said:I see an inconsistency here. They are assuming before hand what is physical and not letting the math guide them to what is physical.
strangerep said:Wiki has a bit more:
http://en.wikipedia.org/wiki/There_is_no_infinite-dimensional_Lebesgue_measure
Let (X, || ||) be an infinite-dimensional, separable Banach space. Then the only locally finite and translation-invariant Borel measure μ on X is the trivial measure, with μ(A) = 0 for every measurable set A. Equivalently, every translation-invariant measure that is not identically zero assigns infinite measure to all open subsets of X.
friend said:I see an inconsistency here. They are assuming before hand what is physical and not letting the math guide them to what is physical.
Fra said:Some more comments.
Part of the measure issue IMO, is exactly that all "mathematically possible paths" aren't necessarily "measureable" from a physical point of view. the question is of course, out of the set of all mathematics, which possibilites are the physical ones (ie the ones we should COUNT)?
/Fredrik
friend said:I see an inconsistency here. They are assuming before hand what is physical and not letting the math guide them to what is physical.
It can make a big difference depending on whether N is "huge but finite" compared to "infinite".friend said:Yes, but we are only talking about processes that allow the dimensionality to approach infinity; the dimensionality is never allowed to actually be infinity. Does that make a difference?strangerep said:[...] it's ill-defined if n is infinite.
jostpuur said:This is not relevant for the topic because path integrals do not rely on translation-invariant measures.
It is possible to define well defined measures on sets such as [itex][0,1]\times [0,1]\times\cdots[/itex] and [itex]\mathbb{R}\times\mathbb{R}\times\cdots[/itex]. Often product topologies are used, and measures are Borel with respect to them.
The last bit is important. I'll write the above in a different way to emphasize somethingfriend said:And as a measure the Dirac delta function satisfies the relation,
[tex]\[
\int_{ - \infty }^{ + \infty } {{\rm{f(x}}_1 {\rm{)\delta (x}}_1 {\rm{ - x}}_0 ){\rm{dx}}_1 } = {\rm{f(x}}_0 )
\]
[/tex]
for all continuous, compactly supported test functions, f(x).
At this point you abandon mathematical rigor because the delta distributionNow, it occurs to me that if we let [tex]\[
{\rm{f(x}}_1 ) = {\rm{\delta (x - x}}_1 )
\]
[/tex]
This only works because you abused some notation. One can of course write insteadthen
[tex]\[
\int_{-\infty }^{+\infty} {{\rm{\delta (x - x}}_1 {\rm{)\delta (x}}_1 {\rm{ - x}}_0 ){\rm{dx}}_1 } = {\rm{\delta (x - x}}_0 )
\]
[/tex]
The above should be rewritten with the integrals separated, and distinct [itex]\Delta t[/itex]'sIf we let
[tex]\[
\Delta = \left[ {\frac{{2i\hbar \left( {t - t_0 } \right)}}{m}} \right]^{1/2}
\]
[/tex]
in the gaussian form of the Dirac delta function, then we can get,
[tex]\[
{\rm{\delta (x - x}}_0 ) = \mathop {\lim }\limits_{t \to t_0 } \left[ {\frac{m}{{2\pi i\hbar (t - t_0 )}}} \right]^{1/2} \exp \left[ {\frac{{im(x - x_0 )^2 }}{{2\hbar (t - t_0 )}}} \right] = \mathop {\lim }\limits_{t \to t_0 } \left[ {\frac{m}{{2\pi i\hbar (t - t_0 )}}} \right]^{1/2} \exp \left[ {\frac{{im}}{{2\hbar }}(\frac{{x - x_0 }}{{t - t_0 }})^2 (t - t_0 )} \right]
\]
[/tex]
[tex]\[
= \mathop {\lim }\limits_{\Delta t \to 0} ({\textstyle{m \over {2\pi i\hbar \Delta t}}})^{1/2} {\mathop{\rm e}\nolimits} ^{(\frac{{im}}{{2\hbar }}(\frac{{\Delta x}}{{\Delta t}})^2 (\Delta t))} = \,\,\,\,\mathop {\lim }\limits_{\Delta t \to 0} ({\textstyle{m \over {2\pi i\hbar \Delta t}}})^{1/2} {\mathop{\rm e}\nolimits} ^{(im(\dot x)^2 \Delta t/2\hbar )}
\]
[/tex]
And when this is substituted in the nth iteration of the recursion relation above, we get,
[tex]\[
\int_{ - \infty }^{ + \infty } {\int_{ - \infty }^{ + \infty } {...\int_{ - \infty }^{ + \infty } {({\textstyle{m \over {2\pi i\hbar \Delta t}}})^{1/2} {\mathop{\rm e}\nolimits} ^{(im(\dot x_n )^2 \Delta t/2\hbar )} ({\textstyle{m \over {2\pi i\hbar \Delta t}}})^{1/2} {\mathop{\rm e}\nolimits} ^{(im(\dot x_{n - 1} )^2 \Delta t/2\hbar )} ...{\rm{ }}({\textstyle{m \over {2\pi i\hbar \Delta t}}})^{1/2} {\mathop{\rm e}\nolimits} ^{(im(\dot x_1 )^2 \Delta t/2\hbar )} {\rm{dx}}_n {\rm{dx}}_{n - 1} ...{\rm{dx}}_1 } } }
\][/tex]
No. What you've done is mathematically incorrect.And when the terms in the exponents are added up, we get,
[tex]\[
\int_{ - \infty }^{ + \infty } {\int_{ - \infty }^{ + \infty } {...\int_{ - \infty }^{ + \infty } {({\textstyle{m \over {2\pi i\hbar \Delta t}}})^{n/2} {\mathop{\rm e}\nolimits} ^{{\textstyle{i \over \hbar }}\int_{t_0 }^t {(m(\dot x^2 (t))/2)} dt} {\rm{dx}}_n {\rm{dx}}_{n - 1} ...{\rm{dx}}_1 } } }
\]
[/tex]
This is Feynman's path integral for the quantum mechanical wavefunction for a free particle. The notation is usually shortened to,
[tex]\[
\int {Dx \cdot {\mathop{\rm e}\nolimits} ^{{\textstyle{i \over \hbar }}S} }
\]
[/tex]
Since this is just an interation of the recursion relation for the Dirac delta measure which results in a Dirac delta which is a well defined measure, then this would mean that the measure of the path integral is well defined as long as we include the exponential integrand as part of the measure, right?
strangerep said:At this point you abandon mathematical rigor because the delta distribution
is not an element in the space of test functions.
strangerep said:The above should be rewritten with the integrals separated, and distinct [itex]\Delta t[/itex]'s
(and distinct limits) for each term. You've assumed you can blithely change the order
of taking the limits and performing the integrals, but that's only valid if everything is
well-behaved (uniform convergence, etc). Since you eventually want to take another limit as
[itex]n\to\infty[/itex], it's incorrect to ignore such details.
No, that's not debatable. See, e.g.,friend said:I'm not at all convinced that the language of distribution theory, functionals, test functions or generalized functions is even relevant to the conversation. The Dirac delta function appears in many applications. Just because the Dirac delta does not conform to the language of distributions and generalized functions does not mean that the integral of two Dirac deltas is not a legitemate integral in and of itself. All it means is that it does not qualify as a test function in the language of distributions. And even that might be debatable.
The theory of generalized functions and distribution theory subsumes the theory ofSo the question is which is more fundamental, integration and the limits involved with the delta function, or the abstract language of generalized functions and distribution theory?
Such questions of when limits can/can't be interchanged are dealt with in ordinary analysis.Does the abstract math of distributions or generalized functions reveal the necessity of which limit to take first, the limit involved with integration or the limits of the parameters that
send the delta to infinity?
That's not quite right. It's defined as a mapping from a space of functions to a space ofI've read a little bit about distribution theory. I'm no expert by any means. But from what I've seen it seems the Dirac delta is always defined in terms of its integration that always equals one
Distributions are defined as linear mappings from functions to scalars. There's anno matter what the value is of the limiting parameter of the delta. That means that the integration is always done first before the limit of the delta function goes to zero. For I cannot imagine how one would even do the integral if the delta parameter were allowed to go to zero first. Is this order of which limit to do first addressed in distribution theory?
The problems arise when you try to extend this naive intuition from the finite caseIt seems the only issue with the integral of two dirac delta functions is which limiting parameter of which delta function to take first. I've briefly browsed through my analysis books, and I did not see anything in general about which limit to do first when multiple limiting processes are involved. I have read that in general this could result in an undefined value since the answer may depend on which limit is done first. But in the case of integrating two dirac delta functions it's clear that you would get the same value no matter which delta's limiting parameter you considered first. I can write this out in mathematical terms if you'd like. So this makes the integral of two deltas well defined, right? If it doesn't matter in which order you take the limiting parameters in the integration of two dirac deltas, then for the same reason it doesn't matter in which order you take the limits in n integrations of n+1 deltas.
And I can only repeat that the important issue here is whether such a sequence has a[...]
And I'd also like to note that since I derived the path integral through a process of iteration, we never have an infinite number of integrations, only a number which approaches infinity in the limit.
strangerep said:No, that's not debatable. See, e.g.,
http://en.wikipedia.org/wiki/Test_function
A test function must be infinitely differentiable. It must also be locally integrable, but a Dirac
delta is nonzero only at a single point, therefore only nonzero on a "set of Lebesgue measure
zero", which means it's not Lebesgue-integrable in any mathematically sensible way.
Then you have not understood a word I've said. I am sorry.friend said:I think this proves my point.strangerep said:See, e.g., http://en.wikipedia.org/wiki/Test_function
A test function must be infinitely differentiable. It must also be locally integrable, but a Dirac
delta is nonzero only at a single point, therefore only nonzero on a "set of Lebesgue measure
zero", which means it's not Lebesgue-integrable in any mathematically sensible way.
A dirac delta is a distribution.It must be differentiable and integrable to qualify as an
element for consideration in distribution theory.
That's correct, but not in the way you're thinking of it. Notions suchThis pretty much states that matters of limits and convergence are more
fundamental than those of distribution theory.
I ask you to give me one example, and you ask me to read 400 pages? Which one is harder? From the pages I have read, nothing seems to prevent the evaluation of the Chapman-Kolmogorov equation in my previous post. Doesn't that equation pretty much establish the recursion relation for at least the gaussian form of the Dirac deltas function?strangerep said:I'm sorry, but I can't spend any more time on this. You really do need
to study a textbook on distributions and generalized functions.
DarMM said:The measure of the path integral not being defined usually refers to the [tex]\mathcal{D}\phi[/tex] measure in the path integral.
DarMM said:This object is basically the infinite dimensional Lesbesgue measure, which you can prove is undefined.
No, the space of paths is a space of functions. [tex]\mathcal{D}\phi[/tex] is a measure on that space. You can prove that this measure doesn't exist, however the space of paths does exist.friend said:I have read excerpts that complain that no measure can be given to the space of "paths". Is the space of paths just an interpretation of the [tex]\mathcal{D}\phi[/tex]?
Well [tex]\mathcal{D}\phi[/tex] exists at finite dimensions. However it doesn't exist in infinite dimensions. This doesn't matter, because the object you should by considering is [tex]e^{-S[\phi]}\mathcal{D}\phi[/tex] which exists in finite and more importantly infinite dimensions.friend said:From what I've seen the proof contains an operation that divides by the dimensionality, or some number to the power of the dimensionality, which gives zero for the measure of infinite dimension. I wonder if it is possible to talk about the measure for finite but approaching infinite for dimensionality?
As you are probably aware, I'm only now just starting to acquaint myself with all this advanced measure theory stuff as it applies to the path integral. It seems really quite complicated, and I would rather not delve too deeply into it unless I have to.DarMM said:No, the space of paths is a space of functions. [tex]\mathcal{D}\phi[/tex] is a measure on that space. You can prove that this measure doesn't exist, however the space of paths does exist.
DarMM said:Well [tex]\mathcal{D}\phi[/tex] exists at finite dimensions. However it doesn't exist in infinite dimensions. This doesn't matter, because the object you should by considering is [tex]e^{-S[\phi]}\mathcal{D}\phi[/tex] which exists in finite and more importantly infinite dimensions.
friend said:But perhaps you know of a resouce where these issues are explained more clearly.
DarMM said:However the object [tex]e^{-S[\phi]}\mathcal{D}\phi[/tex], which is the euclidean path integral, is well defined in the case of quantum mechanics.
DarMM said:In the case of QFT some care is needed in defining [tex]e^{-S[\phi]}\mathcal{D}\phi[/tex],...
DarMM said:However the Minkowski path integral can not be defined, proving this is actually an exercise in Reed and Simon Volume II.
No, although the heuristics work for most calculations.friend said:Is this the "Wiener measure"? Can it be analytically continued to the imaginary plane to provide a measure for the Feynman path integral?
It's best to view [tex]e^{-S[\phi]}[/tex] and [tex]\mathcal{D}\phi[/tex] together, rather than looking for conditions for one to make sense on its own.Is it sufficient to require that the [tex]\phi[/tex] be square integrable so that the exponent exists?
No, the Minkowski space path integral (Feynman path integral) does not have a well defined measure.So the question still remains for me: does the Feynman path integral have a well defined measure for QFT?