[SOLVED] Renormalization

drnobes

Arnold Neumaier wrote:
> Matthew Nobes wrote:
> > On 2005-04-26, Arnold Neumaier <Arnold.Neumaier@univie.ac.at>
wrote:
[lattice QCD]
> >>But you can't go to the limit,
> >
> > Yes you can, in perturbation theory you can take the a->0 limit.
>
> But not on a small or even a big computer.

I think we may be talking past each other. In peturbation theory, I
can compute some quantity to n-loops. I will have some counterterms,
which I can then use to get rid of infinities. If I used dim-reg I get
poles in \epsilon, if I use a lattice I get things like \log(a) and
1/a. Are you claiming I can't adjust counterterms to get rid of the
divergences in the lattice case?

> > The renormalization program works the same way, order by order, you
tune
> > counterterms, etc. It is more technically involved (there are more
> > diagrams, more, and more complicated, Feynman rules) but is
possible.
> >
> > There are some theorems, which prove this, due to Reisz IIRC.
>
> Interesting; I didn't know this.
[snip]
> but the abstract says it is for SU(N) gauge theory. In particular,
> it does not apply to QED.

Okay, but for SU(N) the point stands. As it does for massive QED.
[snip]
> >>while you can in dimensional
> >>regularization at fixed loop order. With lattice calculations,
> >>you'd never come close in accuracy to the experimental value
> >>of the Lamb shift for hydrogen.
> >
> > I don't know about the Lamb shift calculations, but it's worth
pointing
> > out that Tom Kinoshita doesn't use dim-reg for his four- and
five-loop
> > computations of the magnetic moment anomoly.
>
> Yes. He uses NRQED, which is, however, specially adapted to QED.

Huh? For the g-2 he uses standard QED, not NRQED. See hep-ph/0402206,
he uses standard (relativistic) fermion propagators (A3).

Matthew

Arnold Neumaier

jsolomon@mail.com wrote:
>
> Please consider the example of my last post which I will repeat below.
> Let,
>
> F1(n) = \int_1^{\infty} x^n dx (integration in the interval [1,
> oo])
>
> An integral is defined as an infinite sum. For example at n=0 we would
> have,
>
> F1(0) = 1 + 1 + 1 + 1 + 1 + etc. (1)
>
> Now define another function,
>
> F2(n) = -1/(n+1) (2)
>
> Now F1(n) = F2(n) for n<-1. If we replace n by the complex variable z
> we have,
>
> F2(z) = -1/(z+1) (3)
>
> We would say that F2(z) is the analytic continuation of F1(z) to the
> entire complex plane (except z = -1). Now at z = 0 we have F2(0) = -1.
> If F1(z) and F2(z) are equal at z=0 then we would have,
>
> 1 + 1 + 1 +1 + etc. = -1 (4)
>
> which doesn't make sense.
>
> Now suppose I have a mathematical model of a physical process. My
> model predicts that under certain conditions I will get the result R1
> which is given by,
>
> R1 = 1/(F1(0) + 1) (5)
>
> Now when I examine my expression for F1(0) I conclude it is infinite.
> If I use this fact in the above expression for R1 I get,
>
> R1 = 1/(oo + 1) = 0 (6)
>
> This is a perfectly reasonable result for a physical process. However
> by your reasoning I should replace F1(0) by F2(0). In this case,
>
> R1 = 1/(-1 + 1) = 1/0 = oo (7)
>
> Therefore I get a nonsensical result. This example shows that you
> cannot arbitrarily replace a function with its analytical continuation.
> They are not equal everywhere.

They are equal where both make sense, and the analytic continuation
is the unique way to kae sense elsewhere.

It is like (x^2-1)/(x-1)=x+1. Equality holds only for x not 1, but
no one doubts that the right hand side is the right way to extend
the function where it wasn't defined before.


Arnold Neumaier

Arnold Neumaier

Dan Piponi wrote:
> A Bergman wrote:
>
>
>>Sure it does (up to monodromies).
>
>
> Dan Solomon's criticisms are entirely justified. There is no reason to
> think f1(z)=f2(z) outside of C1 because he made no assumption that f1
> was analytic outside of C1. And even if f1 is analytic everywhere in its
> domain it might not actually be defined outside of C1 even though f2 is.

Of course this is true from a fully rigorous point of view, since
there is no rigorous definition of a covariant 4D quantum field theory.
Thus one needs to make somewhere some additional reasonable assumptions,
nd analyticity is _the_ reasonable assumption since it is known to hold
in the nonrelativistic case.


> This replacement of functions with their analytic continuations is just
> one of those things that physicists do, but rarely explicitly state they
> are doing, that make life difficult for mathematicians who try to follow
> their work.

Yes, one needs to learn to read between the lines.
Most of physics is written for physicists, not for mathematicians.


Arnold Neumaier

Arnold Neumaier

C. M. Heard wrote:

> I remember puzzling over dimensional regularization as a graduate student,
> and the issue I had was this: to the extent that a d-dimensional version
> of a quantum field theory is defined at all, it is for non-negative integral
> values of d only, and if the theory is non-trivial, it is usually for some
> subset of the integral values less than four. That is a finite set of
> points, and you cannot make an analytic continuation from that. Sure, the
> integrals that we get as terms in the perturbation series have a form that
> admits definition for a continuum of values, and that will admit a
> continuation; but surely this must be seen as a fortuitous accident, since
> the underlying QFT is itself undefined except for a finite number of points
> in that continuum.

For the traditional physicist's setting of dimensional regularization,
this is indeed true. But there are variants that are better justified.


> Does anyone who does rigorous work in contructive QFT use dimensional
> regularization? Can it actually be made rigorous? (The only
> regularization I've seen discussed in the context of constructive QFT
> is lattice regularization, but that is probably because of my ignorance.)

See the section ''Dimensional regularization'' in my
theoretical physics FAQ at
http://www.mat.univie.ac.at/~neum/physics-faq.txt


Arnold Neumaier

Arnold Neumaier

Matthew Nobes wrote:
> On 2005-04-27, Arnold Neumaier <Arnold.Neumaier@univie.ac.at> wrote:
>
>>jsolomon@mail.com wrote:
>
> [snip]
>
>>>>Arnold Neumaier
>>>
>>>So Dr. Neumaier says yes and you say no,no, no. Maybe you two should
>>>talk to each other.
>>
>>It was a matter of emphasis. We both agree on that _some_
>>regularization procedure must be part of the basic postulates.
>>The question is which one.
>>
>>I recommended dimensional regularization, because it is most flexible,
>>but allowed in my qualifying statement for less flexible
>>alternatives. Matthew Nobes http://www.lns.cornell.edu/~nobes/
>>works with lattice gauge theories and hence emphasizes the freedom
>>to use lattice regulators instead.
>
>
> Not to start an arguement, but it should be pointed out that "most
> flexible" strongly depends on the problem you want to solve. For a
> numerical evaluation of the path integral, a lattice regulator is the
> "most flexible".
>
> As I understand it (and I'm willing to be corrected)
> dim-reg is defined only at the level of perturbation theory.

But perturbation theory in the widest sense, including renormalization
group improved computations and effective 1PI and 2PI theories, which go
far beyond simple perturbation theory. (For example, one can even
consider phase transitions.)


> For non-abelian
> gauge theories, dim-reg is certainly the most flexible scheme for
> perturbative calculations.
>
> [snip]
>
>>The proof of absence of infinities for all 4D renormalizable theories
>>in the limit of removed cutoff has been given only for dimensional
>>regularization, I believe.
>>
>>Thus dimensional regularization is safe, while the situation is
>>less clear for lattice regularization.
>
> The theorems of Reisz apply here. For certain classes of lattice
> actions (including some, but not all of the popular actions used in
> simulations) you can show that the a -> 0 limit in perturbation theory
> gives you the same results as the eps -> 0 limit in dim-reg.

I believe this statement is at best true for asymptotically free
theories.

Phi^4 is nointrivial in perturbation theory, but appears to be trivial
in lattice regularization, which means that one does _not_ get the same
results.


> This is certainly true for the Wilson gauge and quark actions, so a
> lattice regulated theory is equivalent to using dim-reg, from a
> perturbative standpoint.

I'd like to see references substantiating this.


Arnold Neumaier

Matthew Nobes

On 2005-05-02, Arnold Neumaier <Arnold.Neumaier@univie.ac.at> wrote:
> drnobes wrote:
>
[snip]
>>
>> I think we may be talking past each other.
>
> Yes, it seems so.
>
>
>> In perturbation theory, I
>> can compute some quantity to n-loops. I will have some counterterms,
>> which I can then use to get rid of infinities. If I used dim-reg I get
>> poles in \epsilon, if I use a lattice I get things like \log(a) and
>> 1/a. Are you claiming I can't adjust counterterms to get rid of the
>> divergences in the lattice case?
>
> No; only that you can't do the limit on a computer.
> (Though one perhaps can, with today's symbolic packages?)

Okay, we were talking past each other. I was refering to ordinary PT.
And no, you can't take the limit as a->0. You can run at several
spacings and fit, which is good enough for me, but is not taking the
limit.

[snip]
>>>but the abstract says it is for SU(N) gauge theory. In particular,
>>>it does not apply to QED.
>>
>> Okay, but for SU(N) the point stands. As it does for massive QED.
>
>
> The latter is not clear to me, since QED contains fermions.
> Is there any work on proving that lattice regularization works
> with fermions? I thought there were problems with fermion doubling!?

In another post I mentioned that the Reisz theorems apply to Wilson
quarks, which don't have doublers. Far better are the "new"
formulations of Lattice Fermions, which go by the names "domain wall" or
"overlap". The have an exact chiral symmetry (they're also equivalent).

Unfortunately they're very expensive to simulate with.

>>>>out that Tom Kinoshita doesn't use dim-reg for his four- and five-loop
>>>>computations of the magnetic moment anomaly.
>
> Yes, because, as he remarks in reference [62] of the paper quoted below,
> it is not known how to implement it in a numerically stable way at high
> orders. (But using lattice regularization would be even more
> inaccurate...)

Yes, which was my point, regulators should be tuned to match the problem
at hand. I'm still not sure what you mean that the lattice regulator
would be more inaccurate. If the calculation was done to four-loops
with any regulator, it would be the same. (again, I'm *not* talking
about Monte-Carlo simulation, just useing a lattice regulator).

>>>Yes. He uses NRQED, which is, however, specially adapted to QED.
>>
>> Huh? For the g-2 he uses standard QED, not NRQED. See hep-ph/0402206,
>> he uses standard (relativistic) fermion propagators (A3).
>
> I didn't know this recent paper. Earlier work was done with NRQED.
> (Perhaps by Lepage and not Kinoshita?)

Peter (Lepage, my boss) did the Lamb shift with NRQED, and lots of other
bound state problems. I don't think he did the MMA though. The MMA in
NRQED would require you to match the \sigma \dot B operator anyway, which
I suspect would amount to doing the relativistic calculation. I
certainly can't see how you'd get an accurate MMA without an accurate
matching for that operator (and probably at least one more order in v/c
matched to at least one-loop).

--
Matthew Nobes | email: nobes@lepp.cornell.edu
Newman Lab, Cornell University | web: http://lepp.cornell.edu/~nobes/
Ithaca NY 14853, USA |

Dan Piponi

Aaroin Bergman points out:
> Weinberg explicitly states above (11.A.2)

Yes, Weinberg states he is making an analytic continuation, but he
doesn't justify it mathematically. He's simply taking something he
can't compute and replacing it with something he can compute. The
unstated hidden assumption in physics publications is that you can make
this substitution whenever you feel like it. Occasionally physicists do
make these assumptions explicit: eg. I seem to remember one school of
thought explicitly stating, as an axiom of physics, that the S-matrix
is analytic, but this is the exception.

This isn't meant to be a criticism of what physicists do, I'm just
trying to give one reason why physicists and mathematicians can
sometimes speak at cross-purposes, something that has been happening
here.

Arnold Neumaier says:
> Yes, one needs to learn to read between the lines.

But physicists may also sometimes need to recognise when they're doing
that - especially when talking to mathematicians.

Chris Oakley

> >>>At present I believe that there is no such thing as an interacting
> >>>QFT based on Feynman-Dyson perturbation theory in 3+1 dimensions.
> >>>Dr. Neumaier tells me that if I sacrifice Lorentz invariance then
> >>>there is a way of developing the theory from first principles
> >>>without sacrificing consistency. The lack of Lorentz invariance
> >>>makes the theory ugly compared to what I worked out myself, but if
> >>>it does indeed work, I would be forced to accept it, and I would
> >>>qualify my criticisms of QFT (as currently practised) accordingly.
> >>
> >>QFT without Lorentz invariance is not ugly at all. It is simply a
> >>more general approach, from which the Lorentz invariant theories are a
> >>special case.
> >
> > Any situation where a fudge factor is introduced to conceal some basic
> > failure is ugly,
>
> You are shifting grounds. Above you said that the lack of Lorentz
> invariance makes the theory ugly, and I showed that this is not
> the case, just as differential geometry is not ugly though it
> does not assume any symmetries.

I did not say that every generalisation of a theory is ugly. But the
generalisation of QFT to the case where it is only Lorentz invariant with a
particular choice of the generalising parameter certainly is. Especially as
it flies in the face of some of the most copious experimental evidence we
have (i.e. that which supports Special Relativity). If one is going to
generalise, it should be to align with experimental evidence rather than to
intentionally contradict it (Newton's generalisation of the terrestrial
force that makes apples fall to understand planetary motions is a nice
example).

> > whether is is a non-Lorentz-invariant cutoff or
> > (meaningless) complex number of spacetime dimensions in RQFT, or an
extra
> > fictitious amount introduced into a company's accounts to make the books
> > balance. These are all examples of the same phenomenon - dishonesty, and
> > dishonesty is ugly.
>
> There is nothing dishonest in renormalization since nothing is
> concealed. Everything happens completely open, for anyone to check.
> And many understand what is there to understand.

I have seen upright, respectable members of the HEP community write
equations on blackboards where they equate the difference of two divergent
integrals to the number they want to get. I do not believe that such things
would happen had not the floodgates of mathematical impropriety been opened
by advocates of renormalization.

> >>I have more important things to do than to write papers specifically
> >>for someone like you who needs every detail spelled out and gives up
> >>at the slightest inconvenience. My students can do much better;
> >>they give up only when they are really lost, and then come with the
> >>part they completed successfully to show that they deserve further
> >>supervision.
> >
> > If we follow up your earlier suggestion of working with the action in
terms
> > of Fourier transform fields we just get the same inconsistency as
before.
> >
> > The \phi^4 action in position space is
> >
> > I = \int d^4x 1/2 (\partial \phi(x))^2 - m^2/2 \phi(x)^2 - \gamma/4!
> > \phi(x)^4
> >
> > If we write \phi(x) = \int d^4p e^{ip.x} \phi(p) then we get an action
in
> > terms of the Fourier transform fields thus:
> >
> > I = (2\phi)^4 ( 1/2 \int d^4p (p^2-m^2) \phi(p)\phi(-p) -
> > \gamma/4! \int d^4p_1 d^4p_2 d^4 p_3 \phi(p_1) \phi(p_2)
\phi(p_3)
> > \phi(p - p_1 - p_2 - p_3) )
> >
> > I think that it is reasonably clear that the variational equations are
in
> > agreement with those obtained by varying the position space action.
>
> Yes. But this only gives the field equations for the unrenormalized
> theory.
>
>
> > Now let
> > us make \phi(p) vanish outside some given momentum space region, which
using
> > the terminology I developed earlier is
> >
> > \phi(p) = f(p, \Lambda) \chi(p) (*)
> >
> > where \chi(p) is operator-valued and f(p, \Lambda) is a c-number
function
> > that vanishes outside a region whose bounds are defined by \Lambda.
>
> You need to substitute (*) into the above expression for I and get
> a modified action I(Lambda,chi). Then you need to vary chi in this
> action to get the modified field equations. These are consistent with
> this action, and do not include any \chi(p) at momenta beyond the
> cutoff.
>
> But they are different from what you'd get from inserting (8)
> in the unregularized field equations. Which gave you the seeming
> inconsistency.

Why does this make a difference? One just applies the Chain Rule when
varying the action and ends up with the same equations.

Arnold Neumaier

Chris Oakley wrote:


>>>If we follow up your earlier suggestion of working with the action in terms
>>>of Fourier transform fields we just get the same inconsistency as before.
>>>The \phi^4 action in position space is
>>>
>>>I = \int d^4x 1/2 (\partial \phi(x))^2 - m^2/2 \phi(x)^2 - \gamma/4!
>>>\phi(x)^4
>>>
>>>If we write \phi(x) = \int d^4p e^{ip.x} \phi(p) then we get an action
>>>in terms of the Fourier transform fields thus:
>>>
>>>I = (2\phi)^4 ( 1/2 \int d^4p (p^2-m^2) \phi(p)\phi(-p) -
>>> \gamma/4! \int d^4p_1 d^4p_2 d^4 p_3 \phi(p_1) \phi(p_2) \phi(p_3)
>
>>>\phi(p - p_1 - p_2 - p_3) )
>>>
>>>I think that it is reasonably clear that the variational equations are
>>>in agreement with those obtained by varying the position space action.
>>
>>Yes. But this only gives the field equations for the unrenormalized
>>theory.
>>
>>
>>
>>>Now let
>>>us make \phi(p) vanish outside some given momentum space region, which
>>> using the terminology I developed earlier is
>>>
>>>\phi(p) = f(p, \Lambda) \chi(p) (*)
>>>
>>>where \chi(p) is operator-valued and f(p, \Lambda) is a c-number function
>>>that vanishes outside a region whose bounds are defined by \Lambda.
>>
>>You need to substitute (*) into the above expression for I and get
>>a modified action I(Lambda,chi). Then you need to vary chi in this
>>action to get the modified field equations. These are consistent with
>>this action, and do not include any \chi(p) at momenta beyond the
>>cutoff.
>>
>>But they are different from what you'd get from inserting (8)
>>in the unregularized field equations. Which gave you the seeming
>>inconsistency.
>
> Why does this make a difference? One just applies the Chain Rule when
> varying the action and ends up with the same equations.

No. It makes a difference because the variation is over a different
set of allowed changes. If you vary a field which has no fast
Fourier modes you cannot get equations involving the latter!

Do the calculations and you'll see.
Talking without doing is not enough to understand.


Arnold Neumaier

jsolomon@mail.com

Aaron Bergman wrote:
> In article <d59rq4$2q5h$1@godfrey.mcc.ac.uk>,
> "Dan Piponi" <google01@sigfpe.com> wrote:
>
> > Aaroin Bergman points out:
> > > Weinberg explicitly states above (11.A.2)
> >
> > Yes, Weinberg states he is making an analytic continuation, but he
> > doesn't justify it mathematically.
>
> I wouldn't know what it would mean to justify it mathematically.
>

I am not sure from your response if you are saying that it is not
required to justify this procedure mathematicllay or you simply don't
know how to do so. Irregardless I think that Weinbergs explanation of
dimensional regularization on page 477 of his book cannot be justified
mathematically.

Suppose we have the following relationship,

P = A + B (1)

If we have another relationship,

A = C (2)

we can rewrite (1) as,

P = C + B (3)

Now what does the equal sign mean in equation 2. It means two
quantities, A and C, are numerically equal. That is we have some
process of turning A and C into a number. And when that process is
used A and C turn into the same number.

Now what Weinberg does is this. He starts with an expression for the
polarization tensor which is of the form of (1). (See page 477 of his
book). He rewrites (1) in terms of a parameter "d" so he has,

P(d) = A(d) + B(d) (4)

The original expression is recovered when d=4.

He then discovers that for sufficiently small d (I think d<2) he has
the relationships given by 11.2.11 and 11.2.12 of his book. Consider
11.2.12. We write this as,

A(d) = A1(d) (5)

where A(d) is the expression on the left side of 11.2.12 and A1(d) is
the expression on the right side. However the eqaulity is only valid
for d<2. However he assumes the equality is valid for values of d near
4. However this is not correct. You cannot turn A(d) and A1(d) into
the same number for d>2. They are not numberically equal. An integral
is a short hand way of writing an infinite sum. For values of d near 4
A(d) is a divergent sum of positive numbers. A1(d), on the other hand
is finite (except at d=4) and, in fact, can be negative. So we can
find values of d near 4 where we end up with something like this,

1 + 2 + 3 + 5 + 10 + etc. = -3 (6)

(Note - this is for illustrative purposes only)

Yet what Weinberg does is to substitute A1(d) for A(d) in his
expression for the polarization tensor and then evaluates the
expression for values of d near 4 where the two quantities, A(d) and
A1(d), are certainly not numerically equal. This is not a
mathematically corrrect step.

Now it has been argued that this step is justified by analytic
continuation. However this justifies nothing since the analytic
continuation of a function does not necessarily equal the original
function in the entire complex plane.

This is what bothers me about this procedure. It seems to seriously
lack mathematical rigor. I know that this procedure is required to make
QFT work and agrees with experiment. However aren't physicists
interested in mathematical rigor also? In a seperate post Dr. Neumaier
recommended some books that have a more rigorous approach. I have
ordered the one by G. Scharf and I interested in reading how he handles
the problem of regularization.

Dan Solomon

Aaron Bergman

In article <d5b9c4$obl$1@nwrdmz01.dmz.ncs.ea.ibs-infra.bt.com>,
Chris Oakley <coakley@cgoakley.demon.co.uk> wrote:

> > What you really do
> > is to compute everything regularized, with only finite quantities, and
> > then limit away the regulator, keeping the physical couplings constant.

>
> This is not how I would put it. What I think you do is entirely dependent on
> what you want the theory to look like after it has been renormalized. If you
> can tie this down precisely enough then I agree that the renormalization
> scheme is irrelevant. But it is these requirements that in reality define
> the theory - not the quantum field theory you started with.

The quantum theory we started with was nonsense. The only hope is that
it can be derived as an effective theory for some theory which is not
nonsense. That's the trick we pull off with renormalization; we can
compute things that don't depend on the UV completion, just assuming
that such a thing exists.

Aaron

C. M. Heard

Arnold Neumaier wrote:
> C. M. Heard wrote:
> > Does anyone who does rigorous work in contructive QFT use dimensional
> > regularization? Can it actually be made rigorous? (The only
> > regularization I've seen discussed in the context of constructive QFT
> > is lattice regularization, but that is probably because of my ignorance.)
>
> See the section ''Dimensional regularization'' in my
> theoretical physics FAQ at
> http://www.mat.univie.ac.at/~neum/physics-faq.txt

This is a nice exposition on how to regularize integral expressions that
arise in a perturbative computation of the S matrix, but it does not answer
my question of whether it is possible to use dimensional regularization to
actually construct a field theory in a non-perturbative way.

//cmh

Chris Oakley

> This is what bothers me about this procedure. It seems to seriously
> lack mathematical rigor. I know that this procedure is required to make
> QFT work and agrees with experiment. However aren't physicists
> interested in mathematical rigor also? In a seperate post Dr. Neumaier
> recommended some books that have a more rigorous approach. I have
> ordered the one by G. Scharf and I interested in reading how he handles
> the problem of regularization.
>
> Dan Solomon

I too purchased Scharf's book. The title "Finite Quantum Electrodynamics"
certainly sounds promising, but I was disappointed. Superficially it all
looks similar to the good old meaningless renormalized perturbation theory
that we inherited from Feynman, et al, except that it seems to take much
more work to do a lot less. A little less superficially, it seems to be
about introducing extra functions to multiply the fields in time-ordered
products for which certain limits can be taken to extract finite amplitudes.
The principle, though, is the same as every other renormalization scheme:
generalize the theory in some way, set up some kind of differencing with the
parameters and then take a limit in a prescribed way to get finite
amplitudes. It may be finite, but it is still contrived. Their argument is,
however, significantly more involved than any I have seen, so authoritative
comments (from me, at least) will have to wait until I can find time to
study it properly. Their methods are based on a paper by Epstein & Glaser,
Ann. Inst. Poincare A 19 (1973) 211.

Aaron Bergman

In article <1115226335.445472.224850@z14g2000cwz.googlegroups.com>,
jsolomon@mail.com wrote:

> Aaron Bergman wrote:
> > In article <d59rq4$2q5h$1@godfrey.mcc.ac.uk>,
> > "Dan Piponi" <google01@sigfpe.com> wrote:
> >
> > > Aaroin Bergman points out:
> > > > Weinberg explicitly states above (11.A.2)
> > >
> > > Yes, Weinberg states he is making an analytic continuation, but he
> > > doesn't justify it mathematically.
> >
> > I wouldn't know what it would mean to justify it mathematically.
> >
>
> I am not sure from your response if you are saying that it is not
> required to justify this procedure mathematicllay or you simply don't
> know how to do so.

Neither. I'm saying that I'm not sure what it would mean. We don't have
an axiomatic definition of QFT in any useful sense. What we have is a
recipe that's well defined and works. What more do you want?

I've explained, I think, a number of times, why we use analytic
continuation to get sensible answers. You can complain all you want that
the integral is not equal to its analytic continuation everywhere, but
that doesn't mean that the procedure of replacing an integral with its
analytic continuation is ill-defined in any way. It's perfectly
rigorous. Its justification is more complicated, but the basic answer is
that it works. Other cutoffs can be more easily justified, but since
they give the same answer, we're happy with them all.

Aaron

Arnold Neumaier

C. M. Heard wrote:

> Arnold Neumaier wrote:
>
>>C. M. Heard wrote:
>>
>>>Does anyone who does rigorous work in contructive QFT use dimensional
>>>regularization? Can it actually be made rigorous? (The only
>>>regularization I've seen discussed in the context of constructive QFT
>>>is lattice regularization, but that is probably because of my ignorance.)
>>
>>See the section ''Dimensional regularization'' in my
>>theoretical physics FAQ at
>> http://www.mat.univie.ac.at/~neum/physics-faq.txt
>
> This is a nice exposition on how to regularize integral expressions that
> arise in a perturbative computation of the S matrix, but it does not answer
> my question of whether it is possible to use dimensional regularization to
> actually construct a field theory in a non-perturbative way.

I think this is an open research problem.


Arnold Neumaier

Dan Piponi

> It's perfectly rigorous.
said Aaron Bergman

I think rigourous isn't quite the right word here. It's a well defined
procedure (well, I'm not 100% sure if it is but I'll give it the
benefit of the doubt for now) that tells you how to transform one
expression into another one. That transformation isn't one of the usual
ones of mathematics such as substituting A for B in C when we know A=B.
So it's not a mathematical derivation in the usual sense. The situation
is probably similar to Newton working with infinitesimals or Heaviside
with differential operators before either of these things were fully
formalised.

Aaron Bergman

In article <1115415129.605969.117150@z14g2000cwz.googlegroups.com>,
"Dan Piponi" <google01@sigfpe.com> wrote:

> > It's perfectly rigorous.
> said Aaron Bergman
>
> I think rigourous isn't quite the right word here. It's a well defined
> procedure (well, I'm not 100% sure if it is but I'll give it the
> benefit of the doubt for now) that tells you how to transform one
> expression into another one. That transformation isn't one of the usual
> ones of mathematics such as substituting A for B in C when we know A=B.
> So it's not a mathematical derivation in the usual sense. The situation
> is probably similar to Newton working with infinitesimals or Heaviside
> with differential operators before either of these things were fully
> formalised.

If you want me to fully define a quantum field theory for you, then I'm
obviously not going to be able to do it. But, we do understand,
physically, what's going on. This gives us a well-defined procedure to
get a formal power series from the QFT, but that's far from a definition.

Aaron