Path integrals and supposed sum over paths

  • Thread starter pellman
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  • #1
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If I understand correctly--a big if--the path integration method, at least when applied to plain old QM, is described as (1) every possible path the particle could take is assigned an amplitude, (2) sum up (integrate over) these amplitudes for all possible paths.

The problem I have with this is when you look at the actual math, there seem to me be contributions from discontinuous paths. And I don't just mean pointwise continuous but radically discontinuous.

To actually derive the path integral expression, you slice up the time between the initial position and final position into N time intervals, and then integrate over all possible positions at each time interval. As you increase the number of slices, you increase the number of position integrations, until the number of position integrations supposedly "goes over" into a continuum limit as N goes to infinity (which doesn't sound all that convincing to me).

But for each N, the integrations over position are all independent of each other. Of course, for any wild set of N positions, we can always connect them with a continuous function q(t). But as we increase N, we increase the number of positions being integrated over, which necessarily include contributions which are wilder and wilder. I can't see any reason why they would all settle down into continuous functions as N goes to infinity.

I am not just talking about very wild continuous paths. What I mean is if we look at two close time-slices [tex]q_j,t_j[/tex] and [tex]q_{j+1},t_{j}+\delta t[/tex] and we integrate over [tex]q_j[/tex] and [tex]q_{j+1}[/tex] independently, over the whole real line, then there is a lot of contributions from amplitudes associated with large values of [tex]|q_{j+1}-q_j|[/tex]. That's not going to settle down into anything like continuous paths. [tex]|q_{j+1}-q_j|[/tex] is not going to get any smaller as [tex]\delta t[/tex] goes to zero.

However, the author I am reading (MacKenzie http://xxx.lanl.gov/abs/quant-ph/0004090 ) uses arguments based on variations in the action integral from the particle's classical path (cf section 2.2.2 in which MacKenzie calculates the harmonic oscillator propagator) which, I would think, would only work if we are restricted to nice continous variations in the path.

I don't have any problem with the strict definition of the path integral in terms of a limit. But after that the conceptual exposition sounds pretty shaky.

What's my question? I guess it is, do I understand the above correctly?
 
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Answers and Replies

  • #2
strangerep
Science Advisor
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I don't know if this will help, but, AFAIK, the path integral is
mathematically ill-defined - because the measure in an
infinite-dimensional space does not converge (in the absence
of additional assumptions). The issues you raise are part of
the reason why a careful person doesn't say that the
path integral is "derived" in the way you describe. Rather,
they just write down the final thing involving the
Lagrangian and definition, and say that the theory is
"defined" from this starting point. (Personally, I find that
a bit unsatisfying, however.)

Regarding non-convergence of measure in an inf-dim space...
In QFT, this is often handled by extracting the quadratic part
of the Hamiltonian (gives a Gaussian when exponentiated and
analytically continued to imaginary time), absorbing the
Gaussian into the measure (thus guaranteeing that the new
measure does converge), and then perform an
"inverse" analytic continuation back to real time after
performing the integral. A low-brow version of all this
is basically what's done in many physics textbooks
(e.g., Zee). Other books like Glimme & Jaffe do it
more rigorously.

To split the Hamiltonian into kinetic (quadratic)
and interaction parts, one relies on a theorem of
Trotter which says under what conditions this is
allowable. This is also discussed in G&J.
 
  • #3
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I am not just talking about very wild continuous paths. What I mean is if we look at two close time-slices [tex]q_j,t_j[/tex] and [tex]q_{j+1},t_{j}+\delta t[/tex] and we integrate over [tex]q_j[/tex] and [tex]q_{j+1}[/tex] independently, over the whole real line, then there is a lot of contributions from amplitudes associated with large values of [tex]|q_{j+1}-q_j|[/tex]. That's not going to settle down into anything like continuous paths. [tex]|q_{j+1}-q_j|[/tex] is not going to get any smaller as [tex]\delta t[/tex] goes to zero.

Hello,

do we accept only continuous path with finite first derivative or all continues path (like Brownian motion) ?
 
  • #4
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4
Hello,

do we accept only continuous path with finite first derivative or all continues path (like Brownian motion) ?

hi. I really don't know.

I'm just bothered by some rather casual conceptual arguments which seem too rely too much on the idea that path integral is summing up all paths a point particle could take from A to B. But the math makes me wonder if there are contributions from what would be discontinous paths. And would that affect these conceptual arguments? I don' know.
 
  • #5
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I'm just bothered by some rather casual conceptual arguments which seem too rely too much on the idea that path integral is summing up all paths a point particle could take from A to B. But the math makes me wonder if there are contributions from what would be discontinous paths. And would that affect these conceptual arguments? I don' know.

I have not looked at Path integrals from so long time...
I think one of the approach of Path Integral is to consider it as a Wiener Integral with a Wiener process. So in some way, it can be related to random walk where you can expect very strange path happening, they are just unfavored.

http://en.wikipedia.org/wiki/Wiener_process
 
  • #6
1,452
9
I don't know if this will help, but, AFAIK, the path integral is
mathematically ill-defined - because the measure in an
infinite-dimensional space does not converge (in the absence
of additional assumptions).

Actually, it seems the path integral can be derived from only the definition of the Dirac delta function. See the quick and easy derivation at:

http://hook.sirus.com/users/mjake/delta_physics.htm [Broken]

And I am told by Wikipedia that the Dirac delta IS a well defined measure. So if the Dirac delta is well defined, then so is the path integral, right?

I claim no responsibility for anything else on this site. But at least this derivation seems right.
 
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  • #7
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I have not looked at Path integrals from so long time...
I think one of the approach of Path Integral is to consider it as a Wiener Integral with a Wiener process. So in some way, it can be related to random walk where you can expect very strange path happening, they are just unfavored.

Thanks for the info, Barmecides.
 
  • #8
strangerep
Science Advisor
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1,226
Actually, it seems the path integral can be derived from only the definition of the Dirac delta function. See the quick and easy derivation at:

http://hook.sirus.com/users/mjake/delta_physics.htm [Broken]

And I am told by Wikipedia that the Dirac delta IS a well defined measure. So if the Dirac
delta is well defined, then so is the path integral, right?

Not necessarily (at least, not in an infinite-dimensional space without additional tricks).

The derivation looks a bit naive, and does not mention the extra subtleties that
arise in inf-dim spaces. At one point, it assumes that the following formula:
[tex]
\int\int\dots\int \delta(x-x_n)\delta(x_n-x_{n-1})\dots\delta(x_1-x_0)dx_n\dots dx_1=\delta(x-x_0)
[/tex]
remains well-defined in the limit [tex]n\to\infty[/tex]. Later, there's a similar formula
where the delta fns have been replaced by limits of exponentials of the
form
[tex]
f(x) := \exp(im (\dot x)^2 \Delta t \dots)
[/tex]
(modulo some factors which are not important here).

The naive bit is that , whereas an integral like
[tex]
\int_{-N}^N\dots\int_{-N}^N dx_n\dots dx_1 ~~~~(1)
[/tex]
is finite for any finite n, it diverges for [tex]n\to\infty[/tex]. In the ordinary theory of
integration, one uses this fact to define the integral of a multi-dimensional
function f(...), for n,N both finite, as
[tex]
\int_{-N}^N\dots\int_{-N}^N ~ f(x_1,\dots,x_n) ~ dx_n\dots dx_1 ~~~~(2)
[/tex]
and then we take a limit as N goes to infinity. However, (1) diverges in the case
of infinite n, and therefore we cannot validly make the next step to (2), nor
take the limit of infinite N.

To make the derivation a bit more rigorous, one would have to perform a Wick rotation
to imaginary time (such that each exponential becomes a well-behaved Gaussian).
Then it can be shown that (2) does indeed converge as N goes to infinity (and f(...) is
now a product of Gaussians). This is all an elementary version of what I mentioned
in my earlier post about how one absorbs the quadratic part of the Hamiltonian into
the measure to obtain a convergent expression.

I.e., one is relying on the Gaussians to tame the wild ill-defined behaviour.
 
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  • #9
1,452
9
Not necessarily (at least, not in an infinite-dimensional space without additional tricks).

The derivation looks a bit naive, and does not mention the extra subtleties that
arise in inf-dim spaces. At one point, it assumes that the following formula:
[tex]
\int\int\dots\int \delta(x-x_n)\delta(x_n-x_{n-1})\dots\delta(x_1-x_0)dx_n\dots dx_1=\delta(x-x_0)
[/tex]
remains well-defined in the limit [tex]n\to\infty[/tex]. Later, there's a similar formula
where the delta fns have been replaced by limits of exponentials of the
form
[tex]
f(x) := \exp(im (\dot x)^2 \Delta t \dots)
[/tex]
(modulo some factors which are not important here).

The naive bit is that , whereas an integral like
[tex]
\int_{-N}^N\dots\int_{-N}^N dx_n\dots dx_1 ~~~~(1)
[/tex]
is finite for any finite n, it diverges for [tex]n\to\infty[/tex]. In the ordinary theory of
integration, one uses this fact to define the integral of a multi-dimensional
function f(...), for n,N both finite, as
[tex]
\int_{-N}^N\dots\int_{-N}^N ~ f(x_1,\dots,x_n) ~ dx_n\dots dx_1 ~~~~(2)
[/tex]
and then we take a limit as N goes to infinity. However, (1) diverges in the case
of infinite n, and therefore we cannot validly make the next step to (2), nor
take the limit of infinite N.

To make the derivation a bit more rigorous, one would have to perform a Wick rotation
to imaginary time (such that each exponential becomes a well-behaved Gaussian).
Then it can be shown that (2) does indeed converge as N goes to infinity (and f(...) is
now a product of Gaussians). This is all an elementary version of what I mentioned
in my earlier post about how one absorbs the quadratic part of the Hamiltonian into
the measure to obtain a convergent expression.

I.e., one is relying on the Gaussians to tame the wild ill-defined behaviour.


I'm not so sure. You can always break up the complex exponential gaussian into its equivalent sin and cos components on both sides of the equation. Then separately equate the imaginary part and the real parts. Then you are no longer dealing with exponentials, complex numbers, nor gaussing functions. But if I remember correctly, you would still be dealing with delta functions, only in terms of sin and cos and not in terms of e^i... Does anyone have any confirmation on this? Thanks.
 
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  • #10
675
4
Section 1.9 in Quantum Field Theory by Lowell S. Brown covers the question of this thread and provides an alternative definition of the path integral which emphasizes continuous paths rather than the time-slice definition given in MacKenzie http://xxx.lanl.gov/abs/quant-ph/0004090

Thanks to friend, Barmecides and strangerep for your replies.

Todd
 
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