unsolved problems in QED
Melroy
Is there anything besides Casimir effect on which there is no
consensus or is there any prediction which does not agree with experiment?
Melroy
[Moderator's note: one unsolved problem in QED is whether
this theory is mathematically consistent. Most physicists
think not, but nobody has been able to prove it. - jb]
Charles Francis
<melroy...@hotmail.com> writes:
>Are there any? The only thing I can think of is Casimir effect.
>Is there anything besides Casimir effect on which there is no
>consensus or is there any prediction which does not agree with experiment?
>[Moderator's note: one unsolved problem in QED is whether
>this theory is mathematically consistent. Most physicists
>think not, but nobody has been able to prove it. - jb]
There is nothing inconsistent about Scharf's treatment in Finite QED,
and even if one does not like an S-matrix approach there is still
nothing inconsistent about deriving the perturbation expansion from the
interacting Dirac equation, PROVIDING you do not abuse Wick's theorem,
(as happens in treatments which require renormalisation). QED is
demonstrated consistent up to the Landau pole. There is certainly no
consensus about the resolution of the Landau pole. I think many
physicists believe that the Landau pole will be resolved by a short
range modification to QED, and I have been proposing a model with a
short range modification, but it is certainly not an accepted
modification.
Regards
--
Charles Francis
Kevin A. Scaldeferri
In article <b4ocht$h7h$1...@panther.uwo.ca>,
Charles Francis <cha...@clef.demon.co.uk> wrote:
>John Baez wrote in a moderator's note on a message by <melroy...@hotmail.com>:
>
>>[Moderator's note: one unsolved problem in QED is whether
>>this theory is mathematically consistent. Most physicists
>>think not, but nobody has been able to prove it. - jb]
>
>There is nothing inconsistent about Scharf's treatment in Finite QED,
>QED is demonstrated consistent up to the Landau pole. There is certainly no
>consensus about the resolution of the Landau pole. I think many
>physicists believe that the Landau pole will be resolved by a short
>range modification to QED
This is all true, but shouldn't be taken by anyone to contradict what
jb said. The question of the consistency of QED is not about
renormalizability, but about what comes after renormalization: the
flow of the coupling constant and the Landau pole. There may be
short-range physics which renders this a moot point for real life, but
the question still remains about pure QED as a mathematical exercise.
--
======================================================================
Kevin Scaldeferri Calif. Institute of Technology
The INTJ's Prayer:
Lord keep me open to others' ideas, WRONG though they may be.
Arnold Neumaier
> In message <1b7c3dda.03031...@posting.google.com>, Melroy
> <melroy...@hotmail.com> writes:
> >Are there any? The only thing I can think of is Casimir effect.
> >Is there anything besides Casimir effect on which there is no
> >consensus or is there any prediction which does not agree with experiment?
> >[Moderator's note: one unsolved problem in QED is whether
> >this theory is mathematically consistent. Most physicists
> >think not, but nobody has been able to prove it. - jb]
> There is nothing inconsistent about Scharf's treatment in Finite QED,
> and even if one does not like an S-matrix approach there is still
> nothing inconsistent about deriving the perturbation expansion from the
> interacting Dirac equation, PROVIDING you do not abuse Wick's theorem,
> (as happens in treatments which require renormalisation). QED is
> demonstrated consistent up to the Landau pole.
Perturbative QED is only a rudimentary version of the 'real QED';
which can be seen that Scharf's results on the external field case are much
stronger (he constructs in his book the S-matrix)
than those for QED proper (where he only shows the existence of
the power series in alpha, but not their convergence).
The quest for 'existence' of QED is the quest for a framework
where the formulas make sense nonperturbatively, and where the
power series in alpha is a Taylor expansion of a (presumably
nonanalytic) function of alpha that is mathematically well-defined
for alpha around 1/137. And this is still open.
Arnold Neumaier
Charles Francis
<ke...@its.caltech.edu> writes
>In article <b4ocht$h7h$1...@panther.uwo.ca>,
>Charles Francis <cha...@clef.demon.co.uk> wrote:
>...
>>QED is demonstrated consistent up to the Landau pole. There is certainly no
>>consensus about the resolution of the Landau pole. I think many
>>physicists believe that the Landau pole will be resolved by a short
>>range modification to QED
>This is all true, but shouldn't be taken by anyone to contradict what
>jb said. The question of the consistency of QED is not about
>renormalizability, but about what comes after renormalization: the
>flow of the coupling constant and the Landau pole. There may be
>short-range physics which renders this a moot point for real life, but
>the question still remains about pure QED as a mathematical exercise.
Yes. I think some people are still upset about the indefinability of the
equal point multiplication between fields. Indeed the equal point
multiplication between fields is indefinable. However a careful analysis
of the derivation of Feynman rules from the interacting Dirac equation
shows that the equal point multiplication does not appear, and must not
be introduced by careless analysis of the improper integrals and misuse
of Wick's theorem. If not only the equal point multiplication, but also
the short range multiplication between fields which should be removed,
that would eliminate the Landau pole.
Regards
--
Charles Francis
John Baez
Kevin A. Scaldeferri <ke...@its.caltech.edu> wrote:
>Charles Francis wrote:
>>John Baez wrote:
>>>[Moderator's note: one unsolved problem in QED is whether
>>>this theory is mathematically consistent. Most physicists
>>>think not, but nobody has been able to prove it. - jb]
>>There is nothing inconsistent about Scharf's treatment in Finite QED,
>...
>>QED is demonstrated consistent up to the Landau pole.
>This is all true, but shouldn't be taken by anyone to contradict what
>jb said. The question of the consistency of QED is not about
>renormalizability, but about what comes after renormalization: the
>flow of the coupling constant and the Landau pole.
Right. In a bit more detail:
Scharf constructs the S-matrix as a formal power series in the
fine structure constant in QED with an infrared cutoff. He was
by no means the first to do it, but he does it in a specially nice
way.
However, nonexperts should realize that this is a far cry from
proving that QED is a consistent quantum field theory. As always,
you need to read the fine print:
1) "constructs the S-matrix as a formal power series".
In other words, if you use Scharf's result to try to compute the
probability that some collision of particles produces some collection
of outgoing particles with momenta in certain ranges, you get the
answer as an infinite series - with no assurance that this series
converges!
2) "with an infrared cutoff".
In other words, Scharf's work assumes the electron charge becomes
zero outside a certain bounded region of spacetime. This artificial
assumption is made to avoid the problem that a collision could emit
an infinite number of photons having smaller and smaller energies -
the "infraparticle problem".
Of the two problems, 1) is the most important. Conventional
wisdom - based on simpler examples - asserts that the formal power
series does *not* converge. This is not a disaster, because in
many cases one can still extract a finite answer by tricks such
as "Borel summation". However, nobody has proved that these tricks
work for QED. The conventional wisdom - based this time on guesses
about the "Landau pole" - is that they do not.
In short, most physicists who have thought hard about these problems
believe QED is inconsistent (except as a recipe for generating divergent
formal power series). The same holds for the Standard Model,
which also has a Landau pole. However, nobody has *proven* that
these theories are inconsistent! The question remains open.
The Landau pole is a problem one sees, not by simply trying to
sum the formal power series for the S-matrix, but by playing other
games: namely, using a 1-loop or 2-loop calculation to guess the
renormalization group flow of the fine structure constant, and
noticing that it diverges at a finite (but absurdly high) momentum.
One can think of this as a sneaky indirect attempt to gain information
about the behavior of infinitely many Feynman diagrams in QED by
doing calculations that only involve finitely many diagrams. It
involves some guesswork, so it doesn't really *prove* anything about
the consistency or inconsistency of QED.
Ali Soleimani
news:<b54soq$4t5$1...@glue.ucr.edu>...
> Of the two problems, 1) is the most important. Conventional
> wisdom - based on simpler examples - asserts that the formal power
> series does *not* converge. This is not a disaster, because in
> many cases one can still extract a finite answer by tricks such
> as "Borel summation". However, nobody has proved that these tricks
> work for QED. The conventional wisdom - based this time on guesses
> about the "Landau pole" - is that they do not.
>
> In short, most physicists who have thought hard about these problems
> believe QED is inconsistent (except as a recipe for generating divergent
> formal power series). The same holds for the Standard Model,
> which also has a Landau pole. However, nobody has *proven* that
> these theories are inconsistent! The question remains open.
So what is the convential wisdom as to why we can compute the first
few terms of this series, and get answers that look like they're
converging and agree with experiment? There's at least a suspected
mathematical reason we can get away with this, right?
Also, are there any semi-practical QED calculations that "look
divergent" for even just a few terms?
--Ali Soleimani
Charles Francis
writes:
>>Charles Francis wrote:
>>>There is nothing inconsistent about Scharf's treatment in Finite QED,
>>>QED is demonstrated consistent up to the Landau pole.
>Scharf constructs the S-matrix as a formal power series in the
>fine structure constant in QED with an infrared cutoff.
>However, nonexperts should realize that this is a far cry from
>proving that QED is a consistent quantum field theory. As always,
>you need to read the fine print:
>1) "constructs the S-matrix as a formal power series".
Actually he finds it from the Dyson expansion for the U matrix, which he
shows converges in operator norm.
>In other words, if you use Scharf's result to try to compute the
>probability that some collision of particles produces some collection
>of outgoing particles with momenta in certain ranges, you get the
>answer as an infinite series - with no assurance that this series
>converges!
We know the series does not converge at the Landau Pole. However this
may not be relevant. For scattering experiments it is not unreasonable
to apply a cut off to the potential, as is required to show that the
Dyson expansion for the S matrix also converges in operator norm.
>2) "with an infrared cutoff".
I agree that this is an artificial assumption, but an infrared cut-off
is of little direct relevance to the ultraviolet divergence. In fact in
scattering experiments we should strictly be using the U matrix for
large -t1, t2, anyway, and we already know it converges in operator
norm.
In Scharf's treatment the ultraviolet divergence is not treated by an
infrared cut-off, but by the switching function which is zero at the
equal point multiplication. Essentially the switching function is
equivalent to taking the limit after applying an ultraviolet cut-off
(only here we don't have a sharp cut-off, but one controlled by a
continuous function), and then showing that the theory is cut-off
independent.
The important thing is that the ultraviolet divergence is caused by the
abuse of Wick's theorem, and is controlled by the order of taking
limits, not by renormalisation.
>Of the two problems, 1) is the most important. Conventional
>wisdom - based on simpler examples - asserts that the formal power
>series does *not* converge. This is not a disaster
No it is not a disaster. The forced harmonic oscillator also diverges at
its resonant frequency at t=infinity. That doesn't bother us, and there
is no reason to think this is any different. But it does indicate that
the mathematical description of physics breaks down before the
divergence.
>, because in
>many cases one can still extract a finite answer by tricks such
>as "Borel summation". However, nobody has proved that these tricks
>work for QED. The conventional wisdom - based this time on guesses
>about the "Landau pole" - is that they do not.
I think we are more or less agreed that there is an issue concerning the
Landau pole.
Once upon a time (in the fifties I think) physicists use to speculate
that the theory required a short range modification, and that this short
range modification has something to do with gravity. I find that quite
natural, because although gravity is a weak force (sic) it becomes
strong (twice as sick) at short range. What happened to such
speculations? As far as I can tell no one made much headway with them,
which is fine and probably not surprising, but what I don't understand
is why physicists react with hostility to the notion now.
Regards
--
Charles Francis
Arnold Neumaier
> In message <b54soq$4t5$1...@glue.ucr.edu>, John Baez <ba...@galaxy.ucr.edu>
> writes:
> >>Charles Francis wrote:
> >>>There is nothing inconsistent about Scharf's treatment in Finite QED,
> >>>QED is demonstrated consistent up to the Landau pole.
> >Scharf constructs the S-matrix as a formal power series in the
> >fine structure constant in QED with an infrared cutoff.
> >However, nonexperts should realize that this is a far cry from
> >proving that QED is a consistent quantum field theory. As always,
> >you need to read the fine print:
> >1) "constructs the S-matrix as a formal power series".
>
> Actually he finds it from the Dyson expansion for the U matrix, which he
> shows converges in operator norm.
I thought (speaking from memory) he shows convergence only in
the case of a single particle in an external field, not for QED.
The former is a significantly easier problem.
Strong arguments (which haven't lost in half a century
their persuasive power) supporting the view that one should
expect the divergence of the QED power series for all values of
alpha>0 (and independent of energy) are given in
F.J. Dyson,
Divergence of perturbation theory in quantum electrodynamics,
Phys. Rev. 85 (1952), 613--632.
Arnold Neumaier
Squark
> ...Conventional
> wisdom - based on simpler examples - asserts that the formal power
> series does *not* converge. This is not a disaster, because in
> many cases one can still extract a finite answer by tricks such
> as "Borel summation".
What I still fail to understand is why the sum of a finite
number of terms from the beginning of a Borel summable series
approximates the answer. Obviously it doesn't do so for any number
of terms, but it is usually assumed it does so in the beginning,
to some extent. Why is that?
Best regards,
Squark
------------------------------------------------------------------
Write to me using the following e-mail:
Skvark_N...@excite.exe
(just spell the particle name correctly and change the
extension in the obvious way)
Arnold Neumaier
> ba...@galaxy.ucr.edu (John Baez) wrote in message news:<b54soq$4t5$1...@glue.ucr.edu>...
> > ...Conventional
> > wisdom - based on simpler examples - asserts that the formal power
> > series does *not* converge. This is not a disaster, because in
> > many cases one can still extract a finite answer by tricks such
> > as "Borel summation".
> What I still fail to understand is why the sum of a finite
> number of terms from the beginning of a Borel summable series
> approximates the answer. Obviously it doesn't do so for any number
> of terms, but it is usually assumed it does so in the beginning,
> to some extent. Why is that?
if you have a function f(x) which is infinitely often
differentiable, you can apply Taylor's theorem with error term
and get the approximation
f(x)=f(0)+f'(0)x+f''(0)x^2/2 + R(x)
with an error that is provably bounded by C_3x^3 for small x>0;
similarly for order k approximations, with a different
coefficient C_k.
If this function is not analytic (but still infinitely often
differentiable), the coefficients in the power series
f(x)= sum_k f_k x^k, f_k=f^{(k)}/k^!
grow so fast that the convergence radius of the power series
equals zero - thus the series eventually diverges for any nonzero
value of x.
This is no contradiction to the finite order result since
the constant C_k grows in such a case enormously, so that at
any fixed value of x>0, the error bound C_kx^k diverges as k
goes to infinity.
Now take for x the fine structure constant, and for f(x) some
S-matrix element, and you'll have exactly this situation.
Arnold Neumaier
John Baez
In article <c05daf0b.0303...@posting.google.com>,
Ali Soleimani <al...@ugcs.caltech.edu> wrote:
>> Of the two problems, 1) is the most important. Conventional
>> wisdom - based on simpler examples - asserts that the formal power
>> series does *not* converge. This is not a disaster, because in
>> many cases one can still extract a finite answer by tricks such
>> as "Borel summation". However, nobody has proved that these tricks
>> work for QED.
>So what is the convential wisdom as to why we can compute the first
>few terms of this series, and get answers that look like they're
>converging and agree with experiment? There's at least a suspected
>mathematical reason we can get away with this, right?
Even if a formal power series doesn't converge, and even if
it's not Borel summable, we can get away with this as long as
it's "asymptotic". When you have an series
sum a_n x^n,
that's "asymptotic" to the function F(x), the first few terms
give a answer close to F(x) when x is small, even though the
series diverges.
Many physics problems give rise to asymptotic series, and
it's a lot of fun to study examples and see how this works.
I'll describe one below.
If you don't know what an "asymptotic series" is, see
http://mathworld.wolfram.com/AsymptoticSeries.html
for a definition. But beware: the definition here considers
a series that's asymptotic for LARGE values of a parameter x.
So, you need to replace x by 1/x to get the definition relevant
to quantum electrodynamics, where we are interested in x being
the fine structure constant, and interested in series that may
(or may not) be asymptotic for SMALL values of x.
>Also, are there any semi-practical QED calculations that "look
>divergent" for even just a few terms?
No. For a rough idea of what's going on, you should guess
the number of Feynman diagrams you can draw with n vertices.
When n is large this becomes bigger than (137)^n, and roughly
around this point we expect the series to start "looking divergent".
But n must be rather large for this to occur.
I forget how large, but some people have good guesses. Maybe
it's around n = 137, or n = sqrt(137), or something. In any
event, it's big enough that people don't usually calculate such
big Feynman diagrams!
The above simplified analysis ignores the fact that each
Feynman diagram has its own value which may be large, or may
cancel with other diagrams. These complications are the reason
nobody knows whether or not the series in QED are asymptotic.
But here's a much simpler example:
Let's consider quantum field theory in 0+1 dimensions, i.e.,
quantum mechanics. Take the anharmonic oscillator: the system
with Hamiltonian
H = p^2 + q^2 + lambda q^4
Take something like the ground state energy and calculate it perturbatively
as a power series in lambda. You can prove the series diverges, but
when lambda is small you can also prove that adding up the first bunch
of terms (not too many, not too few!) gives you pretty accurate answers.
The smaller lambda is, the more terms you should use, and the better
answers you get.
For example, when lambda is .2, the exact ground state energy is
1.118292...
But when you work out the perturbation expansion you get these partial
sums:
1.15
1.0975
1.15375
1.105372
1.176999
1.049024
1.314970
0.686006
2.353090
and so on. It gets better up to the 5th term and then it gets worse.
I got this example out of page 27 of Reed and Simon's book "Analysis of
Operators". It's a good place to read about this stuff!
John Baez
In article <3e7a2b03$0$9620$3b21...@news.univie.ac.at>,
Arnold Neumaier <Arnold....@univie.ac.at> wrote:
>Charles Francis wrote:
>> In message <b54soq$4t5$1...@glue.ucr.edu>, John Baez <ba...@galaxy.ucr.edu>
>> writes:
>> >Scharf constructs the S-matrix as a formal power series in the
>> >fine structure constant in QED with an infrared cutoff.
>> >However, nonexperts should realize that this is a far cry from
>> >proving that QED is a consistent quantum field theory.
>> Actually he finds it from the Dyson expansion for the U matrix, which he
>> shows converges in operator norm.
>I thought (speaking from memory) he shows convergence only in
>the case of a single particle in an external field, not for QED.
>The former is a significantly easier problem.
You're close, but not quite. Scharf considers an arbitrary
variable number of charged particles, and he considers processes
involving annihilation and pair production... but he shows convergence
of the S-matrix only in the approximation where the electromagnetic
field is treated as an "external field", neglecting the effect of the
charged particles on the field.
He does this in Theorem 5.1 on page 109 of the second edition of
his book. Later he considers full-fledged QED, where the charged
particles affect the electromagnetic field. For this problem
he gets an S-matrix after imposing an infrared cutoff... but he
gets this S-matrix as a formal power series, and does NOT prove
convergence.
In short, the S-matrix for which Scharf shows convergence is *not*
the S-matrix for full-fledged QED. It's just a pale imitation of
the real thing. The upshot is that nobody knows if the S-matrix
for QED makes sense, even with an infrared cutoff.
The reason why Scharf gets further when he limits himself to
an external electromagnetic field is pretty obvious. Full-
fledged QED has nonlinear field equations. Electrons and
positrons in an external electromagnetic result are described
by an inhomogeneous *linear* field equation, which is infinitely
easier to study.
>Strong arguments (which haven't lost in half a century
>their persuasive power) supporting the view that one should
>expect the divergence of the QED power series for all values of
>alpha>0 (and independent of energy) are given in
>
> F.J. Dyson,
> Divergence of perturbation theory in quantum electrodynamics,
> Phys. Rev. 85 (1952), 613--632.
Right. This is why nobody in their right mind should expect
the formal power series in QED to converge! At best they might
be Borel-summable, but the conventional wisdom is that they're
not even that good.
>> We know the series does not converge at the Landau Pole.
This does not make literal sense. The series we are talking
about now are power series in the fine structure constant.
The Landau pole is not a pole in some analytic function of the
fine structure constant. It's a pole in some other sort of
function.
Squark
Arnold Neumaier <Arnold....@univie.ac.at> wrote in message news:<3e7b0e96$0$9620$3b21...@news.univie.ac.at>...
> if you have a function f(x) which is infinitely often
> differentiable, you can apply Taylor's theorem with error term
> and get the approximation
> f(x)=f(0)+f'(0)x+f''(0)x^2/2 + R(x)
> with an error that is provably bounded by C_3x^3 for small x>0;
> similarly for order k approximations, with a different
> coefficient C_k.
>
> If this function is not analytic (but still infinitely often
> differentiable), the coefficients in the power series
> f(x)= sum_k f_k x^k, f_k=f^{(k)}/k^!
> grow so fast that the convergence radius of the power series
> equals zero - thus the series eventually diverges for any nonzero
> value of x.
>
> This is no contradiction to the finite order result since
> the constant C_k grows in such a case enormously, so that at
> any fixed value of x>0, the error bound C_kx^k diverges as k
> goes to infinity.
Sounds great. Two more question arise:
1) Can one prove that if the Taylor series is Borel summable the
result approximates f(x) (up to a function with all order
derivatives vanishing at the expansion point 0, presumably)? I
suppose the answer is yes, but I'd like to be clear on this point.
2) How does one estimate C_k in real life situations? What you
described implies it is possible that the error becomes lower with
the addition of more terms up to some point where it starts
climbing back. How does one know when this point happens?
Arnold Neumaier
> Arnold Neumaier <Arnold....@univie.ac.at> wrote in message
news:<3e7b0e96$0$9620$3b21...@news.univie.ac.at>...
> > if you have a function f(x) which is infinitely often
> > differentiable, you can apply Taylor's theorem with error term
> > and get the approximation
> > f(x)=f(0)+f'(0)x+f''(0)x^2/2 + R(x)
> > with an error that is provably bounded by C_3x^3 for small x>0;
> > similarly for order k approximations, with a different
> > coefficient C_k.
> 1) Can one prove that if the Taylor series is Borel summable the
> result approximates f(x) (up to a function with all order
> derivatives vanishing at the expansion point 0, presumably)? I
> suppose the answer is yes, but I'd like to be clear on this point.
This requires conditions on the function f. The problem is that there
are nonzero functions whose Taylor series vanishes identically
(for example f_0(x)=exp(-1/x^2) with imputed zero at x=0); hence the
coefficients of a Taylor series cannot determine the function,
and the value at a fixed x can be changed to an arbitrary value by adding a
suitable multiple of the above f_0.
But 'nice' functions are Borel summable, and there are precise
mathematical conditions for what it means to be nice.
Of course, proving that some f satisfies these conditions implies
having proved that f(x) exists. This means that for QED nothing is known
about it.
> 2) How does one estimate C_k in real life situations? What you
> described implies it is possible that the error becomes lower with
> the addition of more terms up to some point where it starts
> climbing back. How does one know when this point happens?
This is a problem already present in numerical analysis of simple
problems like computing a limit or an infinite sum from a finite
(and approximate) sequence. For example, Romberg integration
and other extrapolation methods (such as Pade approximation)
must cope with this problem.
What is being done in practice is that you sum the terms until
the change to the current value is bigger than the previous change was.
This works in most cases, though it is not foolproof. From finitely
many terms you can never predict with certainty the reamining ones,
which may change the limit completely. But usually, sequences are
nice, and frequently one has checks to see when something went wrong.
So, in real life, one does n o t estimate the C_k. Getting foolproof
estimates requires a lot of analysis, and in particular, requires that
one knows enough about the function in question to have characterizations
of the global behavior of the derivatives. In particular, one must know
existence of the function and much more. Thus for quantum field theories,
it is far outside the range of current methods.
This is like in many other fields.
For example, physicists 'know' that most dynamical
systems are chaotic, and you can see apparent chaos easily in plots.
But the estimates for a mathematical proof in a specific system
(even in small dimensions) are highly nontrivial, and chaos has been
proved for perhaps a dozen simple systems only. See, e.g.,
http://www.mat.univie.ac.at/~neum/papers/physpapers.html#poinc
[~ is tilde] or
W. Tucker, A Rigorous ODE Solver and Smale's 14th Problem,
Found. Comput. Math. 2, 53-117, 2002.
http://www.cs.utep.edu/interval-comp/tucker02.html
Arnold Neumaier
Arnold Neumaier
Squark wrote:
> 1) Can one prove that if the Taylor series is Borel summable the
> result approximates f(x) (up to a function with all order
> derivatives vanishing at the expansion point 0, presumably)?
For full mathematical details about Borel summability see, e.g.,
B.Yu. Sternin and V.E. Shatalov,
Borel-Laplace transform and asymptotic theory,
CRC Press, Boca Raton 1996.
But it is nontrivial reading...
Arnold Neumaier
Jim Graber
ba...@galaxy.ucr.edu (John Baez) wrote in message
news:<b5fbgd$ki3$1...@glue.ucr.edu>...
[Snip almost everything]
> Even if a formal power series doesn't converge, and even if
> it's not Borel summable, we can get away with this as long as
> it's "asymptotic".
[Snip the rest]
When Good Theories Go Bad
I am sure glad you explained that bit about the probably asymptotic
nature of QED.
When I read your first post about how nothing ever converges in QED, I
began to wonder how Feynman, Schwinger and Tomonaga managed to get a
Nobel prize for such a terrible theory, or why it is often touted as
our most accurate theory, with accuracies to ten decimal places or more.
But now I see we are apparently nowhere near the point where the
theory starts to diverge.
I have some questions about another theory that apparently only goes
so far before it diverges.
I am thinking about the Feynman-Weinberg-Boulware-Deser result that
the only consistent low energy limit of a spin 2 field theory is
general relativity at least "to the second order" or to all orders in
ladder theory, but diverges if loops are included, or something like
that.
For a long time I have wondered what happens at third order. At
first, I thought the answer was that third order is so complex that no
one had ever successfully computed it. Then, I got the idea that at
third order, the first loops appear and the theory gives non
renormalizable infinities. Now I am wondering if it might just be
starting to diverge at third order. If anyone can tell me what the
real breakdown is, I would appreciate it. TIA.
Anyhow, somehow or other the above approach has apparently reached the
end of its useful limits, and people have gone on to string theory and
loop quantum gravity instead. There again, they manage to get to an
approximation to GR, but with "logarithmic corrections to the Hawking
radiation", or something like that.
Apparently both string theory and LQG can make at least some
calculations about blackholes or things that resemble blackholes.
According to Smolin's recent long review, string theory is mostly
limited to extremal black holes, but LQG can deal with non extremal
ones as well. However, most of what I have heard recently seems to
deal with quantum things like Hawking radiation and entropy.
Apparently, long ago Gross published a paper about a "Stringy black
hole" which differs slightly from the standard Schwarzschild GR one.
What is the status of the Stringy black hole and or its LQG
counterpart? Do people still believe they are relevant? To what
extent do they diverge from the GR equivalents? Does any of this
divergence show up in the metric, however minutely, or is it strictly
confined to quantum effects like entropy and Hawking radiation?
Again, TIA for any enlightenment you can give me.
(For purposes of answering these questions, you may assume I have a
moderate understanding of classical GR, but you should assume my
understanding of the quantum effects is near zero.)
Jim Graber
Arnold Neumaier
>
> When I read your first post about how nothing ever converges in QED, I
> began to wonder how Feynman, Schwinger and Tomonaga managed to get a
> Nobel prize for such a terrible theory, or why it is often touted as
> our most accurate theory, with accuracies to ten decimal places or more.
> But now I see we are apparently nowhere near the point where the
> theory starts to diverge.
>
> I have some questions about another theory that apparently only goes
> so far before it diverges.
>
> I am thinking about the Feynman-Weinberg-Boulware-Deser result that
> the only consistent low energy limit of a spin 2 field theory is
> general relativity at least "to the second order" or to all orders in
> ladder theory, but diverges if loops are included, or something like
> that.
>
> For a long time I have wondered what happens at third order. At
> first, I thought the answer was that third order is so complex that no
> one had ever successfully computed it. Then, I got the idea that at
> third order, the first loops appear and the theory gives non
> renormalizable infinities. Now I am wondering if it might just be
> starting to diverge at third order. If anyone can tell me what the
> real breakdown is, I would appreciate it. TIA.
It means that no one found a way to make the third order contribution finite;
the standard recipes that work for renormalizable theories fail.
Note that renormalizable means that the power series exists, although its
convergence is dubious. For calculations one uses only the first few terms.
In quantum gravity, it seems as if the power series cannot even be defined.
It is probably like trying to expand a function like x^{x^3}*e^x as a power
series around x=0. Try and be enlightened.
Arnold Neumaier
Arnold Neumaier
Jim Graber wrote:
>
> I am thinking about the Feynman-Weinberg-Boulware-Deser result that
> the only consistent low energy limit of a spin 2 field theory is
> general relativity at least "to the second order" or to all orders in
> ladder theory, but diverges if loops are included, or something like
> that.
I replied
> It is probably like trying to expand a function like x^{x^3}*e^x as a power
> series around x=0. Try and be enlightened.
Sorry, I chose the wrong function. Try (1+x)/(1+ x^3 log x).
Arnold Neumaier
Alfred Einstead
> There is nothing inconsistent about Scharf's treatment in Finite QED,
... possibly except for the S-matrix, itself.
Squark
> Squark wrote:
> > 1) Can one prove that if the Taylor series is Borel summable the
> > result approximates f(x) (up to a function with all order
> > derivatives vanishing at the expansion point 0, presumably)? I
> > suppose the answer is yes, but I'd like to be clear on this point.
>
> This requires conditions on the function f. The problem is that there
> are nonzero functions whose Taylor series vanishes identically
> (for example f_0(x)=exp(-1/x^2) with imputed zero at x=0); hence the
> coefficients of a Taylor series cannot determine the function,
> and the value at a fixed x can be changed to an arbitrary value by adding a
> suitable multiple of the above f_0.
If you re-read what I wrote and you quote, you'll find the phrase
"up to a function with all order derivatives vanishing at the
expansion point 0", and I wasn't referring only to the zero
function, as you might have guessed. :-)
> > 2) How does one estimate C_k in real life situations? What you
> > described implies it is possible that the error becomes lower with
> > the addition of more terms up to some point where it starts
> > climbing back. How does one know when this point happens?
> So, in real life, one does n o t estimate the C_k. Getting foolproof
> estimates requires a lot of analysis, and in particular, requires that
> one knows enough about the function in question to have characterizations
> of the global behavior of the derivatives. In particular, one must know
> existence of the function and much more. Thus for quantum field theories,
> it is far outside the range of current methods.
Well, John Baez indicated a few methods, though apparently some
guesswork is involved in the estimations there. I suppose I should
read the reference you've given in your other post.
Best regards,
Squark
[Moderator's note: Also read Michael Reed and Barry Simon's _Analysis
of Operators_, especially chapter XII.5, "Summability Methods in
Perturbation Theory". This book is the fourth volume of their series
_Methods of Modern Mathematical Physics_, and it's quite good. - jb]
Charles Francis
In message <b5fd42$kjr$1...@glue.ucr.edu>, John Baez <ba...@galaxy.ucr.edu>
writes:
>In article <3e7a2b03$0$9620$3b21...@news.univie.ac.at>,
>Arnold Neumaier <Arnold....@univie.ac.at> wrote:
>>Charles Francis wrote:
>>> In message <b54soq$4t5$1...@glue.ucr.edu>, John Baez <ba...@galaxy.ucr.edu>
>>> writes:
>>> >Scharf constructs the S-matrix as a formal power series in the
>>> >fine structure constant in QED with an infrared cutoff.
>>> >However, nonexperts should realize that this is a far cry from
>>> >proving that QED is a consistent quantum field theory.
>>> Actually he finds it from the Dyson expansion for the U matrix, which he
>>> shows converges in operator norm.
>>I thought (speaking from memory) he shows convergence only in
>>the case of a single particle in an external field, not for QED.
>>The former is a significantly easier problem.
>You're close, but not quite. Scharf considers an arbitrary
>variable number of charged particles, and he considers processes
>involving annihilation and pair production... but he shows convergence
>of the S-matrix only in the approximation where the electromagnetic
>field is treated as an "external field", neglecting the effect of the
>charged particles on the field.
>He does this in Theorem 5.1 on page 109 of the second edition of
>his book.
I had actually been referring to the Dyson expansion, eq 0.3.18, in
which the integrals are still time ordered. In his argument V is not a
classical field, but an operator. Although he hasn't yet introduced
them, as far as I can see the argument applies to general states in Fock
space, and to field operators, so that it can apply to QED, but with the
important and extremely artificial restriction that the particles behave
as free particles in the in and out states, so that V can be given a cut
off. I don't see how full qed can be given such a cut-off (so I am not
saying I am satisfied) but I don't see a problem in scattering.
In the para following theorem 5.1 Scharf says, more or less as a
corollary, "Since 2.5.28 implies that the interaction H_1(t) is a
bounded operator it follows from 2.5.29 that the perturbation series
2.5.1 converges in operator norm". Again, if H is an operator then
surely the argument applies to field operators (albeit with an
artificial cut-off) which would mean that it applies when we include the
effects of charged particles, not just to external fields?
>>Strong arguments (which haven't lost in half a century
>>their persuasive power) supporting the view that one should
>>expect the divergence of the QED power series for all values of
>>alpha>0 (and independent of energy) are given in
>> F.J. Dyson,
>> Divergence of perturbation theory in quantum electrodynamics,
>> Phys. Rev. 85 (1952), 613--632.
I don't have access to the paper, but I don't see how arguments given
then can apply to modern versions of qed which are supposed to be
regularised when the theory is set up, e.g. by the inclusion of counter
terms in the interaction Hamiltonian, or in Scharf's case, by the use of
a short range switching function. Such modifications effectively
mean that these days we are working with a different theory.
>>> We know the series does not converge at the Landau Pole.
>This does not make literal sense. The series we are talking
>about now are power series in the fine structure constant.
>The Landau pole is not a pole in some analytic function of the
>fine structure constant. It's a pole in some other sort of
>function.
Point taken. But I think you know what I meant; rather than pick hairs
with my poor choice of phrase, why not break the habit of a life time
and do something helpful, like tell me which phrase I should have used?
Regards
--
Charles Francis
Arnold Neumaier
> >>Strong arguments (which haven't lost in half a century
> >>their persuasive power) supporting the view that one should
> >>expect the divergence of the QED power series for all values of
> >>alpha>0 (and independent of energy) are given in
> >> F.J. Dyson,
> >> Divergence of perturbation theory in quantum electrodynamics,
> >> Phys. Rev. 85 (1952), 613--632.
> I don't have access to the paper, but I don't see how arguments given
> then can apply to modern versions of qed which are supposed to be
> regularised when the theory is set up, e.g. by the inclusion of counter
> terms in the interaction Hamiltonian, or in Scharf's case, by the use of
> a short range switching function. Such modifications effectively
> mean that these days we are working with a different theory.
Dyson wrote this when renormalization of QED was already invented.
You can also look at the book by Glimm and Jaffe; he discusses renormalized
QFT in 2 dimensions, and shows that, in the renormalized case, one
needs the Borel transform. There is no reason to think that 4 dimensions
and gauge freedom would make things simpler.
Arnold Neumaier
Arnold Neumaier
>
> Arnold Neumaier <Arnold....@univie.ac.at> wrote in message
> news:<3e80aeb8$0$13674$3b21...@news.univie.ac.at>...
>
> > are nonzero functions whose Taylor series vanishes identically
> > (for example f_0(x)=exp(-1/x^2) with imputed zero at x=0); hence the
> > coefficients of a Taylor series cannot determine the function,
> > and the value at a fixed x can be changed to an arbitrary value by adding a
> > suitable multiple of the above f_0.
>
> If you re-read what I wrote and you quote, you'll find the phrase
> "up to a function with all order derivatives vanishing at the
> expansion point 0", and I wasn't referring only to the zero
> function, as you might have guessed.
What I had meant is that you need conditions on f that single out one
particular function from the infinitude. This is needed since if you know
a function only up to a function with all order derivatives vanishing
at the expansion point 0, and have no other information, you cannot
say _anything_ about f(alpha) at the physical value of the fine structure
constant alpha.
> > > 2) How does one estimate C_k in real life situations? What you
> > > described implies it is possible that the error becomes lower with
> > > the addition of more terms up to some point where it starts
> > > climbing back. How does one know when this point happens?
>
> > So, in real life, one does n o t estimate the C_k. Getting foolproof
> > estimates requires a lot of analysis, and in particular, requires that
> > one knows enough about the function in question to have characterizations
> > of the global behavior of the derivatives. In particular, one must know
> > existence of the function and much more. Thus for quantum field theories,
> > it is far outside the range of current methods.
>
> Well, John Baez indicated a few methods, though apparently some
> guesswork is involved in the estimations there.
Explicit C_k imply an existence proof, which exists only for
some theories in 2 or 3 spacetime dimensions. Nothing exists in
this direction for any field theory physicists are actually
comparing to experiments.
Arnold Neumaier
Mark
From Jim Graber:
> When I read your first post about how nothing ever converges in QED, I
> began to wonder how Feynman, Schwinger and Tomonaga managed to get a
> Nobel prize for such a terrible theory, or why it is often touted as
> our most accurate theory, with accuracies to ten decimal places or more.
One has to make a distinction between the perturbation seris for the
S-matrix
S = S0 + S1 + S2 + S3 + ...
Sn = sum of (renormalized) contributions at nth order; S0 = 1
converging versus things that are functions of this converging. For
instance, the series for the gyromagnetic ratio
g = g0 + g1 + g2 + g3 + ...
gn = total contribution from nth order
may very well converge and converge rapidly.
The situation is parallel to that for the Hamiltonian. Though the
field-theoretic expression for the Hamiltonian
H = integral d^3k h-bar omega(k)/2 (ak ak* + ak* ak)
is ill-defined as an operator in the Fock space representation, the
time-generator Leibnitz derivation operator
[H,_]
is well-defined as
[H,_] = integral d^3k h-bar omega(k)/2 [ak ak* + ak* ak, _]
and in fact:
[H, _] = [:H:, _].
In fact, the conditions that [H,_] = [:H:,_] and that :H: be an operator
serves as a way to uniquely characterize :H: as a Fock Space operator,
up to plus or minus finite constant, in virtue of Schur's Lemma.
Gordon D. Pusch
> Charles Francis wrote:
[Arnold Neumaier <Arnold....@univie.ac.at> wrote:]
It should also be noted that Dyson also provides a much simpler, purely
classical motivating argument for why the S-matrix should NOT be an
analytic function of the fine structure constant: For \alpha > 0, the
classical equilibrium state is a uniform mixture of positive and negative
particles, whereas for negative \alpha, like particles attract and unlike
particles repel, causing the system to break up into positive and negative
domains. Hence, even classically, one expects the point \alpha = 0 to be
a branch point, not a regular point, and that a series expansion in \alpha
will not converge about \alpha = 0. There is no reason to believe that
quantization should change this conclusion.
-- Gordon D. Pusch
perl -e '$_ = "gdpusch\@NO.xnet.SPAM.com\n"; s/NO\.//; s/SPAM\.//; print;'
John Baez
In article <b6beqt$107$1...@panther.uwo.ca>,
Charles Francis <cha...@clef.demon.co.uk> wrote:
>In message <b5fd42$kjr$1...@glue.ucr.edu>, John Baez <ba...@galaxy.ucr.edu>
>writes:
>>Scharf considers an arbitrary
>>variable number of charged particles, and he considers processes
>>involving annihilation and pair production... but he shows convergence
>>of the S-matrix only in the approximation where the electromagnetic
>>field is treated as an "external field", neglecting the effect of the
>>charged particles on the field.
>
>>He does this in Theorem 5.1 on page 109 of the second edition of
>>his book.
>In the para following theorem 5.1 Scharf says, more or less as a
>corollary, "Since 2.5.28 implies that the interaction H_1(t) is a
>bounded operator it follows from 2.5.29 that the perturbation series
>2.5.1 converges in operator norm". Again, if H is an operator then
>surely the argument applies to field operators (albeit with an
>artificial cut-off) which would mean that it applies when we include
>the effects of charged particles, not just to external fields?
No: the problem is that in full-fledged QED the interaction
Hamiltonian is *not a densely defined operator on Fock space*,
even when we include an infrared cutoff. And it's most certainly
not a bounded operator!
As Scharf points out, you need the interaction Hamiltonian to
be a bounded operator whose norm goes to zero pretty fast as
t -> +-infinity for his argument to yield a convergent
perturbative expansion for the S-matrix. You can see this
from equation 0.3.17, which he cites in the paragraph you quote.
In the case of charged particles in an external electromagnetic
field, he shows that the interaction Hamiltonian is bounded and
that its norm goes to zero as t -> +-infinity... *if* we put
conditions on the electromagnetic field saying that it's smooth
enough and goes to zero fast enough in all directions.
But in the case of full-fledged QED, even with an infrared cutoff,
the interaction Hamiltonian is a much more nasty entity. It's not
a bounded operator... it's not even a densely defined operator!
In fact, this problem is one of the main reasons constructive
quantum field theory is a difficult subject. I spent my grad
school years battling this problem, without much success. So,
my jaw dropped when you suggested that Scharf had effortlessly
gotten around it. He hasn't. And he doesn't claim to have done so.
Squark
> But in the case of full-fledged QED, even with an infrared cutoff,
> the interaction Hamiltonian is a much more nasty entity. It's not
> a bounded operator... it's not even a densely defined operator!
Well, hopefully _on the right state space_ it is (densly defined),
isn't it? It should be if QED can be described by AQFT, not so?
Best regards,
Squark
John Baez
In article <939044f.03041...@posting.google.com>,
Squark <fii...@yahoo.com> wrote:
>ba...@galaxy.ucr.edu (John Baez) wrote in message
>news:<b7kbpv$n55$1...@glue.ucr.edu>...
>> But in the case of full-fledged QED, even with an infrared cutoff,
>> the interaction Hamiltonian is a much more nasty entity. It's not
>> a bounded operator... it's not even a densely defined operator!
>
>Well, hopefully _on the right state space_ it is (densely defined),
>isn't it? It should be if QED can be described by AQFT, not so?
If quantum electrodynamics is described by any of the usual
axiom systems, the *actual* Hamiltonian of this theory should
be a densely defined operator on some Hilbert space.
But I wasn't talking about the actual Hamiltonian! I was talking
about the *interaction* Hamiltonian, which is the difference
between the actual Hamiltonian and the free Hamiltonian (the
Hamiltonian for a "free theory" where the charged particles
don't interact with electromagnetism).
Some naive treatments of QED write down a formula for this interaction
Hamiltonian as an operator on the Fock space for the free theory,
and proceed to manipulate it, ignoring the fact that it's not
a densely defined operator. Technically, it's just a densely defined
sesquilinear form. This gives rise to infinities in the Dyson series
for the S-matrix, which must then be removed by renormalization.
Scharf does not make this mistake.
Aaron Bergman
In article <939044f.03041...@posting.google.com>,
fii...@yahoo.com (Squark) wrote:
> ba...@galaxy.ucr.edu (John Baez) wrote in message
> news:<b7kbpv$n55$1...@glue.ucr.edu>...
>
> > But in the case of full-fledged QED, even with an infrared cutoff,
> > the interaction Hamiltonian is a much more nasty entity. It's not
> > a bounded operator... it's not even a densely defined operator!
>
> Well, hopefully _on the right state space_ it is (densly defined),
> isn't it? It should be if QED can be described by AQFT, not so?
I think most people (myself included) doubt QED even exists.
Aaron
--
Aaron Bergman
<http://www.princeton.edu/~abergman/>
<http://aleph.blogspot.com>
Charles Francis
writes
>In article <939044f.03041...@posting.google.com>,
>Squark <fii...@yahoo.com> wrote:
>>ba...@galaxy.ucr.edu (John Baez) wrote in message
>>news:<b7kbpv$n55$1...@glue.ucr.edu>...
>>> But in the case of full-fledged QED, even with an infrared cutoff,
>>> the interaction Hamiltonian is a much more nasty entity. It's not
>>> a bounded operator... it's not even a densely defined operator!
>>Well, hopefully _on the right state space_ it is (densely defined),
>>isn't it? It should be if QED can be described by AQFT, not so?
>If quantum electrodynamics is described by any of the usual
>axiom systems, the *actual* Hamiltonian of this theory should
>be a densely defined operator on some Hilbert space.
But aren't we (almost) certain that it isn't?
>But I wasn't talking about the actual Hamiltonian! I was talking
>about the *interaction* Hamiltonian, which is the difference
>between the actual Hamiltonian and the free Hamiltonian (the
>Hamiltonian for a "free theory" where the charged particles
>don't interact with electromagnetism).
And aren't we (almost) certain that the problem is with the interaction
Hamiltonian, and that the free Hamiltonian is fine?
Doesn't this strongly suggest that the condition
>If quantum electrodynamics is described by any of the usual
>axiom systems
...
can't be satisfied, and that we should look for physical reasons to
modify the axioms systems?
Regards
--
Charles Francis
Charles Francis
In message <b7kbpv$n55$1...@glue.ucr.edu>, John Baez <ba...@galaxy.ucr.edu>
writes
>But in the case of full-fledged QED, even with an infrared cutoff,
>the interaction Hamiltonian is a much more nasty entity. It's not
>a bounded operator... it's not even a densely defined operator!
Ok, that makes sense. I've been overlooking the difference that in
discrete qed H is bounded unless you take the continuum limit, which
usually I don't. Scharf's inclusion of a switching function does the
same job in the limit, but in the limit boundedness goes out the window.
Thanks for this post btw. It has helped me fit a few things together.
>In fact, this problem is one of the main reasons constructive
>quantum field theory is a difficult subject. I spent my grad
>school years battling this problem, without much success. So,
>my jaw dropped when you suggested that Scharf had effortlessly
>gotten around it.
I don't think we can actually solve this problem. What I have done is
solve an easier problem instead. Then the issue is whether the easier
problem models physics.
Regards
--
Charles Francis
Arnold Neumaier
Charles Francis wrote:
>
> In message <b7nmba$rga$1...@glue.ucr.edu>, John Baez <ba...@galaxy.ucr.edu>
> writes
>
> >In article <939044f.03041...@posting.google.com>,
> >Squark <fii...@yahoo.com> wrote:
>
> >If quantum electrodynamics is described by any of the usual
> >axiom systems, the *actual* Hamiltonian of this theory should
> >be a densely defined operator on some Hilbert space.
>
> But aren't we (almost) certain that it isn't?
In all approximations to QED people actually calculate highly
nontrivial things (such as the color of gold, for example),
the Hamiltonian of this theory _is_ a densely defined operator
on some Hilbert space. It would be very strange if this were no
longer the case in full QED. And it would also be very strange if
full QED does not exist while its approximations give reliable
answers to many questions and extremely accurate ones to some.
There are only very fuzzy arguments which suggest that QED may not
exist, and I bet they'll turn out to be unfounded.
Arnold Neumaier
John Baez
In article <b8478q$dqf$1...@panther.uwo.ca>,
Charles Francis <cha...@clef.demon.co.uk> wrote:
>In message <b7nmba$rga$1...@glue.ucr.edu>, John Baez <ba...@galaxy.ucr.edu>
>writes:
>>If quantum electrodynamics is described by any of the usual
>>axiom systems, the *actual* Hamiltonian of this theory should
>>be a densely defined operator on some Hilbert space.
>But aren't we (almost) certain that it isn't?
I feel the need to start by repeating a point you know.
Nobody is really "certain" about anything concerning QED,
except for the mathematics of perturbation theory - which
gives formal power series that may or may not converge.
Beyond that, all we have is guesses and intuition.
Some people feel so sure about their intuition that I suppose
they'd claim to be "certain" that QED admits no rigorous
formulation fitting into the usual axioms for quantum field
theory: e.g., the Garding-Wightman axioms, or the Haag-Kastler
axioms.
I wouldn't be surprised if they are right.
But, there are no theorems along these lines - and it's hard
to even know how one would *state* such theorems, much less
prove them! After all, to prove that "there exists no version
of QED that satisfies the Haag-Kastler axioms", one would need
to state precisely when a theory satisfying the Haag-Kastler
axioms counts as "a version of QED". This is a bit delicate,
and I haven't heard of much work on this.
In general, people working on "constructive quantum field theory"
focus on showing that quantum field theories *do* satisfy various
nice axioms, not on showing they *don't*. For the latter, I guess
we need experts in "destructive quantum field theory". :-)
>>But I wasn't talking about the actual Hamiltonian! I was talking
>>about the *interaction* Hamiltonian, which is the difference
>>between the actual Hamiltonian and the free Hamiltonian (the
>>Hamiltonian for a "free theory" where the charged particles
>>don't interact with electromagnetism).
>And aren't we (almost) certain that the problem is with the interaction
>Hamiltonian, and that the free Hamiltonian is fine?
Free quantum field theories can be rigorously constructed - no problem.
In particular, the free part of the Hamiltonian for QED is a well-defined
self-adjoint operator on the photon/electron-positron Fock space.
As for the rest of your question, it's very important to study
examples of interacting quantum field theories that people have
*succeeded* in rigorously constructing. For example, the phi^4
theory in 2-dimensional spacetime. It's these examples that give
us intuition about how things should work when they *do* work.
In the phi^4 theory in 2d, the free Hamiltonian is a perfectly nice
self-adjoint operator on Fock space. The Hamiltonian of the interacting
theory is a perfectly nice self-adjoint operator on some *other* Hilbert
space. But the supposed "difference" of these two, the "interaction
Hamiltonian", is not a densely defined operator on either space! After
all, to form this "difference", you're trying to subtract operators that
are defined on different spaces! It turns out to exist as a "sesquilinear
form", but not as an operator.
In short, one problem with constructing interacting quantum field
theories is that you can't really write the Hamiltonian of the
interacting theory as an operator on the space of states of the free
theory.
Haag's theorem gives good evidence that this is how things
*should be*. The phi^4 theory in 2d gives an example where this
is how things *are* - and where the problem is surmountable.
All this stuff was known by the mid-1970s.
As for whether the problem is surmountable for QED, nobody
knows for sure. The conventional wisdom is that it's not.
But, conventional wisdom is quite different from certain knowledge.
>Doesn't this strongly suggest that the condition
>
>>If quantum electrodynamics is described by any of the usual
>>axiom systems
>
>can't be satisfied, and that we should look for physical reasons to
>modify the axioms systems?
That's one option. The more usual option is to discard QED as
a "bad" quantum field theory, and focus on "good" ones that we
believe make sense - namely, asymptotically free theories like
nonabelian Yang-Mills theory. That's why the Clay Mathematics
Institute million-dollar prize in constructive quantum field theory
concerns Yang-Mills theory, not QED.
Of course, it's worth remembering that by this criterion, the
Standard Model is a "bad" quantum field theory, since like QED
it has a Landau pole. If we're interested in the real world,
problems with the Standard Model are more interesting than problems
with QED.
Aaron Bergman
In article <3EA6A9E4...@univie.ac.at>,
Arnold Neumaier <Arnold....@univie.ac.at> wrote:
> And it would also be very strange if
> full QED does not exist while its approximations give reliable
> answers to many questions and extremely accurate ones to some.
Not at all. This is standard operating procedure. QED exists as an
effective field theory. Thus, it gives extremely accurate positions
below the cutoff scale. Piece of cake.
> There are only very fuzzy arguments which suggest that QED may not
> exist, and I bet they'll turn out to be unfounded.
The arguments are that the coupling appears to blow up at a finite
scale. This is generally considered bad. Now, no one has proven that
some weird thing might happen to get rid of the pole, but it seems
unlikely.
A.J. Tolland
On Fri, 25 Apr 2003, Arnold Neumaier wrote:
> There are only very fuzzy arguments which suggest that QED may not
> exist, and I bet they'll turn out to be unfounded.
There's apparently some evidence from lattice gauge theory. We
discussed Kim, S., et al., _On the Triviality of Textbook Quantum
Electrodynamics_ (hep-lat/0009029) here once upon a time. You might enjoy
reading it.
--A.J.
Arnold Neumaier
Aaron Bergman wrote:
> In article <3EA6A9E4...@univie.ac.at>,
> Arnold Neumaier <Arnold....@univie.ac.at> wrote:
> > And it would also be very strange if
> > full QED does not exist while its approximations give reliable
> > answers to many questions and extremely accurate ones to some.
> Not at all. This is standard operating procedure. QED exists as an
> effective field theory. Thus, it gives extremely accurate positions
> below the cutoff scale. Piece of cake.
I meant 'exist' in a mathematically well-defined, nonperturbative sense.
Such an existence has not be shown for a single 4D interacting theory,
and therefore not for QED. But the quality of the computed approximations
are a strong indication that there should be a consistent mathematical
foundation (for not too high energies), although it hasn't been found yet.
[Moderator's note: Aaron appears to be claiming that QED
exists in a rigorous way when we put in both an infrared
and an ultraviolet cutoff. This is true. But I don't know
if anyone has tried to rigorously estimate the Lamb shift
or the magnetic moment of an electron as predicted by this
cutoff theory, to demonstrate that it gives "extremely
accurate predictions". - jb]
> > There are only very fuzzy arguments which suggest that QED may not
> > exist, and I bet they'll turn out to be unfounded.
> The arguments are that the coupling appears to blow up at a finite
> scale. This is generally considered bad. Now, no one has proven that
> some weird thing might happen to get rid of the pole, but it seems
> unlikely.
This only shows that QED is likely to have problems at large energy.
But I'd expect QED to be a consistent theory at each fixed energy below
the Landau pole.
Moreover since the Landau pole has been computed by perturbation theory,
it is not even completely clear that it has to be there in a
nonperturbative version. Its position changes with the order
of approximation, and although the value at next order is smaller,
it might well go to infinity as the order grows indefinitely.
The analysis given in hep-lat/9801004 and hep-th/9712244 can also
be interpreted as pointing in this direction, although from the
point of view of a lattice limit.
Arnold Neumaier
Aaron Bergman
> Aaron Bergman wrote:
>> In article <3EA6A9E4...@univie.ac.at>,
>> Arnold Neumaier <Arnold....@univie.ac.at> wrote:
>> > And it would also be very strange if
>> > full QED does not exist while its approximations give reliable
>> > answers to many questions and extremely accurate ones to some.
>> Not at all. This is standard operating procedure. QED exists as an
>> effective field theory. Thus, it gives extremely accurate positions
>> below the cutoff scale. Piece of cake.
> I meant 'exist' in a mathematically well-defined, nonperturbative sense.
> Such an existence has not be shown for a single 4D interacting theory,
> and therefore not for QED. But the quality of the computed approximations
> are a strong indication that there should be a consistent mathematical
> foundation (for not too high energies), although it hasn't been found yet.
>
> [Moderator's note: Aaron appears to be claiming that QED
> exists in a rigorous way when we put in both an infrared
> and an ultraviolet cutoff. This is true. But I don't know
> if anyone has tried to rigorously estimate the Lamb shift
> or the magnetic moment of an electron as predicted by this
> cutoff theory, to demonstrate that it gives "extremely
> accurate predictions". - jb]
I don't know about rigor, but the standard RG type arguments give
you that the predictions of the cutoff theory and what we
normally write down are the same.
>> > There are only very fuzzy arguments which suggest that QED may not
>> > exist, and I bet they'll turn out to be unfounded.
>> The arguments are that the coupling appears to blow up at a finite
>> scale. This is generally considered bad. Now, no one has proven that
>> some weird thing might happen to get rid of the pole, but it seems
>> unlikely.
> This only shows that QED is likely to have problems at large energy.
> But I'd expect QED to be a consistent theory at each fixed energy below
> the Landau pole.
I don't know what you mean by this.
> Moreover since the Landau pole has been computed by perturbation theory,
> it is not even completely clear that it has to be there in a
> nonperturbative version. Its position changes with the order
> of approximation, and although the value at next order is smaller,
> it might well go to infinity as the order grows indefinitely.
It might, but is seems unlikely from what I've heard.
Aaron
Charles Francis
writes:
>In article <b8478q$dqf$1...@panther.uwo.ca>,
>Charles Francis <cha...@clef.demon.co.uk> wrote:
>>In message <b7nmba$rga$1...@glue.ucr.edu>, John Baez <ba...@galaxy.ucr.edu>
>>writes:
>Some people feel so sure about their intuition that I suppose
>they'd claim to be "certain" that QED admits no rigorous
>formulation fitting into the usual axioms for quantum field
>theory: e.g., the Garding-Wightman axioms, or the Haag-Kastler
>axioms.
I would agree with them about the Garding-Wightman axioms. The
Haag-Kastler axioms are more subtle.
>But, there are no theorems along these lines - and it's hard
>to even know how one would *state* such theorems, much less
>prove them! After all, to prove that "there exists no version
>of QED that satisfies the Haag-Kastler axioms", one would need
>to state precisely when a theory satisfying the Haag-Kastler
>axioms counts as "a version of QED". This is a bit delicate,
>and I haven't heard of much work on this.
Actually you have, but in so far as I can tell you are "certain" that it
is misguided. Actually there are two issues that you could argue over,
one is whether Discrete QED counts as "a version of QED", the other is
whether it satisfies an interpretation of the Haag-Kastler axioms. And
you are right, it is a bit delicate and I must admit I have found it
extremely difficult to describe in an acceptable manner. However, if it
was thought to work I believe such semantic issues would cease to be of
concern.
>In short, one problem with constructing interacting quantum field
>theories is that you can't really write the Hamiltonian of the
>interacting theory as an operator on the space of states of the free
>theory.
I think there are other ways around that. If you want to treat the
interacting Hamiltonian as a map from the "free space" to the "free
space" such that it models an interaction then you have to be quite
precise about what you are doing. First you can only identify the "free
space" with the "interacting space" at definite time, so that means you
have to lose manifest covariance. Second an interaction means that you
must be going forward in time. So the map isn't really from H to H, it
is from H(t1) to H(t2) where t2>t1. I don't see a problem in defining a
map to do that until you try taking the limit t2->t1. Of course H(t2) is
isomorphic to H(t1), and usually identified with it. But the moment that
happens the interaction Hamiltonian ceases to be properly defined.
...
>> we should look for physical reasons to
>>modify the axioms systems?
>That's one option.
> If we're interested in the real world,
>problems with the Standard Model are more interesting than problems
>with QED.
As far as I can tell, it is the same problem. If it is possible to
develop a version of qed in the manner described above, and if it is
possible to show that it physically obeys whatever covariance condition
we need to describe physics (apart from manifest covariance) then the
same solution will work for the standard model.
Regards
--
Charles Francis
Arnold Neumaier
>
> In the phi^4 theory in 2d, the free Hamiltonian is a perfectly nice
> self-adjoint operator on Fock space. The Hamiltonian of the interacting
> theory is a perfectly nice self-adjoint operator on some *other* Hilbert
> space. But the supposed "difference" of these two, the "interaction
> Hamiltonian", is not a densely defined operator on either space! After
> all, to form this "difference", you're trying to subtract operators that
> are defined on different spaces! It turns out to exist as a "sesquilinear
> form", but not as an operator.
This means that there is a dense, 'nuclear' subspace W of the Hilbert space such
that the interaction Hamiltonian is a linear mapping from W to its dual space
W^*.
In fact both the free Hamiltonian H_0 and the physical Hamiltonian H_phys
are such mappings. But the inner product needed for H_phys is different from
that
for H_0, which means that the Hilbert spaces W_0 and W_phys (which would make
W < W_0 < W^* and W < W_phys < W^* Gelfand triples = rigged Hilbert spaces)
are different but contain a common nuclear space W.
A different way to say this is that the canonical commutation
relations (CCR) are represented as an algebra of linear operators on W.
The difficulty is that the CCR admit many inequivalent representations.
In particular, the cyclic representation of the free field operators
on the free vacuum state generates Fock space, while the cyclic representation
of the physical field operators on the physical vacuum state generates
a Hilbert space which is _not_ isomprphic to a Fock space..
Another way of looking at this is to realize that the *-algebra defined by the
CCR contains formally unitary operators U with U^*U=UU^*=1 which, however,
don't map Fock space into itself. The Moeller operators used in the interaction
picture would be such operators, which is an intuitive reason for why things
go wrong. Unrenormalized perturbation theory tries to take these U (and all the
terms of the form Texp(iZ) occuring in the derivations) as true
Hilbert space operators and therefore gets lots of terms involving
inner products of states from different Hilbert spaces, which do not make
sense.
This is the reason for the infinities. By using cutoffs or other
renormalization tools, one restores a well-defined inner product at the cost
of approximating things, and then suitable limits make perturbation theory
work. But a mathematically well-defined theory must make sense
nonperturbatively.
When working with nuclear spaces instead of Hilbert spaces, these difficulties
are mathematically absent. In this context, Haag's theorem simply reduces to the
assertion that one cannot use the Fock inner product to get an interacting
theory.
But of course there are lots of other inner products consistent with the
adjoint in the *-algebra, and some of them give good interacting QFTs.
But in 4D, nobody so far has been successful in making all things fit together.
Arnold Neumaier
Mike Mowbray
> Free quantum field theories can be rigorously constructed
> - no problem. In particular, the free part of the Hamiltonian
> for QED is a well-defined self-adjoint operator on the
> photon/electron-positron Fock space.
>
> [...]
>
> In the phi^4 theory in 2d, the free Hamiltonian is a perfectly
> nice self-adjoint operator on Fock space. The Hamiltonian of
> the interacting theory is a perfectly nice self-adjoint
> operator on some *other* Hilbert space. But the supposed
> "difference" of these two, the "interaction Hamiltonian", is
> not a densely defined operator on either space! After all,
> to form this "difference", you're trying to subtract operators
> that are defined on different spaces It turns out to exist
> as a "sesquilinear form", but not as an operator.
>
> In short, one problem with constructing interacting quantum
> field theories is that you can't really write the Hamiltonian
> of the interacting theory as an operator on the space of states
> of the free theory.
>
> Haag's theorem gives good evidence that this is how things
> *should be* [...]
Speaking of Haag's theorem, could you give us your opinion of
the paper mentioned in the "Haag's Theorem for Bears..." thread:
http://ls.poly.edu/~jbain/papers/lsz.pdf
?
The author (J. Bain) seems to claim that Haag's theorem is a
bugbear only because we demand that the full field operators
approach free field operators at asymptotic times (strong
asymptotic convergence). He then advocates adopting instead
a weak asymptotic convergence requirement that only the matrix
elements of the full field operators need converge to the matrix
elements of the free field operators asymptotically. He then
claims that this avoids Haag's theorem.
I smell a rat, but can't pin it down. Can you spot any holes
in his arguments?
- MikeM.
Charles Francis
In message <b8jvfp$qns$1...@panther.uwo.ca>, Arnold Neumaier
<Arnold....@univie.ac.at> writes:
>Aaron Bergman wrote:
>> QED exists as an
>> effective field theory. Thus, it gives extremely accurate positions
>> below the cutoff scale. Piece of cake.
>[Moderator's note: Aaron appears to be claiming that QED
>exists in a rigorous way when we put in both an infrared
>and an ultraviolet cutoff. This is true. But I don't know
>if anyone has tried to rigorously estimate the Lamb shift
>or the magnetic moment of an electron as predicted by this
>cutoff theory, to demonstrate that it gives "extremely
>accurate predictions". - jb]
Isn't that precisely what is done when it is shown that the theory is
cut-off independent to first order, as has been known for years?
>> > There are only very fuzzy arguments which suggest that QED may not
>> > exist, and I bet they'll turn out to be unfounded.
>> The arguments are that the coupling appears to blow up at a finite
>> scale. This is generally considered bad. Now, no one has proven that
>> some weird thing might happen to get rid of the pole, but it seems
>> unlikely.
I should have said it is extremely unlikely that nothing weird happens a
very short distance scales. In particular gravity should kick in, and
curvature should have an effect, or at least some form of quantum
analogue of curvature. After all if a particle could be confined at a
point there really should be some sort of event horizon.
>This only shows that QED is likely to have problems at large energy.
>But I'd expect QED to be a consistent theory at each fixed energy below
>the Landau pole.
Is this actually good enough, or does the Landau pole invalidate loop
integrals? Or don't we even know that, for sure?
>Moreover since the Landau pole has been computed by perturbation theory,
>it is not even completely clear that it has to be there in a
>nonperturbative version.
Having worked on a non-perturbative version, I agree.
Regards
--
Charles Francis
John Baez
Aaron Bergman <aber...@princeton.edu> wrote:
>In article <b8jvfp$qns$1...@panther.uwo.ca>, Arnold Neumaier wrote:
>> Aaron Bergman wrote:
>>> In article <3EA6A9E4...@univie.ac.at>,
>>> Arnold Neumaier <Arnold....@univie.ac.at> wrote:
>>> > And it would also be very strange if
>>> > full QED does not exist while its approximations give reliable
>>> > answers to many questions and extremely accurate ones to some.
>>> Not at all. This is standard operating procedure. QED exists as an
>>> effective field theory. Thus, it gives extremely accurate positions
>>> below the cutoff scale. Piece of cake.
>> [Moderator's note: Aaron appears to be claiming that QED
>> exists in a rigorous way when we put in both an infrared
>> and an ultraviolet cutoff. This is true. But I don't know
>> if anyone has tried to rigorously estimate the Lamb shift
>> or the magnetic moment of an electron as predicted by this
>> cutoff theory, to demonstrate that it gives "extremely
>> accurate predictions". - jb]
>I don't know about rigor, but the standard RG type arguments
>give you that the predictions of the cutoff theory and what we
>normally write down are the same.
Sorry to butt in there like that, but ... right!
This is the conventional wisdom, and it has the
potential to be made rigorous, because the cutoff
theory can be rigorously constructed - so
what you're saying, unlike most claims about
interacting quantum field theories, is a
claim about a well-defined mathematical structure!
It's not just a murky claim about something that
may not exist. It means something fairly definite,
so I could explain it to a mathematician, and they
could set out to actually prove it!
However, I don't think I've seen anyone actually
try to prove it.
And, it's probably not too easy to prove. It's easy
to understand how the cutoff theory works perturbatively,
but it's a lot harder to get control over it at the
nonperturbative level - i.e., to understand how well the
"sum over all Feynman diagrams" is approximated by perturbation
theory, where you sum over finitely many diagrams. If
we could do this, we'd be a lot closer to understanding
what happens when we remove the cutoff!
(I put "sum over all Feynman diagrams" in quotes, because
this sum probably diverges even with a cutoff, requiring
Borel summation. Also, the only cutoff theory I know how
to rigorously construct in a nonperturbative way uses
canonical quantization, not a path integral.)
Matthew Nobes
[snip]
> (I put "sum over all Feynman diagrams" in quotes, because
> this sum probably diverges even with a cutoff, requiring
> Borel summation. Also, the only cutoff theory I know how
> to rigorously construct in a nonperturbative way uses
> canonical quantization, not a path integral.)
What about field theories with lattice cuttoffs, confined to
finite volume?
In this case the path integral reduces to a finite (but large)
number of sums over allowed field modes. I would think that
that theory is rigorously constructable, though I have no idea
if somebody has gone all out and done it.
--
Matthew Nobes
c/o Physics Dept. Simon Fraser University, 8888 University
Drive Burnaby, B.C., Canada
http://normland.phys.sfu.ca
Aaron Bergman
In article <b93n9l$ml5$1...@glue.ucr.edu>, ba...@galaxy.ucr.edu (John Baez)
wrote:
> (I put "sum over all Feynman diagrams" in quotes, because
> this sum probably diverges even with a cutoff, requiring
> Borel summation. Also, the only cutoff theory I know how
> to rigorously construct in a nonperturbative way uses
> canonical quantization, not a path integral.)
Something wrong with lattice gauge theories?
John Baez
Matthew Nobes <man...@sfu.ca> wrote:
>John Baez <ba...@galaxy.ucr.edu> wrote:
>> (I put "sum over all Feynman diagrams" in quotes, because
>> this sum probably diverges even with a cutoff, requiring
>> Borel summation. Also, the only cutoff theory I know how
>> to rigorously construct in a nonperturbative way uses
>> canonical quantization, not a path integral.)
>What about field theories with lattice cuttoffs, confined to
>finite volume?
Good point... I was just experiencing a senior moment.
As far as I know there's no rigorous lattice formulation of
theories with chiral fermions, like the Standard Model, thanks
to the fermion doubling problem. But, this problem does not
afflict QED.
>In this case the path integral reduces to a finite (but large)
>number of sums over allowed field modes. I would think that
>that theory is rigorously constructable, though I have no idea
>if somebody has gone all out and done it.
It's easy to rigorously construct the Euclidean path integral
for gauge fields on a finite-volume lattice. In some cases
people have gone on and done something much harder: proved
the estimates needed to show there's a well-defined continuum
theory in the limit as you shrink the lattice spacing to zero
and increase the lattice volume to infinity. This amounts to
removing the "ultraviolet cutoff" and "infrared cutoff", respectively.
If you can do this, then with some extra work you'll be able
to use the Osterwalder-Schrader theorem to rigorously obtain
a quantum field theory on Minkowski spacetime.
In a famous piece of unpublished work, Balaban is supposed to
have taken this approach to SU(2) Yang-Mills theory in 4 dimensions -
the "pure" Yang-Mills theory, not coupled to fermions. He was
able to remove the ultraviolet cutoff, but only if the infrared
cutoff was maintained below a certain *fixed* length scale.
This means he couldn't go further and try to remove the infrared
cutoff.
So they say, anyway. Apparently it took Balaban a huge amount of
work to get this far... and frustrated by getting stuck at this point,
he never wrote up his work for publication!
Of course, just because Balaban couldn't go further doesn't mean
nobody else could. Constructive quantum field theory relies
on what mathematicians call "hard analysis": that is, lots of
clever inequalities. In hard analysis, it usually takes a series
of papers by different authors before the best results are obtained,
and then even more papers to find the really slick proofs. That's
why it's a real pity Balaban didn't publish his work. Especially
since there's now a million-dollar prize for constructing SU(2)
Yang-Mills theory in 4d spacetime, and proving that it has a mass gap!
Folks interested in this stuff should read this:
Vincent Rivasseau, From Perturbative to Constructive Renormalization,
Princeton, N.J., Princeton University Press, 1991.
Charles Francis
writes
>And, it's probably not too easy to prove. It's easy
>to understand how the cutoff theory works perturbatively,
>but it's a lot harder to get control over it at the
>nonperturbative level - i.e., to understand how well the
>"sum over all Feynman diagrams" is approximated by perturbation
>theory, where you sum over finitely many diagrams. If
>we could do this, we'd be a lot closer to understanding
>what happens when we remove the cutoff!
>(I put "sum over all Feynman diagrams" in quotes, because
>this sum probably diverges even with a cutoff, requiring
>Borel summation.
I dare say how easy it is to prove depends how you put the cut-off in.
If you use a cut-off, delta, in time, it is natural to think that the
rate of interaction is bounded. One way to interpret this is to make the
theory discrete in time. Then in each iteration of the Dyson expansion
the range of integration has to be modified. Instead of
t_n > t_(n-1)
we have to have
t_n > t_(n-1) + delta
Then in any finite time period T the number of iterations is bounded by
T/delta. Since we can find Feynman rules from the Dyson expansion the
"sum over all Feynman diagrams" becomes a finite sum.
An interesting consequence is that whatever scale of lattice the
appropriate difference equation is associated with the same differential
equation, so you end up showing the that the theory is independent of
the cut-off (to first order) and with the equations of qed anyway.
> Also, the only cutoff theory I know how
>to rigorously construct in a nonperturbative way uses
>canonical quantization, not a path integral.)
It is fairly straight forward to formally avoid canonical quantisation
(either first or second) by starting from a Hilbert space of states and
constructing Fock space. It's a long time since I read about it, but as
I recall the theorems about the equivalence of the path integral to
other formulations of qm do apply in this approach.
Regards
--
Charles Francis
Charles Francis
>And, it's probably not too easy to prove. It's easy
>to understand how the cutoff theory works perturbatively,
>but it's a lot harder to get control over it at the
>nonperturbative level - i.e., to understand how well the
>"sum over all Feynman diagrams" is approximated by perturbation
>theory, where you sum over finitely many diagrams. If
>we could do this, we'd be a lot closer to understanding
>what happens when we remove the cutoff!
>(I put "sum over all Feynman diagrams" in quotes, because
>this sum probably diverges even with a cutoff, requiring
>Borel summation.
I dare say how easy it is to prove depends how you put the cut-off in.
If you use a cut-off, delta, in time, such that the rate of interaction
is bounded, then in each iteration of the Dyson expansion the range of
integration has to be modified. Instead of
t_n > t_(n-1)
we have to have
t_n > t_(n-1) + delta
Then in any finite time period T the number of iterations is bounded by
T/delta. Since we can find Feynman rules from the Dyson expansion the
"sum over all Feynman diagrams" becomes a finite sum.
> Also, the only cutoff theory I know how
>to rigorously construct in a nonperturbative way uses
>canonical quantization, not a path integral.)
It is fairly straight forward to avoid canonical quantisation (either
Bill Hobba
Aaron Bergman wrote:
> > Not at all. This is standard operating procedure. QED exists as an
> > effective field theory. Thus, it gives extremely accurate positions
> > below the cutoff scale. Piece of cake.
Arnold Neumaier replied:
> I meant 'exist' in a mathematically well-defined, nonperturbative sense.
> Such an existence has not be shown for a single 4D interacting theory,
> and therefore not for QED. But the quality of the computed approximations
> are a strong indication that there should be a consistent mathematical
> foundation (for not too high energies), although it hasn't been found yet.
The moderator added:
> [Moderator's note: Aaron appears to be claiming that QED
> exists in a rigorous way when we put in both an infrared
> and an ultraviolet cutoff. This is true. But I don't know
> if anyone has tried to rigorously estimate the Lamb shift
> or the magnetic moment of an electron as predicted by this
> cutoff theory, to demonstrate that it gives "extremely
> accurate predictions". - jb]
I seem to recall something of relevance in Dirac's Principles of
Quantum Mechanics. I had a peek at page 311 and it seems (of course I
could be misinterpreting what Dirac said) that he claims to have done
such in his Lectures on Quantum Field Theory.
As an aside I have just finished Weinberg's volume 1 on QFT (a proper
study - not just a cursory read - but please don't ask me any difficult
questions about it - the study was not as deep as I would have liked). The
impression I was left with was that there MAY still be some problems with
the renormalization procedure but exactly what they were he was a bit coy
on. Still I rather like Weinberg's idea of an effective field theory and I
am not so uneasy about such things these days.
Thanks
Bill
Arnold Neumaier
>
> Aaron Bergman wrote:
>
> > > Not at all. This is standard operating procedure. QED exists as an
> > > effective field theory. Thus, it gives extremely accurate positions
> > > below the cutoff scale. Piece of cake.
>
> Arnold Neumaier replied:
>
> > I meant 'exist' in a mathematically well-defined, nonperturbative sense.
> > Such an existence has not be shown for a single 4D interacting theory,
> > and therefore not for QED. But the quality of the computed approximations
> > are a strong indication that there should be a consistent mathematical
> > foundation (for not too high energies), although it hasn't been found yet.
>
> The moderator added:
>
> > [Moderator's note: Aaron appears to be claiming that QED
> > exists in a rigorous way when we put in both an infrared
> > and an ultraviolet cutoff. This is true. But I don't know
> > if anyone has tried to rigorously estimate the Lamb shift
> > or the magnetic moment of an electron as predicted by this
> > cutoff theory, to demonstrate that it gives "extremely
> > accurate predictions". - jb]
>
> I seem to recall something of relevance in Dirac's Principles of
> Quantum Mechanics. I had a peek at page 311 and it seems (of course I
> could be misinterpreting what Dirac said) that he claims to have done
> such in his Lectures on Quantum Field Theory.
The 4th edition I have ends on p.310...
But I doubt that later editions contain what jb alluded to.
The high accuracy predictions of the Lamb shift etc I have seen
are done with lots of uncontrolled approximations.
A recent survey is hep-ph/0002158.
Note that perturbative QED (even in Scharf's rigorous treatment)
has nothing at all to say about how to model bound states.
Bound states don't exist perturbatively: the poles in the S-matrix
can arise only by summing infinitely many Feynman diagrams,
and I haven't seen a single rigorous treatment of such an issue in
4D field theory.
> As an aside I have just finished Weinberg's volume 1 on QFT (a proper
> study - not just a cursory read - but please don't ask me any difficult
> questions about it - the study was not as deep as I would have liked). The
> impression I was left with was that there MAY still be some problems with
> the renormalization procedure but exactly what they were he was a bit coy
> on. Still I rather like Weinberg's idea of an effective field theory and I
> am not so uneasy about such things these days.
It is interesting that he states repeatedly (I don't have the book here
but could find out where if there is interest) that bound state problems
(and this includes the Lamb shift) are still very poorly understood.
Arnold Neumaier
Bill Hobba
> > > [Moderator's note: Aaron appears to be claiming that QED
> > > exists in a rigorous way when we put in both an infrared
> > > and an ultraviolet cutoff. This is true. But I don't know
> > > if anyone has tried to rigorously estimate the Lamb shift
> > > or the magnetic moment of an electron as predicted by this
> > > cutoff theory, to demonstrate that it gives "extremely
> > > accurate predictions". - jb]
Bill Hobba wrote:
> > I seem to recall something of relevance in Dirac's Principles of
> > Quantum Mechanics. I had a peek at page 311 and it seems (of course I
> > could be misinterpreting what Dirac said) that he claims to have done
> > such in his Lectures on Quantum Field Theory.
Arnold Neumaier replied:
> The 4th edition I have ends on p.310...
> But I doubt that later editions contain what jb alluded to.
Strange my copy is also the 4th edition and goes up to page 312. What it
says on 311 is:
'The calculations of the Lamb shift and anomalous magnetic moment are
rather complicated. They are given in detail working from the
Hamiltonian H, in the authors work Lectures in Quantum Field Theory
(Academic Press 1966). The results are in good agreement with
experiment and provide a confirmation of the theory.
The calculations were made in terms of the Heisenber picture
throughout. One may tackle Quantum Electrodynamics on the Schrodenger
picture, looking for the solution to the Schrodenger equation by
taking the no particle ket, or a ket corresponding to just a few
particles present, as the initial ket of a perturbation procedure and
apply the standard probation technique. One finds that the later
terms are large and depend strongly on the cutoff, or are infinite if
the there is no cutoff. The pertubation procedure is not logically
valid under these conditions.'
Thus I would say your interpretation is probably correct. He is
merely alluding to the well known divergences that appear in the
perturbation method.
> The high accuracy predictions of the Lamb shift etc I have seen
> are done with lots of uncontrolled approximations.
> A recent survey is hep-ph/0002158.
Thanks for the reference. It will probably help with my constant on
and off investigation of QFT. Why is this subject so bloody hard yet
once it gets a grip on you you can't let go?
Arnold Neumaier replied:
> Note that perturbative QED (even in Scharf's rigorous treatment)
> has nothing at all to say about how to model bound states.
> Bound states don't exist perturbatively: the poles in the S-matrix
> can arise only by summing infinitely many Feynman diagrams,
> and I haven't seen a single rigorous treatment of such an issue in
> 4D field theory.
Bill Hobba wrote:
> > As an aside I have just finished Weinberg's volume 1 on QFT (a
> > proper study - not just a cursory read - but please don't ask me
> > any difficult questions about it - the study was not as deep as I
> > would have liked). The impression I was left with was that there
> > MAY still be some problems with the renormalization procedure but
> > exactly what they were he was a bit coy on. Still I rather like
> > Weinberg's idea of an effective field theory and I am not so
> > uneasy about such things these days.
Arnold Neumaier replied:
> It is interesting that he states repeatedly (I don't have the book here
> but could find out where if there is interest) that bound state problems
> (and this includes the Lamb shift) are still very poorly understood.
I can't find what your referring to but he confirms what you said
previously about the perturbative procedure not being valid. On page
462 (regarding composite states) Weinberg states:
'Unfortunately it has not been possible to implement this procedure in
quantum field theories because as we have seen Z=0 means the particle
couples as strongly as possible to its constituents and this rules out the
use of perturbation theory'
Thanks
Bill
Arnold Neumaier
Bill Hobba wrote:
> > > As an aside I have just finished Weinberg's volume 1 on QFT (a
> > > proper study - not just a cursory read - but please don't ask me
> > > any difficult questions about it - the study was not as deep as I
> > > would have liked). The impression I was left with was that there
> > > MAY still be some problems with the renormalization procedure but
> > > exactly what they were he was a bit coy on. Still I rather like
> > > Weinberg's idea of an effective field theory and I am not so
> > > uneasy about such things these days.
> Arnold Neumaier replied:
> > It is interesting that he states repeatedly (I don't have the book here
> > but could find out where if there is interest) that bound state problems
> > (and this includes the Lamb shift) are still very poorly understood.
> I can't find what your referring to but he confirms what you said
> previously about the perturbative procedure not being valid. On page
> 462 (regarding composite states) Weinberg states:
>
> 'Unfortunately it has not been possible to implement this procedure in
> quantum field theories because as we have seen Z=0 means the particle
> couples as strongly as possible to its constituents and this rules out the
> use of perturbation theory'
Yes, this is on p.462. And on p.564 he says,
'These problems are those involving bound states [...]
such problems necessarily involve a breakdown of ordinary
perturbation theory. [...] The pole therefore can only arise
from a divergence of the sum of all diagrams [...]'
On p.560, he writes,
'It must be said that the theory of relativistic effects
and radiative corrections in bound states is not yet in an
entirely satisfactory shape.'
This remark suggests that he seems to think that, in contrast,
for scattering problems, the theory is in an entirely satisfactory
state, as given in the rest of his book. Thus 'satisfactory'
does not mean 'mathematically rigorous', but only
'well understood from a physical, approximate point of view'.
Thus there is still scope for interesting new work on bound states
on both the physical and mathematical side.
Arnold Neumaier