Spatial representation of field commutator

  • #1

Main Question or Discussion Point

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
 

Answers and Replies

  • #2
410
0
Consider doing the integral in terms of E instead of p?
 
  • #3
Yes, sadly... same result :(
 
  • #4
410
0
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.
 
  • #5
410
0
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].
 
Last edited:
  • #6
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
 
  • #7
strangerep
Science Advisor
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Last edited:
  • #8
Hans de Vries
Science Advisor
Gold Member
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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 http://functions.wolfram.com/webMathematica/FunctionEvaluation.jsp?name=BesselK". 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
 
Last edited by a moderator:
  • #9
Hans de Vries
Science Advisor
Gold Member
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23
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 http://functions.wolfram.com/webMathematica/FunctionEvaluation.jsp?name=BesselK". 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]
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
 
Last edited by a moderator:
  • #10
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!
 
  • #11
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  • #12
strangerep
Science Advisor
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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.
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...)
 
  • #13
strangerep
Science Advisor
3,038
835
It's using the method of "stationary phase approximation".
[...]
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]).
 

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