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From Feynman diagrams to potential 
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#1
Jan2414, 08:28 AM

P: 9

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)=[itex]\frac{f^{2}}{\mu^{2}}(1P_{ex})(\tau_1\cdot\tau_2)(\sigma_1\cdot\nabla_1)(\sigma_2\cdot\nabla_1 )\frac{e^{\mur_1r_{2}}}{r_1r_{2}}[/itex] (10.51) from the amplitude [itex]S_{fi}=\frac{(ig_0)^2M^2}{(2\pi)^2\sqrt{E_1E_2E'_1E'_2}}(2\pi)^4\delta^4(p_1+p_2p'_1p'_2){[\chi^{+}_1\bar{u}(p'_1)i\gamma^5\tau u(p_1)\chi_1]\frac{i}{(p'_1p_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'_2p_1)^2\mu^2}\cdot[\chi^{+}_1\bar{u}(p'_1)i\gamma^5\tau u(p_2)\chi_2]}[/itex] (10.45) I can get the formula 10.50 [itex] \bar{u}(p'_1,s_1)\gamma^5 u(p_1,s_1)=u^{+}(s'_1)\frac{\sigma\cdot(p_1p'_1)}{2M}u(s_1)[/itex] but I can't get the 10.51, please give me an idea or suggestion, or any information, thank you! 


#2
Jan2414, 11:15 AM

P: 1,020

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 ( x_{1} (σ_{1}.p/2m)x_{1})



#3
Jan2414, 11:43 PM

P: 9

One is FoldyWouthuysen transform which need the complete Hamilton in Chapter 4, the other is 'bigsmall' 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} [itex][\chi^+_1\bar{u}(p'_1)i\gamma^5\tau u_1(p_1)\chi_1]\frac{i}{(p'_1p_1)^2\mu^2}\cdot[\chi^+_2\bar{u}(p'_2)i\gamma^5\tau u_1(p_2)\chi_2][/itex] =[itex](\tau_1\cdot\tau_2)\chi^+_1 \frac{\sigma_1\cdot(p_1p'_1)}{2m}\chi_1 \chi^+_2 \frac{\sigma_2\cdot(p_2p'_2)}{2m}\chi_2\frac{1}{(p'_1p_1)^2\mu^2}[/itex] Before Fourier transform, how to deal with the [itex]\chi[/itex] 


#4
Jan2514, 03:36 AM

P: 1,020

From Feynman diagrams to potential
Yes,this is what I have said you to do.Those [itex]\chi [/itex] 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
[itex](\tau_1\cdot\tau_2) \frac{\sigma_1\cdot(p_1p'_1)}{2m} \frac{\sigma_2\cdot(p_2p'_2)}{2m}\frac{1}{(p'_1p_1)^2\mu^2}[/itex] 


#5
Jan2614, 11:16 AM

P: 9

then [itex]\int^{\infty}_{\infty}d^3p\frac{1}{\vec{p}^2+\mu^2}e^{i\vec{p}\cdot\vec{r}}[/itex] =[itex]\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}[/itex] =4[itex]\pi^2\frac{e^{\mu r}}{r}[/itex] so, [itex]\sigma_1\cdot(p_1p'_1)[/itex]and[itex]\sigma_2\cdot(p_2p'_2)[/itex] don't involved in integration, instead of being the operator [itex]p=i\hbar \nabla[/itex]. Although, the [itex]p_2p'_2[/itex] is replaced by [itex]p_1p'_1[/itex] for the delta funtion. But the formula of the integrate should be [itex](\tau_1\cdot\tau_2) \sigma_1\cdot(\nabla_1\nabla'_1) \sigma_2\cdot(\nabla_1\nabla'_1) \frac{e^{\mur_1r_2}}{r_1r_2}[/itex] 


#6
Jan3014, 06:52 AM

P: 1,020

Hey,sorry for being late but the point is that you have to take the conservation principle also into account from p_{1}+p_{2}=p_{1}'+p_{2}' which gives p_{1}p_{1}'=p_{2}'p_{2}=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.



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