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Mike James · Guest

I>m in the process of making sure that the maths I>m using is 100% clear
to me and this means I keep on finding that I don>t understand things
that I thought I once did.

Give the position space propagator expressed as a Fourier transform of
1/(p^2-m^2)the integral is clearly a problem because of the pole at p^2=m^2.

My questions is:
If the integral is to have any meaning surely it should have a single
form and be the unique result of a limiting process - i.e. the solution
shouldn>t depend on how the limit is taken.

So why do we get different results depending on how the poles are avoided?

Also is there a sense in which the contour integrals relate to any real
limit e.g. an integral with the poles removes by an interval which we
then take to the limit.
mikej

Hendrik van Hees · Guest

Mike James wrote:

[quote]Give the position space propagator expressed as a Fourier transform
of 1/(p^2-m^2)the integral is clearly a problem because of the pole
at p^2=m^2.

My questions is:
If the integral is to have any meaning surely it should have a
single form and be the unique result of a limiting process - i.e.
the solution shouldn>t depend on how the limit is taken.
[/quote]
Indeed, that>s what you *must* do to give the above expression a
definite sense in physics.
[quote]
So why do we get different results depending on how the poles are
avoided?
[/quote]
The reason is that it depends on the physical context which propagator
you have to use. If you do vacuum quantum-field theory what you
usually want is the Feynman propagator which in this case coincides
with the time-ordered propagator which for charged spin-0 fields in
position-time representation is defined as

G(x,y)=<0|T \phi(x) \phi^{\dagger}(y)|0>,

where T is the time-ordering operator, ordering the field operators
from right to left in increasing time-order.

To achieve this, you can put an i \epsilon as follows

G(p)=1/(p^2-m^2+i \epsilon)

with the usual west-coast metric for the Minkowski product.
[quote]
Also is there a sense in which the contour integrals relate to any
real limit e.g. an integral with the poles removes by an interval
which we then take to the limit.
[/quote]
In a sense yes. You can use the following identity

1/(x+i epsilon)=P 1/x -i \pi \delta(x),

where P is the principial value, to be used when integrating a test
function times this expression over x \in R.

For more information on different Green>s functions in realtivistic
QFT and in various contexts of many-body theory, see my notes

http://theorie.physik.uni-giessen.de/~hees/publ/lect.pdf
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 Gießen
http://theory.gsi.de/~vanhees/faq/

Tom Roberts · Guest

Mike James wrote:
[quote]Give the position space propagator expressed as a Fourier transform of
1/(p^2-m^2)the integral is clearly a problem because of the pole at p^2=m^2.
My questions is:
If the integral is to have any meaning surely it should have a single
form and be the unique result of a limiting process - i.e. the solution
shouldn>t depend on how the limit is taken.
So why do we get different results depending on how the poles are avoided?
[/quote]
This is a common property of integrals in physics which arise as real
integrals from -infinity to +infinity, with poles on the real axis, or
as complex Fourier transforms. One applies analytic continuation to the
(real) integrand to perform the integral on the complex plane instead of
the real line. This is useful because usually the integrand is zero at
either +i*infinity or -i*infinity (sometimes both) -- that permits one
to close the contour "at complex infinity" in either the +i or -i sense.
One can only do that when the integrand is zero along the added path at
"complex infinity", so the original integral along the real line has the
same value as the integral on the complex plane around the closed path.
Once one has justified the closed path, one can use the residue theorem
to easily compute the value of the integral by looking at the poles of
the integrand inside the (now closed) path.

Now it is clear that since the integrand is analytic, one can distort
the portion of the path along the real axis to avoid the pole(s) there.
Whether one distorts in the +i or the -i direction affects the value of
the integral, as that determines whether or not the pole is included
inside or outside the (closed) path of integration. This is the crux of
your question -- how can such an arbitrary choice affect the value of a
physically-meaningful integral?

The answer is usually contained in the sign of the result, and/or in its
complex phase. These depend on the direction of the closure at complex
infinity -- closed at +i*infinity the boundary is traversed
counter-clockwise, while closed at -i*infinity the boundary is traversed
clockwise, and the latter direction changes the sign of the result. The
integrand can also have a dependence on that direction that affects the
complex phase of the result. One must then apply physical reasoning to
the result, and be able to show that one or the other signs or phases is
unphysical. For instance, such integrals often arise in computing
Green>s functions, and one closure is advanced and the other is retarded
-- advanced Green>s functions are unphysical (they violate causality).

Advanced and retarded Green>s functions both give valid
solutions to the mathematical equations, but physics is
not math, and physically we can reject advanced solutions.
But without such physical reasoning, one must accept both
solutions.

Another possibility is that the original integral is not really defined
along the real axis, but rather along (x+i*\epslion) for x real and very
small (real) \epsilon. This then defines explicitly how the path avoids
poles on the real axis (usually it must be closed in the -i*infinity
direction due to the sign in exp(-2pi*i*z) of the integrand, thus
enclosing poles on the real axis). This is the case for a careful
treatment of Fourier transforms.

Yes, mathematical rigor is completely lacking above. My
discussion is at the level common in physics. It can be
cleaned up and made rigorous, at the expense of clarity.
I didn>t try because it is already complex enough <groan>.

[quote]Also is there a sense in which the contour integrals relate to any real
limit e.g. an integral with the poles removes by an interval which we
then take to the limit.
[/quote]
I>m not sure what you are asking. Note in the above procedure it is
necessary that the original real integral has limits -infinity to
+infinity, and that along the added boundary at complex infinity the
integrand be zero. When making the above discussion rigorous, one starts
with finite limits of the original integral, closes the boundary at
finite radius, and then takes a limit as the path goes to complex
infinity everywhere along the added path. Perhaps that answers your
question....

Tom Roberts

cokonov · Guest

Following Mike>s question,

So different choises of the integral contour leads to different
propagators.
It seem to me that all these propagator will give the same physical
solution.
However is there any physical difference for the choice of
propagators?
or it just try to get a time-ordered form?

Carlos

[quote]Mike James wrote:
Give the position space propagator expressed as a Fourier transform
of 1/(p^2-m^2)the integral is clearly a problem because of the pole
at p^2=m^2.

My questions is:
If the integral is to have any meaning surely it should have a
single form and be the unique result of a limiting process - i.e.
the solution shouldn>t depend on how the limit is taken.

Indeed, that>s what you *must* do to give the above expression a
definite sense in physics.

So why do we get different results depending on how the poles are
avoided?

The reason is that it depends on the physical context which propagator
you have to use. If you do vacuum quantum-field theory what you
usually want is the Feynman propagator which in this case coincides
with the time-ordered propagator which for charged spin-0 fields in
position-time representation is defined as

G(x,y)=<0|T \phi(x) \phi^{\dagger}(y)|0>,

where T is the time-ordering operator, ordering the field operators
from right to left in increasing time-order.

To achieve this, you can put an i \epsilon as follows

G(p)=1/(p^2-m^2+i \epsilon)

with the usual west-coast metric for the Minkowski product.

Also is there a sense in which the contour integrals relate to any
real limit e.g. an integral with the poles removes by an interval
which we then take to the limit.

In a sense yes. You can use the following identity

1/(x+i epsilon)=P 1/x -i \pi \delta(x),

where P is the principial value, to be used when integrating a test
function times this expression over x \in R.

For more information on different Green>s functions in realtivistic
QFT and in various contexts of many-body theory, see my notes

http://theorie.physik.uni-giessen.de/~hees/publ/lect.pdfhttp://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]

mikej · Guest

On 31 Oct, 05:15, cokonov <bogoliu...@gmail.com> wrote:
[quote]Following Mike>s question,

So different choises of the integral contour leads to different
propagators.
[/quote]
That seems to be exactly the case.

[quote]It seem to me that all these propagator will give the same physical
solution.
[/quote]
As long as you allow the "usual" formulation of adding a solution to
the homogenous equation to give a solution that satisifes any boundary
conditions.
If you don>t do this then the raw Green>s functions give you solutions
with different boundary conditions.

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

[/quote]
If you just want to use them as Green>s functions that satisfy
particular boundary conditions they DO give you different solutions.
mikej

mikej · Guest

On 30 Oct, 17:48, Tom Roberts <tjroberts...@sbcglobal.net> wrote:
[quote]This is a common property of integrals in physics which arise as real
integrals from -infinity to +infinity, with poles on the real axis, or
as complex Fourier transforms. One applies analytic continuation to the
(real) integrand to perform the integral on the complex plane instead of
the real line. This is useful because usually the integrand is zero at
either +i*infinity or -i*infinity (sometimes both) -- that permits one
to close the contour "at complex infinity" in either the +i or -i sense.
One can only do that when the integrand is zero along the added path at
"complex infinity", so the original integral along the real line has the
same value as the integral on the complex plane around the closed path.
Once one has justified the closed path, one can use the residue theorem
to easily compute the value of the integral by looking at the poles of
the integrand inside the (now closed) path.
[/quote]
I>m happy with the idea of contour integration but only where the
poles are not on any of the paths. If the poles are not on any of the
paths then all of the integrals are well defined and using residues to
compute the integral of any section of the parth is simple.
However if the pole is on a path - lets stick with the real axis then
that part of the path integral is improper and might not be defined.
I can see that moving the path to "miss" the pole might well give a
well defined integral but clearly the residues you have to include
depend on how you miss the pole i.e. pole inside or outside contour. I
can also see that the real integral defined by the contour can be
regarded as the limit of the "avoiding the pole" i.e. let ie got to
zero. However as there is more than one contour and more than one
result there is more than one limit - and this surely means that the
limit is ill defined.

[quote]Now it is clear that since the integrand is analytic, one can distort
the portion of the path along the real axis to avoid the pole(s) there.
Whether one distorts in the +i or the -i direction affects the value of
the integral, as that determines whether or not the pole is included
inside or outside the (closed) path of integration. This is the crux of
your question -- how can such an arbitrary choice affect the value of a
physically-meaningful integral?
[/quote]
Yes this is indeed my question - with perhaps a sideways glance toward
what the maths is all about.

[quote]The answer is usually contained in the sign of the result, and/or in its
complex phase. These depend on the direction of the closure at complex
infinity -- closed at +i*infinity the boundary is traversed
counter-clockwise, while closed at -i*infinity the boundary is traversed
clockwise, and the latter direction changes the sign of the result. The
integrand can also have a dependence on that direction that affects the
complex phase of the result. One must then apply physical reasoning to
the result, and be able to show that one or the other signs or phases is
unphysical. For instance, such integrals often arise in computing
Green>s functions, and one closure is advanced and the other is retarded
-- advanced Green>s functions are unphysical (they violate causality).
[/quote]
It is this connection that I really don>t get.
We want position Green>s functions but have derived momentum Green>s
functions and have the usual connection between momentum and position
space.
What I can>t follow is why the prescriptions for missing the poles
give you anything that behaves like a position space Green>s function
and why the different contours give different physical
interpretations.
Bascially the Green>s momentum space function isn>t a bounded operator
and so strictly isn>t in the space that the fourier transform applies
to.

[quote]Â Â Â Â Advanced and retarded Green>s functions both give valid
Â Â Â Â solutions to the mathematical equations, but physics is[/quote]

Oh No · Guest

Thus spake Mike James <mike.james@infomaxgroup.co.uk>
[quote]I>m in the process of making sure that the maths I>m using is 100% clear
to me and this means I keep on finding that I don>t understand things
that I thought I once did.

Give the position space propagator expressed as a Fourier transform of
1/(p^2-m^2)the integral is clearly a problem because of the pole at p^2=m^2.

My questions is:
If the integral is to have any meaning surely it should have a single
form and be the unique result of a limiting process - i.e. the solution
shouldn>t depend on how the limit is taken.

So why do we get different results depending on how the poles are avoided?

Also is there a sense in which the contour integrals relate to any real
limit e.g. an integral with the poles removes by an interval which we
then take to the limit.
[/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

Make sure you don>t miss the subpage where the propagators are actually
evaluated

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

Note the notation for creation and annihilation operators defined at

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

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

Arnold Neumaier · Guest

mikej schrieb:

[quote]I>m happy with the idea of contour integration but only where the
poles are not on any of the paths. If the poles are not on any of the
paths then all of the integrals are well defined and using residues to
compute the integral of any section of the parth is simple.
However if the pole is on a path - lets stick with the real axis then
that part of the path integral is improper and might not be defined.
I can see that moving the path to "miss" the pole might well give a
well defined integral but clearly the residues you have to include
depend on how you miss the pole i.e. pole inside or outside contour. I
can also see that the real integral defined by the contour can be
regarded as the limit of the "avoiding the pole" i.e. let ie got to
zero. However as there is more than one contour and more than one
result there is more than one limit - and this surely means that the
limit is ill defined.
[/quote]
Tere is no limit. when there are two poles, there are 4 different
values of the integral, defiend by the 4 topologically different ways
of avoiding the poles.

Corresponding to this, there are 4 different propagators, which have
different physical meanings: the advanced, retarded, Feynman propagator,
and another one. Theory tells when which propagator is needed.

Arnold Neumaier

======= Moderator>s note ============
quotations trimmed

Hendrik van Hees · Guest

mikej wrote:

[quote]I>m happy with the idea of contour integration but only where the
poles are not on any of the paths. If the poles are not on any of
the paths then all of the integrals are well defined and using
residues to compute the integral of any section of the parth is
simple. However if the pole is on a path - lets stick with the real
axis then that part of the path integral is improper and might not
be defined. I can see that moving the path to "miss" the pole might
well give a well defined integral but clearly the residues you have
to include depend on how you miss the pole i.e. pole inside or
outside contour. I can also see that the real integral defined by
the contour can be regarded as the limit of the "avoiding the pole"
i.e. let ie got to zero. However as there is more than one contour
and more than one result there is more than one limit - and this
surely means that the limit is ill defined.
[/quote]
Right. The physicist>s way of doing calculations is often a bit
sloppy, and particularly in this case of propagators one has to be
very careful to define it in the right way.

An alternative is not to do the full Fourier transformation from
time-position to energy-momentum space but just go from position to
momentum space. Then you get the socalled "Mills representation",
which however has the disadvantage of being not manifestly Poincare
covariant.

Then you can impose the boundary conditions for the various
propagators just wrt. the time variable as they are formulated
physically in the beginning of the calculation.

Then you obtain usually some analytic expressions multiplied with
Theta functions in time. These Theta functions are then the origin
for the various i epsilon descriptions which give the energy-Fourier
integral a proper meaning by telling you how to deform the
integration contour from the real axis to avoid the poles in the
right way to find the propagator for the problem in question.
[quote]
Now it is clear that since the integrand is analytic, one can
distort the portion of the path along the real axis to avoid the
pole(s) there. Whether one distorts in the +i or the -i direction
affects the value of the integral, as that determines whether or
not the pole is included inside or outside the (closed) path of
integration. This is the crux of your question -- how can such an
arbitrary choice affect the value of a physically-meaningful
integral?

Yes this is indeed my question - with perhaps a sideways glance
toward what the maths is all about.
[/quote]
As I said before, you have to decide for each physics question, which
propagator to choose. E.g., in classical electrodynamics usually you
need the retarded propagator. Then the boundary condition is

G(t,x)=0 for t<0,

i.e.

G(t,x)=Theta(t) g(t,x),

You can determine g(t,x) most easily with the Fourier method applied
only to x and keep time as fourth independent variable.

[quote]It is this connection that I really don>t get.
We want position Green>s functions but have derived momentum Green>s
functions and have the usual connection between momentum and
position space.
[/quote]
Mathematically it>s simple to understand: The propagator for the wave
equation is

\Box G(t,x):=(\partial_t^2-\Delta_x) G(t,x)=-\delta(t) \delta^{(3)}(x)

If you have two solutions, G1 and G2, of this equation their
difference fulfills the homogeneous (free) wave equation

\Box (G1-G2)=0,

and the free-wave equation has a lot of solutions. So the Green>s
function must be determined by imposing appropriate initial (and
perhaps boundary) conditions, depending on the physical problem at
hand.

[quote]What I can>t follow is why the prescriptions for missing the poles
give you anything that behaves like a position space Green>s
function and why the different contours give different physical
interpretations.
[/quote]
You can easily check that the different contours give solutions of the
Green>s function equation which differ by exactly a solution of the
homogeneous equation.

If you are interested in a somewhat old-fashioned but very concise
discussion of these issues, see Vol. 6 of the Pauli Lectures (Dover).

[quote]Bascially the Green>s momentum space function isn>t a bounded
operator and so strictly isn>t in the space that the fourier
transform applies to.
[/quote]
The Green>s function (as all N-point vertex functions of QFT) are not
functions but distributions.

--
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/

Hendrik van Hees · Guest

cokonov wrote:

[quote]Following Mike>s question,

So different choises of the integral contour leads to different
propagators.
[/quote]
Exactly.

[quote]It seem to me that all these propagator will give the same physical
solution.
[/quote]
That>s not right! It is very important to make sure to use the right
propagator for the purpose of your application.

[quote]However is there any physical difference for the choice of
propagators?
or it just try to get a time-ordered form?
[/quote]
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 Gießen
http://theory.gsi.de/~vanhees/faq/

Mike James · Guest

Arnold Neumaier wrote:
[quote]
Tere is no limit. when there are two poles, there are 4 different
values of the integral, defiend by the 4 topologically different ways
of avoiding the poles.
[/quote]
Yes I follow the 4 topologically different ways of avoiding the poles
and the way that this gives 4 different propagators.
But can you clarify "there is no limit"?
It looks like at least a suggestion of a limit in the use of epsilon and
appealing to the idea that the deformed contour is very like the
original path along the real axis...

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?

I>m not sure I follow this part of your statement at all.
mikej

mikej · Guest

On 29 Oct, 18:21, Hendrik van Hees <Hendrik.vanH...@theo.physik.uni-
giessen.de> wrote:
Snip
[quote]
Also is there a sense in which the contour integrals relate to any
real limit e.g. an integral with the poles removes by an interval
which we then take to the limit.

In a sense yes. You can use the following identity

1/(x+i epsilon)=P 1/x -i \pi \delta(x),

where P is the principial value, to be used when integrating a test
function times this expression over x \in R.
--
Hendrik van Hees Â Â Â Â Â Â Â Â Â Â Â Â Institut fÃ¼r Theoretische Physik
[/quote]
I>ve looked up principle value and related topics but can>t find
anything like the expression quoted above - but I can see it is
relevant to what I want to know.
Can you point me in the right direction to find out more?
mikej

Mike James · Guest

Hendrik van Hees wrote:
[quote]
Right. The physicist>s way of doing calculations is often a bit
sloppy, and particularly in this case of propagators one has to be
very careful to define it in the right way.
[/quote]
I>m happy with sloppy but only when I can see how it works and
preferably how it might be "cleaned up".
In this case I really can>t see the reasoning.

[quote]An alternative is not to do the full Fourier transformation from
time-position to energy-momentum space but just go from position to
momentum space. Then you get the socalled "Mills representation",
which however has the disadvantage of being not manifestly Poincare
covariant.
[/quote]
Sorry I didn>t mean to bring in this sort of split by using the terms
position and momentum space. I was being sloppy :-)
I should have said time-position and energy-momentum space.

[quote]As I said before, you have to decide for each physics question, which
propagator to choose. E.g., ...
snip[/quote]

Happy with all of this but to ask again..

[quote]
What I can>t follow is why the prescriptions for missing the poles
give you anything that behaves like a position space Green>s
function and why the different contours give different physical
interpretations.

You can easily check that the different contours give solutions of the
Green>s function equation which differ by exactly a solution of the
homogeneous equation.
[/quote]
Check yes - but predict?
That is I can see that we might just be lucky and the different contours
give Green>s functions with different properties (i.e. boundary
conditions etc.) but its a bit remarkable that we get these particular
physically meaningful (?) propagators....
There must be something going on deeper?

[quote]If you are interested in a somewhat old-fashioned but very concise
discussion of these issues, see Vol. 6 of the Pauli Lectures (Dover).
[/quote]
I will look it up as soon as I can (think I have a copy).

[quote]Bascially the Green>s momentum spacfunction isn>t a bounded
operator and so strictly isn>t in the space that the fourier
transform applies to.

The Green>s function (as all N-point vertex functions of QFT) are not
functions but distributions.
[/quote]

Yes I think I have deduced this even though it is rarely pointed out in
the literature - but this only makes me even more uneasy about the
procedure! :-)
mikej

Arnold Neumaier · Guest

Mike James schrieb:
[quote]Arnold Neumaier wrote:
Tere is no limit. when there are two poles, there are 4 different
values of the integral, defiend by the 4 topologically different ways
of avoiding the poles.

Yes I follow the 4 topologically different ways of avoiding the poles
and the way that this gives 4 different propagators.
But can you clarify "there is no limit"?
It looks like at least a suggestion of a limit in the use of epsilon and
appealing to the idea that the deformed contour is very like the
original path along the real axis...
[/quote]
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.

[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?
[/quote]
The real integral is only a shorthand for the complex integral.
It has no meaning in itself.

Arnold Neumaier

=========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.

It>s much simpler to think of the Green>s functions as analytically
continued meromorphic functions in the complex p0 plane. Then in vacuo

G_c(z,p)=1/(z-p^2-m^2),

and one can define the "standard Green>s functions" by certain limits,
e.g.,

G_ret(p0,p)=G_c(p0+i 0^+,p)
G_adv(p0,p)=G_c(p0-i 0^+,p)
G_{Feyn}(p0,p)=G_c(p0+i 0^ sign(p0),p)

Mike James · Guest

Arnold Neumaier wrote:
[quote]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.
[/quote]
I can see that you could think about it like this but then you end up
with a function of epsilion.

[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.
[/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?
Any references I can turn to?

[quote]Arnold Neumaier

=========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.
[/quote]
Thank you anonymous moderator :-)
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?
If so can any one provide a reference?

[quote]It>s much simpler to think of the Green>s functions as analytically
continued meromorphic functions in the complex p0 plane. Then in vacuo

G_c(z,p)=1/(z-p^2-m^2),
and one can define the "standard Green>s functions" by certain limits,
e.g.,

G_ret(p0,p)=G_c(p0+i 0^+,p)
G_adv(p0,p)=G_c(p0-i 0^+,p)
G_{Feyn}(p0,p)=G_c(p0+i 0^ sign(p0),p)

[/quote]
I think I can see what you are getting at but it is still a long way
from being clear to me.
Again can you give me a reference to read?
mikej


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