Contour integration in propagator

[ianhoolihan]

This is exactly what I don't like with Peskin Schroeder! First he writes some sloppy undefined integral and then pulls some contour magic out of the hat. To the contrary the important point is not to find the integrand of the Fourier transform, which is quite simple algebra, but to derive the correct contour from the beginning.

In vacuum qft, what you usually have to calculate in perturbation theory, is the time-ordered propagator, which is defined for a Klein-Gordan field, by the vacuum-expectation value
\mathrm{i} \Delta_c(x,y)=\langle \Omega | \mathcal{T}_c \hat{\phi}(x) \hat{\phi}^{\dagger}(y) | \Omega \rangle.
Then you write out the time-ordering symbol with Heaviside-unitstep functions, and those precisely tell you how you have to choose your path to circumvent the poles in the p^0-Fourier integral.

You find many details about these issues in my writeup on Green's functions for relativistic many-body theory in the real-time Schwinger-Keldysh formalism:

http://fias.uni-frankfurt.de/~hees/publ/green.pdf

I'm glad someone else finds PS a little fishy sometimes. To be honest, I'm working through PS as it the usual basis for courses (including those that I'll be taking this year), and not because I think it is well written. Hopefully I'll gain the main ideas, and then can progress to Weinberg...

I had a brief skim over the pdf you linked, but at this stage, it's a little too far beyond me!

Sorry, the numerator in the integrand should have been $e^{-ikz}$ (or else specified instead that $k<0$).

I picked that up a while after I'd posted the solutions too. Don't think it changes things.

That's only true if the total contribution from the small detouring semicircles around the real poles is zero. In the partial solutions so far, that has not been proven.

I think what's been said is that we are not trying to find the integral in the limit of smaller semicircles. That is the path, and they are meant to be of finite radius. (I'm still waiting to hear if they can be any anticlockwise contour surrounding the poles.) Hence, the residue theorem works fine.

 

[strangerep]

Regarding the point about defining the integral correctly... of course we intend
the Cauchy principal value interpretation of the integral. I.e., the ill-defined integral
$$
I_1 ~:=~ \int_{-\infty}^{+\infty} \!\! dx \; \frac{e^{-ikx}}{x^2 - a^2}
$$
is tacitly interpreted as
$$
\def\eps{\epsilon}
I_1 ~:=~ P.V. \int_{-\infty}^{+\infty} \!\! dx \; \frac{e^{-ikx}}{x^2 - a^2}
~\equiv~
\lim_{\eps\to 0} \left(\,
\int_{-\infty}^{-a-\eps}\!\!\!\!\dots
~+~ \int_{-a+\eps}^{a-\eps} \!\!\!\!\dots
~+~ \int_{a+\eps}^{+\infty}\!\!\!\!\dots \;\right) ~,
$$
to be performed by contour integration in the complex $x$-plane.
Cf. http://en.wikipedia.org/wiki/Cauchy_principal_value

Any use of such complex energies, etc, is only a technical device, since complex
energies (in this context, at least) are unphysical. To be physically useful, the
result for $I_1$ must not depend on which technical device we choose.
The Cauchy PV interpretation, combined with contour integration satisfies
this criterion, since it doesn't matter which way we choose to form a closed
contour (provided the total contour integral exists, of course, and that any extra
detours in the complex plane are accounted for when isolating the
contribution from $I_1$ which is what we want).

 

[strangerep]

I'm glad someone else finds PS a little fishy sometimes.

Do you have a copy of the errata list for your edition of PS? They can be found at:
http://www.slac.stanford.edu/~mpeskin/QFT.html

 

[ianhoolihan]

Regarding the point about defining the integral correctly... of course we intend
the Cauchy principal value interpretation of the integral. I.e., the ill-defined integral
$$
I_1 ~:=~ \int_{-\infty}^{+\infty} \!\! dx \; \frac{e^{-ikx}}{x^2 - a^2}
$$
is tacitly interpreted as
$$
\def\eps{\epsilon}
I_1 ~:=~ P.V. \int_{-\infty}^{+\infty} \!\! dx \; \frac{e^{-ikx}}{x^2 - a^2}
~\equiv~
\lim_{\eps\to 0} \left(\,
\int_{-\infty}^{-a-\eps}\!\!\!\!\dots
~+~ \int_{-a+\eps}^{a-\eps} \!\!\!\!\dots
~+~ \int_{a+\eps}^{+\infty}\!\!\!\!\dots \;\right) ~,
$$
to be performed by contour integration in the complex $x$-plane.
Cf. http://en.wikipedia.org/wiki/Cauchy_principal_value

Any use of such complex energies, etc, is only a technical device, since complex
energies (in this context, at least) are unphysical. To be physically useful, the
result for $I_1$ must not depend on which technical device we choose.
The Cauchy PV interpretation, combined with contour integration satisfies
this criterion, since it doesn't matter which way we choose to form a closed
contour (provided the total contour integral exists, of course, and that any extra
detours in the complex plane are accounted for when isolating the
contribution from $I_1$ which is what we want).

OK, I'm getting confused by the opposing views presented, but I think posts #23 and #24 clarify things. The principal value approach would be correct if attempting to evaluate along the real axis (principal value is and is not (?) the same as the real integral, from what I've been told). However, PS are not talking of this integral. They are talking of $\int_C$ where $C$ is the given (or possibly more general) contour around the poles.

strangerep, thanks for the list of errata --- most useful!

 

[strangerep]

In the second case both poles are within the contour, and then you have according to the residue theorem (I take a&gt;0 to make the notation simple)
\int_{\mathcal{C}_w} \mathrm{d} z \frac{\exp(\mathrm{i} z k)}{k^2-a^2}=2 \pi \mathrm{i} \left [\frac{\exp(\mathrm{i} k a)}{2a}-\frac{\exp(-\mathrm{i} k a)}{2 a} \right ]=-\frac{2 \pi}{a} \sin(k a).
So here you have a factor 2 wrong.

Your second expression is the same as
$$\frac{\pi i}{a} \Big( e^{ika} - e^{-ika} \Big)$$
which is the same as Ian's result (except for a minus sign at the front which I think is due to the different contour orientations). But I don't see a "factor of 2" error here. (?)

 

[strangerep]

OK, I'm getting confused by the opposing views presented, [...]

Indeed -- I'm hoping this thread will (eventually) clarify all these matters precisely, and in full detail.

The principal value approach would be correct if attempting to evaluate along the real axis (principal value is and is not (?) the same as the real integral, from what I've been told). However, PS are not talking of this integral. They are talking of $\int_C$ where $C$ is the given (or possibly more general) contour around the poles.

I don't yet concede that this is indeed what they intended. Probably I'll have to email Michael Peskin or Dan Schroeder eventually and ask them.

But in the meantime, perhaps you'd like to have a go at evaluating some
variations on that other integral I mentioned. You could start with this one:
$$
\int\!\! d^4p\;\delta(p^2 - m^2) e^{-ipx} ~,
$$
where the integral is over all of 4D momentum space.
Just do the $p_0$ integration so that you end up with a 3D momentum integral.
I think you'll find the result interesting when you compare it to what we've been
discussing in PS.

 

[ianhoolihan]

Indeed -- I'm hoping this thread will (eventually) clarify all these matters precisely, and in full detail.

I don't yet concede that this is indeed what they intended. Probably I'll have to email Michael Peskin or Dan Schroeder eventually and ask them.

But in the meantime, perhaps you'd like to have a go at evaluating some
variations on that other integral I mentioned. You could start with this one:
$$
\int\!\! d^4p\;\delta(p^2 - m^2) e^{-ipx} ~,
$$
where the integral is over all of 4D momentum space.
Just do the $p_0$ integration so that you end up with a 3D momentum integral.
I think you'll find the result interesting when you compare it to what we've been
discussing in PS.

$$\int dp^0 \delta(p^2-m^2) = \int dp^0 \frac{1}{2E_p}\left[\delta(p^0 + E_p) + \delta(p^0-E_p)\right] = \frac{1}{2E_p}\left[\theta(p^0 + E_p) + \theta(p^0 - E_p)\right]$$
though I'm not 100% about the last step. I think this is (2.40) of PS, though I'm not sure why it's relevant...?

 

[strangerep]

$$\int dp^0 \delta(p^2-m^2) = \int dp^0 \frac{1}{2E_p}\left[\delta(p^0 + E_p) + \delta(p^0-E_p)\right] = \frac{1}{2E_p}\left[\theta(p^0 + E_p) + \theta(p^0 - E_p)\right]$$
though I'm not 100% about the last step.

You forgot the $e^{ipx}$.

I think this is (2.40) of PS, though I'm not sure why it's relevant...?

It gets more interesting (imho) when you start adding factors of $\pm\Theta(p_0)$ to the original integrand,
and think about time-ordered versions...

But, on second thoughts, maybe it's better if you get further into PS before
exploring this sidetrack.

 

[ianhoolihan]

You forgot the $e^{ipx}$.

$$\int dp^0 \delta(p^2-m^2)e^{-ip^0x^0} = \int dp^0 \frac{1}{2E_p}\left[\delta(p^0 + E_p) + \delta(p^0-E_p)\right]e^{-ip^0x^0} = \frac{1}{2E_p}\left[e^{iE_p x^0} + e^{-iE_p x^0}\right]\ldots ??$$

It gets more interesting (imho) when you start adding factors of $\pm\Theta(p_0)$ to the original integrand,
and think about time-ordered versions...

But, on second thoughts, maybe it's better if you get further into PS before
exploring this sidetrack.

I agree --- I think I'll steer clear of this for now.

 

[qbert]

I don't yet concede that this is indeed what they intended. Probably I'll have to email Michael Peskin or Dan Schroeder eventually and ask them.

Look at the structure of the presentation of eq. 2.54

[\phi(x), \phi(y)] is a number so that
[\phi(x), \phi(y)] = \left&lt; 0\right| [\phi(x), \phi(y)] \left| 0 \right&gt;
which is (by eqn 2.53)
= \int \frac{d^3p}{(2 \pi)^3} \, \frac{1}{2 E_{\bf p}}<br /> \left( e^{-i p \cdot (x-y)} - e^{i p \cdot (x-y)} \right)

rewrite the second integral and take p→-p. and
use E_p=E_-p.
= \int \frac{d^3p}{(2 \pi)^3} \, <br /> \left( <br /> \left. \frac{1}{2 E_{\bf p}} e^{-i p \cdot (x-y)} \right|_{p^0 = E_{\bf p}} <br /> - \left. \frac{1}{2 E_{\bf p}} e^{-i p \cdot (x-y)} \right|_{p^0 = -E_{\bf p}} <br /> \right)

From here we factor the integrand
= \int \frac{d^3p}{(2 \pi)^3} \, e^{i {\bf p}\cdot ({\bf x} - {\bf y})} <br /> \left( <br /> \frac{1}{2 E_{\bf p}} e^{-i E_{\bf p} (x_0-y_0)}<br /> - \frac{1}{2 E_{\bf p}} e^{-i (-E_{\bf p})(x_0-y_0)} <br /> \right)

Now we can easily show
\int_C \frac{e^{-i k z}}{z^2 - a^2} dz<br /> = - 2 \pi i \left( \frac{e^{-ika}}{2a} - \frac{e^{ika}}{2a} \right)<br />
where C is any contour which circles both a, and -a clockwise.

noticing due to the exponential (and k>0) we can add a loop at infinity
for free. we pick the contour along the real axis and just skirt above
the poles and close at ∞, clockwise.

= \int \frac{d^3p}{(2 \pi)^3} \, e^{i {\bf p}\cdot ({\bf x} - {\bf y})} <br /> \left( <br /> \frac{-1}{2 \pi i} \int_C dp^0 \frac{e^{-i p^0 (x_0-y_0)}}{(p^0)^2 - E_{\bf p}^2} <br /> \right)

Now use p \cdot (x-y) = p^0(x_0-y_0) - {\bf p}\cdot ({\bf x} -{\bf y}),
p^2 = (p^0)^2 - {\bf p}^2, and E_{\bf p} = \sqrt{ {\bf p}^2 - m^2},

= \int \frac{d^3p}{(2 \pi)^3} \int_C \frac{dp^0}{2 \pi i} \, <br /> \frac{-1}{p^2 - m^2} e^{-i p\cdot (x-y)} <br />

Which is exactly what they have, including pole prescription (which was forced
on us).

 

[ianhoolihan]

rewrite the second integral and take p→-p. and
use E_p=E_-p.
= \int \frac{d^3p}{(2 \pi)^3} \, <br /> \left( <br /> \left. \frac{1}{2 E_{\bf p}} e^{-i p \cdot (x-y)} \right|_{p^0 = E_{\bf p}} <br /> - \left. \frac{1}{2 E_{\bf p}} e^{-i p \cdot (x-y)} \right|_{p^0 = -E_{\bf p}} <br /> \right)

I agree with what you've done. However, to clarify, if $\tilde{p}=-p$, does $d^3\tilde{p}=d^3p$? I've seen this fleetingly before, but couldn't find anything about it online. Why is this the case?

 

[kloptok]

The p\rightarrow -p substitution is something which is easy to understand when you sit down and do it but it looks strange the first time you see it and is usually not explained at all in textbooks. We have that d^3 p \rightarrow -d^3 p since dp \rightarrow -dp. But the minus sign cancels out since the limits of the integral also change sign, i.e. for p\rightarrow -p, \int^{\infty}_{-\infty} dp \rightarrow \int^{-\infty}_{+\infty} (-dp) =- \int^{+\infty}_{-\infty} (-dp)=\int^{\infty}_{-\infty} dp

 

[ianhoolihan]

The p\rightarrow -p substitution is something which is easy to understand when you sit down and do it but it looks strange the first time you see it and is usually not explained at all in textbooks. We have that d^3 p \rightarrow -d^3 p since dp \rightarrow -dp. But the minus sign cancels out since the limits of the integral also change sign, i.e. for p\rightarrow -p, \int^{\infty}_{-\infty} dp \rightarrow \int^{-\infty}_{+\infty} (-dp) =- \int^{+\infty}_{-\infty} (-dp)=\int^{\infty}_{-\infty} dp

Silly me!

 

[strangerep]

[...]we pick the contour along the real axis and just skirt above
the poles and close at ∞, clockwise.

Do you mean a contour where the part near the real line is shifted slightly upwards? If so, that's equivalent (or so I think) to shifting the poles down slightly into the lower halfplane, and keeping the contour on the real axis. This would indeed give the result you obtained. I also now think this is what PS intend, since they refer to it as $D_R$, the retarded propagator.

[...]Which is exactly what they have, including pole prescription (which was forced
on us).

But it's not exactly what they've done. They use a contour on the real axis, and take little detours around the poles. The contributions from those little semicircular detours must be taken into account, since (afaict) the result from this calculation is not the same as shifting the poles slightly into the lower halfplane.

However, since the difference between the two results is apparently only a factor of 2, perhaps this could all be resolved by a simple field redefinition.

(BTW, thanks for the effort of typing it out.)

 

[qbert]

Do you mean a contour where the part near the real line is shifted slightly upwards? If so, that's equivalent (or so I think) to shifting the poles down slightly into the lower halfplane, and keeping the contour on the real axis. This would indeed give the result you obtained. I also now think this is what PS intend, since they refer to it as $D_R$, the retarded propagator.

I meant the contour given with little half-circles avoiding the poles, and a large
return route in the negative half plane. This is exactly equivalent to pushing the poles down a little bit
and integrating along the real axis.

But it's not exactly what they've done. They use a contour on the real axis, and take little detours around the poles. The contributions from those little semicircular detours must be taken into account, since (afaict) the result from this calculation is not the same as shifting the poles slightly into the lower halfplane.

However, since the difference between the two results is apparently only a factor of 2, perhaps this could all be resolved by a simple field redefinition.

(BTW, thanks for the effort of typing it out.)

The easiest way to evaluate the integral is the residue theorem.

If you want to calculate the contribution of the little circles then what you'll get is
I = Pv.∫ + ∫_{C_1}+∫_{C_2}, which will just add up to
the sum of the residues inside (with a minus sign for going the wrong way.)
with a little work you can show Pv.∫ is exactly 1/2 the sum of residues
and the integral over the circles adds the missing 1/2.

 

[qbert]

also this business of PV + half circle = push pole is exactly
the content of the Sokhotski-Plemelj theorem
(which for some reason i call a Dirac Identity)

http://en.wikipedia.org/wiki/Sokhotski–Plemelj_theorem

 

[ianhoolihan]

But it's not exactly what they've done. They use a contour on the real axis, and take little detours around the poles. The contributions from those little semicircular detours must be taken into account, since (afaict) the result from this calculation is not the same as shifting the poles slightly into the lower halfplane.

If I asked you to calculate the integral around the contour shown (the semicircles, the real line bits, and the arc in the lower half plane), you'd use the residue theorem alone. If I asked you to find the principal value along the real line, you'd do what we'd been doing and subtract of semicircle contributions. PS are asking the first case.

 

[strangerep]

If I asked you to calculate the integral around the contour shown (the semicircles, the real line bits, and the arc in the lower half plane), you'd use the residue theorem alone. If I asked you to find the principal value along the real line, you'd do what we'd been doing and subtract of semicircle contributions. PS are asking the first case.

OK, I agree that one must read PS that way to avoid it being "wrong". I would have preferred less magical formulation in terms of P.V. integrals, but probably that's just a personal taste thing.

 

[ianhoolihan]

OK, I agree that one must read PS that way to avoid it being "wrong". I would have preferred less magical formulation in terms of P.V. integrals, but probably that's just a personal taste thing.

Until I'm told otherwise, I think it is just a sneaky way to simplify the expression of the formula. My main argument for this is that (as far as I know) it doesn't depend on the contour, as long as it is anticlockwise and includes both poles. So it's deceptive to draw it along the real axis, with small semicircles around the poles. Not magical though. If a different contour was drawn not on the real axis, a P.V. integral wouldn't even make sense.

If I am wrong, please inform me, as I suspect I am --- I'm not sure why such specific contours are the only ones I ever see in textbooks.

 

[strangerep]

[...] Not magical though. [...]

I meant "magical" in the sense that particular contours, including the little detours into the unphysical complex plane, are used. I think one could insist on P.V. integrals along the real axis and adjust the formulation accordingly (e.g., adjust factors of 2 in the present case).

Perhaps this is purism vs shortcut-ism and I should just leave it alone.
In any case, I need to think it through further for the more difficult cases.

 

[vanhees71]

Sure, you can use the Dirac lemma, when you formulate the propagator as an p^0 integral along the real axis and define the integrand by shifting the poles infinitesimally up and/or down away from the real axis. E.g., for the free time-ordered propagator, needed in vacuum-qft perturbation theory
\Delta(p)=\frac{1}{p^2-m^2+\mathrm{i} 0^+}.
This you can reformulate as
\Delta(p)=\mathrm{PV} \frac{1}{p^2-m^2} - \mathrm{i} \pi \delta(p^2-m^2).
The proof is simpler for the statement
\frac{1}{p^0-z+\mathrm{i} 0^+}=\mathrm{PV}\frac{1}{p_0-z}-\mathrm{i} \pi \delta(p^0-z)
and using the residue theorm on an arbitrary holomorphic test function.