r d 3 r e ikvectore r \u03b8\u03c6 vector r T vector r vector r e i vector k vector r 1 2

R d 3 r e ikvectore r θφ vector r t vector r vector

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r d 3 r ′′ e ikvectore r ( θ,φ ) · vector r T ( vector r ,vector r ′′ ) e i vector k · vector r ′′ = 1 (2 π ) 3 / 2 bracketleftbigg e i vector k · vector r + e ikr r f ( θ,φ | vector k ) bracketrightbigg , (65) where we have introduced the scattering amplitude, f ( θ,φ | vector k ), 2 defined as f ( θ,φ | vector k ) = M 2 π planckover2pi1 2 integraldisplay d 3 r d 3 r ′′ e ikvectore r ( θ,φ ) · vector r T ( vector r ,vector r ′′ ) e i vector k · vector r ′′ . (66) With the definitions vector k = kvector e r ( θ,φ ) , (67) and f ( vector k | vector k ) = f ( θφ | vector k ) , (68) the scattering amplitude becomes f ( vector k | vector k ) = M 2 π planckover2pi1 2 integraldisplay d 3 r d 3 r ′′ e i vector k · vector r T ( vector r ,vector r ′′ ) e i vector k · vector r ′′ = (2 π ) 2 M planckover2pi1 2 integraldisplay d 3 r d 3 r ′′ ( vector k | vector r )( vector r | T | vector r ′′ )( vector r ′′ | vector k ) = (2 π ) 2 M planckover2pi1 2 ( vector k | T | vector k ) . (69) This shows that the scattering amplitude is the Fourier transform of the T-matrix. In optics it is well known that the image in the far-field is the Fourier transform of the near-field image. This is because propagation over a large distance allows the Fourier components of the field to spatially separate. This is exactly what we are seeing here. The T-matrix describes the exact scattered field over all space, including inside the interaction region. The Fourier transform of the T-matrix therefore describes the same scattered field in momentum space. The scattering amplitude, on the other hand, describes the scattered field only in the far-field region ( r → ∞ ), hence it follows that the different momentum components would spatially separate, with the result that the scattering amplitude should reflect the Fourier transform of the scattered field. 6 The Scattering Cross-Section Up to now we have taken the incident wave to be a plane-wave | ψ 0 ) = | vector k ) , where ( vector r | vector k ) = e i vector k · vector r (2 π ) 3 / 2 . (70) This state is delta-normalized so that ( vector k | vector k ) = δ 3 ( vector k vector k ). We note that this choice of normalization has the drawback that that | ψ 0 ( vector r ) | 2 does not have the right units to be a probability density. The units of a probability density are 1 / [ volume ], while Eq. ( ?? ) shows that | ψ 0 ( r ) | 2 = |( vector r | vector k )| 2 is clearly dimensionless. One possible way to give a probabilistic interpretation to | ψ 0 ( vector r ) | 2 would be to assume a finite quantiza- tion volume V , so that the probability density would be defined as ρ ( vector r ) = | ψ 0 ( vector r ) | 2 (2 π ) 3 V , chosen so that 2 Note that f ( x, y | a ) should be read as “f of x and y given a”. 9
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integraltext V d 3 ( vector r ) = 1. For a plane-wave, the probability density is uniform, giving ρ ( vector r ) = ρ = 1 /V . A plane wave corresponds to a uniform probability flow at speed v = planckover2pi1 v/M , so that the current density is j = ρv = v/V .
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  • Spring '10
  • MichaelMoore
  • mechanics, Angular Momentum, SCATTERING, Fundamental physics concepts, plane wave, scattering amplitude

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