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- - **Fourier transform of scattering hamiltonian**
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Fourier transform of scattering hamiltonianHey,
I am looking at the coupling hamiltonian for electrons in an EM field. In particular I'm interested in the inelastic scattering (this isn't the dominant part for inelastic scattering but it's confusing me). The part of the hamiltonian in the time/space domain that I'm interested in is [itex]H_A = (\frac{e^2}{2mc^2})\sum_j A(r_j,t) \cdot A(r_j,t)[/itex] Where [itex]j[/itex] sum across all the electrons in this many particle problem. [itex]A[/itex] is the vector potential of EM field. Now I have another paper which fourier transforms this part of the Hamiltonian as [itex]H_A = (\frac{e^2}{2mc^2})\sum_{k_1,\omega_1} \sum_{k_2,\omega_2} N(-k_1 + k_2) A(k_1,\omega_1) \cdot A^*(k_2,\omega_2)[/itex] where [itex] N(-k_1 + k_2) = \Sigma_j e^{i(k_1 - k_2)} \cdot r_j [/itex] which is the fourier transform of the many particle number operator for the electrons. The questionHow exactly is the hamilton fourier transformed in this way? It seems to imply that [itex] A(k_1,w_1) = \Sigma_{k_2,\omega_2} N(-k_1 + k_2) A^*(k_2,\omega_2) [/itex] assuming that the fourier transformed hamiltonian can be written as [itex]H_A = (\frac{e^2}{2mc^2})\sum_{k_1,\omega_1} A(k_1,\omega_1) \cdot (k_1,\omega_1)[/itex] which I am not certain of.. |

Re: Fourier transform of scattering hamiltonianHi dkin,
I'm not familiar with that usage but to be able to evaluate the Fourier transform you need to express the Hamiltonian in such a way that every derivative (with respect to time or space) is explicitly shown. The potentials you write are apparently dependent on some other expression of what they consist of and which presumably contain derivative operators. If the potentials are defined using the Dirac delta function that needs to be explicitly shown also because it will have a term in the Fourier transform. |

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