Quantum Field Theory: The Big, Simple Picture?
Jay R. Yablon
and for all the drama and encyclopedic tomes about quantum theory, it
seems to me that there is really a very simple and clear line between
classical and quantum theory, at least when one views QFT through the
lens of Path Integration.
For classical theory, one has a field equation, or "equation of
motion," such as, but not limited to, Maxwell's
J^u=d_vF^vu (1)
or Einstein's
-kT^uv=R^uv-.5g^uvR (2)
One then obtains a Lagrangian density L which replicates the classical
field equation through the Euler-Lagrange prescription. In the former
case, we have the Maxwell Lagrangian density and in the latter the
Einstein-Hilbert Lagrangian where L_matter is a constant (I think -.5)
times the trace T=T^s_s.
For QFT, setting aside the fact that some path integrals are very
hard to calculate exactly, the conceptual prescription is really very
simple: Take L for the field psi, and put it in the path integral:
Z = $dpsi exp i[d^4x sqrt(-g) L] (3)
Then, calculate the intergal. Period.
When one does the calculation, and if one can do the calculation exactly
and preferably keep the boundary terms too, we then obtain an equation
for transition amplitudes which tells what our classical field theory
turns into once it is "path integral" quantized.
Yes, there is a lot of detail which emerges from all of this, and
sometimes calculating (3) exactly to all orders and leaving out nothing
as an approximation or simplification can be intractably difficult as a
mathematical challenge (and you have to add a few things along the way
such as +i epsilon to avoid poles when doing an inverse Fourier
transformation of the propagator and you may run into renormalization
problems). But fundamentally, conceptually, it seems to me that that is
it, and that the boundary between and the transition from classical to
quantum theory is just that simple, and that everything else is doing
calculation and overcoming obstacles in doing the calculations. I know
that I am totally ignoring canonical quantization, though from what I
understand, that and path integral quantization are two alternative
approaches (and the only two known approaches), so if you have one, you
implicitly have both. Am I being naive or missing something big?
Thanks,
Jay
____________________________
Jay R. Yablon
Email: jya...@nycap.rr.com
co-moderator: sci.physics.foundations
Weblog: http://jayryablon.wordpress.com/
Web Site: http://home.roadrunner.com/~jry/FermionMass.htm
Igor Khavkine
> I have been studying quantum field theory for about two years now,
> and for all the drama and encyclopedic tomes about quantum theory, it
> seems to me that there is really a very simple and clear line between
> classical and quantum theory, at least when one views QFT through the
> lens of Path Integration.
The line between classical and quantum theory is the same, independent
of whether you talk of a point particle or a continuous field. On the
one side, you have a classical phase space and functions on it, on the
other side, you have a Hilbert space of states, operators on it and
their expectation values. However, quantized field systems do have
many unique and interesting properties, and that's what the tomes are
written about.
> For QFT, setting aside the fact that some path integrals are very
> hard to calculate exactly, the conceptual prescription is really very
> simple: Take L for the field psi, and put it in the path integral:
>
> Z = $dpsi exp i[d^4x sqrt(-g) L] (3)
>
> Then, calculate the intergal. Period.
> [...] Am I being naive or missing something big?
It really depends on what you are looking for. One could say that all
there is to understanding turbulence is solving the Navier-Stokes
equations. The equations have been known over a hundred years, yet the
problem of turbulence in many senses remains unsolved still. In this
analogy, one could replace the Navier-Stokes equations by the path
integral and turbulence by any of the many open and interesting
questions in field theory. It must also be remarked, that, just as in
hydrodynamics, many interesting and non-trivial problems of field
theory have been posed and solved.
However, there are deeper issues still. The number Z by itself tells
you nothing. In fact, unless you take care to arrange things
otherwise, it is not even a real number; it is just infinity. And what
has proven very challenging is to define precisely what a path
integral is. So, the question of computing it "exactly" is rather ill
posed, since the "it" you are computing is not well defined to begin
with. What does exist, are well defined "approximation" schemes (most
notably perturbation theory and lattice discretization). I put
"approximation" in quotes because each scheme gives a well defined
sequence of answers and the limit of such a sequence (if it exists) is
by fiat declared to be the definition of the path integral.
Unfortunately, in many cases of interest, this limit is not known to
exist (and sometimes even known not to exist).
And even further, a path integral by itself is simply a computational
tool: it is a convenient way to express expectation values of certain
products of operators, called time-ordered vacuum n-point functions.
One still has to connect these expectation values to experimental
measurements, such as the output of particle counters, spectrometers,
photo-multipliers, thermometers, etc. Time ordered n-point functions
have an intimate relation to scattering amplitudes (as should be
explained in any good QFT book), that is why the path integral is of
immediate use in particle physics, where most experiments are reduced
to observing the results of scattering events. However, for other
kinds of observables, one would have to work somewhat harder and the
path integral tool is not necessarily the best one to use.
If you are interested in any of these issues, you really do need to
read through at least some of the many tomes written on QFT.
Igor
Jay R. Yablon
news:53a0fced-b6e9-48ae...@a32g2000yqm.googlegroups.com...
Thank you Igor,
I both appreciate and am somewhat amused by your reply. I am
appreciative because of the time you took and, as always, the good
information you provided. I am amused because you put in a great deal
of beating around the bush to avoid a direct answer to my question,
which answer, when you boil it down, appears to be "yes."
Taking some of what you said, it seems to me that if one has a classical
field theory characterized by a given Lagrangian density L, and if one
churns it through the path integral which is in many senses just a
"machine" for processing a classical Lagrangian and turning it into
quantum amplitudes, and if by hook or by crook or by whatever method one
needs to resort to (renormalization, perturbation theory, lattice gauge,
Fadeev-Popov, everything else in the tomes, etc., etc.), one manages to
derive a well-defined "answer" to the path integral, which "answer"
mathematically amounts to taking the definite integral from negative
infinity to positive infinity of a complicated mathematical integrand
and getting a finite, non-divergent expression out the other end, then
this "answer" is, in fact, the "quantum" version of the classical field
theory. Am I, or am I not, correct in this view? Obviously, what I
just said presupposes that you have a classical Lagrangian density to
begin with, but that caveat aside, I do believe that this is a question
which does have a simple, yes or no answer: Is the path integral,
however difficult it can be to calculate with, and however many tricks
on needs to use to get a good answer, essentially a mathematical machine
for quantizing a classical field theory?
I do not mean by this reply to give short shrift to all of the "tomes"
which have been written about this subject. But if anything, physics
today suffers from an over-abundance of information, and it is important
to be able to step back and put it all into context in terms of a few,
overarching conceptual principles from which everything else emanates by
clever, persistent deduction. And one of those big principles, in my
view, if that if you have a classical field theory represented by a
Lagrangian density, and you want to know what the associated quantum
field theory looks like, you plug it into the path integral and then do
whatever you have to do to get a sensible, finite answer out the other
end.
As a specific instance of all of this, suppose one starts with the
Einstein Hilbert action in a form which restates the Einstein field
equation via the Euler-Lagrange prescription, inserts that into the path
integral, properly selects the "field of integration" (possibly,
sqrt(-g)g^uv), evaluates the definite integral from -oo to +oo, and
finds some way to get a mathematically correct, finite answer. I know I
am understating how difficult that is, which is why I am purposely
posing this as a hypothetical. IF one was able to do that -- never
mind the difficulty of actually doing the exercise -- would this
hypothetical "answer," in principle, be a quantum field theory of
gravitation which corresponds with the classical -kappaT^uv=G^uv?
> However, there are deeper issues still. The number Z by itself tells
> you nothing. In fact, unless you take care to arrange things
> otherwise, it is not even a real number; it is just infinity. And what
> has proven very challenging is to define precisely what a path
> integral is. So, the question of computing it "exactly" is rather ill
> posed, since the "it" you are computing is not well defined to begin
> with. What does exist, are well defined "approximation" schemes (most
> notably perturbation theory and lattice discretization). I put
> "approximation" in quotes because each scheme gives a well defined
> sequence of answers and the limit of such a sequence (if it exists) is
> by fiat declared to be the definition of the path integral.
> Unfortunately, in many cases of interest, this limit is not known to
> exist (and sometimes even known not to exist).
All you have told me is that it is very often extremely difficult to do
sensible calculations with the path integral, and that one often needs
to develop what has turned out to be an abundant variety of
"'approximation' schemes" to get good answers many of which are
assembled into the "tomes," and that sometimes we either cannot in
principle, or do not know how to in practice, get good answer. But you
have not at any point said that the path integral really isn't a way to
quantize a classical field theory, which is what I asked.
> And even further, a path integral by itself is simply a computational
> tool: it is a convenient way to express expectation values of certain
> products of operators, called time-ordered vacuum n-point functions.
> One still has to connect these expectation values to experimental
> measurements, such as the output of particle counters, spectrometers,
> photo-multipliers, thermometers, etc. Time ordered n-point functions
> have an intimate relation to scattering amplitudes (as should be
> explained in any good QFT book), that is why the path integral is of
> immediate use in particle physics, where most experiments are reduced
> to observing the results of scattering events. However, for other
> kinds of observables, one would have to work somewhat harder and the
> path integral tool is not necessarily the best one to use.
That I do not dispute. All you are saying there is that sometimes,
there are better tools than path integration to quantize the theory.
Sort of like telling me that multiplying is not as easy as converting
into and then adding logarithms -- that still does not make it wrong to
multiply. You have not said at any point that the path integral, in
principle, is not a quantization machine, you have only articulated that
it is an unwieldy and sometime intractable machine which requires some
clever approaches to make it work properly.
> If you are interested in any of these issues, you really do need to
> read through at least some of the many tomes written on QFT.
I have read quite a bit. But, after looking at trees, I always try to
see the forest in the end. IMHO, we need more, not less of that these
days.
What I am specifically interested in at this time, is understanding how
to apply the path integral to a) the Maxwell action, b) the Yang-Mills
action, and c) the Einstein Hilbert action; select the proper
integration variable; do the definite integration; and get a finite,
sensible answer in closed form out the other end. I am not interested
in becoming a physics or mathematics encyclopedia. I will happily study
and have studied any and all aspects of the tomes, and any and all
mathematics, which gives me insight into how to better skin those three
particular cats. And, IMHO there is nothing like actually doing the
exercise of trying to calculate these things through, because actually
doing the math and overcoming the calculation obstacles along the way
teaches you things that just reading a text can never teach. All the
texts teach, are the ingredients for doing a calculation, and one always
needs to take care that one starts out with the right ingredients. The
path integral, it seems to me, is the central ingredient when the goal
is to bake a quantum field theory from a classical one.
Jay.
> Igor
>
X-Phy
> As a specific instance of all of this, suppose one starts with the
> Einstein Hilbert action in a form which restates the Einstein field
> equation via the Euler-Lagrange prescription, inserts that into the path
> integral, properly selects the "field of integration" (possibly,
> sqrt(-g)g^uv), evaluates the definite integral from -oo to +oo, and
> finds some way to get a mathematically correct, finite answer. I know I
> am understating how difficult that is, which is why I am purposely
> posing this as a hypothetical. IF one was able to do that -- never
> mind the difficulty of actually doing the exercise -- would this
> hypothetical "answer," in principle, be a quantum field theory of
> gravitation which corresponds with the classical -kappaT^uv=G^uv?
You'll have a quantized version of the field theory of gravitation,
but even if cooked by Feynman, and whatever the number of tomes, the
paths integral quantization hasn't that magical power of reconciling
two theories (QM and GR) that initially contradict one another.
Indeed, you'll have the field only part of general relativity, without
matter. The unified theory isn't yet in textbooks, and it'll be
necessary to depart from them if some progress is wanted.
Jay R. Yablon
news:fb3de115-a79f-47b9...@x31g2000yqx.googlegroups.com...
X-Phy:
Why do you say "without matter"? If a non-zero stress energy tensor
trace is in the Einstein Hilbert action, then to get the Ricci curvature
independent of the field one would apply the functional variation
delta/delta T^uv. Then, one after moving what is effectively the Ricci
curvature R outside the integral by turning it into a series of
delta/delta T^uv operators with parallel underlying structure to R, one
would be left having to integrate the remaining sqrt(g) g_uv T^uv term
over the field variable of integration which I believe needs to be
sqrt(-g)g^uv) (Weinberg, Chapter 10 as I recall, I asked about this in
some other threads and have now got past my initial misfiring on this).
This appears to be of the form of a Fourier transform between conjugate
"field space" and "source space" which parallels Fourier transforms
between conjugate spacetime and momentum space. And, BTW, this is why I
asked about the Fourier transform of F(x^-(10/3)), because boiled down
to the underlying "skeletal" mathematical structure, that is what needs
to be transformed to do the E-H path integral. (There is an x^-2
originating from the g_uv, and another x^-4/3 originating from the
metric determinant g. This has a parallel role to the underlying
Gaussian integral that shows up as the skeleton for QED path
integration.) I am working up an exercise and am a few days from having
something I'd feel comfortable posting, with some other patent work that
I have to get done first. But it seems to me so long as you retain a
non-zero trace energy in the E-H action, the opportunity will exist to
have both matter and fields, defining conjugate spaces, with T^uv
specifying the Fourier-transformed space conjugate to the field space of
g_uv. And -- the answer that comes out of this is completely finite,
with no special tricks other than good deductive mathematical elbow
grease.
For as much of the the specific math as I can convey in summary fashion,
the stress energy part of the path integral which remains after you get
out the R terms using delta/delta T^uv and remove the $d^4x to simplify
the math is:
exp(-.5i sqrt(-g)g_uv T_uv) (1)
By the substitution of variables:
x_uv==sgrt(g)g_uv (2)
w^uv==.5iT^uv (3)
this turns into:
exp(-ix_ux w^uv) (4)
You should be able to see this, just thinking about what you already
know about the matter part of the EH action. That, of course, is a
variant of the Fourier kernel.
The origin of F(x^-(10/3)) is a bit more involved and I can't do it full
justice here, but because of that substitution of variables (2),(3),
other terms sitting next to the exponential (1) but inside the integral
also need to reflect this substitution, and so after that is also taken
care of, one ends up having to integrate a term, with the indexes all
suppressed, which is:
$ x^-(10/3) exp(-ix w) dx (5)
and that integral is just a Fourier transform F(x^-(10/3)), which is
#310 in the table in
http://74.125.95.132/search?q=cache:fmiTYnSjDbgJ:en.wikipedia.org/wiki/Fourier_transform+fourier+transform&cd=1&hl=en&ct=clnk&gl=us,
and which I have been able to confirm does work for fractional n using
fractional factorials generated by gamma functions, and which is fully
finite.
Jay
Gerard Westendorp
[..]
>
> For QFT, setting aside the fact that some path integrals are very
hard to calculate exactly, the conceptual prescription is really very
simple: Take L for the field psi, and put it in the path integral:
>
> Z = $dpsi exp i[d^4x sqrt(-g) L] (3)
>
> Then, calculate the intergal. Period.
Do you get classical field theory if Plank's constant goes to zero?
If so, can you see this from eq (3)?
Gerard
Jay R. Yablon
news:fb3de115-a79f-47b9...@x31g2000yqx.googlegroups.com...
X-Phy:
I posted a reply three days ago which has still not arrived here. In
it, I asked why you said "you'll have the field only part of general
relativity, without matter." I have since realized after a parallel
reply from Dr. Carlip at
http://groups.google.com/group/sci.physics.research/browse_frm/thread/d15e35b971697965#
that a particularly simple formulation of the Einstein-Hilbert action,
directly including the stress energy trace T:
S=$[(1/2k)R-(1/2)T]sqrt(-g)d^4x (1)
is apparently not known (and thought to be incorrect.)
Consequently, I have written up a proof of (1), in the first ~3 pages in
the file linked below:
http://jayryablon.files.wordpress.com/2009/11/einstein-hilbert-action-4.pdf
Once the relationship (1) is shown to be correct and known, the
possibilities for selecting a suitable "measure" of field integration
for the path integral become much more clearly defined. Section 2 in
the above link lays this out, and if you think section 1 is correct, I'd
ask you to consider looking at section 2, and letting me know if you
think this is on the right track.
Thanks,
Jay
Jay R. Yablon
news:fb3de115-a79f-47b9...@x31g2000yqx.googlegroups.com...
After several false starts where I tried to mess around with the L_m
in the Einstein-Hilbert action, I think I may finally be able to show,
mathematically, how to evaluate the path integral:
Z=${-oo to +oo}Dg exp[i $d^4x sqrt(-g)((1/2k)R+L_m)] (1)
for the E-H action, without any messing around. This is at the link
below:
I have indeed had to rely on an approach not in the textbooks, because
the usual tricks that are employed in gauge theory mightily resist any
useful application to gravitation for reasons outlined in the above.
In section 1, we briefly review the Einstein-Hilbert action and insert
this into the standard Feynman path integral. That is, in section 1, we
simply obtain (1), while laying out the standard, known background
material which supports this action and relates it to the Einstein
equation. Nothing in section 1 is new.
Then, in section 2, we show how to *reparameterize* the matter
Lagrangian in such a way as to allow the path integral (1) to be
mathematically evaluated over the definite field-density range from
negative to positive infinity, using Fourier analysis.
The end result, deduced in (2.25) and generalized for various choices of
"measure" in (2.33), is the *mathematical* deduction that the evaluation
of (2) above over the definite range -oo to +oo is as follows:
Z=${-oo to +oo} D(sqrt(-g)g_ab) exp [i $d^4x sqrt(-g) ((1/2k)R + L_m)]
= delta^(4xoo)[.25 g_ab ((1/2k)R + L_m) ] (2)
where delta^(4xoo) is a 4 x infinite dimensional Dirac delta (Fourier
impulse) function.
The E-H Lagrangian density:
L = (1/2k)R + L_m (3)
therefore moves through the path integration intact, and ends up inside
an impulse function in the space conjugate to spacetime. In this
formulation, the traceless EM field (L=.25 F^uv F_uv) clearly would
contribute, as would any trace matter kT=R.
I want to be very clear on one point: I am saying nothing here or in
this paper about the *physics* that is associated with (2). I am simply
addressing the *mathematical* problem of evaluating the definite
integral (1), and getting a real answer, which, according to what is
deduced here, the mathematics tells us is (2). The only place where
anything other than pure mathematical calculation comes into play, is in
selecting the "measure" for the integration field. Rather than "choose"
one measure over another, I have used what seem to be the four most
reasonable possibilities for the measure, and, as it turns out, they all
lead to the same mathematical outcome.
I will keep my fingers crossed that after several trials that were
errors, perhaps I finally have figured out the right way to do this.
Thanks.
Jay R. Yablon
news:7mrv4rF...@mid.individual.net...
> "X-Phy" <xphys...@gmail.com> wrote in message
> news:fb3de115-a79f-47b9...@x31g2000yqx.googlegroups.com...
>> On 17 nov, 22:04, "Jay R. Yablon" <jyab...@nycap.rr.com> wrote:
>>
>> You'll have a quantized version of the field theory of gravitation,
>> but even if cooked by Feynman, and whatever the number of tomes, the
>> paths integral quantization hasn't that magical power of reconciling
>> two theories (QM and GR) that initially contradict one another.
>> Indeed, you'll have the field only part of general relativity,
>> without
>> matter. The unified theory isn't yet in textbooks, and it'll be
>> necessary to depart from them if some progress is wanted.
>
> After several false starts where I tried to mess around with the
> L_m
> in the Einstein-Hilbert action, I think I may finally be able to
> show,
> mathematically, how to evaluate the path integral:
>
> Z=${-oo to +oo}Dg exp[i $d^4x sqrt(-g)((1/2k)R+L_m)] (1)
>
> for the E-H action, without any messing around. This is at the link
> below:
>
> http://jayryablon.files.wordpress.com/2009/11/fourier-path-integration-of-the-einstein-hilbert-action.pdf
>
Overnight, it became apparent to me that I should specifically discuss
the "measure problem" which Dr. Carlip raised in one of his recent posts
in a parallel thread.
I have therefore posted an update at:
What the above now contains, just added, is a detailed discussion of how
to select the measure so that the overall path integral evaluates to
something that is invariant under general coordinate transformations. I
begin to introduce this discussion at equation (2.13), and the upshot is
that the only measure which meet this criteria of a generally-invariant
evaluation of the path integral is sqrt(-g), alone.
In brief, the "measure problem" is resolved by distinguishing between
local coordinate invariance and global coordinate invariance. The goal
is to end up with a result that is globally-invariant, even if the
variable of integration is itself not locally-invariant. One may think
of this approach as one of "hidden invariance." ***This all works,
because in taking a definite integral, even over a coordinate dependent
measure, the measure drops out from the overall expression that results
from evaluating the definite integral.***
In essence, the way this works is that when we do path integrals, we are
forced to select a particular system of coordinates, perform the
definite path integration in this system of coordinates, and obtain a
result. Then, we transform into a different system of coordinates,
again do the path integration, and obtain a second result. The goal is
to find that the second result is the same as the first result.
Repeating this ad infinitum, we choose system after system of
coordinates until we have exhausted every possible coordinate system,
and each time we do the integration, it is the goal to end up with the
same, invariant result. That is, no matter what system of coordinates
we choose, our path integral should yield the same invariant result.
Again, even though any particular choice of measure is a non-covariant
choice, the path integral should evaluate invariantly no matter what
coordinates we choose to represent the measure. This is then a
"globally" covariant result. This is what we mean by "hidden
covariance." This updated post demonstrates that sqrt(-g) is the only
measure which meets this goal and is the only truly invariant choice.
The final result of this effort, is in equation (3.19):
Z = $(-oo to +oo) Dsqrt(-g) exp[i S_EH] = delta^(4xoo)(L_EH) (1)
where the Einstein Hilbert action:
S_EH = $(-oo to +oo) sqrt(-g) L_EH d^4x (2)
and the Einstein Hilbert Lagrangian density is:
L_EH = (1/2 kappa) R + L_matter (3)
If you look closely at (3.19), and think about the Fourier
transformation:
$(-oo to +oo) dx exp[i x w] = delta (w) (4)
it should be self-evident that (1) (which is (3.19) in the linked file)
evaluates to a Dirac delta. In fact, in retrospect, one can omit just
about everything leading up to (3.19), stare at (3.19) closely for a few
moments, and realize that (3.19) is just a disguised form of (4), with a
coordinate-dependent measure that "washes out" no matter what coordinate
system one uses for the integration.
I just realized I am missing some sqrt(2pi) factors on the deltas. I'll
put this in a next draft, but those are non-essential to the main
result.
Jay.
[[Mod. note -- 63 very-excessively-quoted lines snipped. -- jt]]
Arnold Neumaier
This "answer" is much less well-defined than you think. At present,
in 4 dimensions, only the case of free fields has a well-defined path
integral.
In the interacting case, one gets (for renormalizable theories)
finite answers only for Taylor coefficients of path integrals,
and only when the parameters in the Lagrangian diverge in the
particular manner determined by renormalization theory.
For unrenormalizable theories, one does not even get this much.
> Obviously, what I
> just said presupposes that you have a classical Lagrangian density to
> begin with, but that caveat aside, I do believe that this is a question
> which does have a simple, yes or no answer:
The only simple answer can then be: No, you are not correct.
> Is the path integral,
> however difficult it can be to calculate with, and however many tricks
> on needs to use to get a good answer, essentially a mathematical machine
> for quantizing a classical field theory?
No. It is an ill-defined, nonmathmatical machine for approximately
quantizing a classical theory. Mathematically, the path integral has
never been shown to exist for interactive 4D theories.
> I have read quite a bit. But, after looking at trees, I always try to
> see the forest in the end. IMHO, we need more, not less of that these
> days.
After having looked at the elementary particles (the trees), looking
at the material (the forest) made of them, already moves you from
the quantum realm to the classical realm. Statistical mechanics (or
thermodynamics) is the forest you are trying to see - but in your eyes,
things seem to come out quite differently....
Jay R. Yablon
"Arnold Neumaier" <Arnold....@univie.ac.at> wrote in message
news:4B0BE6E1...@univie.ac.at...
With all due respect, you are beating around the bush in the same way as
Igor. You tip your hand when you say "*At present*, in 4 dimensions,
only the case of free fields has a well-defined path integral." That
does not mean that I am incorrect per your assertion below, but only
that at the moment the jury is still out, because nobody has yet figured
out how to use the path integral to get finite answers for other than
free fields. That does not mean that this is not possible, nor, I grant
you, does this mean that this *is* possible. We just do not know.
Using the analogy of the path integral as a "machine," I would maintain
that this is a "machine" that we human beings have not yet figured out
how to properly operate. That in no way means that path integration is
an inherently defective as a machine.
>
>> Obviously, what I
>> just said presupposes that you have a classical Lagrangian density to
>> begin with, but that caveat aside, I do believe that this is a
>> question
>> which does have a simple, yes or no answer:
>
> The only simple answer can then be: No, you are not correct.
Per above, I think that the real answer is that we just do not know,
because we human operators still don't quite know how to "operate" the
path integral machine except in certain limited cases.
>
>> Is the path integral,
>> however difficult it can be to calculate with, and however many
>> tricks
>> on needs to use to get a good answer, essentially a mathematical
>> machine
>> for quantizing a classical field theory?
>
> No. It is an ill-defined, nonmathmatical machine for approximately
> quantizing a classical theory. Mathematically, the path integral has
> never been shown to exist for interactive 4D theories.
The key here: "*has never been shown to exist* for interactive 4D
theories." This does not mean it does not exist. Nor does this means
that it does exist. It means that the jury is still out -- unless
someone has proved a "no-go theorem" to show that the path integral
*cannot* exist for interactive 4D theories. If such a theorem has been
proved, I would appreciate a reference to such proof, and would assume
that you would have already said "has been shown to NOT exist" rather
than the weaker "has never been shown to exist."
I have tried to make some headway on these questions, in the newly
posted SPR thread "The Mathematical Substructure Underlying Quantum
Field Theory, and Why I Like Anthony Zee's Book so Much."
>
>> I have read quite a bit. But, after looking at trees, I always try
>> to
>> see the forest in the end. IMHO, we need more, not less of that
>> these
>> days.
>
> After having looked at the elementary particles (the trees), looking
> at the material (the forest) made of them, already moves you from
> the quantum realm to the classical realm. Statistical mechanics (or
> thermodynamics) is the forest you are trying to see - but in your
> eyes,
> things seem to come out quite differently....
Yes, it is the human condition that different people see the same things
through different eyes. That is what keep life interesting, and one
should generally exercise caution in making value judgments about
different sets of eyes. ;-)
You interpreted my "forest and trees" remark way too literally. I am
simply talking in a non-technical, colloquial manner about stepping back
out of the all the gory details to try to see a big picture. And, in
that big picture, I maintain that the path integral does take a
classical theory and quantize that theory, and that the problems we have
using path integrals for interacting 4-D theories have less to do with
the path integral itself, than with limitations in our ability as humans
beings to operate that path integral machine, at least to date. Absent
some no-go theorem about applying the path integral to 4-D interacting
theories, I would think that you and I and maybe even Igor can come to
terms on at least agreeing that the jury is still out on these
questions.
Then, perhaps we can look at how to advance the envelope of knowledge by
perhaps learning how to better "operate" the path integral machine.
Jay
Igor Khavkine
> With all due respect, you are beating around the bush in the same way as
> Igor.
I don't see how either of us is beating around the bush. But let me
try once again, and I'll try to be as clear as possible. Here's the
information you need to know, in bullet form.
- Except for isolated situations, the path integral is not well
defined. In fact, forget the "well", it's just *not defined*.
- The path integral is a tool. All of QFT can be discussed without it.
- When it is defined, it is sometimes useful.
- However, it is not universal. Some questions cannot be answered
using the path integral alone.
- Many people can spend their entire careers using the path integral
in the domains where it is sort of well defined and be perfectly happy
with it. (Hence, books like Zee's.)
- However, you've indicated an interest in problems such as some
properties of gauge theories and gravity where the simple approach,
after many decades of trying, has yet to work and where hand waving
arguments are not expected to help much.
Just to be clear, discussing the existence of something that is not
defined is akin to discussing the problem of how many angles fit on
the head of a pin. The problem is simply ill posed because the term
"angel" is never defined.
Igor
Igor Khavkine
> The final result of this effort, is in equation (3.19):
>
> Z = $(-oo to +oo) Dsqrt(-g) exp[i S_EH] = delta^(4xoo)(L_EH) (1)
Jay, as I pointed out in another post. This result is wrong. No matter
what steps you believe lead you to this result, it is incorrect. This
equality can only hold if S_EH were linear in the integration
variable, be it g_uv, sqrt(-g)g_uv, or just sqrt(-g). And it should be
quite clear that S_EH is not linear in any of these variables.
Moreover, if you only write Dsqrt(-g), you are not even doing a path
integral: (1) you are not integrating over all degrees of freedom (10
components of g) and (2) sqrt(-g) is not algebraically independent of
g_uv.
Igor
Jay R. Yablon
"Gerard Westendorp" <wes...@xs4all.nl> wrote in message
news:4b065b14$0$22919$e4fe...@news.xs4all.nl...
http://jayryablon.files.wordpress.com/2008/11/zee-field-equations-and-inverses-markup.pdf
Jay
Arnold Neumaier
>
> "Arnold Neumaier" <Arnold....@univie.ac.at> wrote in message
> news:4B0BE6E1...@univie.ac.at...
>>> Is the path integral,
>>> however difficult it can be to calculate with, and however many tricks
>>> on needs to use to get a good answer, essentially a mathematical machine
>>> for quantizing a classical field theory?
>>
>> No. It is an ill-defined, nonmathmatical machine for approximately
>> quantizing a classical theory. Mathematically, the path integral has
>> never been shown to exist for interactive 4D theories.
>
> The key here: "*has never been shown to exist* for interactive 4D
> theories."
As long as it has not been shown to be well-defined, it _is_
ill-defined.
> This does not mean it does not exist. Nor does this means
> that it does exist. It means that the jury is still out
No. it means that, at present, it is ill-defined, and that therefore
the question of its existence is, strictly speaking, meaningless.
What I meant in precise terms is, that nobody has been able to figure
out a definition of the path integral that makes logically sense.
At present, the path integral is just a convenient intuition without
logical content, just like Dirac's delta function before it was given
a logically defensible meaning as a distribution. The hope is that
someone comes along to turn it into something respectable. But this is
unlikely to happen by looking at big pictures -- it needs the most
careful attention to the small details. The devil is, as always, in
the details, and can be cast out only by understanding these.
> You interpreted my "forest and trees" remark way too literally. I am
> simply talking in a non-technical, colloquial manner about stepping back
> out of the all the gory details to try to see a big picture. And, in
> that big picture, I maintain that the path integral does take a
> classical theory and quantize that theory,
It does it only if one can give a clear meaning to it, which one cannot
at present. Thus it is premature to conclude that the path integral
gives the correct big picture.
My big picture looks very different (see my FAQ), and does not feature
the (fundamentally flawed) path integral as anything fundamental.
Juan R.
I will add two more bullets:
- The path-integral formalism is not fundamental. Indeed, textbooks
explain how to obtain it from the canonical formalism.
- For some important theories the simplest version of the path-integral
method is, in Weinberg own words, "simply wrong". Weinberg cites the
case of the nonlinear sigma model. This is the reason which Weinberg
prefers to *obtain* path integrals (including vertices corrections to
Feynmans original version) from the canonical formalism.
--
http://www.canonicalscience.org/
BLOG:
http://www.canonicalscience.org/en/publicationzone/canonicalsciencetoday/canonicalsciencetoday.html
X-Phy
> > You interpreted my "forest and trees" remark way too literally. �I am
> > simply talking in a non-technical, colloquial manner about stepping back
> > out of the all the gory details to try to see a big picture. �And, in
> > that big picture, I maintain that the path integral does take a
> > classical theory and quantize that theory,
On 27 nov, 22:20, Arnold Neumaier <Arnold.Neuma...@univie.ac.at>
wrote:
> It does it only if one can give a clear meaning to it, which one cannot
> at present. Thus it is premature to conclude that the path integral
> gives the correct big picture.
Being ill-defined and having no clear meaning are two very different
things. The Feynman's paths integral has a clear meaning since it is
the integral form of the Huygens principle. It can be given a
heuristic definition, which is very insightful. What it says is that
quantization occurs by the introduction of some wave function in some
space, and that the least action principle can be reduced to an
interference phenomena through the Fermat theorem. In few words, it
makes the connexion between the Lagrangian and the wave propagation.
The Lagrangian formalism can be shown to be equivalent to a shortest
path problem in a space ad hoc. After that, the paths integral's work
is done and is no longer needed in building physics, though it could
be called again to duty for some calculations.
Canonical quantization is just another way to introduce a wave
function, and is well-defined. I say that for those who would say I
only wave my hands. Then through the Huygens principle (differential
form,) all the above concepts can be written down mathematically. The
path's integral was first of all invoked as an interpretation, for
what I think it failed altogether, since Schr�dinger took just the
opposit route in deriving his equation.
Wow, that makes a God damned number of great names, and I don't think
they would appreciate being called non clear sensical. In conlusion,
the paths integral give the big picture, since it is no more than an
"artist view" of all other quantization approaches. To the collection
of great names, I would add Georges Seurat, who didn't made "well-
defined" paintings, but on which everybody see what it is about, that
is, they have a clear meaning.
Phillip Helbig---undress to reply
<cd78e060-4921-4d3c...@m38g2000yqd.googlegroups.com>,
X-Phy <xphys...@gmail.com> writes:
> Being ill-defined and having no clear meaning are two very different
> things.
My maths professor once remarked that "well defined" is not well
defined.
Question: Is this actually true, in a mathematical-formal sense? What
about people using the terms "well defined" and "ill defined" in this
discussion? Is there some algorithm, or some test, which determines
whether a concept is ill defined or well defined?
Jay R. Yablon
Igor, you are correct. Thank you. After two days of playing around
with the Ricci scalar R in terms of g_uv, I made the very careless dumb
mistake of ignoring the g_uv-dependency of R which is in S_EH. I saw a
supposed quick route to the integration and was off and running way too
fast. Mea culpa.
Back on planet earth, let me ask a question or two before I head off on
on another wild goose chase. (Not that I do not learn some things along
the way while chasing geese, but I was particularly inattentive on this
one.)
In QED we use the functional derivative d/dJ^u (d=delta, the functional
derivative, I will use d here just for conciseness of notation). Then
we apply this to the generating functional F(J^u), so that we can
replace a function of the gauge field V(A^u) with V(d/dJ^u) and remove
some terms involving the gauge field from the integral by making them
independent of the variable / field of integration. I may not have the
words just right, but you know what I am saying. What makes this work
is that J^u is independent of A^u, as is V(d/dJ^u).
In gravitational theory, the EH action is (k=kappa, L_m=L_matter):
S_EH = $(-oo to +oo) sqrt(-g) {(1/2 k) R + L_m} d^4x (1)
My question is this: IF for gravitational theory one were to try to
specify a functional derivative analogous to d/dJ^u which is completely
independent of g_uv (which is tricky because g_uv connects all sorts of
scalar products together and raises and lowers every vector and tensor
index), what would that d/dJ^u analog be?
Am I correct in believing that it would have to be d/dT^ab using the
*upper indexed* energy source tensor T^ab?
If so, then because the Einstein equation in relation to the classical
variation of (1) requires, being careful with upper and lower indexes
that:
T^ab = 2 g^au g^bv (1/sqrt(-g)) d(sqrt(-g)L_m)/dg^uv, (2)
then would not the functional derivative that is totally free of g_uv
be:
d/dT^ab
=d/d{2 g^au g^bv (1/sqrt(-g)) d(sqrt(-g)L_m)/dg^uv} (3)
using T^ab in (2)?
And, although it *looks like* d/dT^ab is highly dependent on g_uv, am I
correct to look at (3) as saying that *it is L_m which has the
g_uv-dependence*, and that T^ab is the only thing in (3) which is
g_uv=independent? And, is it correct to look at (2) as a functional
differential equation defining a field-dependent L_m in terms of a
field-independent source T^ab?
Thanks,
Jay
Arnold Neumaier
> Jay R. Yablon wrote:
>
>>> You interpreted my "forest and trees" remark way too literally. I am
>>> simply talking in a non-technical, colloquial manner about stepping back
>>> out of the all the gory details to try to see a big picture. And, in
>>> that big picture, I maintain that the path integral does take a
>>> classical theory and quantize that theory,
>
> On 27 nov, 22:20, Arnold Neumaier <Arnold.Neuma...@univie.ac.at>
> wrote:
>
>> It does it only if one can give a clear meaning to it, which one cannot
>> at present. Thus it is premature to conclude that the path integral
>> gives the correct big picture.
>
> Being ill-defined and having no clear meaning are two very different
> things.
We clearly have different interpretations of the words. In my view,
an ill-defined concept cannot have a clear meaning. Its meaning is
intrinsically vague (strictly speaking, it is even meaningless)
since clarity requires a logically intelligible definition.
> The Feynman's paths integral has a clear meaning since it is
> the integral form of the Huygens principle.
but with an ill-defined and hennce meaningless integral.
> It can be given a
> heuristic definition, which is very insightful.
I agree that the path integral gives insight.
But insight is something very different from clarity.
> What it says is that
> quantization occurs by the introduction of some wave function in some
> space,
No. Path integrals are numbers, not wave functions!
The wave functions are introduced in addition to the path integrals,
to heuristically relate the formalism to the Schroedinger equation,
and thus recover the standard picture of quantum mechanics.
Formally, the path integral appears as an intuitive but ill-defined
integral expression for the kernel in the position representation of
an (in a canonical approach well-defined) unitary propagation operator.
The latter formulation is more general, more versatile, more precise,
and hence more basic than the former, which gives only one possible
form of intuition about the propagation.
> and that the least action principle can be reduced to an
> interference phenomena through the Fermat theorem. In few words, it
> makes the connexion between the Lagrangian and the wave propagation.
> The Lagrangian formalism can be shown to be equivalent to a shortest
> path problem in a space ad hoc.
... and only with ad hoc handwaving arguments that cannot be made
precise.
> After that, the paths integral's work
> is done and is no longer needed in building physics, though it could
> be called again to duty for some calculations.
>
> Canonical quantization is just another way to introduce a wave
> function, and is well-defined.
More precisely:
The path integral is well-defined only for free systems, while
canonical quantization is well-defined for all classical systems
with finitely many degrees of freedom. For non-free field theories,
both are ill-defined.
> I say that for those who would say I
> only wave my hands. Then through the Huygens principle (differential
> form,) all the above concepts can be written down mathematically.
This shows that the path integral is a logically dubious heuristics
only, while canonical quantization is the logically clear foundation.
> Wow, that makes a God damned number of great names, and I don't think
> they would appreciate being called non clear sensical.
I don't understand why the innocent number 5 (counting Feynman, Huygens,
Fermat, Lagrange, and Schroedinger) should be damned, or what it should
have to do with God, who presumably has a much clearer picture of the
natural laws than we can ever have.
But being clear, insightful, or useful are very different attributes.
Great physicists get their fame for being innovative and useful,
not for clarity. Thus your arguments support only that the path integral
was innovative when feynman introduced it and was found useful,
but says nothing about its clarity.
> In conlusion,
> the paths integral give the big picture, since it is no more than an
> "artist view" of all other quantization approaches.
Path integrals give _a_ big picture, but not _the_ big picture.
Pictures in visual art have often no clear meaning, and there may
be many big pictures of the same object, clear to very varying degrees.
The most famous pictures of an object are usually not the clearest ones.
The same is true for the pictures of physical theories.
In conclusion, the path integral provides a simple and intuitive big
picture (one of several alternatives) which is sometimes illuminating,
at other times thoroughly misleading.
It lacks the clarity needed for understanding quantum field theory.
Remember that it was Feynman - who introduced the path integral -
who said that nobody understands quantum mechanics!
Arnold Neumaier
Arnold Neumaier
> In article
> <cd78e060-4921-4d3c...@m38g2000yqd.googlegroups.com>,
> X-Phy <xphys...@gmail.com> writes:
>
>> Being ill-defined and having no clear meaning are two very different
>> things.
>
> My maths professor once remarked that "well defined" is not well
> defined.
If it were so, then nothing would be well-defined, and the term would
have lost its meaning. (Having no clear meaning would then be not
well-defined either.)
> Question: Is this actually true, in a mathematical-formal sense?
No; this must have been meant to be a joke.
In mathematics, the statement ``... is well-defined'' is usually
used to express the fact that although a definition given seems to
depend on the way other things are chosen, the result is in fact
independent of these choices. This is something any competent reader can
check, and hence something well-defined.
> What
> about people using the terms "well defined" and "ill defined" in this
> discussion? Is there some algorithm, or some test, which determines
> whether a concept is ill defined or well defined?
I call (in this discussion and otherwise) formal concepts in mathematics
or theoretical physics well-defined if, given a definition of the
concepts, one can say unambiguously, in formal terms already known,
what each phrase in the definition means.
In case of the path integral, this means specifying either the
measure or the limit in a formal, unambiguous way from which the
properties used later can be derived. This can be done
in function spaces for path integrals of the Wiener type,
but has not been done for those of the Feynman type.
Thus Wiener path integrals are well-defined, while Feynman
path integrals are not. This is why numerical work in lattice
gauge theories is based upon Euclidean field theory with
imaginary times. This amounts to replacing the Feynman integrals
by Wiener integrals, a necessity for getting a reliable numerics.
One cannot compute reliably with ill-defined stuff.
Arnold Neumaier
Igor Khavkine
> In gravitational theory, the EH action is (k=kappa, L_m=L_matter):
>
> S_EH = $(-oo to +oo) sqrt(-g) {(1/2 k) R + L_m} d^4x � (1)
>
> My question is this: �IF for gravitational theory one were to try to
> specify a functional derivative analogous to d/dJ^u which is completely
> independent of g_uv (which is tricky because g_uv connects all sorts of
> scalar products together and raises and lowers every vector and tensor
> index), what would that �d/dJ^u analog be?
>
> Am I correct in believing that it would have to be d/dT^ab using the
> *upper indexed* energy source tensor T^ab?
No. The simple reason is that neither T^ab nor T_ab appear as
independent parameters in the Lagrangian. T is always a function of
both the metric and the matter fields.
The rest of your questions rest of the premise that you could be
taking functional derivatives with respect to the stress energy
tensor. Without this premise, they no longer make sense.
Igor
Jay R. Yablon
> On Nov 30, 6:27 pm, "Jay R. Yablon" <jyab...@nycap.rr.com> wrote:
>
>> In gravitational theory, the EH action is (k=kappa, L_m=L_matter):
>>
>> S_EH = $(-oo to +oo) sqrt(-g) {(1/2 k) R + L_m} d^4x (1)
>>
>> My question is this: IF for gravitational theory one were to try to
>> specify a functional derivative analogous to d/dJ^u which is
>> completely
>> independent of g_uv (which is tricky because g_uv connects all sorts
>> of
>> scalar products together and raises and lowers every vector and
>> tensor
>> index), what would that d/dJ^u analog be?
>>
>> Am I correct in believing that it would have to be d/dT^ab using the
>> *upper indexed* energy source tensor T^ab?
>
> No. The simple reason is that neither T^ab nor T_ab appear as
> independent parameters in the Lagrangian. T is always a function of
> both the metric and the matter fields.
I want to be crystal clear what you are saying.
What is in the Lagrangian density are the Ricci curvature R, and
L_matter. The upper-indexed energy tensor must be:
T^ab = 2 g^au g^bv (1/sqrt(-g)) d(sqrt(-g)L_m)/dg^uv (1)
in order to reproduce the Einstein equation. Neither T^ab nor T_uv nor
the trace T=g_ab T^ab are in the Lagrangian. All that is there is R,
which is a field-dependent as can be (indeed it is all field), and
L_matter which has a function of both T^ab and g^ab, through equation
(1). As I understand it, T_ab IS the one thing that is field
independent (all matter, no field) (IS THIS SO?), but since T^ab it does
not appear in the Lagrangian density except via (1), there is no
apparent opportunity to exploit this through a functional derivative in
the same way that we can do in QED by postulating a J.A source/field and
then taking d/dJ to get A.
Anything amiss or missing?
Thanks,
Jay
Jay R. Yablon
> Jay R. Yablon wrote:
>>
>> "Arnold Neumaier" <Arnold....@univie.ac.at> wrote in message
>> news:4B0BE6E1...@univie.ac.at...
>
>>>> however difficult it can be to calculate with, and however many
>>>> tricks
>>>> on needs to use to get a good answer, essentially a mathematical
>>>> machine
>>>> for quantizing a classical field theory?
>>>
>>> No. It is an ill-defined, nonmathmatical machine for approximately
>>> quantizing a classical theory. Mathematically, the path integral has
>>> never been shown to exist for interactive 4D theories.
>>
>> The key here: "*has never been shown to exist* for interactive 4D
>> theories."
>
> ill-defined.
>
>
>> that it does exist. It means that the jury is still out
>
> the question of its existence is, strictly speaking, meaningless.
> What I meant in precise terms is, that nobody has been able to figure
> out a definition of the path integral that makes logically sense.
You keep proving my point and citing your proof as a refutation.
You say "at present" is it ill-defined. No quarrel there. But you are
inherently making a definite statement about the present state of human
knowledge, and as well as an uncertain statement about the ultimate fate
of the path integral... I do not see what "ill-defined" implies
"meaningless." It is also consistent with "not well understood yet."
> At present, the path integral is just a convenient intuition without
> logical content, just like Dirac's delta function before it was given
> a logically defensible meaning as a distribution.
Agreed. But the Dirac delta still had meaning as distribution before
its meaning as a distribution as discovered and proved. Just like
lightening was still governed by Maxwell's equations back when the
ancients thought the Gods we angry.
> The hope is that
> someone comes along to turn it into something respectable.
I think "respectable" is a poor choice of words. I would suggest that
"better defined" is a better choice unless your intention was to achieve
dramatic effect.
> But this is
> unlikely to happen by looking at big pictures -- it needs the most
> careful attention to the small details. The devil is, as always, in
> the details, and can be cast out only by understanding these.
I am curious, because the discussion here is still in generalities. Can
you please state, with some precision, exactly what in your view makes
the path integral in our present state of knowledge, "ill-defined"? Is
it a convergence issue? Is it all of the infinite products? A
combination of things? What, to the best you can articulate it, makes
this "ill-defined" "at present"? (And I will look at your FAQs to see
what is there and be back to you further.)
>
>> You interpreted my "forest and trees" remark way too literally. I am
>> simply talking in a non-technical, colloquial manner about stepping
>> back
>> out of the all the gory details to try to see a big picture. And, in
>> that big picture, I maintain that the path integral does take a
>> classical theory and quantize that theory,
>
> It does it only if one can give a clear meaning to it, which one
> cannot
> at present. Thus it is premature to conclude that the path integral
> gives the correct big picture.
>
> My big picture looks very different (see my FAQ), and does not feature
> the (fundamentally flawed) path integral as anything fundamental.
I do not perceive that we are at a state of scientific development when
one can state conclusively that path integral will, or will not be, of
lasting value. As I said before, the jury is still out on this, and it
seems to me to be a subject of debate in scientific circles. We are
simply dealing with a question of what presumptions ought to be
operative about what will ultimately happen when the jury turns in its
verdict. If we want to have some fun on the side, we can post wagers as
to whether history will eventually show the path integral to have been a
passing tool or a fundamental viewpoint. Then, we can pray that God
will leave us all here for another generation or so, and we can look
back to today and perhaps see which intuitions about the eventual state
of path integral turn out to have been vindicated by the passage of
time. You seem to be of the view that the path integral will fade in
importance over time; I tend to see this in the opposite way. x-phy
points out in reply that path integration is the "artists view" of QFT,
Maxwell's equations and GR are certainly earlier vindications of
artistic viewpoints.
I will take a good look thorough your FAQs, as I am curious about why
you believe this to be path integration to be "fundamentally flawed" as
opposed to presenting what I might couch as "transitory challenges." I
suppose you do believe that there is already a "no-go" theorem about the
long-term prognosis of the path integral?
Jay
Arnold Neumaier
Any statement anyone makes can only be about the present state of human
knowledge.
But you take your point already proven if there is a slight option for
it to be possibly true. I'd not call this a proven point but only a hope.
> and as well as an uncertain statement about the ultimate fate
> of the path integral... I do not see what "ill-defined" implies
> "meaningless." It is also consistent with "not well understood yet."
>
>> At present, the path integral is just a convenient intuition without
>> logical content, just like Dirac's delta function before it was given
>> a logically defensible meaning as a distribution.
>
> Agreed. But the Dirac delta still had meaning as distribution before
> its meaning as a distribution as discovered and proved.
There is a big difference, though. Before Schwarz, nobody bothered to
find a logically meaningful definition of the delta function.
But many have tried in vain to find a logical interpretation of
the Feynman path integral. Thus in my assessment the locical status
of the latter is much more in doubt.
Usually, the impossibility of something shows much earlier through
failed attempts before it can be proved to be impossible.
Think of trisecting the angle or squaring the circle.
> Just like
> lightening was still governed by Maxwell's equations back when the
> ancients thought the Gods we angry.
Taking your analogy seriously, this would mean that when the right
meaning has been found, the intuition the path integral provides
about quantum field theory will be like the intuition angry Gods
provide about lightnings. Very intuitive, but...
>> The hope is that
>> someone comes along to turn it into something respectable.
>
> I think "respectable" is a poor choice of words. I would suggest that
> "better defined" is a better choice unless your intention was to achieve
> dramatic effect.
I meant ``logically respectable''.
>> But this is
>> unlikely to happen by looking at big pictures -- it needs the most
>> careful attention to the small details. The devil is, as always, in
>> the details, and can be cast out only by understanding these.
>
> I am curious, because the discussion here is still in generalities. Can
> you please state, with some precision, exactly what in your view makes
> the path integral in our present state of knowledge, "ill-defined"?
There exists no prescription how the limit in the ``definition'' of
the path integral is to be taken to yield (in theory - independent
of the difficulty of computing them) numbers that have the properties
commonly ascribed to the path integral.
> Is
> it a convergence issue? Is it all of the infinite products? A
> combination of things? What, to the best you can articulate it, makes
> this "ill-defined" "at present"? (And I will look at your FAQs to see
> what is there and be back to you further.)
My FAQ at http://www.mat.univie.ac.at/~neum/physics-faq.txt
contains mostly a path-integral-free approach to QFT,
just because it is mathematically dubious.
Only the sections 6b/c and S9c contain a bit about path integrals,
>>> You interpreted my "forest and trees" remark way too literally. I am
>>> simply talking in a non-technical, colloquial manner about stepping
>>> back
>>> out of the all the gory details to try to see a big picture. And, in
>>> that big picture, I maintain that the path integral does take a
>>> classical theory and quantize that theory,
>> It does it only if one can give a clear meaning to it, which one
>> cannot
>> at present. Thus it is premature to conclude that the path integral
>> gives the correct big picture.
>>
>> My big picture looks very different (see my FAQ), and does not feature
>> the (fundamentally flawed) path integral as anything fundamental.
>
> I do not perceive that we are at a state of scientific development when
> one can state conclusively that path integral will, or will not be, of
> lasting value. As I said before, the jury is still out on this,
... and may well be for another 1000 years -- remember the squaring of
the circle...
That's why I don't wait with my assessment till then. I live now!
> and it
> seems to me to be a subject of debate in scientific circles. We are
> simply dealing with a question of what presumptions ought to be
> operative about what will ultimately happen when the jury turns in its
> verdict.
You and I speculate the very opposite in this respect.
The path integral lives only because of its connection to
the propagator in canonical quantum mechanics. Few in quantum
mechanics use it since the canonical approach is so much more
powerful.
This shows that the path integral is a crutch. It is used in QFT
only because we do not yet have a nonpertubative view of the latter.
But physics is moving towards the nonperturbative - it is needed to
understand all quantum phenomena apart from scattering.
Lattice QFT and its variants will be the computational tools of
the future. They are nonperturbative, and have at least in 2D a good
contnuum limit. The 2D path integral can be well-defined only through
the lattice approach, in a quite indirect manner. Thus nothing at all
suggests that the path integral is anything fundamental.
> I will take a good look thorough your FAQs, as I am curious about why
> you believe this to be path integration to be "fundamentally flawed" as
> opposed to presenting what I might couch as "transitory challenges." I
> suppose you do believe that there is already a "no-go" theorem about the
> long-term prognosis of the path integral?
60 years of failed attempts by some of the best mathematicians and
physicists speak with a loud voice.
Arnold Neumaier
Igor Khavkine
> I want to be crystal clear what you are saying.
>
> What is in the Lagrangian density are the Ricci curvature R, and
> L_matter. The upper-indexed energy tensor must be:
>
> T^ab = 2 g^au g^bv (1/sqrt(-g)) d(sqrt(-g)L_m)/dg^uv (1)
>
> in order to reproduce the Einstein equation. Neither T^ab nor T_uv nor
> the trace T=g_ab T^ab are in the Lagrangian. All that is there is R,
> which is a field-dependent as can be (indeed it is all field),
So far so good.
> and
> L_matter which has a function of both T^ab and g^ab, through equation
> (1).
And this is definitely wrong. In fact, I'm at a loss as to how you
could read this from what I wrote in this thread or other previous
discussions.
The matter Lagrangian L_m is a function of the matter fields, which
I'll collectively call phi. The matter fields are all fields that are
not the metric (EM field, Dirac field, Yang-Mills field, and so on).
The stress energy tensor T (no matter where you put the indices) is a
function of phi and g. The Lagrangian L_m is also a function of phi
and g, but not a function of T (since T is constructed from L_m by
varying with respect to g). Anything else you want to say must be
consistent with the above, and so far very little has been.
Igor
Igor Khavkine
> "Arnold Neumaier" <Arnold.Neuma...@univie.ac.at> wrote in message
> news:4B0DA97C...@univie.ac.at...
> > Jay R. Yablon wrote:
> > As long as it has not been shown to be well-defined, it _is_
> > ill-defined.
>
> >> This does not mean it does not exist. Nor does this means
> >> that it does exist. It means that the jury is still out
>
> > No. it means that, at present, it is ill-defined, and that therefore
> > the question of its existence is, strictly speaking, meaningless.
> > What I meant in precise terms is, that nobody has been able to figure
> > out a definition of the path integral that makes logically sense.
>
> You keep proving my point and citing your proof as a refutation.
>
> You say "at present" is it ill-defined. No quarrel there. But you are
> inherently making a definite statement about the present state of human
> knowledge, and as well as an uncertain statement about the ultimate fate
> of the path integral... I do not see what "ill-defined" implies
> "meaningless." It is also consistent with "not well understood yet."
If you've not quite grasped the use of the term "meaningless" in this
context, I'll kindly ask you provide an example of a calculation of a
wigamobuzom.
What's a wigamobuzom and how do you even know that it exists to be
calculated, you might ask? Well, it's not defined. From a mathematical
point of view, there is no difference between a wigamobuzom and a path
integral (in the extremely general context you want it that term to be
applicable in). Neither term has a mathematical definition. To speak
of calculating either one or the other is hence meaningless.
Igor
X-Phy
> After several false starts where I tried to mess around with the L_m
> in the Einstein-Hilbert action, I think I may finally be able to show,
> mathematically, how to evaluate the path integral:
>
> Z=${-oo to +oo}Dg exp[i $d^4x sqrt(-g)((1/2k)R+L_m)] (1)
>
> for the E-H action, without any messing around. This is at the link
> below:
>
> http://jayryablon.files.wordpress.com/2009/11/fourier-path-integratio...
You overlooked than in the matter Lagrangian density of matter, the
metric tensor appears since the motion of matter depends of it. In
the quantized version of GR, it no longer have a determinate value.
That not the same as if you started from a given energy-momentum
density.
Jay R. Yablon
> On Dec 1, 1:10 am, "Jay R. Yablon" <jyab...@nycap.rr.com> wrote:
>
>> I want to be crystal clear what you are saying.
>>
>> What is in the Lagrangian density are the Ricci curvature R, and
>> L_matter. The upper-indexed energy tensor must be:
>>
>> T^ab = 2 g^au g^bv (1/sqrt(-g)) d(sqrt(-g)L_m)/dg^uv (1)
>>
>> in order to reproduce the Einstein equation. Neither T^ab nor T_uv
>> nor
>> the trace T=g_ab T^ab are in the Lagrangian. All that is there is R,
>> which is a field-dependent as can be (indeed it is all field),
>
> So far so good.
>
>> and
>> L_matter which has a function of both T^ab and g^ab, through equation
>> (1).
>
> And this is definitely wrong. In fact, I'm at a loss as to how you
> could read this from what I wrote in this thread or other previous
> discussions.
>
> The matter Lagrangian L_m is a function of the matter fields, which
> I'll collectively call phi. The matter fields are all fields that are
> not the metric (EM field, Dirac field, Yang-Mills field, and so on).
Let me see if I have this straight now. Thanks for patiently bearing
with me.
In other words, L_m at least contains the usual standard model
Lagrangian (D_u=d^u-igA_u, the gauge-covariant derivative):
L_m=psi-bar(D_u gamma^u - m)psi-.25F_uvF^uv+|iD_u phi|^2 -V(phi)+...
(2)
where psi is the Dirac field and phi is a scalar field and +... means
anything else that one might put there which is not a metric tensor
field. For Yang-Mills, we treat many of the objects in the above as NxN
Hermitian matrices for any SU(N). And, g_uv is implied in (2) because
it sits between any two contracted indexes.
Then, with the Einstein-Hilbert Lagrangian density, we simply take the
standard model Lagrangian with all of the non-metric fields, and then
add in the (1/2k)R for the Ricci scalar thereby introducing the metric
fields through the curvature scalar. That is:
L_EH = sqrt(-g)((1/2k)R+L_m) (3)
with L_m containing all of (2).
> The stress energy tensor T (no matter where you put the indices) is a
> function of phi and g.
Because it is constructed from (2) by variation with respect to g^uv
according to
T^ab = 2 g^au g^bv (1/sqrt(-g)) d(sqrt(-g)L_m)/dg^uv (1)?
> The Lagrangian L_m is also a function of phi
> and g,
as in (2) above?
> but not a function of T (since T is constructed from L_m by
> varying with respect to g).
as in (1) above?
> Anything else you want to say must be
> consistent with the above, and so far very little has been.
Does all of the above work? If so, that is very simple, and I read way
too much into everything. All you do is add a term with R into the
standard model Lagrangian as in (3). If you want the energy tensor, you
vary as in (1). And, if you do the same variation on the R term, you
get R_uv-.5g_uvR. And both variations together, give you the Einstein
field equation.
Jay.
> Igor
X-Phy
> Usually, the impossibility of something shows much earlier through
> failed attempts before it can be proved to be impossible.
> Think of trisecting the angle or squaring the circle.
Your logical assessment is just plain wrong. If something is proven
impossible, of course there were failed attemps before, but the
converse isn't true. Many things have been successful after many
failed attemps. The big Fermat theorem is a famous example. Then the
moto of the Shadoks: "Plus �a rate, plus �a marche." [The more it
fails, the more it succeeds.]
X-Phy
wrote:
"Arnold Neumaier" <Arnold....@univie.ac.at> a �crit dans le
message de
news:4B119000...@univie.ac.at...
>> Being ill-defined and having no clear meaning are two very different
>> things.
> We clearly have different interpretations of the words. In my view,
> an ill-defined concept cannot have a clear meaning. Its meaning is
> intrinsically vague (strictly speaking, it is even meaningless)
> since clarity requires a logically intelligible definition.
Having a clear meaning isn't the same as being understood by
everybody.
>> The Feynman's paths integral has a clear meaning since it is
>> the integral form of the Huygens principle.
> but with an ill-defined and hennce meaningless integral.
You can't make a logical deduction between concepts that are
fundamentally different.
>> It can be given a
>> heuristic definition, which is very insightful.
> I agree that the path integral gives insight.
> But insight is something very different from clarity.
>> What it says is that
>> quantization occurs by the introduction of some wave function in some
>> space,
> No. Path integrals are numbers, not wave functions!
I disagree totally, wave functions are numbers too. The final
evaluation of the paths integral, if taken to every points in the
configuration space for a given time slice, gives a field of complex
number that is the wave function by definition. Indeed, the transition
amplitude in the position representation is proportional to the
probability amplitude of detecting the system at that point, aka the
wave function.
The paths integral was intended to put that wave function under the
carpet in order to gain an interpretation of QM. The same is true with
the canonical quantization. When p becomes -i@x, that operator must act
upon a function, even if the later is never introduced explicitly. The
purpose of both approaches is indeed to hide that wave function that we
don't want to see, because it put to light the very difficulty of QM:
the non locality of the wave function collapse.
> The wave functions are introduced in addition to the path integrals,
> to heuristically relate the formalism to the Schroedinger equation,
> and thus recover the standard picture of quantum mechanics.
Are introduced, and not added, because they were there from the
beginning.
> Formally, the path integral appears as an intuitive but ill-defined
> integral expression for the kernel in the position representation of
> an (in a canonical approach well-defined) unitary propagation operator.
>
> The latter formulation is more general, more versatile, more precise,
> and hence more basic than the former, which gives only one possible
> form of intuition about the propagation.
We know that there are systems without a Lagrangian formulation, that's
not what we are discussing. As for basicness and all fundamentalities,
it is only a matter of taste, like in mathematics some theorem can be
taken in the place of an axiom. They are different from generality.
What is only required is consistency and completeness.
That said, you should be aware that different mathematical equivalent
formulations are available for the same process. The paths integral is
just another (very convoluted) representation of the unitary propagation
operator, taking benefice of the superposition principle, just like the
Huygens principle. Mathematically, the propagation operator is made to
act on a delta function, and the result is decomposed into delta
functions. The integral makes the final superposition. The central
difficulty is to properly wield the set of all trajectories, which is a
mathematical issue only, and thus doesn't impair the meaning.
>> and that the least action principle can be reduced to an
>> interference phenomena through the Fermat principle. In few words, it
>> makes the connexion between the Lagrangian and the wave propagation.
>> The Lagrangian formalism can be shown to be equivalent to a shortest
>> path problem in a space ad hoc.
> ... and only with ad hoc handwaving arguments that cannot be made
> precise.
I don't understand why you say they can't be made precise since that
have been done in the past, as soon as with Schr�dinger. I myself
derived the shortest path problem, it's elementary algebra.
>> Canonical quantization is just another way to introduce a wave
>> function, and is well-defined.
> More precisely:
>
> The path integral is well-defined only for free systems,
No, it works very well for one point particle in a potential, therefore
not free.
> while canonical quantization is well-defined for all classical systems
> with finitely many degrees of freedom. For non-free field theories,
> both are ill-defined.
>> I say that for those who would say I
>> only wave my hands. Then through the Huygens principle (differential
>> form,) all the above concepts can be written down mathematically.
> This shows that the path integral is a logically dubious heuristics
> only, while canonical quantization is the logically clear foundation.
???
>> Wow, that makes a God damned number of great names, and I don't think
>> they would appreciate being called non clear sensical.
> But being clear, insightful, or useful are very different attributes.
> Great physicists get their fame for being innovative and useful,
> not for clarity. Thus your arguments support only that the path integral
> was innovative when feynman introduced it and was found useful,
> but says nothing about its clarity.
It wasn't innovative, precisely. Fermat and Schr�dingen were
innovative, mathematically rigorous and crystal clear.
>> In conlusion,
>> the paths integral give the big picture, since it is no more than an
>> "artist view" of all other quantization approaches.
> Path integrals give _a_ big picture, but not _the_ big picture.
> Pictures in visual art have often no clear meaning, and there may
> be many big pictures of the same object, clear to very varying degrees.
> The most famous pictures of an object are usually not the clearest ones.
Ok, _a_ big picture. But it isn't because there are other pictures that
it isn't _one_.
> The same is true for the pictures of physical theories.
>
>
> In conclusion, the path integral provides a simple and intuitive big
> picture (one of several alternatives) which is sometimes illuminating,
> at other times thoroughly misleading.
> It lacks the clarity needed for understanding quantum field theory.
> Remember that it was Feynman - who introduced the path integral -
> who said that nobody understands quantum mechanics!
I don't know what you are after here, but it's not the same thing as
me. Indeed, there are two components in QM:
- The mere wave mechanics, which is nothing more than a classical
theory, even in second quantification. It is perfectly well depicted by
the paths integral or other points of view. Included here are all the
issues of infinities, renormalization, time ordering and all that.
- Proper QM, with the probabilistic interpretation and the projection
postulate. It is counter intuitive and no picture has yet been found
for it.
If you are attempting to fudge them in some obscure formalism or
something, you are misled. Even canonical quantization or any other
phrase ending in 'quantization', whatever well defined, fundamental, or
general they are, can't provide a clear meaning to proper QM. They
aren't interpretations but reformulations. The paths integral and all
other approaches preferred by you mean one a an only thing: mere wave
mechanics. Indeed, they don't make understand proper QM.
X-Phy
> What's a wigamobuzom and how do you even know that it exists to be
> calculated, you might ask? Well, it's not defined. From a mathematical
> point of view, there is no difference between a wigamobuzom and a path
> integral (in the extremely general context you want it that term to be
> applicable in). Neither term has a mathematical definition. To speak
> of calculating either one or the other is hence meaningless.
A wigamobuzom is a device to kill kangaroos in Mexico. What? There is
no kangaroo in Mexico? Then it isn't a wigamobuzom.
Ill defined is different from not defined at all. So in this example
a wigamobuzon is ill defined, but has a clear meaning. In every day
life, there are many not well defined things: beauty, smartness,
highness, yet they have a clear meaning, everybody know what it is
about, because the meaning always comes before the defintition.
[[Mod. note -- I would like to hope that, by virtue of (among other
things) using precisely-defined terminology, we can reach a higher
degree of consensus in physics discussions than people can in debating
whether or not (say) a given object is "beautiful" or "smart" or "high".
-- jt]]
Arnold Neumaier
> On 30 nov, 18:31, Arnold Neumaier <Arnold.Neuma...@univie.ac.at>
> wrote:
>
> "Arnold Neumaier" <Arnold....@univie.ac.at> a �crit dans le
> message de
> news:4B119000...@univie.ac.at...
>
>>> Being ill-defined and having no clear meaning are two very different
>>> things.
>
>> We clearly have different interpretations of the words. In my view,
>> an ill-defined concept cannot have a clear meaning. Its meaning is
>> intrinsically vague (strictly speaking, it is even meaningless)
>> since clarity requires a logically intelligible definition.
>
> Having a clear meaning isn't the same as being understood by
> everybody.
I agree. But this does not affect the truth of what I sis.
>>> The Feynman's paths integral has a clear meaning since it is
>>> the integral form of the Huygens principle.
>
>> but with an ill-defined and hennce meaningless integral.
>
> You can't make a logical deduction between concepts that are
> fundamentally different.
You can't make a logical deduction based on Feu=ynman integrals
that have never given a logically sound definition.
>>> It can be given a
>>> heuristic definition, which is very insightful.
>
>> I agree that the path integral gives insight.
>> But insight is something very different from clarity.
>
>>> What it says is that
>>> quantization occurs by the introduction of some wave function in some
>>> space,
>
>> No. Path integrals are numbers, not wave functions!
>
> I disagree totally, wave functions are numbers too. The final
> evaluation of the paths integral, if taken to every points in the
> configuration space for a given time slice, gives a field of complex
> number that is the wave function by definition.
By whose definition?
A wave function is generally defined simply as an arbitrary square
integrable function in L^2(R^3). Your definition is, in contrast,
very complicated (and I doubt very much that it is correct).
And even if it is correct, what you describe only says that the
wave function is something that you compute by evaluating a family
of path integrals all all values of a position parameter.
> Indeed, the transition
> amplitude in the position representation is proportional to the
> probability amplitude of detecting the system at that point, aka the
> wave function.
You are far too genorous with your use of the word ``is''.
A transition amplitude is not a wave function, but the inner product
of an ingoing wave function and an outgoing wave function.
>> The wave functions are introduced in addition to the path integrals,
>> to heuristically relate the formalism to the Schroedinger equation,
>> and thus recover the standard picture of quantum mechanics.
>
> Are introduced, and not added, because they were there from the
> beginning.
So where is the Hilbert space? Is it there before the path integral,
or constructed from it? Your presentation is very vague, far too vague
for something claimed to be fundamental.
>> Formally, the path integral appears as an intuitive but ill-defined
>> integral expression for the kernel in the position representation of
>> an (in a canonical approach well-defined) unitary propagation operator.
>>
>> The latter formulation is more general, more versatile, more precise,
>> and hence more basic than the former, which gives only one possible
>> form of intuition about the propagation.
>
> We know that there are systems without a Lagrangian formulation, that's
> not what we are discussing. As for basicness and all fundamentalities,
> it is only a matter of taste, like in mathematics some theorem can be
> taken in the place of an axiom. They are different from geneality.
> What is only required is consistency and completeness.
The Feynam path integral has never been given a consistent formulation.
Thus one of your basic requirements is not satisfied.
>>> and that the least action principle can be reduced to an
>>> interference phenomena through the Fermat principle. In few words, it
>>> makes the connexion between the Lagrangian and the wave propagation.
>>> The Lagrangian formalism can be shown to be equivalent to a shortest
>>> path problem in a space ad hoc.
>
>> ... and only with ad hoc handwaving arguments that cannot be made
>> precise.
>
> I don't understand why you say they can't be made precise since that
> have been done in the past, as soon as with Schr�dinger.
Schroedinger didn't use the Feynamn path integral.
Nobody has made the latter precise, except for free particles.
>>> Canonical quantization is just another way to introduce a wave
>>> function, and is well-defined.
>
>> More precisely:
>>
>> The path integral is well-defined only for free systems,
>
> No, it works very well for one point particle in a potential, therefore
> not free.
It works well, in a heuristic fashion.
But you demanded for a foundation that it has a consistent formulation.
The latter is known only for free systems.
>> This shows that the path integral is a logically dubious heuristics
>> only, while canonical quantization is the logically clear foundation.
>>> Wow, that makes a God damned number of great names, and I don't think
>>> they would appreciate being called non clear sensical.
>
>> But being clear, insightful, or useful are very different attributes.
>> Great physicists get their fame for being innovative and useful,
>> not for clarity. Thus your arguments support only that the path integral
>> was innovative when feynman introduced it and was found useful,
>> but says nothing about its clarity.
>
> It wasn't innovative, precisely. Fermat and Schr�dingen were
> innovative, mathematically rigorous and crystal clear.
But they didn't use the Feynman path integral att all.
They used only canonical quantization.
>>> In conlusion,
>>> the paths integral give the big picture, since it is no more than an
>>> "artist view" of all other quantization approaches.
>
>> Path integrals give _a_ big picture, but not _the_ big picture.
>> Pictures in visual art have often no clear meaning, and there may
>> be many big pictures of the same object, clear to very varying degrees.
>> The most famous pictures of an object are usually not the clearest ones.
>
> Ok, _a_ big picture. But it isn't because there are other pictures that
> it isn't _one_.
It is a big, heuristic, and useful picture.
But one that couldn't be made logically consistent, in spite of
many attempts by some of the best mathematicians and physicists.
Let us not confuse usefulness and logical consistency.
Arnold Neumaier