From: israel@math.ubc.ca (Robert Israel) Subject: Re: Analytic Function? Date: 18 Aug 2000 17:10:44 GMT Newsgroups: sci.math,sci.physics.research Summary: Using Borel summation to deal with divergent series e.g. in quantum field theory In article <8nhmg3$lgu@edrn.newsguy.com>, Daryl McCullough wrote: >As a toy example, I imagined that some field theory gave >a power series expansion for some observable of the form: > > 1. O(z) = sum of (-z)^j j! >where z is some coupling constant. >Wildly divergent. However, if we use the integral representation >of the factorial, we get: > 2. O(z) = sum of (-z)^j (integral of exp(-t) t^j dt) >Taking the summation inside gives: > 3. O(z) = integral of exp(-t) (sum of (-zt)^j) dt >But the summation is the formal power series for 1/(1+et). >Therefore, > 4. O(z) = integral of exp(-t)/(1+zt) >which is much better behaved as a function of z. So, maybe when >we are faced with a power series like 1, it's a hint that we >are really dealing with a case like 4. >Would any of the summation tricks for divergent series give you >something like 4 if you start with something like 1? Basically this is Borel summability. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ============================================================================== From: daryl@cogentex.com (Daryl McCullough) Subject: Re: Analytic Function? Date: 18 Aug 2000 23:14:51 GMT Newsgroups: sci.math,sci.physics.research daryl@cogentex.com says... >As a toy example, I imagined that some field theory gave >a power series expansion for some observable of the form: > > 1. O(z) = sum of (-z)^j j! [stuff deleted] >Therefore, > > 4. O(z) = integral of exp(-t)/(1+zt) Robert Israel informed me that this is a special case of Borel summability. The general case works like this: Suppose that we have a formal power series for f(z): f(z) = sum c_n z^n We can improve the convergence by replacing c_n by c_n/n!, but that changes the answer: F(z) = sum c_n/n! z^n However, we can cleverly write f(z) in terms of F(z) as follows: First, multiply and divide by n! 1. f(z) = sum (c_n/n!) * n! z^n Now, rewrite the second n! as integral exp(-t) t^n dt: 2. f(z) = sum c_n/n! z^n integral exp(-t) t^n dt Bring the summation inside: 3. f(z) = integral exp(-t) (sum c_n/n! (zt)^n) dt Now, rembering the definition of F(z), we rewrite this as: 4. f(z) = integral exp(-t) F(zt) My example was a special case in which c_n = (-1)^n n! and F(z) = 1/(1+z) Daryl McCullough CoGenTex, Inc. Ithaca, NY ============================================================================== From: baez@galaxy.ucr.edu (John Baez) Subject: Re: Analytic Function? Date: 23 Aug 2000 15:34:27 GMT Newsgroups: sci.math,sci.physics.research In article <8nhmg3$lgu@edrn.newsguy.com>, Daryl McCullough wrote: >QED perturbation rules give us a prediction >for observables as a formal power series in the coupling constant. >However, this power series may not converge, but perhaps there >is information in the power series, nevertheless. Right! >As a toy example, I imagined that some field theory gave >a power series expansion for some observable of the form: > > 1. O(z) = sum of (-z)^j j! > >where z is some coupling constant. [sneaky resummation trick deleted] >Therefore, > > 4. O(z) = integral of exp(-t)/(1+zt) > >which is much better behaved as a function of z. So, maybe when >we are faced with a power series like 1, it's a hint that we >are really dealing with a case like 4. >Would any of the summation tricks for divergent series give you >something like 4 if you start with something like 1? As Robert Israel pointed out, you've reinvented Borel summation! Borel summation is very useful in quantum mechanics and quantum field theory. For example, consider the anharmonic oscillator, which is the quantum mechanical system with Hamiltonian H = (p^2 + q^2)/2 + c q^4 When the coupling constant c is greater than zero, the ground state energy of this system is an analytic function of c. Unfortunately, right at c = 0 the ground state energy is not analytic. So when you use perturbation theory to express the ground state energy as a power series in c, you get a series with zero radius of convergence. Fortunately, this series is asymptotic - and you can use Borel summation when c > 0 to sum it to the right answer. Very similar remarks hold for a wide variety of quantum field theory problems, like the phi^4 theory with a spatial cutoff but no ultraviolet cutoff in spacetime dimensions 2 and 3. In this example (and others), the n-point functions can be correctly computed by doing perturbation theory and then using Borel summation.