## Re: QFT, divergent power series, and all that...

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no,location=no, scrollbars=yes,resize=yes,status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>In article <e7f834be.0404020428.14843d78@posting.google.com>, Frank\nHellmann <C.i.m@gmx.net> writes\n>\n>How should a poor student make sense of this? We start with a\n>lagrangian, apply some more or less arbitrary prescriptions to get a\n>qunatum theory, which we can not solve, and then when we develop it\n>the predictions are divergent, but the lowest order terms are in\n>excellent agreement with experiment. WTF?\n>\n>Can anybody shed some light on this?\n\nScharf sheds, Finite QED, some light on quantisation. The arbitrary\nprescriptions for producing operators are not so bad if you work through\nFock space and construct them.\n>\n>To ask some direct questions:\n>Are those powerseries divergent in the sense that they run of to\n>infinity, or non convergent but finite (like summing over an infininte\n>series of -2 and +2, evidently bounded yet not convergent)?\n\nThey are divergent.\n\n>What is one to make of this divergence physically?\n\nThere isn\'t a generally accepted answer. My own view is that qed breaks\ndown on very small time/distance scales, and as a result the series\nbecomes less and less accurate for higher order terms.\n\n\n\n\nRegards\n\n--\nCharles Francis\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">&nbsp;&nbsp;View this Usenet post in original ASCII form </a></div><P></jabberwocky>In article <e7f834be.0404020428.14843d78@posting.google.com>, Frank
Hellmann <C.i.m@gmx.net> writes
>
>How should a poor student make sense of this? We start with a
>lagrangian, apply some more or less arbitrary prescriptions to get a
>qunatum theory, which we can not solve, and then when we develop it
>the predictions are divergent, but the lowest order terms are in
>excellent agreement with experiment. WTF?
>
>Can anybody shed some light on this?

Scharf sheds, Finite QED, some light on quantisation. The arbitrary
prescriptions for producing operators are not so bad if you work through
Fock space and construct them.
>
>Are those powerseries divergent in the sense that they run of to
>infinity, or non convergent but finite (like summing over an infininte
>series of -2 and +2, evidently bounded yet not convergent)?

They are divergent.

>What is one to make of this divergence physically?

There isn't a generally accepted answer. My own view is that qed breaks
down on very small time/distance scales, and as a result the series
becomes less and less accurate for higher order terms.

Regards

--
Charles Francis

 PhysOrg.com physics news on PhysOrg.com >> Promising doped zirconia>> New X-ray method shows how frog embryos could help thwart disease>> Bringing life into focus


Charles Francis wrote in message news:... > In article , Frank > Hellmann writes ...... > > >What is one to make of this divergence physically? > > There isn't a generally accepted answer. My own view is that qed breaks > down on very small time/distance scales, and as a result the series > becomes less and less accurate for higher order terms. > > In his "Dialog on quantum gravity" Rovelli states that "the weak field approximation fails for GR" because it "is based on Feynman integrals that sum over infinite momenta, namely over regions of arbitrarily small volume". A result of QLG is that "there is literally no volume smaller than the Planck volume" so that "it makes no sense to integrate over degrees of freedom far smaller than the Planck length" since such regions "literally do not exist". Now, according to your view above, if "there is literally no volume smaller than the Planck volume" that would impact divergences in QFT too. Right? Regards, IV



Frank Hellmann wrote: > Ok, here is a poor undergrad student struggling with his first courses > on QFT... Overall an frustrating experience. After the beautifully > concise and mathematically (or at least formulistically) precise > treatment of mechanics and quantum mechanics (particularly in the > books of Landau Lifschitz and Dirac respectively) this appears > horribly complicated and obscure (I'm mostly working with Weinbergs > book, any other recommendations?). > > One particular aspect is really driving me mad though, particularly > because none of the books I'm reading appears to have much to say on > it. > > It appears that the pertubation series does generally not converge. > Now my prof in advanced stat phys had to say something on this, but it > was very hand wavy and included various incantations of a mystic > concept called Borel summation. > > How should a poor student make sense of this? We start with a > lagrangian, apply some more or less arbitrary prescriptions to get a > qunatum theory, which we can not solve, and then when we develop it > the predictions are divergent, but the lowest order terms are in > excellent agreement with experiment. WTF? > > Can anybody shed some light on this? > What is one to make of this divergence physically? > What is Borel summation? (I have found applications and simple > definitions on the web, and in earlier threads but it remains > mysterious to me, particularly it is said that the original value of > the sum can be extracted from the Borel sum, how would one even define > the "original value of the sum" if this sum is divergent?) See http://www.lns.cornell.edu/spr/2003-05/msg0051062.html and context. > How is it's use justified in QFT? (in particular, is it just a > mathematical operation to get something out or is there physics > involved at that step) In certain cases, where nonperturbative QM applies, one can show that the nonperturbative result satisfies the properties needed to show that Borel summation of the perturbative expansion reproduces the nonperturbative result. Arnold Neumaier

## Re: QFT, divergent power series, and all that...

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no,location=no, scrollbars=yes,resizable=yes,status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>In article &lt;61789046.0404061053.6a1a619d@posting.google.com&gt;, Italo\nVecchi &lt;vecchi@weirdtech.com&gt; writes\n&gt;Charles Francis &lt;charles@clef.demon.co.uk&gt; wrote in message news:&lt;c4sap\n&gt;f\\$c99\\$1@lfa222122.richmond.edu&gt;...\n&gt;&gt; In article &lt;e7f834be.0404020428.14843d78@posting.google.com&gt;, Frank\n&gt;&gt; Hellmann &lt;C.i.m@gmx.net&gt; writes\n&gt;.....\n&gt;&gt;\n&gt;&gt; &gt;What is one to make of this divergence physically?\n&gt;&gt;\n&gt;&gt; There isn\'t a generally accepted answer. My own view is that qed breaks\n&gt;&gt; down on very small time/distance scales, and as a result the series\n&gt;&gt; becomes less and less accurate for higher order terms.\n&gt;&gt;\n&gt;&gt;\n&gt;In his "Dialog on quantum gravity" Rovelli states that "the weak field\n&gt;approximation fails for GR" because it "is based on Feynman integrals\n&gt;that sum over infinite momenta, namely over regions of arbitrarily\n&gt;small volume". A result of QLG is that "there is literally no volume\n&gt;smaller than the Planck volume" so that "it makes no sense to\n&gt;integrate over degrees of freedom far smaller than the Planck length"\n&gt;since such regions "literally do not exist".\n\nFor my own reasons I think the limiting distance is much smaller, the\nSchwarzschild radius of an electron. But I agree with the principle.\n&gt;\n&gt;Now, according to your view above, if "there is literally no volume\n&gt;smaller than the Planck volume" that would impact divergences in QFT\n&gt;too. Right?\n\nSpecifically in would impact the Landau Pole, which is the last\nremaining divergence which does not already have a resolution. If there\nis a lower bound on time/distance there will be also be a bound on the\nnumber of vertices possible in a Feynman diagram, and the series will\nterminate.\n\n\n\n\nRegards\n\n--\nCharles Francis\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">&nbsp;&nbsp;View this Usenet post in original ASCII form </a></div><P></jabberwocky>In article <61789046.0404061053.6a1a619d@posting.google.com>, Italo
Vecchi <vecchi@weirdtech.com> writes
>Charles Francis <charles@clef.demon.co.uk> wrote in message news:<c4sap

>f$c99$1@lfa222122.richmond.edu>...
>> Hellmann <C.i.m@gmx.net> writes

>.....
>>
>> >What is one to make of this divergence physically?

>>
>> There isn't a generally accepted answer. My own view is that qed breaks
>> down on very small time/distance scales, and as a result the series
>> becomes less and less accurate for higher order terms.
>>
>>

>In his "Dialog on quantum gravity" Rovelli states that "the weak field
>approximation fails for GR" because it "is based on Feynman integrals
>that sum over infinite momenta, namely over regions of arbitrarily
>small volume". A result of QLG is that "there is literally no volume
>smaller than the Planck volume" so that "it makes no sense to
>integrate over degrees of freedom far smaller than the Planck length"
>since such regions "literally do not exist".

For my own reasons I think the limiting distance is much smaller, the
Schwarzschild radius of an electron. But I agree with the principle.
>
>Now, according to your view above, if "there is literally no volume
>smaller than the Planck volume" that would impact divergences in QFT
>too. Right?

Specifically in would impact the Landau Pole, which is the last
remaining divergence which does not already have a resolution. If there
is a lower bound on time/distance there will be also be a bound on the
number of vertices possible in a Feynman diagram, and the series will
terminate.

Regards

--
Charles Francis



Charles Francis wrote in message news:... > In article <61789046.0404061053.6a1a619d@posting.google.com>, Italo > Vecchi writes .... > >In his "Dialog on quantum gravity" Rovelli states that "the weak field > >approximation fails for GR" because it "is based on Feynman integrals > >that sum over infinite momenta, namely over regions of arbitrarily > >small volume". A result of QLG is that "there is literally no volume > >smaller than the Planck volume" so that "it makes no sense to > >integrate over degrees of freedom far smaller than the Planck length" > >since such regions "literally do not exist". > > For my own reasons I think the limiting distance is much smaller, the > Schwarzschild radius of an electron. But I agree with the principle. > > > >Now, according to your view above, if "there is literally no volume > >smaller than the Planck volume" that would impact divergences in QFT > >too. Right? > > Specifically in would impact the Landau Pole, which is the last > remaining divergence which does not already have a resolution. If there > is a lower bound on time/distance there will be also be a bound on the > number of vertices possible in a Feynman diagram, and the series will > terminate. > Thanks, very interesting. I wonder what's so special about the Schwarzschild radius of (of all things) the electron. I also wonder at which scales QFT series start to diverge from experimental results ,i.e. which terms do you actually have to throw away since they ruin your predictions instead of enhancing them. I guess that could provide an answer to the Schwarzschild electron radius vs Planck length issue. Regards, IV



In article <61789046.0404090027.3c4943c2@posting.google.com>, Italo Vecchi wrote: > I also wonder at which scales QFT series start to diverge from > experimental results ,i.e. which terms do you actually have to throw > away since they ruin your predictions instead of enhancing them. The rough guide for asymtotic series is that they get better until you exceed the inverse of the expansion parameter number of terms. So in QED, the expansion parameter is $1/137,$ so you might expect the approximation to get better for 100 terms or so. In QCD it's more like $1/10,$ and there are examples of series which show very poor convergence (though they still provide reasonable apporximations). I do lattice QCD perturbation theory, and here the coupling is more like .2-0.3 so we don't expect much beyond 3-4 terms would be any use. Matthew Nobes -- manobes@sdf.lonestar.org SDF Public Access UNIX System - http://sdf.lonestar.org



Thank you a lot for the pointers. I particularly like Weinbergs book so far because it appears to keep the physics in the foreground, while linking them extensively to the mathematics (at least in the first few chapters which are all I have studied yet...), but for the time being I don't need to understand just pass exams, so I'm gonna tackle this again later, including the examples by Mr Baez which shed quite some light on these issues. I have one additional mathematical question though, every now and then the term operator valued distribution crops up. Now for ordinary QM we have distributions, where we identify the Hilbert/Schwartz Space used to construct them in the first place naturally through the integral representation of the distributions: $f_d(g) = \int dx (f(x)*g(x))$ with f element of the Hilbert Space (which contains the Schwartz space) and g element of the Schwartz space and $f_d$ the distribution naturally assosciated with f. This assosciation of course preserves the vector space structure of the function space, and allows us to define a large number of nice tools on the Hilbert space in a very naturall fashion. Now for operator valued distributions, how does this construction proceed? How does one identify operator valued functions of a particular space with the distributions? Is there a similarily naturall way to do it? cheers, Frank Hellmann



Frank Hellmann wrote: > I have one additional mathematical question though, every now and then > the term operator valued distribution crops up. Now for ordinary QM we > have distributions, where we identify the Hilbert/Schwartz Space used > to construct them in the first place naturally through the integral > representation of the distributions: > $f_d(g) = \int dx (f(x)*g(x))$ > with f element of the Hilbert Space (which contains the Schwartz > space) and g element of the Schwartz space and $f_d$ the distribution > naturally assosciated with f. This assosciation of course preserves > the vector space structure of the function space, and allows us to > define a large number of nice tools on the Hilbert space in a very > naturall fashion. > > Now for operator valued distributions, how does this construction > proceed? How does one identify operator valued functions of a > particular space with the distributions? Is there a similarily > natural way to do it? If $\phi(x)$ is the operator valued distribution, and g a Schwartz test function then $\phi(g) = \int dx \phi(x) g(x)$ is a good operator. Arnold Neumaier



Matthew A. Nobes wrote: > In article <61789046.0404090027.3c4943c2@posting.google.com>, > Italo Vecchi wrote: > > >>I also wonder at which scales QFT series start to diverge from >>experimental results ,i.e. which terms do you actually have to throw >>away since they ruin your predictions instead of enhancing them. > > > The rough guide for asymtotic series is that they get better until > you exceed the inverse of the expansion parameter number of terms. This is a very poor guide. It very much depends on the magnitude of the terms in the series. Generally, an asymptotic series can be trusted as long as the first discarded term is smaller than the last used term, though even this rule may be fallacious. One can study these things quite well with functions which have known asymptotic expansions (Watson's lemma, etc.) Arnold Neumaier



Arnold Neumaier wrote in message news:<407BE7B2.6070206@univie.ac.at>... > Matthew A. Nobes wrote: .... > > The rough guide for asymtotic series is that they get better until > > you exceed the inverse of the expansion parameter number of terms. > > This is a very poor guide. > It very much depends on the magnitude of the terms in the > series. Generally, an asymptotic series can be trusted as long as the > first discarded term is smaller than the last used term, though even > this rule may be fallacious. > > One can study these things quite well with functions which have > known asymptotic expansions (Watson's lemma, etc.) > The problem here is that we do not have a mathematical function, only experimental results. I was suggesting that the onset of divergence from measured data may correspond to the lower bound on physically meaningful time/distance. IV



> If $\phi(x)$ is the operator valued distribution, and g a Schwartz test > function then > $\phi(g) = \int dx \phi(x) g(x)$ > is a good operator. > Hu? I don't see what you are doing there... Is this to imply that the dual space is taken as the space of mapings from the operaor valued functions to the operators instead of to the reals? And if not: Assuming $\phi(x)$ is an operator valued function, and ovd $(\phi)$ is a distribution maping from the space of operator valued functions into the reals, then for any test function g (x) we could get the function $\phi(x)g(x),$ that is $\phi$ operating on g, which if we integrate gives a function in the reals that depends on linearly on $\phi$ and in some way on g. Thus at least naivly it appears that there is no naturall way to identify the distribution ovd $(\phi,g) := \int dx \phi(x) g(x)$ with the operator valued function $\phi (x),$ as there is an additional dependency on g. Does every possible function g then furnish an equivalent identification for the function space in the dual space? Is the g dependency therefore irrelevant? cheers, Frank.



Italo Vecchi wrote: > Arnold Neumaier wrote in message news:<407BE7B2.6070206@univie.ac.at>... > >>Matthew A. Nobes wrote: > > ... > >>>The rough guide for asymtotic series is that they get better until >>>you exceed the inverse of the expansion parameter number of terms. >> >>This is a very poor guide. >>It very much depends on the magnitude of the terms in the >>series. Generally, an asymptotic series can be trusted as long as the >>first discarded term is smaller than the last used term, though even >>this rule may be fallacious. >> >>One can study these things quite well with functions which have >>known asymptotic expansions (Watson's lemma, etc.) >> > > > The problem here is that we do not have a mathematical function, only > experimental results. I was suggesting that the onset of divergence > from measured data may correspond to the lower bound on physically > meaningful time/distance. With experimental results you just have numbers, and not infinite series, so questions of convergence do not occur. On the other hand, if you know of an infinite series a finite number of terms only, the result can be, strictly speaking, anything. But usually one applies some extrapolation algorithm (e.g., the $\epsilon$ or $\eta$ algorithm) to get a meaningful guess for the limit, and estimates the error by doing the same several times, keeping a variable number of terms. The difference between consecutive results can count as a reasonable (though not foolproof) error estimate of these results. But to have reliable bounds you need to know an exact definition of what you are approximating, and work from there. Arnold Neumaier



John Baez wrote: > Exercise: Show that $1^k - 2^k + 3^k - 4^k + .$.. is Abel summable > and compute its Abel sum. > > Exercise: Show that Abel summation dominates (C,k) summation for > all k. > > Exercise: Find a series that is Abel summable but not (C,k) summable > for any k. OK, I eventually blundered my way through the Cesaro exercises, but I don't know how to do the first Abel exercise. Please take pity and give some more hints. $- MM$.



Frank Hellmann wrote: >>If $\phi(x)$ is the operator valued distribution, and g a Schwartz test >>function then >> $\phi(g) = \int dx \phi(x) g(x)$ >>is a good operator. >> > > > Hu? > I don't see what you are doing there... > Is this to imply that the dual space is taken as the space of mapings > from the operaor valued functions to the operators instead of to the > reals? No. Probably I'd have said that phihat(g) $= \int dx \phi(x) g(x)$ is a good operator. If one works rigorously, one assumes the existence of the left hand side as a mapping from test functions to operators, and regards the right hand side just as a different notation for it. > And if not: > > Assuming $\phi(x)$ is an operator valued function, and ovd $(\phi)$ is a > distribution maping from the space of operator valued functions into > the reals, I don't understand how you read that into what I wrote. In my statement, $\phi(x)$ is not a function but a distribution, and g is a mollifier that smears the singularity so that one gets something nice. Arnold Neumaier



> > And if not: > > > > Assuming $\phi(x)$ is an operator valued function, and ovd $(\phi)$ is a > > distribution maping from the space of operator valued functions into > > the reals, > > I don't understand how you read that into what I wrote. > In my statement, $\phi(x)$ is not a function but a distribution, > and g is a mollifier that smears the singularity so that one > gets something nice. > > > Arnold Neumaier Well a distribution is a linear mapping (aka operator) from the test functions into the reals so phihat(g) is certainly a distribution, an $\phi(x)$ it's integral core, as phihat(g) is real valued if $\phi(x)$ and g(x) are both real valued. I don't see where operator valued distributions enter the picture with what you wrote, this seems to be just standard distributions to me. If I'm way of the mark do you know a good textbook that contains an introduction?



In article <38b55e8c.0404152345.219b8d58@posting.google.com>, MM wrote: >John Baez wrote: > > > Exercise: Show that $1^k - 2^k + 3^k - 4^k + .$.. is Abel summable > > and compute its Abel sum. > > > > Exercise: Show that Abel summation dominates (C,k) summation for > > all k. > > > > Exercise: Find a series that is Abel summable but not (C,k) summable > > for any k. > >OK, I eventually blundered my way through the Cesaro exercises, but I >don't know how to do the first Abel exercise. I think I've got this one! Let me add a 0th term and change the signs (so that my solution will be the negative of the original): we want to evaluate $$0^k - 1^k + 2^k - .[/itex].$. = \sum_j (-1)^j j^k$. To do Abel summation, we're supposed to consider the function f defined by f(x) $= \sum_j (-1)^j j^k x^j$. Let D be the operator defined by $Dg = x (dg/dx)$. Then $D(x^j) = j x^j,$ so f(x) $= \sum_j (-1)^j D^k(x^j)= D^k(\sum_j (-x^j))= D^k(1/(1+x))$. Like any good physicist, I'm being cavalier about issues of uniform convergence and the like, switching orders of derivatives and sums with wild abandon. But I think this is all OK for $|x| < 1$. Now we have to evaluate this thing at $x = 1$. If my kung fu were stronger, I'd have been able to do this in some clever way. What I actually did was give it to Maple, evaluate it explicitly for k from 1 to 20, and feed the results into the Encyclopedia of Integer Sequences. [Incidentally, anyone who's read this far into this post would probably like the Encyclopedia of Integer Sequences. If you don't know about it, check out http://www.research.att.com/~njas/sequences/ ] After playing around a bit, I found that the sum is $-(2^(k+1) - 1) B_{k+1} / (k+1),$$ where [itex]B_n$ is the nth Bernoulli number. Remember that I flipped the sign at the beginning, so the solution to the original problem the negative of this. I haven't actually proved this last expression, but it works for k=1 through 20, and I also spot-checked it $at k=100,$ so I'm morally sure it's right. I suppose that it should be easy enough to prove by messing around a bit with the definitions of the Bernoulli numbers. $$-Ted$$ -- [E-mail me at name@domain.edu, as opposed to name@machine.domain.edu.]



In article <38b55e8c.0404152345.219b8d58@posting.google.com>, MM wrote: >John Baez wrote: > > Exercise: Show that $1^k - 2^k + 3^k - 4^k + .$.. is Abel summable > > and compute its Abel sum. > > > > Exercise: Show that Abel summation dominates (C,k) summation for > > all k. >OK, I eventually blundered my way through the Cesaro exercises, but I >don't know how to do the first Abel exercise. > >Please take pity and give some more hints. I apologize for giving such a hard one to start with! How about this: $$1 - 2 + 3 - 4 + .[/itex]... To Abel sum this, you need to compute the limit of sum $(-1)^{n+1} n x^n$$ as x approaches 1 from below. The trick is to find an explicit closed form for this sum, and then take the limit. If you have trouble with the "trick" part, maybe it's good to think a bit more generally. If you know sum [itex]a_n x^n = f(x)$ then how is sum $n a_n x^n$ related to f(x)? This is a nice thing to know. If you know it, you can reduce the Abel sum $$1 - 2 + 3 - 4 + .[/itex]... $:= lim_{x -> 1}$ sum $(-1)^{n+1} n x^n$$ to this easier one: $$1 - 1 + 1 - 1 + .$... $:= lim_{x -> 1}$ sum $(-1)^{n+1} x^n$$ You can do the latter one with a geometric series! Alternatively, you can do it using Cesaro summation, together with the theorem that whenever Cesaro summation works, Abel summation does too - and gives the same answer. If you did the Cesaro summation problems, presumably you know already that Cesaro summation gives $$1 - 1 + 1 - 1 + .$.$.. = 1/2$$ If you get stuck, I can tell you about some webpages of mine that give the answer to the Abel sum [itex]1 - 2 + 3 - 4 + .$.. and more generally $1^k - 2^k + 3^k - 4^k + .$.. This stuff is fun! It's like black magic. ....................................................................... ..... Puzzle 19: As of February 2004, five of the ten richest people in the world had the same last name. What is it? If you get stuck, go here: http://www.math.ucr.edu/home/baez/puzzles/19.html