Contour integration in propagator

[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
$$[\phi(x), \phi(y)] = \left< 0\right| [\phi(x), \phi(y)] \left| 0 \right>$$
which is (by eqn 2.53)
$$= \int \frac{d^3p}{(2 \pi)^3} \, \frac{1}{2 E_{\bf p}} \left( e^{-i p \cdot (x-y)} - e^{i p \cdot (x-y)} \right)$$

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

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

Now we can easily show
$$\int_C \frac{e^{-i k z}}{z^2 - a^2} dz = - 2 \pi i \left( \frac{e^{-ika}}{2a} - \frac{e^{ika}}{2a} \right)$$
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.

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

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 $E_{\bf p} = \sqrt{ {\bf p}^2 - m^2}$,

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

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.
$$= \int \frac{d^3p}{(2 \pi)^3} \, \left( \left. \frac{1}{2 E_{\bf p}} e^{-i p \cdot (x-y)} \right|_{p^0 = E_{\bf p}} - \left. \frac{1}{2 E_{\bf p}} e^{-i p \cdot (x-y)} \right|_{p^0 = -E_{\bf p}} \right)$$

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.