Fourier Transform quartic interaction

[sbh77]

Hi all,

This might be simple but I haven't figured out a way to do this.

Basically I have the result in coordinate space and 3+1 spacedimensions, to order lambda^2,

\frac{(-i \lambda)^2}{2!} \int dx dy (i D_F(x-y))^2 (i D_F(x1-x)) (i D_F(x2-x)) (i D_F(x3-y)) (i D_F(x4-x))

which is just a bubble with 4 legs, 2 on each side. But now I want to put it into momentum space using Fourier transforms. Maybe I am getting lost in the details but I know that one of the legs transforms as,

i D_F(x1-x) ---> \int \frac{d^4p}{(2\pi)^4} \exp^{-i p * (x1-x)} \frac{i}{p^2-m^2+i\epsilon}

I also know that some delta functions will come out to knock out some of the integrals, but I can't figure it all out. Does someone know how to do this and can show me explicitly?

Thanks much,
Brett

 

[AndyW]

Hello Brett,

Thank you for your question. Transforming from coordinate space to momentum space can be a tricky process, but it is an important step in many calculations in physics. In this case, you are trying to transform a bubble with 4 legs from coordinate space to momentum space.

To do this, we can start by writing out the integral in momentum space. Using the Fourier transform, we can write the bubble as:

\frac{(-i \lambda)^2}{2!} \int \frac{d^4p_1}{(2\pi)^4} \frac{d^4p_2}{(2\pi)^4} \frac{d^4p_3}{(2\pi)^4} \frac{d^4p_4}{(2\pi)^4} \tilde{D}_F(p_1) \tilde{D}_F(p_2) \tilde{D}_F(p_3) \tilde{D}_F(p_4) (2\pi)^4 \delta^{(4)}(p_1+p_2-p_3-p_4)

where \tilde{D}_F(p) is the Fourier transform of the propagator iD_F(x), and the delta function comes from the momentum conservation at each vertex.

To evaluate this integral, we can first simplify it by using the Feynman rules for the propagator and the vertex. This gives us:

\frac{(-i \lambda)^2}{2!} \int \frac{d^4p_1}{(2\pi)^4} \frac{d^4p_2}{(2\pi)^4} \frac{d^4p_3}{(2\pi)^4} \frac{d^4p_4}{(2\pi)^4} \frac{i}{p_1^2-m^2+i\epsilon} \frac{i}{p_2^2-m^2+i\epsilon} \frac{i}{p_3^2-m^2+i\epsilon} \frac{i}{p_4^2-m^2+i\epsilon} (2\pi)^4 \delta^{(4)}(p_1+p_2-p_3-p_4)

Next, we can use the delta function to eliminate one of the integrals. For example, we can set p

 

[sir-pinski]

Hi Brett,

The Fourier transform of a quartic interaction in momentum space can be solved using the following steps:

1. First, we can rewrite the quartic interaction in the following form:

\frac{(-i \lambda)^2}{2!} \int dx dy (i D_F(x-y))^2 (i D_F(x1-x)) (i D_F(x2-x)) (i D_F(x3-y)) (i D_F(x4-x))

= \frac{(-i \lambda)^2}{2!} \int d^4x_1 d^4x_2 d^4x_3 d^4x_4 (i D_F(x_1-x_2))^2 (i D_F(x_3-x_4))^2

2. Next, we can use the Fourier transform of the Feynman propagator in momentum space, given by:

i D_F(x-y) = \int \frac{d^4p}{(2\pi)^4} \exp^{-i p * (x-y)} \frac{i}{p^2-m^2+i\epsilon}

3. By substituting this expression into the quartic interaction, we get:

\frac{(-i \lambda)^2}{2!} \int d^4x_1 d^4x_2 d^4x_3 d^4x_4 \left( \int \frac{d^4p_1}{(2\pi)^4} \exp^{-i p_1 * (x_1-x_2)} \frac{i}{p_1^2-m^2+i\epsilon} \right)^2 \left( \int \frac{d^4p_2}{(2\pi)^4} \exp^{-i p_2 * (x_3-x_4)} \frac{i}{p_2^2-m^2+i\epsilon} \right)^2

4. We can then use the convolution theorem to simplify this expression, which states that the Fourier transform of a product of two functions is equal to the convolution of their individual Fourier transforms. This gives us:

\frac{(-i \lambda)^2}{2!} \int d^4p_1 d^4p_2 \left( \int d^4x_1 d^4x_