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From Feynman diagrams to potential

  1. Jan 24, 2014 #1
    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}}(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}|}[/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_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]}[/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_1-p'_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. jcsd
  3. Jan 24, 2014 #2
    The basic idea is to take the non relativistic limit of the matrix element Mfi,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 ( x11.p/2m)x1)
     
  4. Jan 24, 2014 #3
    Thank you for your advice. Actually, I know the exponential term (Yukawa potential [itex]\frac{e^{-\mu r}}{r}[/itex]) come from the Fourier transform of the propagator [itex]\frac{1}{(p'_1-p_1)^2-\mu^2}[/itex]. But I can understand how to get the term [itex](\sigma_1\cdot\nabla_1)(\sigma_2\cdot\nabla_1)[/itex]. 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 Mfi
    [itex][\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][/itex]
    =[itex](\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}[/itex]
    Before Fourier transform, how to deal with the [itex]\chi[/itex]
     
    Last edited: Jan 24, 2014
  5. Jan 25, 2014 #4
    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_1-p'_1)}{2m} \frac{\sigma_2\cdot(p_2-p'_2)}{2m}\frac{-1}{(p'_1-p_1)^2-\mu^2}[/itex]
     
  6. Jan 26, 2014 #5
    if we define the [itex]\vec{p}'_1-\vec{p}_1[/itex]=[itex]\vec{p}[/itex]and [itex]\vec{r}_1-\vec{r}_2[/itex]=[itex]\vec{r}[/itex]
    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_1-p'_1)[/itex]and[itex]\sigma_2\cdot(p_2-p'_2)[/itex] don't involved in integration, instead of being the operator [itex]p=-i\hbar \nabla[/itex]. Although, the [itex]p_2-p'_2[/itex] is replaced by [itex]p_1-p'_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^{-\mu|r_1-r_2|}}{|r_1-r_2|}[/itex]
     
  7. Jan 30, 2014 #6
    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.
     
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