Derivation of Yukawa potential: quandry

[A/4]

Can someone provide some help with a derivation in Peskin and Schroeder (equation 4.126, p.122):

$$V(\bold{x}) = \int \frac{d^3q}{(2\pi)^3} \frac{-g^2}{|\bold{q}|^2+m^2}e^{i\bold{q}\cdot\bold{r}}$$

$$= \frac{-g^2}{4\pi^2}\int_0^\infty dq\; q^2\; \frac{e^{iqr}-e^{-iqr}}{iqr}} \frac{1}{q^2+m^2}$$

They derive the position-space Yukawa potential by Fourier-transforming the Feynman amplitude for the process. Perhaps I'm just being obtuse, but the simplifications from the first to the second line of the equation (once they've done the angular integration) don't seem clear to me.

In particular, what is the justification for this term: $$\frac{e^{iqr}-e^{-iqr}}{iqr}}$$

Any help would be appreciated.

 

[malawi_glenn]

Use $$\bold{q}\cdot\bold{r} = qrcos(\theta )$$

$$d^3q = q^2sin(\theta ) d\phi dq$$

$$q \text{ from } 0 \text{ to} +\infty$$

$$\theta \text{ from } 0 \text{ to}\pi$$

$$\phi \text{ from } 0 \text{ to}2\pi$$

 

[sir-pinski]

Thank you.

The term \frac{e^{iqr}-e^{-iqr}}{iqr} is known as the Fourier transform of a delta function. In order to understand this, we need to first look at the definition of a delta function. The delta function, denoted as \delta(x), is a mathematical function that has the property of being zero everywhere except at x=0, where it is infinite. Its integral over any interval containing x=0 is equal to 1. Mathematically, we can write this as:

\int_{-\infty}^{\infty} \delta(x) dx = 1

Now, let's consider the Fourier transform of this delta function. The Fourier transform of a function f(x) is given by:

\tilde{f}(k) = \int_{-\infty}^{\infty} f(x) e^{-ikx} dx

Using the properties of the delta function, we can write the Fourier transform of the delta function as:

\tilde{\delta}(k) = \int_{-\infty}^{\infty} \delta(x) e^{-ikx} dx = 1

This means that the Fourier transform of the delta function is a constant function with value 1. Now, coming back to our original equation, we can see that the term \frac{e^{iqr}-e^{-iqr}}{iqr} is essentially the Fourier transform of a delta function at x=q. This is why it appears in the derivation of the Yukawa potential.

I hope this helps clarify the derivation for you. If you have any further questions, please feel free to ask.