# Klein-Gordon propagator ill-defined?

1. Aug 6, 2010

### Šquark

Hi, I've had the following problem in elementary quantum field theory. The propagator for the Klein-Gordon scalar field takes the form

$$D(x-y)=\int\frac{\textrm{d}^3\mathbf{p}}{(2\pi)^3}\frac{1}{2\sqrt{|\mathbf{p}|^2+m^2}}e^{-ip\cdot(x-y)}$$

I was interested what the propagator looks like for space-like separations. Putting $$x^0-y^0=0$$, going to the spherical polars in momentum space, and integrating out the angular coordinates gives

$$D(x-y)=\frac{1}{(2\pi)^2r}\int_0^\infty\frac{p\sin(pr)}{\sqrt{p^2+m^2}}\textrm{d}p$$
but this integral doesn't converge because the function

$$p/\sqrt{p^2+m^2} \rightarrow 1$$ as $$p \rightarrow \infty$$

so for large $$p$$, the integrand is just $$\sin(pr)$$.

Peskin & Schroeder shift the contour in the complex plane and transform this into a convergent integral of a decaying exponential, but their integral is just different from the ill-defined integral above.

I know that it just works if we use the above expression for the Klein-Gordon propagator, but I am lead to conclude that it is nothing but a symbolic prescription, which only works if we are told how to shift the contour in the complex plane, because the shifting process changes the nature of the integral.

Or is there something that I'm missing here?

Last edited: Aug 7, 2010
2. Aug 6, 2010

### ismaili

I think your calculation is correct.
But Peskin's calculation is also correct.
Note that in his eq(2.52)
$$D(x-y) = \frac{1}{4\pi r^2}\int_m^{\infty}d\rho\frac{\rho e^{-\rho r}}{\sqrt{\rho^2-m^2}}$$
(where I think the argument of the square root should be corrected as I wrote)
also suffers a divergence problem, i.e. $$\rho\rightarrow m$$.

And the way Peskin calculated the integral is just by deforming complex contour.
Those two integrals are equivalently the same.
What he shows in eq(2.52) is that when $$r\rightarrow \infty$$, the integral is convergent to $$e^{-mr}$$ which is non-zero.

To preserve the Causality, we need $$D(x-y) - D(y-x)$$ to be vanishing at space-like separation, and this is indeed the case, because they subtract out. Even the integral D(x-y) is divergent at $$x^0 = y^0, \mathbf{x} \neq \mathbf{y}$$, but they subtract out.

3. Aug 7, 2010

### Šquark

Not quite. This is clear from the fact that my expression is not well-defined, whereas theirs is well-defined. The discrepancy appears during the shifting as follows: P&S first write the original integral as (just manipulation of complex exponential)

$$D(x-y) = \frac{1}{8i\pi^2r} \int_{-\infty}^\infty\frac{pe^{ipr}}{\sqrt{p^2+m^2}}\textrm{d}p$$
This is still not convergent because the imaginary part of the integrand behaves like $$\sin(pr)$$ for large $$p$$.

The integral is then rendered convergent by shifting the contour from the real axis to form a `keyhole' around the positive imaginary axis starting at the branch point $$im$$.

However, for this to be allowed, the integral of our complex function over the upper-semicircle must go to $$0$$ as the radius of the circle is increased to $$\infty$$. But it doesn't go to $$0$$ since for large $$p=Re^{i\varphi}$$, the integand doesn't diminish, and Jordan's lemma isn't aplicable.

4. Aug 7, 2010

### ismaili

Why?
For large $$|p|$$, the integrand DO diminish, right?
Due to the exponential factor.

5. Aug 7, 2010

### Šquark

Yes, but only for large enough imaginary part of $$p$$. There are non-vanishing contributions to the semicircle integral from the regions just above the real axis.

For the Jordan's lemma to be applicable, the integrand must have the form
$$f(p)e^{irp}$$
where $$|f(z)|\rightarrow 0$$ as $$|z|\rightarrow \infty$$ uniformly in the upper half-plane. But here
$$|f(p)|=\left|\frac{p}{\sqrt{p^2+m^2}}\right|$$
which goes to $$\pm 1$$ depending whether we're in the right- or left- hand upper quatre-plane.

6. Aug 7, 2010

### weejee

I think the key point here is that the real space propagator should be well behaved, but not necessarily its momentum space (Fourier transformed) counterpart. For example, in the integrand of the following expression,
$$D(x-y)=\frac{1}{(2\pi)^2r}\int_0^\infty\frac{p\sin(pr)}{\sqrt{p^2+m^2}}\textrm{d}p ,$$
you can multiply the integrand by $$e^{-\epsilon p}$$ and the integral converges. Then, you can take the limit $$\epsilon \rightarrow 0^{+}$$ in the final result, and it will give you the desired result.

If you Fourier transform the obtained real space propagator to the momentum space, you will exactly get $$\frac{1}{2\sqrt{|\mathbf{p}|^2+m^2}}e^{-ip\cdot(x-y)}$$, probably without any shady $$\epsilon$$ prescription. However, in an inverse Fourier transformation back to the real space, that seemingly weird $$\epsilon$$ prescription will be inevitable.

It seems that Peskin's closing his eyes on the large circle in the complex plane also amounts to this $$\epsilon$$ prescription. Something like multiplying $$\frac{4e^{\epsilon p}}{(e^{\epsilon p}+1)^{2}}$$ on the integrand of
$$D(x-y) = \frac{1}{8i\pi^2r} \int_{-\infty}^\infty\frac{pe^{ipr}}{\sqrt{p^2+m^2}}\textrm{d}p$$
which makes the contribution from the large circle vanish.

This exponential factor make the integrand vanish only when p has a big positive imaginary part. It is insufficient to kill the contribution from the whole large circle.

Last edited: Aug 7, 2010
7. Aug 7, 2010

### Šquark

Thanks, it makes much more sense now. However, I can't see why different $$\epsilon$$ regularizations should give the same results.

There are certainly many different fast-decaying functions, with which we can multiply the integrand, integrate, and then take the limit $$\epsilon\rightarrow 0$$. But how is it guaranteed that this result is always the same?

8. Aug 7, 2010

### weejee

I'm glad that it helped. :)

I think that we are safe to use any regularization procedure as long as
1) The result of the inverse Fourier transformation is independent of $$\epsilon$$ after we take the limit that it goes to zero(Doesn't behave something like $$1/\epsilon$$),
2) The Fourier transformation is well-defined without any $$\epsilon$$ prescription.

I think it is right, but as I didn't prove it rigorously, I'm not totally sure if it doesn't have loopholes.

9. Aug 7, 2010

### ismaili

Oh yes...You are right.
I forgot the Lemma require the integrand after taken exponential out should approach zero.
Then probably you need some regularization to make $$f(p)$$ goes to zero when $$z\rightarrow\infty$$.

10. Aug 10, 2010

### NanakiXIII

You can remove the $p$ in the numerator by using

$$\frac{\partial}{\partial r} e^{i p r} = i p e^{i p r}.$$

You can take the differentiation with respect to $r$ out and worry about that when you've done the integral.

11. Aug 12, 2010

### Rus Almighty

I think the right propagator should be

$$D(x-y)=\int\frac{\textrm{d}^4\mathbf{p}}{(2\pi)^4}\frac{1}{p^2+m^2}}e^{-ip\cdot(x-y)}$$

Last edited: Aug 12, 2010