# Klein-Gordon propagator derivation

• I

## Main Question or Discussion Point

Basically they are looking for $G$, that solves the equation
$$(\square _{x}+m^{2})G(x,y)=-\delta (x-y).$$
So they take the Fourier transform to get
$$\left(-p^{2}+m^{2}\right)G(p)=-1.$$
Then they say that we need to add an $i\epsilon$ and when we take the inverse Fourier transform we get
$$G(x,y)={\frac {1}{(2\pi )^{4}}}\int d^{4}p\,{\frac {e^{-ip(x-y)}}{p^{2}-m^{2}\pm i\varepsilon }}~.$$
I don't quite follow the derivation. First I would like to ask if the Fourier transform even makes sense when it has poles in it and we can't do an inverse Fourier transform to get back the original function? Also how can we just add $i\epsilon$, maybe it will change the result significantly? In the Wiki article it goes on to talk about how the sign of $\epsilon$ gives different propagators. So $\epsilon$ has a significant effect, how can we just add it?

Basically I am asking for a more rigorous derivation of the propagator, why the Fourier transform still works when poles are involved and where exactly the $\epsilon$ comes from.

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strangerep
First I would like to ask if the Fourier transform even makes sense when it has poles in it and we can't do an inverse Fourier transform to get back the original function?
In general, yes. One treats the integral as a Cauchy Principal Value (CPV) Integral -- meaning that we exclude a small disk centered on the pole and then take a limit as the disk's radius goes to 0.

As for why it works -- well it's because the CPV integral is compatible with Lebesgue integration, which underpins the general theory of the Fourier transform.

Also how can we just add $i\epsilon$, maybe it will change the result significantly? In the Wiki article it goes on to talk about how the sign of $\epsilon$ gives different propagators. So $\epsilon$ has a significant effect, how can we just add it?
The original formulas defining the propagator follow from ordinary Poincare invariance. But for real world physics, we must add in a causality criterion by hand -- since in the real world there is an arrow of time, and energy is bounded below. The $i\epsilon$ prescription excludes one pole while keeping the other. This has the effect of making the propagator "causal".

An alternate way to get there (i.e., to introduce causality) is to simply multiply the position-space propagator by a suitable step function (in the time domain).

• Fervent Freyja and dextercioby
Is there a way of showing that in general as long as I treat the infinities using the CPV, that everything works out correctly, I can do the Fourier transform and the inverse and there are no inconsistencies or other problems? Basically I mean can we show that if I have a function, I do the Fourier transform, get infinities, but when doing the inverse transform I use the CPV, I get back the original function.

As I understand, we want the CPV of the integral $$G(x,y)={\frac {1}{(2\pi )^{4}}}\int d^{4}p\,{\frac {e^{-ip(x-y)}}{p^{2}-m^{2} }}~,$$
but for causality we want the integral to be zero for certain spacetime events $x$ and $y$ (time arrow etc), but otherwise it should give the CPV of the above integral, and adding the $i\varepsilon$ to the denominator does just that?

The original formulas defining the propagator follow from ordinary Poincare invariance.
How do they follow?

Actually, that's an interesting point you raise here. Not sure how to resolve it, but maybe we can figure it out together.

In mathematics, Green's functions are known as fundamental solutions and physicists treat the issue in a generally very sloppy manner, that, however, is very useful for calculating things. Searching for fundamental solutions to KGE on google or in the literature will probably put you on the right track.

As far as I remember from my calculus courses, the proper way to do it is to express everything properly in terms of distributions and then take convolutions to get the general solution. Either way, you're right, the $\epsilon$ looks very fishy and strangerep's answer also seems to suggest that it's not proper mathematics.

strangerep
Basically I mean can we show that if I have a function, I do the Fourier transform, get infinities, but when doing the inverse transform I use the CPV, I get back the original function.
Well, not every function has a Fourier transform. The function must satisfy certain conditions. This would be treated (at different levels) in books on transform theory that maintain at least some mathematical rigor. I'm not sure which book would be best since I don't know your level of formal mathematical education.

But if you do a Fourier transform properly, you don't get infinities (else the transform simply wouldn't exist). The point of the CPV technique is to exclude the poles as a "set of measure zero" (after taking the radius$\to$0 limit I mentioned earlier). Sets of measure zero don't affect Lebesgue integrals, and that why these two kinds of integral are compatible.

As I understand, we want the CPV of the integral $$G(x,y)={\frac {1}{(2\pi )^{4}}}\int d^{4}p\,{\frac {e^{-ip(x-y)}}{p^{2}-m^{2} }}~,$$
but for causality we want the integral to be zero for certain spacetime events $x$ and $y$ (time arrow etc), but otherwise it should give the CPV of the above integral, and adding the $i\varepsilon$ to the denominator does just that?
Well, it fudges the result (imho). [I'll try and say a bit more later -- right now I've got chain saws, etc, being used nearby.]

Which textbook(s) are you working from? (It would be more efficient, and better, to try and follow a textbook presentation than for me to try and reproduce one here.) This subject is more often discussed on PF over in the quantum forum, where you'll find lots of older threads on this topic.

strangerep said:
The original formulas defining the propagator follow from ordinary Poincare invariance.
How do they follow?
It's just a direct translation of the $M^2$ Casimir invariant of the Poincare group across to Hilbert space operators.

PeterDonis
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This subject is more often discussed on PF over in the quantum forum
In view of which, I have moved this thread to the quantum forum.

For my level, I have a undergraduate degree in physics. Some reading material or website which does this kind of thing with rigor and proof would be great (deriving the classical propagator first, before I move on to the quantum field theory stuff).

I am reading John A. Peacock "Cosmological Physics" chapter 7. There they just do the Fourier transform, say the integral is undefined because of the poles, and then they say we'll add $i\varepsilon$ to displace the poles and avoid problems, without explaining why we can do this and still trust the result.

As an example, I'll do a one dimensional inverse transform of $(p-m)^{-1}$, using CPV:
$$\frac{1}{2\pi} \int_{-\infty}^{\infty} dp \ \frac{e^{-ipx}}{p-m} = -\frac{i}{2} e^{-imx}.$$
I closed the contour around the pole and used a semicircle in the lower half plane. This closed contour gives zero, so the above CPV integral must be minus half of the residue of the pole. But now when I do the Fourier transform of $e^{-imx}$, it is a delta function and not $(p-m)^{-1}$, so there seems to be an inconsistency here?

It's just a direct translation of the $M^2$ Casimir invariant of the Poincare group across to Hilbert space operators.
This I don't follow immediately, I will have to read up on it some time.

EDIT:

I forgot that the integral depends on the sign of x, so instead we have
$$\frac{1}{2\pi} \int_{-\infty}^{\infty} dp \ \frac{e^{-ipx}}{p-m} = -\frac{i}{2} e^{-imx} \text{sgn}(x),$$
and checking the Fourier transform of the signum function from a table it would indeed be of the form $(p-m)^{-1}$, so it seems to work out.

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vanhees71
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The problem is that you have to specify which Green's function you want. You define
$$(\Box+m^2) G(x)=-\delta^{(4)}(x).$$
Now given any solution of this inhomogeneous equation, e.g., the retarded Green's function (which you usually need in classical physics or in linear-response theory) $G_{\text{ret}}$, then any solution of the above equation has the form
$$G(x)=G_{\text{ret}}(x)+G_0(x),$$
where $G_0$ is an arbitrary solution of the homogeneous equation,
$$(\Box+m^2) G_0(x)=0.$$
Which Green's function you finally get is determined by how you run around the poles on the real energy axis in the complex energy plane. You have to close the contour for t<0 in the upper and for t>0 in the lower plane with a semi-circle at infinity to be able to apply the theorem of residues. For the retarded Green's function you have to push poth poles slightly below the real axis, i.e.,
$$G_{\text{ret}}(x)=\int_{\mathbb{R}^4} \mathrm{d}^4 p \frac{1}{(2 \pi)^4} \frac{\exp(-\mathrm{i x \cdot p})}{(p^0+\mathrm{i} 0^+)^2-\vec{p}^2-m^2}.$$
It's not so clear to me what you get from just calculating the principal-value part of the integral. Maybe you are able to figure that out, you can use "Dirac's formula"
$$\frac{1}{x+\mathrm{i} 0^+}=\text{PV}\frac{1}{x} -\mathrm{i} \pi \delta(x).$$
The integral you give in #1 leads to the time-ordered (with $+\mathrm{i} \epsilon$) or the anti-time ordered Green's function (with $-\mathrm{i} \epsilon$).

• dextercioby
Thanks, now I can understand how we can change the way we go around the poles and still have a propagator solving the original equation, as changing the way we go around the poles is equivalent to adding a contour that goes in a small circle around the pole and near the pole the operator $(\square + m^2)$ is zero, meaning it solves the homogeneous equation.

I guess I can understand how deriving the propagator works. We can instead solve the equation $$(\square _{x}+m^{2} + i\varepsilon)G(x,y)=-\delta (x-y).$$
And using the solution $G(x,y)$ to find the general solution $f(x)$ to $$(\square _{x}+m^{2})f(x)= h(x),$$
by using $$f(x) = -\int d^4y\ G(x,y) h(y),$$
so that $$(\square _{x}+m^{2})f(x)= \int d^4y\ (\delta (x-y) +i\varepsilon G(x,y)) h(y).$$
So
$$(\square _{x}+m^{2})f(x) = h(x) + i\varepsilon \int d^4y\ G(x,y) h(y).$$
Presumably the last part will go to zero as $\varepsilon\to 0$, but I should check it by explicitly finding the propagator $G(x,y)$ and determining which kind of $h(y)$ are allowed so that it actually will.

This way I haven't introduced an ad hoc way of avoiding the poles and the steps are more justified (i.e. we don't have undefined integrals and then just add something to avoid the divergences).

But I think the CPV would give something different, because they would have different contours than what the propagators actually have (including both of the poles, one of them or none of them), or am I wrong?

strangerep
As an example, I'll do a one dimensional inverse transform of $(p-m)^{-1}$, using CPV:
$$\frac{1}{2\pi} \int_{-\infty}^{\infty} dp \ \frac{e^{-ipx}}{p-m} = -\frac{i}{2} e^{-imx}.$$
I closed the contour around the pole and used a semicircle in the lower half plane. This closed contour gives zero, so the above CPV integral must be minus half of the residue of the pole.
I suspect this is wrong (by a factor of 2), but I can't be sure since you didn't show your details. It sounds like you haven't allowed for a contribution from the small semicircle, centered on $m$, in your overall computation.

A proper CPV integral excludes any contributions from such detours, i.e.,
$$CPV \int_{-\infty}^{+\infty}\!\!\!\!\cdots ~~:=~~ \lim_{\epsilon\to 0} \left( \int_{-\infty}^{m-\epsilon} \!\!\!\!\cdots ~+~ \int_{m+\epsilon}^{+\infty}\!\!\!\!\cdots \right)$$ If you omit the contribution from the small semicircular section of the contour, you don't have a closed contour and hence can't validly use the Cauchy residue theorem. When performed correctly, the CPV integral is independent of which way you avoid the poles.

The $i\epsilon$ stuff has the effect of including one or other of the small semicircular sections in the definition of the integral. Hence one might say it changes the interpretation of the integral from "CPV" to "Causal". To see this, one must return to position space and examine the effect of a $\Theta(x^0)$ multiplier.

[Edit: You asked earlier for a reference that discussed the classical propagator first. Try Nussenzveig's "Causality & Dispersion Relations", ch1, esp. section 1.3.]

(More later, perhaps.)

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• vanhees71
I suspect this is wrong (by a factor of 2), but I can't be sure since you didn't show your details. It sounds like you haven't allowed for a contribution from the small semicircle, centered on $m$, in your overall computation.

A proper CPV integral excludes any contributions from such detours, i.e.,
$$CPV \int_{-\infty}^{+\infty}\!\!\!\!\cdots ~~:=~~ \lim_{\epsilon\to 0} \left( \int_{-\infty}^{m-\epsilon} \!\!\!\!\cdots ~+~ \int_{m+\epsilon}^{+\infty}\!\!\!\!\cdots \right)$$ If you omit the contribution from the small semicircular section of the contour, you don't have a closed contour and hence can't validly use the Cauchy residue theorem. When performed correctly, the CPV integral is independent of which way you avoid the poles.
As I understand the CPV integral excludes a region of length $\varepsilon$ to both sides of the poles. So it covers the whole real axis, except the region around the pole. On the complex plane, if I would close this contour with a big semicircle going from $-\infty$ to $\infty$ on the bottom half of the plane (this semicircle gives a zero contribution, because the exponential term goes very small there), and also use a small semicircle around the pole to avoid it, the whole closed contour would give zero, as it doesn't have any poles in it. As the total contour gives zero and the big semicircle gives zero. the CPV trajectory and the small semicircle around the pole give equal and opposite contributions. The small semicircle gives half the residue (because it is a semicircle) $\int_C dp\ f(p)/(p-m) = i\pi f(m)$ and the CPV therefore gives the negative of that.

If I did this correctly, the CPV integral can be represented by semicircles around the poles, however the propagator has in its trajectory both the real axis and it also goes around the poles, so they would give different results? In other words the CPV integral covers the real axis, except the poles, but the propagator also goes around the poles, so the CPV in fact wouldn't give the propagator?

strangerep
[...] The small semicircle gives half the residue (because it is a semicircle)$\int_C dp\ f(p)/(p-m) = i\pi f(m)$ and the CPV therefore gives the negative of that.
You're relying on a symmetry argument here (wrt $\theta$) without demonstrating the symmetry. But I think it is the correct answer -- it matches what I get when I do the semicircle part carefully as a line integral.

If I did this correctly, the CPV integral can be represented by semicircles around the poles, however the propagator has in its trajectory both the real axis and it also goes around the poles, so they would give different results? In other words the CPV integral covers the real axis, except the poles, but the propagator also goes around the poles, so the CPV in fact wouldn't give the propagator?
Yes, that's the conclusion I reached also. It gets trickier still if there are 2 poles, as in the KG case.

The point is that blithely taking Fourier transforms (whose underlying theory assumes a CPV interpretation of the integrals), and introducing the $i\epsilon$ prescription, is a bit of a fudge when computing a causal propagator.

More later. (I keep saying that, I know, but I feel a need to refresh my memory properly by reviewing Nussenzveig and some other texts.)

vanhees71
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Which propagator you have to use is well-defined by the physical problem at hand. Since in #1 the OP asked about the "classical Klein-Gordon equation" I guess, what is wanted is the retarded propagator, and thus, I'd not work in the energy-momentum representation but in the Mills representation to beginn with, i.e., I'd define
$$G_{\text{ret}}(x)=\int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{k}}{(2 \pi)^3} \tilde{G}_{\text{ret}}(t,\vec{k}) \exp(\mathrm{i} \vec{k} \cdot \vec{x}),$$
and "retarded" means that
$$G_{\text{ret}}(x)=0 \quad \text{for} \quad t<0.$$
We want
$$(\Box+m^2)G_\text{ret}(x)=-\delta^{(4)}(x),$$
i.e.,
$$(\partial_t^2+\omega^2) \tilde{G}_{\text{ret}}(t,\vec{k})=-\delta(t), \quad \omega=+\sqrt{\vec{k}^2+m^2}.$$
This is easily solved
$$\tilde{G}_{\text{ret}}(t,\vec{k})=\begin{cases} 0 &\text{for} \quad t<0, \\ A \exp(-\mathrm{i} \omega t) + B \exp(+\mathrm{i} \omega t) & \text{for} \quad t>0. \end{cases}$$
To find $A$ and $B$ we first need that $\tilde{G}$ is continuous at $t=0$, i.e.,
$$A+B=0 \; \Rightarrow \; B=-A.$$
To also get $A$, we integrate the equation over $t \in (-\epsilon,\epsilon)$ and take $\epsilon \rightarrow 0^+$, which gives that
$$\dot{\tilde{G}}(0^+,\vec{k})=-1 \; \Rightarrow \; -2 A \mathrm{i} \omega =-1 \; \Rightarrow \; A=\frac{1}{2 \mathrm{i} \omega},$$
so that we finally get
$$\tilde{G}_{\text{ret}}(t,\vec{k})=-\Theta(t) \frac{\sin(\omega t)}{\omega}.$$
To get the result for the energy-momentum representation you only have to Fourier transform this with respect to $t$:
$$g_{\text{ret}}(k)=\int_{\mathbb{R}} \mathrm{d} t \tilde{G}_{\text{ret}}(t,\vec{k}) \exp(\mathrm{i} k^0 t).$$
This converges only if $\mathrm{Im} k^0>0$. Then the straight-forward integration yields
$$g_{\text{ret}}(k)=\frac{1}{(k^0+\mathrm{i} 0^+)-\omega^2}.$$

• chingel
Thank you. It seems that using $t$ instead of energy it follows rigorously and we don't have to deal with undefined integrals.

• vanhees71