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Scattering Amplitude for a given potential

  1. Nov 22, 2012 #1
    Hello again,

    My question is on determining the scattering amplitude via time-dependent perturbation theory (first-order) for a given potential - I believe the perturbation potential is modelled due to some interaction between two scalar particles and has form:

    [tex]\delta V=\lambda \Phi_{f'}^{*}\Phi_{i'}[/tex]

    Where I believe the i' state is the initial state of some incoming particle which interacts with another particle in a state i, then both particles rebound in to states f' and f respectively. I believe we use an equation of the following form to determine the scattering amplitude (provided that they are much less than 1):

    [tex]a_{f}(t)=\frac{i}{2E_{f}}\int dt\int d^3x\Phi_{f}^{*}\delta V\Phi_{i}[/tex]

    where I am assuming plane wave solutions to each wave equation, so they have following form where the N's are normalization constants.

    [tex]\Phi_{i}=N_{1}e^{-E_{i}t+i\mathbf{p}_{i}\cdot \mathbf{x}}\; ,\; \Phi_{i'}=N_{3}e^{-E_{i'}t+i\mathbf{p}_{i'}\cdot \mathbf{x}}[/tex]
    [tex]\Phi_{f}^{*}=N_{2}e^{E_{f}t-i\mathbf{p}_{f}\cdot \mathbf{x}}\; ,\; \Phi_{f'}^{*}=N_{4}e^{E_{f'}t-i\mathbf{p}_{f'}\cdot \mathbf{x}}[/tex]

    So I have then substituted these into my scattering amplitude equation to attain:

    [tex]a_{f}(t)=\frac{i\lambda}{2E_{f}}N_{1}N_{2}N_{3}N_{4}\int dt\: e^{i(E_{f}+E_{f'}-E_{i}-E_{i'})t}\int d^3x\: e^{-i(\mathbf{p}_f+\mathbf{p}_{f'}-\mathbf{p}_i-\mathbf{p}_{i'})\cdot \mathbf{x}}[/tex]

    Now, I think if I have done this correctly, then the integral of the exponential with respect to time becomes a delta function describing conservation of energy of the interaction such that:

    [tex]\int_{-\infty}^{\infty} dt\: e^{i(E_{f}+E_{f'}-E_{i}-E_{i'})t}=2\pi\delta(E_f+E_{f'}-E_i-E_{i'})[/tex]

    I think the integral with respects to 'x' does something similar, can someone help me out with this integral?

    Thanks guys!
    SK
     
  2. jcsd
  3. Nov 22, 2012 #2
    Actually is this correct?

    [tex](2\pi)^3\delta^3(\mathbf{p}_i+\mathbf{p}_{i'}-\mathbf{p}_f-\mathbf{p}_{f'})=\int d^3x\: e^{-i(\mathbf{p}_f+\mathbf{p}_{f'}-\mathbf{p}_i-\mathbf{p}_{i'})\cdot \mathbf{x}}[/tex]

    Essentially enforcing conservation of momentum as well?
     
  4. Nov 23, 2012 #3
    yes,it is definitely correct.
     
  5. Nov 26, 2012 #4
    Cheers man!

    Thanks,
    SK
     
  6. Nov 26, 2012 #5
    I have attained a scattering amplitude of form:

    [tex]\frac{i\lambda(2\pi)^4N_1N_2N_3N_4}{2(E_i+E_{i'}-E_{f'})}[/tex]

    from the above, what does this tell me about the interaction? I think we should expect scattering amplitudes much less than 1 though I'm not sure how this equation shows it.

    It seems that all I can say for this scattering amplitude solution is that the scattering of some particle from states i to f by some other particle scattering from states i' to f' is inversely proportional to the energy of state f or the the energies i+i'-f', so that the scattering is not only dependent on one state of the scattered particle f but also on the energy states of all other particle states involved in the interaction.

    Is there anything else this equation says?
     
    Last edited: Nov 26, 2012
  7. Nov 28, 2012 #6
    this is not the way you should do it.you have turned the amplitude into energy denominator form.You should retain those 4 delta functions(momentum+energy) and the absolute square of it shows the probability type thing.The delta function is treated after it.You can see some books lke feynman 'quantum electrodynamics' on how to treat it.you should draw some conclusion from absolute square of it rather than from amplitude itself(other than some analyticity,unitarity etc.)
     
  8. Nov 28, 2012 #7
    Ahh I see what you mean, I may have been doing this calculation incorrectly but I understand what you are saying and it still applies. Thanks Andrien!
     
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