# Amplitude for fermion-fermion Yukawa scattering

1. Sep 24, 2014

### ChrisVer

Suppose that we have that:
$\mathcal{M} = c \bar{u}^{s'}(p') u^s(p) \bar{u}^{r'}(k') u^r(k)$

For a fermion fermion scattering: $f(k,r)+ f(p,s) \rightarrow f(k',r')+ f(p',s')$

Now if I want to calculate the polarization summed and averaged squared amplitude:

$\frac{|c|^2}{4} \sum_{r,r',s,s'} \bar{u}^{s'}(p') u^s(p) \bar{u}^{r'}(k') u^r(k) \bar{u}^{r}(k) u^{r'}(k') \bar{u}^{s}(p) u^{s'}(p')$

In index form I think this can be written:
$\frac{|c|^2}{4} \sum_{r,r',s,s'} \sum_{a,b,c,d} \bar{u}^{s'}(p')_a u^s(p)_a \bar{u}^{r'}(k')_b u^r(k)_b \bar{u}^{r}(k)_c u^{r'}(k')_c \bar{u}^{s}(p)_d u^{s'}(p')_d$
So I can move the spinors around, no? So that I will finally get:
$\frac{|c|^2}{4} Tr \Big( [p+m][p'+m] \Big) Tr \Big( [k+m][k'+m] \Big)$?

where the momenta in the trace are in fact slashed...

2. Sep 24, 2014

### Orodruin

Staff Emeritus
Looks reasonable to me.

3. Sep 24, 2014

### ChrisVer

Also I have one more question concerning this...
When you move the spinors, do you get a - for each commutation?

4. Sep 24, 2014

### arivero

Yukawa?

5. Sep 24, 2014

### ChrisVer

yes Yukawa....
The expression for the amplitude looks like the case for a very low energetic scalar propagator... So that $c = \frac{1}{q^2-m^2} \rightarrow -\frac{1}{m^2}$

6. Sep 24, 2014

### arivero

Nice

I wonder, then... is it not possible to get a repulsive Yukawa core in Quantum Field Theory?

7. Sep 24, 2014

### ChrisVer

I think all the yukawa couplings have attractive potential only....both for fermion-fermion and antifermion-fermion...

8. Sep 24, 2014

### arivero

Yep, even spin seems to imply atractive potential.... what about a massive photon?

9. Sep 25, 2014

### ChrisVer

What about it? It doesn't have a Yukawa coupling because it's a vector boson...?

10. Sep 25, 2014

### arivero

Well, it is not a Yukawa coupling, but can it be approximated by a Yukawa potential? I was wondering how to use QFT to produce a repulsive Yukawa potential. It seems that spin 0 and spin 2 always generate universally attractive potentials when you do the Born approximation.

11. Sep 25, 2014

### ChrisVer

hmmm...I am not entirely sure if that would be the case, but you can always look at what happens with the $W^\pm ,Z^0$ bosons, which are spin 1 and have mass. The reason I say that is that although the Yukawa potential appears always repulsive, the repulsion of the Coulomb potential is a result of the metric $g_{00} = -1$, whereas its attraction is a result of the $\gamma^0$ which exists in the vertex (which doesn't exist for the yukawa). So I think it's because the Lagrangian for the EM (spin-1 particles) comes with the $\gamma^{\mu}$ to keep lorentz invariance (the product $\gamma^{\mu}D_{\mu}$, in contrast to the Yukawa which doesn't need it (you have a scalar propagator).

However I'd try writing:
If you allow a massive [small value] photon, then you can indeed get a Yukawa potential [since the EM potential can be obtained from the Yukawa by sending m to zero]. However instead of using the exponential as it is, you can expand it:
$V= -\frac{g^2}{4 \pi r} e^{-mr} \approx - \frac{g^2}{4 \pi r} + \frac{m g^2}{4 \pi} - \frac{g^2}{8 \pi} m^2 r + \mathcal{O}(m^3)$

The first term can allow for attractive and repulsive forces, depending on the charges of your particles (practically it's the coulomb potential). The middle term is just a constant term, so it can be neglected so you are left appoximately with:
$V= V_{C} + V_{a}$ with $V_a = - \frac{g^2}{8 \pi} m^2 r$
I am not quiet sure if this tells us anything, it's a weird result since it implies some confinement. However I'd say that partially it can. It would denote a mass for the photon if such a term could exist, at least for the scalar part (unphysical/ghost) of the photon. What do you think?

Last edited: Sep 25, 2014
12. Sep 25, 2014

### nrqed

No because at that stage you are not dealing with quantum fields, just with matrices and vectors.

13. Sep 25, 2014