From Feynman diagrams to potential

[Thor Shen]

I met a problem when I read the textbook "Relativistic Quantum Mechanics" by J.D.Bjorken. He said we can get the potential
V(r_1,r_2)=$\frac{f^{2}}{\mu^{2}}(1-P_{ex})(\tau_1\cdot\tau_2)(\sigma_1\cdot\nabla_1)(\sigma_2\cdot\nabla_1)\frac{e^{-\mu|r_1-r_{2}|}}{|r_1-r_{2}|}$
(10.51)
from the amplitude
$S_{fi}=\frac{(-ig_0)^2M^2}{(2\pi)^2\sqrt{E_1E_2E'_1E'_2}}(2\pi)^4\delta^4(p_1+p_2-p'_1-p'_2){[\chi^{+}_1\bar{u}(p'_1)i\gamma^5\tau u(p_1)\chi_1]\frac{i}{(p'_1-p_1)^2-\mu^2}\cdot[\chi^{+}_2\bar{u}(p'_2)i\gamma^5\tau u(p_2)\chi_2] -[\chi^{+}_2\bar{u}(p'_2)i\gamma^5\tau u(p_1)\chi_1]\frac{i}{(p'_2-p_1)^2-\mu^2}\cdot[\chi^{+}_1\bar{u}(p'_1)i\gamma^5\tau u(p_2)\chi_2]}$
(10.45)

I can get the formula 10.50
$\bar{u}(p'_1,s_1)\gamma^5 u(p_1,s_1)=u^{+}(s'_1)\frac{\sigma\cdot(p_1-p'_1)}{2M}u(s_1)$
but I can't get the 10.51, please give me an idea or suggestion, or any information, thank you!

 

[andrien]

The basic idea is to take the non relativistic limit of the matrix element M_fi,the Fourier transform of the corresponding matrix element will give you the potential.you also have to write the four component spinor u in terms of two component spinor like ( x₁ (σ₁.p/2m)x₁)

 

[Thor Shen]

The basic idea is to take the non relativistic limit of the matrix element M_fi,the Fourier transform of the corresponding matrix element will give you the potential.you also have to write the four component spinor u in terms of two component spinors like ( x₁ (σ₁.p/2m)x₁)

Thank you for your advice. Actually, I know the exponential term (Yukawa potential $\frac{e^{-\mu r}}{r}$) come from the Fourier transform of the propagator $\frac{1}{(p'_1-p_1)^2-\mu^2}$. But I can understand how to get the term $(\sigma_1\cdot\nabla_1)(\sigma_2\cdot\nabla_1)$. I know there are two methods to transform the relativistic potential to non-relativistic one.
One is Foldy-Wouthuysen transform which need the complete Hamilton in Chapter 4, the other is 'big-small' components nonrelativistic limit in Chapter 1. I try to use the latter one to simplify the 10.51 and get the matrix element M_fi
$[\chi^+_1\bar{u}(p'_1)i\gamma^5\tau u_1(p_1)\chi_1]\frac{i}{(p'_1-p_1)^2-\mu^2}\cdot[\chi^+_2\bar{u}(p'_2)i\gamma^5\tau u_1(p_2)\chi_2]$
=$(\tau_1\cdot\tau_2)\chi^+_1 \frac{\sigma_1\cdot(p_1-p'_1)}{2m}\chi_1 \chi^+_2 \frac{\sigma_2\cdot(p_2-p'_2)}{2m}\chi_2\frac{-1}{(p'_1-p_1)^2-\mu^2}$
Before Fourier transform, how to deal with the $\chi$

 

[andrien]

Yes,this is what I have said you to do.Those $\chi$ are not problem because they are spinors like (1 0) or (0 1),so you can forget them for the moment.you have to Fourier transform
$(\tau_1\cdot\tau_2) \frac{\sigma_1\cdot(p_1-p'_1)}{2m} \frac{\sigma_2\cdot(p_2-p'_2)}{2m}\frac{-1}{(p'_1-p_1)^2-\mu^2}$

 

[Thor Shen]

Yes,this is what I have said you to do.Those $\chi$ are not problem because they are spinors like (1 0) or (0 1),so you can forget them for the moment.you have to Fourier transform
$(\tau_1\cdot\tau_2) \frac{\sigma_1\cdot(p_1-p'_1)}{2m} \frac{\sigma_2\cdot(p_2-p'_2)}{2m}\frac{-1}{(p'_1-p_1)^2-\mu^2}$

if we define the $\vec{p}'_1-\vec{p}_1$=$\vec{p}$and $\vec{r}_1-\vec{r}_2$=$\vec{r}$
then
$\int^{\infty}_{-\infty}d^3p\frac{1}{\vec{p}^2+\mu^2}e^{-i\vec{p}\cdot\vec{r}}$
=$\int^{2\pi}_{0}d\phi\int^{1}_{-1}dcos\theta\int^{\infty}_{0}dp\frac{p^2}{p^2+\mu^2}e^{-ipcos\vartheta r}$
=4$\pi^2\frac{e^{-\mu r}}{r}$
so,
$\sigma_1\cdot(p_1-p'_1)$and$\sigma_2\cdot(p_2-p'_2)$ don't involved in integration, instead of being the operator $p=-i\hbar \nabla$. Although, the $p_2-p'_2$ is replaced by $p_1-p'_1$ for the delta funtion. But the formula of the integrate should be
$(\tau_1\cdot\tau_2) \sigma_1\cdot(\nabla_1-\nabla'_1) \sigma_2\cdot(\nabla_1-\nabla'_1) \frac{e^{-\mu|r_1-r_2|}}{|r_1-r_2|}$

 

[andrien]

Hey,sorry for being late but the point is that you have to take the conservation principle also into account from p₁+p₂=p₁'+p₂' which gives p₁-p₁'=p₂'-p₂=q say,when you put it into your expression it only depends on q,now you have to just take the Fourier transform with respect to q and the answer falls in it place.