# Measure on Path Integral not defined

• friend
In summary: I guess would be the correct word, is inherent in trying to define a "measure" on an uncountable space of paths. This is because the product of two distributions is not generally well defined. In fact, I think it's safe to say that this problem is at the core of the continuum probability problem.
friend
Where can I find an On-line exposition of the undefined nature of the measure in the Feynman path integral? Thanks.

Just curious, what do you mean by the phrase "the undefined nature of the measure" ?

AEM said:
Just curious, what do you mean by the phrase "the undefined nature of the measure" ?

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. Most of the time "undefined" means it is infinite or has more than one value depending on how you evalute it. I'd like clarification on the meaning of "undefined" as well as you do. For it seems impossible for the path integral formulation to be fundamental if it is based on undefined measures in mathematical analysis.

Just guessing here, but maybe it has something to do with the inabilitiy to form larger measures from the measure of disjoint unions of smaller subsets of paths. Or maybe the infinite dimensions violates the requirement to have only a finite intersection of subsets. Or maybe it has something to do with the product of distributions not generally being well defined. I really don't know.

Last edited:
I think that maybe "ambigous" would be an different word.

Without furhter qualifiers the usual notion
$$\int Df e^{iS[f]}$$
where $$f: X \mapsto f(X)$$, and a measure is defined on X.

is not well defined, because it's only a symbolic notation. It doesn't mean it can't be defined, but it's at minimum ambigous. There is a CHOICE involved to make sense of this.

Then the question is what measure to assign/construct to the "function space" f belongs to so that this intergral makes sense?

Just like before you introduced the riemmann integral in calculus, transition from discrete sum to an infinite sum is ambigous.

It is very clear what "summing over a finite set of discrete path" means, but what does it actually mean when the paths form an infinite uncountable set? You need to define a MEASURE on the now continuous space of paths.

This is actually also at the core of my personal objection for the continuum probability. It's simply because I do not see the physical basis for DISTINGUISHING and continuum of real numbers. This objection is very close at least conceptually to the problem of "defining a measure of physically distinguishable paths" in the path integral, when you are using a language that permitts an uncountable infinite set of paths.

If you see this as a mathematical problem, then it's a matter of constructing these measures for certain classes of functions.

From the physics point of view, it's also a matter of knowing what we are trying to do. What is the action? What is a transition amplitude?

/Fredrik

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.
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.

(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.)
(The last bit is important in the case of the path integral over a nonseparable space of paths.)

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.)

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. 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.

I'm not sure if this helps, but I tend to see the physical perspective and from that perspective, failue or non-separability, means there is no sensible way to find the countable set to "approximate" the uncountable set of paths, and thus no way to perform the "sum over path" as and ordinary sum over natural numbers.

As I see it, the ambigousness, is that to make sense out of this, one constrains the space of paths, to something in order to be able to define the integral by means of a countable sequences. But the CHOICE of countable sequence, implies as far as I understand that we impose and arbitrary constraint on the "space of paths".

Instead of such what I persoanlly think is physically ambigous way, I think we should define that the space of physically distinguishable paths is. And I am sure if will end up managable, since I can't imagine the physical meaning in a physical observer beeing able to distinguish an uncountable set. Even the intinite countable, is suspect, but it could be reached in the large complexity limit, but when it's uncountable things get suspect.

Another way of phrasing the ambigous choice of constraining the spce, is that it amounts to a non-unique choice of ergodic hypothesis with a equiprobable set of microstates over the space of paths. I think this way of solving this problem is strange. I think there may be another way, which implies reconstructing the entire "path integral" instead of trying to invent a way for it to make sense. Even if we invent a mathematical sense, it may be physically ambigous, since we need to not only be able to calculate it, the calculations should also match nature.

I think the set of "physically distinguishable paths" IS the "natural" constraint we need. The question then seems reduce to figuring out what physically distinguishing a path, from the point of view of an observer, really means? What does it mean for an observer to "distinguish" different possibilities?

I think somehow this "distinguishability" i the observer-dependent MEASURE on the space of paths we are looking for.

/Fredrik

friend said:
I'm not see the nonseparable nature of the space of paths. [...]

A space of well-behaved continuous paths is separable.
See, e.g.

http://en.wikipedia.org/wiki/Classical_Wiener_space

Separability follows from the Stone-Weierstrass theorem.
(Any continuous function can be approximated arbitrarily
closely by a polynomial.)

By my previous remark about nonseparability I only meant that even
if one considers a space of more pathological paths, one still has
trouble defining a suitable measure.

So what about the dx1dx2...dxn. Is this a well defined measure, even if n is infinite? Are we talking about an ill-defined measure only when we interpret this as a space of paths?

friend said:
So what about the dx1dx2...dxn. Is this a well defined measure, even if n is infinite?
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.

Are we talking about an ill-defined measure only when we interpret this as a space of paths?
No. A path is just a particular kind of function. If we have a space of such functions which
is also a vector space (ie closed under addition and scalar multiplication) on which a norm is
defined, then we have (almost) a Banach space. (A true Banach space is what you get if the
space contains all its limit points.)

(We work with vectors in a finite dimensional space in the form of ordered tuples.
Similarly, we work with vectors in an infinite dimensional space in the form of functions.)

When gravity is addes, this very issue gets even worse, for reference one could note how Loll writes in

"The Emergence of Spacetime or Quantum Gravity on Your Desktop"
http://arxiv.org/abs/0711.0273

which reasons likes this:

"At the heart of the approach lies an explicit realization of the infamous “Sum over Histories”, also known as the gravitational path integral,
...
Before one has specified the integration space, the integration measure and the conditions under which the integration leads to a meaningful (i.e. non-infinite) result, it should be regarded as a statement of intent rather than a well-defined mathematical quantity."

Their CHOICE is to make a specific construction

"The integration space G is a space of causal, Lorentzian geometries, obtained from a certain limiting process, which will be described below."

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

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

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:
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.

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?

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.

I have to interject here that this is where physicists and mathematicians diverge. Remember it was Hilbert who said "Every school boy in Gottingen knows more about four dimensional geometry than Einstein, yet it was he who created General Relativity not the mathematicians."

The history of physics is full of examples where physical intuition guided the choice of mathematical concepts and how they were to be interpreted.

When it comes to deciphering nature methinks one must keep a balance between one's intuition about the physical and rigor of the mathematicians. A noted mathematician, Rene Thom once quipped "The word rigor often preceeds the word mortis".

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.

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 $[0,1]\times [0,1]\times\cdots$ and $\mathbb{R}\times\mathbb{R}\times\cdots$. Often product topologies are used, and measures are Borel with respect to them.

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.

1) I now know that you are looking to derive physics from logic or math. On that point I think we can agree to disagree, although I partially appreciate your quest :) I've already presented some of my arguments why physics from pure logic is still ambigous. It lies in the choice of the axioms that defines your logic. If you consider the "set of all possible logical systems" then I have a hard time to see how that can be constructive and lead to physical predictions.

2) I do not like CDT in itself, but I just injected that paper as an example of reasoning. But the general idea to lead the physical insight choose guide the choice of mathematical abstractions, instead of the other way around is more natural to me at least.

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

Fra said:

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

I want to voice my agreement with this statement.

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.

If you refer to them assuming too much structure in their sample space, like lorentian causality, then I fully agree. But that's I think a different point since I think the causal structure of spacetime is emergent with spacetime is. So I would like to instead see a different sampling space even without spacetime topology and instead somehow just sample over physical complexions, and hopefully see that spacetime structure including the lorenzian structure is a result of an underlying selforganisation. I also reject the manual choice of the EH-action.

So I really don't defend the CDT paper. But it's because I think they are not radical enough.

/Fredrik

friend said:
strangerep said:
[...] it's ill-defined if n is infinite.
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?
It can make a big difference depending on whether N is "huge but finite" compared to "infinite".
Sometimes, properties which hold for all finite N no longer hold in the limit (at least, not the
way you'd like them to). Such is the case here.

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 $[0,1]\times [0,1]\times\cdots$ and $\mathbb{R}\times\mathbb{R}\times\cdots$. Often product topologies are used, and measures are Borel with respect to them.

Are you talking about Gaussian/Wiener measures? If so, yes they're used in path
integrals of course, but entail some extra assumptions about analytic continuation, etc.

Let me suggest a different approach. It seems the Feynman path integral can also be obtained by interating a recursion relation for the Dirac delta function. And the Dirac delta function is a well defined measure. The Dirac delta function, usually symbolized as $$${\rm{\delta (x - x}}_0 )$$$, is defined as:

$$$\int_{ - \infty }^{ + \infty } {{\rm{\delta (x}}_1 {\rm{ - x}}_0 ){\rm{dx}}_1 } = 1$$$

such that as a parameter approaches zero, the amplitude approaches infinity. There are many ways to represent the Dirac delta function. For example, the gaussian form of the Dirac delta function is,

$$${\rm{\delta (x - x}}_0 ) = \mathop {\lim }\limits_{\Delta \to 0} \frac{1}{{(\pi \Delta ^2 )^{1/2} }}e^{ - (x - x_0 )^2 /\Delta ^2 }$$$

Here, as $$$\Delta$$$ goes to zero, $$${\rm{\delta (x - x}}_0 )$$$ approaches infinity at $$${\rm{x = x}}_0$$$ and is zero for $$${\rm{x}} \ne {\rm{x}}_0$$$. But it integrates to 1 independent of the value of $$$\Delta$$$.

And as a measure the Dirac delta function satisfies the relation,

$$$\int_{ - \infty }^{ + \infty } {{\rm{f(x}}_1 {\rm{)\delta (x}}_1 {\rm{ - x}}_0 ){\rm{dx}}_1 } = {\rm{f(x}}_0 )$$$

for all continuous, compactly supported test functions, f(x). See more details at:

http://en.wikipedia.org/wiki/Dirac_delta_function#As_a_measure

Now, it occurs to me that if we let $$${\rm{f(x}}_1 ) = {\rm{\delta (x - x}}_1 )$$$, then

$$$\int_{ - \infty }^{ + \infty } {{\rm{\delta (x - x}}_1 {\rm{)\delta (x}}_1 {\rm{ - x}}_0 ){\rm{dx}}_1 } = {\rm{\delta (x - x}}_0 )$$$

This is a recursion relation that we can apply as many times as you like. Applying it again gives,

$$$\int_{ - \infty }^{ + \infty } {\int_{ - \infty }^{ + \infty } {{\rm{\delta (x - x}}_2 {\rm{)}}\left( {{\rm{\delta (x}}_2 {\rm{ - x}}_1 {\rm{)\delta (x}}_1 {\rm{ - x}}_0 ){\rm{dx}}_1 } \right){\rm{dx}}_2 } } = \int_{ - \infty }^{ + \infty } {{\rm{\delta (x - x}}_2 {\rm{)\delta (x}}_2 {\rm{ - x}}_0 )} {\rm{dx}}_2 = {\rm{\delta (x - x}}_0 )$$$

And applying it n times gives,

$$$\int_{ - \infty }^{ + \infty } {\int_{ - \infty }^{ + \infty } {...\int_{ - \infty }^{ + \infty } {{\rm{\delta (x - x}}_n {\rm{)\delta (x}}_n {\rm{ - x}}_{n - 1} {\rm{)}}...{\rm{\delta (x}}_1 {\rm{ - x}}_0 ){\rm{dx}}_1 {\rm{dx}}_2 ...{\rm{dx}}_n } } } = {\rm{\delta (x - x}}_0 )$$$

If we let

$$$\Delta = \left[ {\frac{{2i\hbar \left( {t - t_0 } \right)}}{m}} \right]^{1/2}$$$

in the gaussian form of the Dirac delta function, then we can get,

$$${\rm{\delta (x - x}}_0 ) = \mathop {\lim }\limits_{t \to t_0 } \left[ {\frac{m}{{2\pi i\hbar (t - t_0 )}}} \right]^{1/2} \exp \left[ {\frac{{im(x - x_0 )^2 }}{{2\hbar (t - t_0 )}}} \right] = \mathop {\lim }\limits_{t \to t_0 } \left[ {\frac{m}{{2\pi i\hbar (t - t_0 )}}} \right]^{1/2} \exp \left[ {\frac{{im}}{{2\hbar }}(\frac{{x - x_0 }}{{t - t_0 }})^2 (t - t_0 )} \right]$$$

$$$= \mathop {\lim }\limits_{\Delta t \to 0} ({\textstyle{m \over {2\pi i\hbar \Delta t}}})^{1/2} {\mathop{\rm e}\nolimits} ^{(\frac{{im}}{{2\hbar }}(\frac{{\Delta x}}{{\Delta t}})^2 (\Delta t))} = \,\,\,\,\mathop {\lim }\limits_{\Delta t \to 0} ({\textstyle{m \over {2\pi i\hbar \Delta t}}})^{1/2} {\mathop{\rm e}\nolimits} ^{(im(\dot x)^2 \Delta t/2\hbar )}$$$

And when this is substituted in the nth iteration of the recursion relation above, we get,

$$$\int_{ - \infty }^{ + \infty } {\int_{ - \infty }^{ + \infty } {...\int_{ - \infty }^{ + \infty } {({\textstyle{m \over {2\pi i\hbar \Delta t}}})^{1/2} {\mathop{\rm e}\nolimits} ^{(im(\dot x_n )^2 \Delta t/2\hbar )} ({\textstyle{m \over {2\pi i\hbar \Delta t}}})^{1/2} {\mathop{\rm e}\nolimits} ^{(im(\dot x_{n - 1} )^2 \Delta t/2\hbar )} ...{\rm{ }}({\textstyle{m \over {2\pi i\hbar \Delta t}}})^{1/2} {\mathop{\rm e}\nolimits} ^{(im(\dot x_1 )^2 \Delta t/2\hbar )} {\rm{dx}}_n {\rm{dx}}_{n - 1} ...{\rm{dx}}_1 } } }$$$

And when the terms in the exponents are added up, we get,

$$$\int_{ - \infty }^{ + \infty } {\int_{ - \infty }^{ + \infty } {...\int_{ - \infty }^{ + \infty } {({\textstyle{m \over {2\pi i\hbar \Delta t}}})^{n/2} {\mathop{\rm e}\nolimits} ^{{\textstyle{i \over \hbar }}\int_{t_0 }^t {(m(\dot x^2 (t))/2)} dt} {\rm{dx}}_n {\rm{dx}}_{n - 1} ...{\rm{dx}}_1 } } }$$$

This is Feynman's path integral for the quantum mechanical wavefunction for a free particle. The notation is usually shortened to,

$$$\int {Dx \cdot {\mathop{\rm e}\nolimits} ^{{\textstyle{i \over \hbar }}S} }$$$

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?

friend said:
And as a measure the Dirac delta function satisfies the relation,

$$$\int_{ - \infty }^{ + \infty } {{\rm{f(x}}_1 {\rm{)\delta (x}}_1 {\rm{ - x}}_0 ){\rm{dx}}_1 } = {\rm{f(x}}_0 )$$$

for all continuous, compactly supported test functions, f(x).
The last bit is important. I'll write the above in a different way to emphasize something
below.
$$\int_{- \infty}^{+\infty}dx_1 \delta (x_0 - x_1) \left( f(x_1) \right) ~=~ f(x_0)$$
This emphasizes more clearly that the Dirac delta should be considered as a
linear mapping from the space of test functions to the space of scalars.
I.e., in my rewrite above, I've emphasized the delta "acting on" the function f,
yielding the scalar value $f(x_0)$.

Now, it occurs to me that if we let $$${\rm{f(x}}_1 ) = {\rm{\delta (x - x}}_1 )$$$
At this point you abandon mathematical rigor because the delta distribution
is not an element in the space of test functions.

then
$$$\int_{-\infty }^{+\infty} {{\rm{\delta (x - x}}_1 {\rm{)\delta (x}}_1 {\rm{ - x}}_0 ){\rm{dx}}_1 } = {\rm{\delta (x - x}}_0 )$$$
This only works because you abused some notation. One can of course write instead
$$\int_{- \infty}^{+\infty }dx \delta (x - x_0 ) \left( \int_{- \infty}^{+\infty}dx_1 \delta (x - x_1) \left( f(x_1) \right) \right) ~=~ f(x_0)$$
which is equivalent to what you wanted, except that I've been careful not to
move the positions of the integral signs.

Then...

If we let

$$$\Delta = \left[ {\frac{{2i\hbar \left( {t - t_0 } \right)}}{m}} \right]^{1/2}$$$

in the gaussian form of the Dirac delta function, then we can get,

$$${\rm{\delta (x - x}}_0 ) = \mathop {\lim }\limits_{t \to t_0 } \left[ {\frac{m}{{2\pi i\hbar (t - t_0 )}}} \right]^{1/2} \exp \left[ {\frac{{im(x - x_0 )^2 }}{{2\hbar (t - t_0 )}}} \right] = \mathop {\lim }\limits_{t \to t_0 } \left[ {\frac{m}{{2\pi i\hbar (t - t_0 )}}} \right]^{1/2} \exp \left[ {\frac{{im}}{{2\hbar }}(\frac{{x - x_0 }}{{t - t_0 }})^2 (t - t_0 )} \right]$$$

$$$= \mathop {\lim }\limits_{\Delta t \to 0} ({\textstyle{m \over {2\pi i\hbar \Delta t}}})^{1/2} {\mathop{\rm e}\nolimits} ^{(\frac{{im}}{{2\hbar }}(\frac{{\Delta x}}{{\Delta t}})^2 (\Delta t))} = \,\,\,\,\mathop {\lim }\limits_{\Delta t \to 0} ({\textstyle{m \over {2\pi i\hbar \Delta t}}})^{1/2} {\mathop{\rm e}\nolimits} ^{(im(\dot x)^2 \Delta t/2\hbar )}$$$

And when this is substituted in the nth iteration of the recursion relation above, we get,

$$$\int_{ - \infty }^{ + \infty } {\int_{ - \infty }^{ + \infty } {...\int_{ - \infty }^{ + \infty } {({\textstyle{m \over {2\pi i\hbar \Delta t}}})^{1/2} {\mathop{\rm e}\nolimits} ^{(im(\dot x_n )^2 \Delta t/2\hbar )} ({\textstyle{m \over {2\pi i\hbar \Delta t}}})^{1/2} {\mathop{\rm e}\nolimits} ^{(im(\dot x_{n - 1} )^2 \Delta t/2\hbar )} ...{\rm{ }}({\textstyle{m \over {2\pi i\hbar \Delta t}}})^{1/2} {\mathop{\rm e}\nolimits} ^{(im(\dot x_1 )^2 \Delta t/2\hbar )} {\rm{dx}}_n {\rm{dx}}_{n - 1} ...{\rm{dx}}_1 } } }$$$
The above should be rewritten with the integrals separated, and distinct $\Delta t$'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
$n\to\infty$, it's incorrect to ignore such details.

And when the terms in the exponents are added up, we get,

$$$\int_{ - \infty }^{ + \infty } {\int_{ - \infty }^{ + \infty } {...\int_{ - \infty }^{ + \infty } {({\textstyle{m \over {2\pi i\hbar \Delta t}}})^{n/2} {\mathop{\rm e}\nolimits} ^{{\textstyle{i \over \hbar }}\int_{t_0 }^t {(m(\dot x^2 (t))/2)} dt} {\rm{dx}}_n {\rm{dx}}_{n - 1} ...{\rm{dx}}_1 } } }$$$

This is Feynman's path integral for the quantum mechanical wavefunction for a free particle. The notation is usually shortened to,

$$$\int {Dx \cdot {\mathop{\rm e}\nolimits} ^{{\textstyle{i \over \hbar }}S} }$$$

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?
No. What you've done is mathematically incorrect.

However, in modern physics one does not "derive" the path integral in this way.
Rather, one starts from a similar expression involving a Hamiltonian which is quadratic
in the free part, then one tries to perform a Wick rotation to imaginary time so that
the "it" in the exponent becomes "-t", producing a Gaussian function. It can indeed be
shown that if these Gaussian terms are included as part of the measure, the expression
converges. (That's what Gaussian/Wiener measure is all about.) Later, after evaluating
the integral, ones tries to Wick-rotate back to real time. To be rigorous, one must show
that the assumptions about analyticity implicit in all this are indeed valid. This is
difficult if interactions are included in the Hamiltonian.

strangerep said:
At this point you abandon mathematical rigor because the delta distribution
is not an element in the space of test functions.

I really appreciate your comments on this. I hope to get some serious discussion on these matters.

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.

So 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? 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?

I'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 no 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?

It 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?

strangerep said:
The above should be rewritten with the integrals separated, and distinct $\Delta t$'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
$n\to\infty$, it's incorrect to ignore such details.

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.

I'd like to make you aware that I found this equation in the book, "Path Integrals in Quantum Mechanics, Statistics, Polymer Physics, and Financial Markets", by Hagen Kleinert, page 91. You can also see in a book review at:

http://users.physik.fu-berlin.de/~kl...es/pthic04.pdf

It shows how a quantum transition amplitude can be interpreted as a dirac delta function equal to the integration of a great number of products of delta functions.

$$$\left( {x_b t_b |x_a t_a } \right) = \prod\limits_{n = 1}^N {\left[ {\int_{ - \infty }^{ + \infty } {dx_n } } \right]} \prod\limits_{n = 1}^{N + 1} {\left\langle {x_n |x_{n - 1} } \right\rangle } = \prod\limits_{n = 1}^N {\left[ {\int_{ - \infty }^{ + \infty } {dx_n } } \right]} \prod\limits_{n = 1}^{N + 1} {\delta \left( {x_n - x_{n - 1} } \right)} = \delta \left( {x_b - x_a } \right)$$$

The last two equations on the right can be obtained by interating a recursion relation for the dirac delta function. So you can see here that QM can be derived from this recursion relation assuming that it is valid.

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.

I look forward to your responses.

Last edited by a moderator:
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.
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.

Also note that " the language of distributions and generalized functions" was invented
in part to place ill-defined-but-useful things like the Dirac delta into a well-defined context,
i.e., as a distribution.

So 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?
The theory of generalized functions and distribution theory subsumes the theory of
more well-behaved functions.

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?
Such questions of when limits can/can't be interchanged are dealt with in ordinary analysis.

I'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
That's not quite right. It's defined as a mapping from a space of functions to a space of
scalars. The case when the function is constant everywhere is just a special case of this.

no 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?
Distributions are defined as linear mappings from functions to scalars. There's an
entire body of theory in functional analysis (nuclear spaces, dual spaces, and lots of other
stuff). "Doing the integral" involves considering sequences in the nuclear spaces, which
induce meaningful sequences in the dual space, and the analysis proceeds in this way.
But I can't possibly hope to summarize this here.

It 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.
The problems arise when you try to extend this naive intuition from the finite case
to the infinite case. Finite intuition doesn't always remain valid when extended to
infinity. E.g., the rational numbers are a countable set, but the set of the (equivalence
classes of) Cauchy sequences of rationals (i.e., the real numbers) is not.

[...]
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.
And I can only repeat that the important issue here is whether such a sequence has a
good (convergent) limit.

I seem to be repeating myself though. I don't think I can say much more since
I have lots of other things to do. I hope you'll understand that I don't have time
to teach you functional analysis and general topology. Maybe someone over in
the maths forums might have more time for this sort of thing.

Well, it *is* possible (in some limited sense) to do path integrals rigorously. One first performs Wick rotation, and then defines the Wiener measure on the space of nice paths. Then we analytically continue (not always possible) to Minkowski space.

The measure of the path integral not being defined usually refers to the $$\mathcal{D}\phi$$ measure in the path integral. This object is basically the infinite dimensional Lesbesgue measure, which you can prove is undefined. However the object $$e^{-S[\phi]}\mathcal{D}\phi$$, which is the euclidean path integral, is well defined in the case of quantum mechanics. In the case of QFT some care is needed in defining $$e^{-S[\phi]}\mathcal{D}\phi$$, in particular there will appear terms physicists call counter-terms. However at the end of the day you can still define it.
However the Minkowski path integral can not be defined, proving this is actually an exercise in Reed and Simon Volume II.

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.

I think this proves my point. It must be differentiable and integrable to qualify as an element for consideration in distribution theory. This pretty much states that matters of limits and convergence are more fundamental than those of distribution theory. I guess I'm looking for one clear example of where the limit process is defined in terms of the elements of distribution theory such that the limiting process is changed or guided by distribution theory. If there is no such example, then I think that means I'm free to consider how the integral of two gaussian dirac delta functions might be evaluated.

I found this equation in, "The Feynman Integral and Feynman's Operational Calculus", by Gerald W. Johnson and Michael L. Lapidus, page 37, called a Chapman-Kolmogorov equation:

$$$\int_{ - \infty }^{ + \infty } {\left( {\frac{\lambda }{{2\pi \left( {t - s} \right)}}} \right)^{\frac{1}{2}} e^{ - {\textstyle{{\lambda (\omega - \upsilon )^2 } \over {2\left( {t - s} \right)}}}} \left( {\frac{\lambda }{{2\pi \left( {s - r} \right)}}} \right)^{\frac{1}{2}} e^{ - {\textstyle{{\lambda (\upsilon - u)^2 } \over {2\left( {s - r} \right)}}}} {\rm{d}}\upsilon } = \left( {\frac{\lambda }{{2\pi \left( {t - r} \right)}}} \right)^{\frac{1}{2}} e^{ - {\textstyle{{\lambda (\omega - u)^2 } \over {2\left( {t - r} \right)}}}}$$$

This equation is also confirmed in the book review at:

This equation does not consider limits. But it's easy to see that placing limits on (t-s) and (s-r) would lead to the integration of two gaussian form Dirac delta functions. This integral as shown is independent of the value of (t-s) or (s-r). So placing limits on (t-s) and (s-r) would not change its evaluation. That's why I think its well defined. This is the recursion relation I use to develop the path integral. It proves the recursion relation for at least the gaussian form of the Dirac delta. I'm not sure it proves the recursion relation in general. The book does not spell out how they got this equation. Does anyone know how they got this equation? I tried it and got cross-multiplication terms in the exponent; it did not seem obvious how to integrate it.

Last edited:
friend said:
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.
I think this proves my point.
Then you have not understood a word I've said. I am sorry.

It must be differentiable and integrable to qualify as an
element for consideration in distribution theory.
A dirac delta is a distribution.
Distributions act on test functions to yield scalars.
I don't know how I can say this any more clearly.

This pretty much states that matters of limits and convergence are more
fundamental than those of distribution theory.
That's correct, but not in the way you're thinking of it. Notions such
as limits and convergence have a meaning within the subject of
general topology, which deals with more generalized ways of expressing
whether one entity is "nearly" the same as another. Distribution theory
works with limits and convergence expressed in terms of something
called "weak topology", which is different from the standard topology
that you're implicitly thinking of when you say "limits and convergence".

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.

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.
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?

DarMM said:
The measure of the path integral not being defined usually refers to the $$\mathcal{D}\phi$$ measure in the path integral.

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 $$\mathcal{D}\phi$$?

DarMM said:
This object is basically the infinite dimensional Lesbesgue measure, which you can prove is undefined.

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?

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 $$\mathcal{D}\phi$$?
No, the space of paths is a space of functions. $$\mathcal{D}\phi$$ is a measure on that space. You can prove that this measure doesn't exist, however the space of paths does exist.

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?
Well $$\mathcal{D}\phi$$ exists at finite dimensions. However it doesn't exist in infinite dimensions. This doesn't matter, because the object you should by considering is $$e^{-S[\phi]}\mathcal{D}\phi$$ which exists in finite and more importantly infinite dimensions.

DarMM said:
No, the space of paths is a space of functions. $$\mathcal{D}\phi$$ is a measure on that space. You can prove that this measure doesn't exist, however the space of paths does exist.
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.

I'm presently starting to browse through Johnson and Lapidus, "The Feynman Integral and Feynman's Operational Calculus". I get sentences here and there that claim the non-existence of the Feynman measure and that there is no countably additive measure on the space of paths, etc. But I'm not seeing any clear proof of those claims. I probably wouldn't know if I saw it, but I would think if it were clear, then there would be reference to theorems and sections in the book where the proof would be worked out. So I'm getting confusing signals at this point. And I'm not sure any of this is actually relevant to my interests. But perhaps you know of a resouce where these issues are explained more clearly.

DarMM said:
Well $$\mathcal{D}\phi$$ exists at finite dimensions. However it doesn't exist in infinite dimensions. This doesn't matter, because the object you should by considering is $$e^{-S[\phi]}\mathcal{D}\phi$$ which exists in finite and more importantly infinite dimensions.

Yes, I came to the same conclusion just by iterating a recursion relation of the gaussian form of the dirac delta function (post 21 in this thread). It very much appeals to me to think that the measure of the path integral can be derived from the measure associated with the dirac delta. For it seems the dirac delta is the most basic of all measures, counting only those elements in a set. And I wonder more fundamentally if all measures can be described in terms of the dirac delta measure. Have you come across anything like that in your studies?

One interesting philosophical consideration is that the path integral does give physical results. And so one would expect that the path integral would have well defined mathematical underpinnings. How could undefined mathematics give physical results unless reality itself is undefinable?

friend said:
But perhaps you know of a resouce where these issues are explained more clearly.

Are you aware of the book "Mathematical Theory of Feynman Path Integrals: An Introduction" by Sergio A. Albeverio, et. al. ? Perhaps that would be of use to you.

DarMM said:
However the object $$e^{-S[\phi]}\mathcal{D}\phi$$, which is the euclidean path integral, is well defined in the case of quantum mechanics.

Is this the "Wiener measure"? Can it be analytically continued to the imaginary plane to provide a measure for the Feynman path integral?

DarMM said:
In the case of QFT some care is needed in defining $$e^{-S[\phi]}\mathcal{D}\phi$$,...

Is it sufficient to require that the $$\phi$$ be square integrable so that the exponent exists?

DarMM said:
However the Minkowski path integral can not be defined, proving this is actually an exercise in Reed and Simon Volume II.

So the question still remains for me: does the Feynman path integral have a well defined measure for QFT?

Thanks

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?
No, although the heuristics work for most calculations.

Is it sufficient to require that the $$\phi$$ be square integrable so that the exponent exists?
It's best to view $$e^{-S[\phi]}$$ and $$\mathcal{D}\phi$$ together, rather than looking for conditions for one to make sense on its own.

So the question still remains for me: does the Feynman path integral have a well defined measure for QFT?
No, the Minkowski space path integral (Feynman path integral) does not have a well defined measure.

Replies
1
Views
946
Replies
4
Views
608
Replies
6
Views
1K
Replies
13
Views
1K
Replies
14
Views
3K
Replies
3
Views
1K
Replies
2
Views
1K
Replies
3
Views
1K
Replies
1
Views
843
Replies
5
Views
2K