# Homework Help: Zee, Quantum Field Theory in a Nutshell, problem 1.3.1

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1. Jul 2, 2016

### Maurice7510

1. The problem statement, all variables and given/known data
I'm working through Zee for some self study and I'm trying to do all the problems, which is understandably challenging. Problem 1.3.1 is where I'm currently stuck: Verify that D(x) decays exponentially for spacelike separation.

2. Relevant equations
The propagator in question is
$$D(x) = -i \int \frac{d^3k}{2(2\pi)^3} \frac{e^{-i\boldsymbol{k}\cdot\boldsymbol{x}}}{\sqrt{\boldsymbol{k}^2+m^2}}$$

3. The attempt at a solution
Presumably, I would have to solve the integral and show that it decays exponentially (the spacelike aspect has already been taken into account for the above integral) and what I did was switch to spherical coordinates and integrated over the azimuthal:

$$D = \frac{-i}{8\pi^2}\int dr\,d\theta\frac{e^{-irx\cos\theta}}{\sqrt{r^2+m^2}}r\cos\theta$$

This is where I'm stuck. The square root in the denominator suggests this is a branch cut integral but I haven't been able to find a source that explains it sufficiently. If anyone could help me figure this out I'd appreciate it. Thanks.

2. Jul 3, 2016

### stevendaryl

Staff Emeritus
I feel that there is something wrong with this equation. Does $x$ mean a spatial vector, or a 4-vector? If it's a 4-vector, then there is a problem, because on the right-hand side of the equals sign, there is no mention of the time component of $x$. If $x$ is a spatial vector vector, then it doesn't make any sense to talk about spacelike separations.

3. Jul 3, 2016

### Maurice7510

That is curious, though I wrote exactly what he has in the book. Initially, we have
$$D(x) = -i \int \frac{d^3k}{(2\pi)^32\omega_k}[e^{-i(\omega_kt - \boldsymbol{k}\cdot\boldsymbol{k})}\theta(x^0)+e^{i(\omega_kt - \boldsymbol{k}\cdot\boldsymbol{k})}\theta(-x^0)]$$
which he reduces to the above equation in consideration of spacelike separation where $x^0= 0$. So I suppose the notation on the left is a four vector, it just happens to be $x = (0, x^1, x^2, x^3) \equiv \boldsymbol{x}$.

4. Jul 8, 2016

### Maurice7510

I now realize I made a mistake in the spherical coordinate substitution, the integral should be
$$D(x) = \frac{-i}{8\pi^2} int dr\,d\theta \frac{e^{-irx\cos\theta}}{\sqrt{r^2+m^2}} r^2\sin\theta$$
The integral in $\theta$ is fairly straight forward at this point, and I got
$$D(x) = \frac{-i}{4\pi x}\int dr \frac{r\sin(rx)}{\sqrt{r^2+m^2}}$$
which i now have no idea how to solve

5. Jul 9, 2016

### malawi_glenn

That is because that integral has no solution in terms of elementary functions.

6. Jul 26, 2016

### Fred Wright

If you assume that the magnitude of the radial velocity is much less than the speed of light, then you can expand the denominator in the integral to second order in $p_r$.
$\sqrt{p^2 + m^2} = m\sqrt{1 + \frac {p^2} {m^2}} \approx m(1 + \frac {p^2} {m^2}) = \frac {(p - im)(p + im)} {m}$
The integral becomes
$D(x) \approx \frac {1} {2(2\pi)^2} \frac {m} {|x|} \int_{0}^\infty dr \frac {r[(-i) \sin(r|x|)]} {(r - im)(r + im)} = \frac {1} {2(2\pi)^2} \frac {m} {|x|} \mathcal {Im} \int_{0}^\infty dr \frac {r e^{-i|x|r}} {(r - im)(r + im)}$
We have simple poles at $z_0 = \pm im$. We use the Cauchy Integral theorem and the residue theorem, closing a semi-circular contour in the lower half-plane, and noting that the integral along the arc tends to zero as the radius tends to $\infty$ and the integral along the real axis from $-\infty$ to $+\infty$ is twice the integral from 0 to $\infty$. We find,
$D(x) \approx \frac {1} {16\pi}\frac{m} {|x|}e^{-|x|m}$