Spatial representation of field commutator

blue2script

Hi all!

I worked for hours on this simple commutator of real scalar fields in qft:

[tex]\left[\Phi\left(x\right),\Phi\left(y\right) \right] = i\Delta\left( x-y \right)[/tex]

where

[tex]\Delta\left(x\right) = \frac{1}{i}\int {\frac{{d^4 p}}
{{\left( {2\pi } \right)^3 }}\delta \left( {p^2 - m^2 } \right)\operatorname{sgn} \left( {p^0 } \right)e^{ - ip \cdot x} }[/tex]

The task is to solve this integral and to look at the cases [tex]x^2 = 0[/tex] and [tex]m\rightarrow 0[/tex].

But, what I get in my calculations is that for space like x the commutator is zero whereas the integral diverges for time-like x. Normally there should be a Bessel function involved in the end and I am just totally confused now. But maybe I should show the steps I took:

i) Spacelike x

The expression is Lorentz-invariant. For space-like x we go into a reference frame where [tex]x_0[/tex] is zero, then
[tex]\Delta \left( x \right) = \frac{1}
{i}\int {\frac{{d^3 \vec p}}
{{\left( {2\pi } \right)^3 }}\frac{1}
{{2\sqrt {\vec p^2 + m^2 } }}\left( {e^{ - i\vec p \cdot \vec x} - e^{ - i\vec p \cdot \vec x} } \right)}
[/tex]
and this is zero. What argument fails here? Calculating just one term of the bracket leads to a modified Bessel function. I followed Weinberg, p. 202 here - its strange...

ii) Timelike x

Then we go into a reference frame where [tex]\vex x = 0[/tex] and we get
[tex]\begin{gathered}
\Delta \left( x \right) = \frac{1}
{i}\int {\frac{{d^3 \vec p}}
{{\left( {2\pi } \right)^3 }}\frac{1}
{{2\sqrt {\vec p^2 + m^2 } }}\left( {e^{ - iEx_0 } - e^{iEx_0 } } \right)} \hfill \\
= 4\pi \frac{1}
{i}\int {\frac{{dp}}
{{\left( {2\pi } \right)^3 }}\frac{{p^2 }}
{{2\sqrt {p^2 + m^2 } }}\left( {e^{ - iEx_0 } - e^{iEx_0 } } \right)} \hfill \\
\end{gathered}
[/tex]
This integral diverges quadratically.

So there has to be something wrong, otherwise this task wouldn't make any sense... I hope somebody can help me here, I am totally lost with this task. I worked hours on that problem and I don't know what to do any more.

A big thanks in advance to everybody!!

Blue2script

lbrits

Consider doing the integral in terms of E instead of p?

blue2script

Yes, sadly... same result :(

lbrits

Peskin says
[tex]D(x-y) = \frac{1}{4\pi^2} \int_m^\infty\!dE\, \sqrt{E^2 - m^2} e^{-i E t}[/tex]
from which I suspect you'll find your answer. It's in the lower bound of the E integral.

lbrits

My bad. This might not be exactly your expression, which is due to the pole prescription in the Feynman propagator. In any event, I think you'll be able to get something that goes like [tex]e^{-m t}[/tex].

blue2script

ah, ok, I see that on p. 26 is exactly what I wanted. Thanks for the hint! But then again: I can't see why there the integral 2.51 goes to [tex]e^{-imt][/tex] for t goes to infinity? I mean, sure, for big E the oscillation is too high and gives zero. But then on the lower end the prefactor [tex]\sqrt{E^2 - m^2[/tex] gives zero. On the other hand, if we don't take the m in the exponent too seriously, this is true.

A big Thanks to you, Ibrits! Do you have a better explanation for the limit t->\infinity?

Blue2script

strangerep

It's using the method of "stationary phase approximation".
See http://en.wikipedia.org/wiki/Station..._approximation

BTW, are you still trying to solve the spacelike case? If so, Scharf's
"Finite Quantum Electrodynamics", pp64-69, might help.

Hans de Vries

Notice that Weinberg's results are totally different as those from P&S and Zee.
(Weinberg is right, the others are wrong). The exact definition of the outside
the light-cone behavior of the Feynman propagator is given by:

[tex]\frac{1}{4\pi^2}~\frac{m}{\sqrt{r^2-t^2}}~K_1(m\sqrt{r^2-t^2})[/tex]

This corresponds to Weinberg's equation (5.2.9). For small arguments we can
simply replace the Bessel function [itex]K_1(r)[/itex] with [itex]1/r[/itex], for instance:

[tex]K_1(0.001)~=~999.996238[/tex]

Try it yourself here. Since m is very small we can in general simply replace
Weinberg's (5.2.9) with:

[tex]\frac{1}{4\pi^2}~\frac{1}{r^2-t^2}[/tex]

Which is independent of the mass m (!) and much more serious.

It's hard for me to understand, now it has become more or less routine to make
anti-hydrogen in the laboratory, how one can maintain the idea that anti-matter
propagates backward in time... Going backwards in time we see:

Somehow gamma flashes mysteriously produce anti-protons and positrons at the
inside surface of a vacuum container. These instantaneously pair up to form
anti-hydrogen which turns out to be very cold (amazingly after such high energy
events) After a while the anti-hydrogen spontaneously ionizes (it's instable).
The constituents are then accelerated to ultrarelativistic speed and end their
lives in high energy synchrotron collisions..

Oh well....



Regards, Hans

Hans de Vries

Oops, no more time to edit over-impulsive posts. Rushing back home didn't help...
Of course m should be mc/hbar which is not small. So the 1/r behavior is indeed
below the Compton radius. Later on I want to say a few things more.


Regards, Hans

blue2script

Hi Hans,

I just came back seeing your posts... thanks a lot for your answers!! I am impressed... and will check everyting you said ;). But, well, that won't happen before tuesday, I am on a bike tour the next three days.

But please go on to continue writing, I am very interested!! There are so many mysterious things in qft, especially for a, well somewhat, a beginner (I hear two courses on qft right now and already had courses in qed and qcd).

So, if I may please you, I am interested in everything related with this topic! A big thanks in advance for your couragement!!

Have a nice weekend, I will check back monday evening!

Blue2script

PS: @strangerep Thanks for the hint of the stationary phase method, helped me a lot!

kof9595995

Could you explain how in more details? I don't see how to apply the stationary phase method because in this case the phase factor's 1st derivative is never 0.

strangerep

Sorry, I don't have enough time right now, but I'll try to remember to get back to this sometimes next week. (Since you're asking a necro-question I'll have to refresh my memory...)

strangerep

Well, well. I wrote that 3.5 years ago, and now have the benefit of that extra study time, so I must now post a correction...

The [itex]e^{-imt}[/itex] (for [itex]t\to\infty[/itex] is not obtained by the stationary phase method. Indeed the stationary phase method doesn't seem to work here at all -- neither by a stationary exponent (critical point of the 1st kind) nor the boundary points of the integration (critical points of the 2nd kind).

A much better reference for calculating these propagators is

G. Scharf, "Finite Quantum Electrodynamics -- The Causal Approach".

In the section titled "Discussion of the Commutation Functions", he calculates lots of these propagators exactly.

The answer is that the exact propagator is a Bessel function (actually a Hankel function in our case). The asymptotic behaviour of for large arguments approaches something like this:
[tex]
H^{(2)}_0(z) ~\sim~ \sqrt{\frac{2}{\pi z}} \, e^{-iz} ~~~~~~[|z|\to\infty]
[/tex]
(Ref: Abramowitz & Stegun.)

(In our case, z is mt.)

Imho, P&S really should have given a reference instead of simply quoting the asymptotic result (and should also have mentioned the factor of [itex]1/\sqrt{t}[/itex]).