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Igor Khavkine · Guest

On Nov 1, 6:42 am, Mike James <mike.ja...@infomaxgroup.co.uk> wrote:
[quote]Arnold Neumaier wrote:

The real integral is only a shorthand for the complex integral.
It has no meaning in itself.

Now I>m completely lost.....
How can I regard the complex path integral as the fundamental entity
when I>m trying to obtain a Fourier transform to the x,t space?
Any references I can turn to?
[/quote]
Actually, you don>t want a Fourier transform, you want an integral
representation of a solution to a differential equation (just think of
the time dimention for now, as the spatial dimensions come in without
much hassle). Slightly different integral representations result in
specific different t-space boundary conditions. Here are the details.

You have a linear differential operator with constant coefficients, D.
Replace each t-differential in D with -iw and you get a polynomial in w,
G^(-1). The trick now is to show that any function of the form

f = Int dw exp(-iwt) G

is a solution of Df = 0. Showing that is pretty easy, given that Int dw
is interchangeable with differentiation and gives Int dw (const) = 0.
Both these properties are satisfied if Int dw is taken to be a closed
complex contour inegral, where the contour avoids any singularities. A
related question is whether every solution of Df = 0 can be represented
in this form. At least when the poles of G are simple, the answer is Yes
(as long as you are also allow linear combinations of contour
integrals). A basis for the solutions is given by contours that circle
each pole of G. I>m sure that with a little bit of work this statement
can be generalized when the poles of G are no longer only simple.

So far, there are no mathematical difficulties. All integrals are finite
and well defined. However, this formalism can be easily generalized to
include contours that do not avoid all singularities of the integrand.
However, in these cases, as such integrals do not make sense on their
own, a limiting proceduce must be given as part of the definition. The
limiting proceduce can be shown to correspond to a certain kind of
singularity or boundary condition on f.

Let me illustrate with a particular example. Take the integral over the
whole real line when G has no poles on it. Then the integral

f = Int dw exp(-iwt) G

does not avoid the essential singularity of exp(-iwt) at w -> +-oo. G is
analytic along the integration contour. So, assuming that the integral
converges, it can be computed by the method of residues. If t > 0, then
the integration contour is closed in the lower half of the complex
plane, and in the upper half if t < 0. This prescription creates a
singularity in f at t = 0. This singularity is such that f satisfies the
equation Df = d, where d is some distribution that has support only at t
= 0. This distribution can be evaluated by direct calculation (if anyone
is curious about the details, please ask).

Another example modifies the previous one by stipulating that G does
have a pole on the real axis. For the integral to be well defined, the
pole must be avoided in some way. The most general prescription consists
of a linear combination of contours deformed either above or below the
pole. In particular, you can check that going above the pole excludes
its residue from the integral when t < 0, while going below the pole
excludes the residue when t > 0. In short, these prescriptions impose
the boundary condition on f that a particular independent component of
the solution to Df = 0 vanishes in the limit t -> -oo or t -> +oo,
respectively. Particularly, if G has no poles in the upper half plane,
then going above the pole is equivalent to the advanced boundary
condition (f = 0 in the t -> -oo limit).

As has been pointed out several times in this thread, the only way to
know which is the right integral representation to use is to know the
boundary conditions on the t-space solution. For evaluating Feynman
diagrams, it is instructive to study the space-time amplitude rules
first. The space-time rules make it clear which Green function is being
used. The momentum-space rules can be derived from there, along with the
needed prescription for avoiding propagator poles.

Hope this helps.

Igor

Oh No · Guest

Thus spake Mike James <mike.james@infomaxgroup.co.uk>
[quote]Now I>m completely lost.....
How can I regard the complex path integral as the fundamental entity
when I>m trying to obtain a Fourier transform to the x,t space?
[/quote]
You can>t when there is a pole on the path. As per my earlier post, the
propagator is correctly found by commuting the partial fields as per
Wick>s theorem. The contour integral is introduced so as to have an
integration over energy as well as time. At this point the pole is
placed above or below the real number line, depending on whether the
contour is closed above or below (for whichever the integral on an
infinite semicircle tends to 0). The propagator is then valid in the
limit when epsilon goes to 0. Provided limits are taken in the right
order, this all works fine.

Regards

--
Charles Francis
moderator sci.physics.foundations.
charles (dot) e (dot) h (dot) francis (at) googlemail.com (remove spaces and
braces)

http://www.teleconnection.info/rqg/MainIndex

Mike James · Guest

Oh No wrote:

[quote]The problem here is probably the way in which the propagator has been
introduced in whatever treatment you have seen. When one evaluates the
propagator from Wick>s theorem it is necessary to insert a small value
i*epsilon to take the pole off the path. I have given a derivation at

http://www.teleconnection.info/rqg/FeynmanDiagrams
[/quote]
Not really - I>m familiar with a number of derivations of the
propagators and they fail to explain the way that the i*epsilon
procedure can be justified let alone to give some idea of what is going
on. Your web pages don>t add anything to this situation and simply go
through the same motions.

But thanks never the less as reading the pages was no problem :-)
mikej

Oh No · Guest

Thus spake Mike James <mike.james@infomaxgroup.co.uk>
[quote]Oh No wrote:

The problem here is probably the way in which the propagator has been
introduced in whatever treatment you have seen. When one evaluates the
propagator from Wick>s theorem it is necessary to insert a small value
i*epsilon to take the pole off the path. I have given a derivation at

http://www.teleconnection.info/rqg/FeynmanDiagrams

Not really - I>m familiar with a number of derivations of the
propagators and they fail to explain the way that the i*epsilon
procedure can be justified let alone to give some idea of what is going
on. Your web pages don>t add anything to this situation and simply go
through the same motions.

[/quote]
Have I misunderstood the question? Specifically i*epsilon is introduced
in the lemma at the top of

http://www.teleconnection.info/rqg/Propagators

We have an expression for the propagator which contains a factor given
on the RHS, which does not contain epsilon. This factor is replaced by
the integral on the LHS, which is shown by a straightforward contour
integral and in which the pole is not on the path. This is a
straightforward mathematical identity, which determines the position of
the pole.

Regards

--
Charles Francis
moderator sci.physics.foundations.
charles (dot) e (dot) h (dot) francis (at) googlemail.com (remove spaces and
braces)

http://www.teleconnection.info/rqg/MainIndex

Hendrik van Hees · Guest

Mike James wrote:

[quote]=========Moderator>s note==========

Of course, if you keep the epsilons finite, you obtain functions,
dependent on epsilon. Then, in the limit epsilon->0^{\pm} you
obtain various Green>s functions which differ in solutions to the
source-free wave equation.

Thank you anonymous moderator :-)
[/quote]
You are welcome.

[quote]but my question is why do you get these different solutions using this
procedure? Does it always work?
That is:
Can I take Df=S (D=diff op, f funtion, S source term) and solve by a
fourier transform F to give F(DG)=F(delta) (G= greens function) which
then gives F(G) as a rational function (perhaps a polynomial), and
finally get G (or should that be Gs in the plural) by inverse
transform of the rational function using the contour trick to avoid
any poles I care to?
[/quote]
That should work generally.

[quote]If so can any one provide a reference?
[/quote]
I can only repeat what I said before. Take textbooks on QFT. I also
posted links to my own manuscripts.

Also the question is which physics you like to understand since the
choice of the propagator depends on the physics you like to describe.
Usually, the first time one hears about propagators is classical
electromagnetism. The standard reference is

J.D. Jackson, Classical Electrodynamics

Let>s give the usual arguments for classical electromagnetics as an
example:

[quote]From the Maxwell equations you derive after fixing the gauge to the
Lorenz gauge,[/quote]

\partial_{mu} A^{mu}=0, (1)

you obtain the wave equation for each component of the vector potential:

\Box A^{mu}=j^{mu} (2)

Now you like to solve this with help of a Green>s function:

A^{mu}(x)=\int d^4 x' G(x,x') j^{mu}(x') (3)

This means your G should fulfill the equation

\Box_x G(x,x')=\delta^{(4)}(x-x') (4)

In the classical context you want the retarded propagator which is
defined by

G(x,x')=0 for t<t' (5)

since there should be no effect of the currents on the fields from the
future. With this condition the propagator is uniquely defined, as we
shall see in a moment.

Further it>s clear from space-time homogeneity that (for the here
considered case of free space)

G(x,x')=G(x-x') (6)

Then we write

G(x)=\int d^3 k/(2pi)^3 g(t,k) exp(i k X), (7)

where X denotes the spatial three vector components of the four vector
x.

Then the Fourier transform of Eq. (4) leads to equation for the function
g(t,k)

\partial_t^2 g(t,k) + k^2 g(t,k)= delta(t) (8)

which has to be solved under the retardation condition (5).

For t \neq 0, the equation is solved by

g(t,k)=A exp(-i |k| t)+B exp(+i |k| t) (9)

A and B are different for t<0 and t>0 and have to be choosen such that
(5) is fulfilled and one gets a solution of (8). It>s clar that

A=B=0 for t<0 (10)

is the unique solution for t<0 due to (5).

Further g(t,k) should be continuous at t=0. Which leads to

A+B=0 for t>0. (11)

Now we need a second condition. To find it, we integrate (8) over a
small intervall (-eps,eps) wrt. t which gives

\partial t g(eps,k)+k^2 g(eps,k)=1,

since for t<0 g vanishes. For eps->0^+, we find

\partial t g(0,k)=1.

This means

A-B=i/k for t>0

Together with (11)

we find

A=i/(2|k|), B=-i/(2|k|)

Plugging this in (9) and using g(t,k)=0 for t<0 we finally get

g(t,k)=Theta(t) sin[|k| t]/|k|

Plugging this into (7), you obtain the desired result

G_ret(t,X)=Theta(t)/(4 pi r) delta(t-r) with r=|X| (12)

This can be written in manifestly covariant form:

G_ret(x)=Theta(t)/(2 pi) delta(x^2), (13)

where "covariant" is meant with respect to proper orthochronous Lorentz
transformations, which do not change the direction of time.

You can also easily confirm that you get this Green>s function in
momentum space by putting the following i eps description to the
denominator

\tilde{G}_ret(k)=1/(k^2+i eps sign(k0)), (14)

where epsilon is meant to be positive and let to 0 after the integral

G(x)=int d^4 k \tilde{G}_ret(k) exp(-i k x)

is performed which leads back to (12) or (13).

Here the Theta(t) comes about from the fact that for the k0 integration
one has to close the contour in the upper k0 plane for t<0 and in the
lower one for t>0. Due to the i eps description in (14) both poles are
shifted to the lower k0 plane, and thus the k0 integral is 0 for t<0 as
it should be.

As you see on this most elementary example of the classical case, as
soon as you define precisely your boundary conditions in time you get
uniquely the propagator you want to get.

In the quantum-field theory case it depends on what you like to
calculate. In the vacuum theory from the Dyson-Wick perturbative series
for the S-matrix elements you know that you need the time-ordered
propagator, and then the same line of arguments applied above, using
the decomposition of the fields in annihilation and creation operators,
leads to a unique i eps description, namely

\tilde{G}_{time ord}(k)=1/(k^2+i eps).

For massive particles, it>s

\tilde{G}_{time ord}(p)=1/(p^2-m^2+i eps)

In many-body theory you need to double the time contour or in
equilibrium to use the Matsubara formalism with an imaginary time
running from 1 to -i beta (beta=1/T).

All this you can find in my QFT and Green>s function scripts on my
homepage.

--
Hendrik van Hees Institut für Theoretische Physik
Phone: +49 641 99-33342 Justus-Liebig-Universität Gießen
Fax: +49 641 99-33309 D-35392 Gießen
http://theory.gsi.de/~vanhees/faq/

cokonov · Guest

On Nov 1, 1:11 am, Hendrik van Hees <Hendrik.vanH...@theo.physik.uni-
giessen.de> wrote:
[quote]cokonov wrote:
Following Mike>s question,

So different choises of the integral contour leads to different
propagators.

Exactly.

It seem to me that all these propagator will give the same physical
solution.

That>s not right! It is very important to make sure to use the right
propagator for the purpose of your application.

However is there any physical difference for the choice of
propagators?
or it just try to get a time-ordered form?

In the vacuum quantum field theory where you calculate, e.g., cross
sections for reactions between a few particles, you need the
time-ordered form. For other purposes, e.g., if you want to evaluate
the reaction of a many-body system to an external disturbance (to
obtain transport properties of the corresponding quantities like
viscosity (momentum transport), heat conductivity (energy transport),
conductivity (el. charge transport), etc. you need the retarded
propagator.

All these propagators can also be seen as certain limits of an
analytic propagator, defined in the complex energy plane. For
details, the socalled real-time contour formalism of quantum field
theory is a very convenient concept. It>s invented to get expectation
values of observables or transition probabilities (and not only
transition amplitudes which are sufficient in vacuum QFT, but not in
the many-body realm). Have a look at my Green>s-function manuscript
on my homepage:

http://theorie.physik.uni-giessen.de/~hees/publ/green.pdf

--
Hendrik van Hees Institut für Theoretische Physik
Phone: +49 641 99-33342 Justus-Liebig-Universität Gießen
Fax: +49 641 99-33309 D-35392 Gießenhttp://theory.gsi.de/~vanhees/faq/
[/quote]
Thank you very much Hendrik. My understanding looks like this:
we use different propagator, due to the difference of boundary
conditions according to the physical problem. But here the boundary
condition has not be added yet. So I am really wandering why there are
four Green functions. Hope I can understand this after I read your
notes.
Thank you again.

Arnold Neumaier · Guest

Mike James schrieb:
[quote]Arnold Neumaier wrote:
Arnold Neumaier wrote:
There is no limit since the value of eps does not matter.
The form of the eps prescription simply tells one which of the
four propagators is intended.

I can see that you could think about it like this but then you end up
with a function of epsilion.
[/quote]
The meaning of the epsilons is the following.
Let epsilon>0. This moves the poles from the real axis; so the integral
is well-defined. Now move the integration path into the complex domain
by more than eps near the poles. You can do this in may ways, wothout
altering the meaning of the integral. Then move epsilon back to zero.
Now you have the original integral, independent of epsilon, but
integrated over some complex contour. In this sense, the epsilon
description tells you unambiguously how to place the complex contour
that gives the inegral the desired interpretation.

[quote]
And if there is no limit how can I regard the contour integrals as
having anything to do with the real integral they are supposed to help
us evaluate?
The real integral is only a shorthand for the complex integral.
It has no meaning in itself.

Now I>m completely lost.....
How can I regard the complex path integral as the fundamental entity
[/quote]
Try reading
http://en.wikipedia.org/wiki/Propagator#Scalar_propagator

Arnold Neumaier

Hendrik van Hees · Guest

Arnold Neumaier wrote:

[quote]Try reading
http://en.wikipedia.org/wiki/Propagator#Scalar_propagator
[/quote]
It>s a pretty good article, but one should be aware that all this is
valid in vacuo, not in equilibrium or off-equilibrium field theory,
where the Feynman propagator is usually different from the time
ordered propagator! See, e.g.,

Landsmann, N. P., and van Weert, Ch. G.: Real- and Imaginary-time
Field Theory at Finite Temperature and Density , Physics Reports 145,
141, 1987
http://dx.doi.org/10.1016/0370-1573(87)90121-9

pp. 196

However, one should be aware that there is a mistake in this article
concerning the factorization theorem which holds true only for the
original equilibrium-modified Schwinger-Keldysh contour whose real
part is closed. This is clearly worked out in

Gelis, Francois: The Effect of the vertical part of the path on the
real time Feynman rules in finite temperature field theory , Z. Phys.
C 70, 321, 1996

Gelis, Francois: A new approach for the vertical part of the contour
in thermal field theories, Phys. Lett. B455, 205?212, 1999

This formal mistake, however, doesn>t affect the definition of the
various real-time Green>s functions and their application to
(perturbative) real-time Feynman diagrams, as long as one applies
them to the above mentioned Schwinger-Keldysh contour.

--
Hendrik van Hees Institut für Theoretische Physik
Phone: +49 641 99-33342 Justus-Liebig-Universität Gießen
Fax: +49 641 99-33309 D-35392 Gießen
http://theory.gsi.de/~vanhees/faq/


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