- #1
Sekonda
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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 modeled 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 into 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
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 modeled 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 into 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