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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

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# Scattering Amplitude for a given potential

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