Zee QFT problem I.4.1: inverse square laws in (D+1)-dimensions

[Mirod]

Homework Statement

Calculate the analog of the inverse square law in a (2 + 1)-dimensional universe, and more generally in a (D + 1)-dimensional universe.

Relevant Equations

The potential energy of two sources of a free scalar field:

$$E = - \int \frac{d \vec k}{(2 \pi)^D} \frac{e^{{\vec k} \cdot {\vec r}}}{{\vec k}^2 + m^2}$$

I tried to do it for 2+1 D (3+1 is done in the text, by writing the integral in spherical coordinates and computing it directly). In 2+1 D I wrote it as:
$$E = - \int \frac{d^2 k}{ (2\pi)^2 } \frac{e^{kr cos\theta}}{k^2 + m^2} = - \int_0^{\infty} \int_0^{2\pi} \frac{d k d\theta}{ (2\pi)^2 } \frac{k e^{kr cos\theta}}{k^2 + m^2}$$

The integral over $$\theta$$ is a Bessel function:
$$\int_0^{2\pi} d \theta e^{kr cos\theta} = 2\pi I_0 (ikr)$$

Thus:
$$E = - \int_0^{\infty} \frac{d k }{ 2\pi} \frac{k I_0 (ikr)}{k^2 + m^2}$$
I got stuck here, as I have no idea if this integral is doable (looking at the properties of the Bessel functions it doesn't seem doable). Regarding arbitrary D+1 dimensions, I have no idea where to start.

 

[mark726]

Thank you for sharing your approach to solving the integral in 2+1 dimensions. It is always interesting to see different methods and perspectives in scientific research.

In response to your question about whether the integral is doable, I would like to offer some insights. While the integral may seem daunting at first glance, there are actually some techniques that can be used to evaluate it.

Firstly, let's look at the integral in the form:

E = - \int_0^{\infty} \frac{d k }{ 2\pi} \frac{k I_0 (ikr)}{k^2 + m^2}

We can see that the integrand is a product of two functions, namely k and the Bessel function I_0 (ikr). This form suggests that we can use the technique of integration by parts to evaluate the integral.

Integration by parts states that for two functions f(x) and g(x), the integral of their product can be expressed as:

\int_a^b f(x)g(x)dx = [f(x)G(x)]_a^b - \int_a^b G(x)f'(x)dx

where G(x) is the antiderivative of g(x). Applying this to our integral, we have:

E = - \frac{k I_0 (ikr)}{2\pi} \Big|_0^\infty + \int_0^\infty \frac{d k}{2\pi} \frac{I_0 (ikr)}{k^2 + m^2}

We can see that the first term evaluates to zero as k approaches infinity, leaving us with:

E = \int_0^\infty \frac{d k}{2\pi} \frac{I_0 (ikr)}{k^2 + m^2}

Now, let's focus on the second term. We can rewrite the Bessel function I_0 (ikr) as a series expansion:

I_0 (ikr) = \sum_{n=0}^\infty \frac{(ikr)^{2n}}{(2n)!}

Substituting this into our integral, we have:

E = \int_0^\infty \frac{d k}{2\pi} \frac{1}{k^2 + m^2} \sum_{n=0}^\