# 51 scattering in quantum mechanics we calculate

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5.1 Scattering In quantum mechanics, we calculate amplitudes, h f | i i , and probabilities, |h f | i i| 2 . In field theory, we calculate the same objects. Let us consider the situation where two particles collide, and two or more particles come out. In the Schr¨ odinger picture, we want to calculate h f ; t = ∞| i ; t = -∞i . In this Heisenberg picture, we are trying to measure h f | S | i i . We are interested in this matrix S . Classically, if we throw a beam of particles on a large particle,s we can consider the cross-section area as σ = #particles scattered time × number density of beam × velocity of beam . We may think this as N = , where L is the luminosity. What we want to do now is to talk about quantum mechanics. Here, σ is just a cross-section. Particles have a probability of scattering: P = N scatter /N incident . We are going to let N incident = 1, so we are throwing one particles at a time. Then the flux is Flux = | ~v | V = | ~v 1 - ~v 2 | V . Now our formula for σ is = V T 1 | ~v 1 - ~v 2 | dP, dP = |h f | S | i i| 2 h f | f ih i | i i d Π . The last factor d Π is the density of states. On a line of size L , momenta are p n = 2 π L n and so dp = 2 π L dn . So we have d Π = Y j V (2 π ) 3 d 3 p j . The initial and final states are given by | i i = | p 1 i| p 2 i , | f i = | p 3 i · · · | p n i . Because we are working in a box, we consider | p | p i = 2 E p δ 3 (0) = 2 E p V . Then | i | i i = 2 E 1 2 E 2 V 2 , | f | f i = n Y j =3 (2 E j ) V.
Physics 253a Notes 16 Then we have = V T |h f | S | i i| 2 | ~v 1 - ~v 2 | Q j (2 E i )2 E 1 2 E 2 V n Y n V (2 π ) 3 d 3 p i . We write S = 1 + iT, where T = (2 π ) 4 δ 4 ( p 1 + p 2 - p 3 - · · · - p n ) M , because momentum is conserved. Then |h f | S | i i| 2 f 6 = i = (2 π ) 8 δ 4 ( p ) δ 4 (0) | M | 2 . If we plug this in, we get = 1 (2 E 1 )(2 E 2 ) | ~v 1 - ~v 2 | × | M 2 | Y j d 3 p j (2 π ) 3 1 2 E j (2 π ) 4 δ 4 ( p ) . This second term is also called the Lorentz-invariant phase space, d Π LIPS . So we can write the decay wrate as d Γ = 1 2 E 1 | M | 2 d Π LIPS . There is no flux factor, and no 1 / 2 E 2 . 5.2 Two-to-two scattering Let us look at the example of a 2 2 scattering. Let us call the four particles p 1 , p 2 , p 3 , p 4 . In the center of mass frame, we have | ~ p 1 | = | ~ p 2 | = p i , | ~ p 3 | = | ~ p 4 | = p f . Energy conservation is E 1 + E 2 = E 3 + E 4 = E CM . Now we look at d Π LIPS = (2 π ) 4 δ 4 ( p 1 + p 2 - p 3 - p 4 ) d 3 p 3 (2 π ) 3 1 2 E 3 d 3 p 4 (2 π ) 3 1 2 E 4 . But this has a lot of redundancies, so we can express in terms on the direction. If we integrate over ~ p 4 , we get d Π LIPS = 1 4(2 π ) 2 dp 3 E 3 E 4 δ ( E 3 + E 4 - E CM ) = d Ω 16 π 2 Z p 2 f dp f 1 E 3 E 4 δ ( q m 2 3 + p 2 f + q m 2 4 + p 2 f - E CM ) = d Ω 16 π 2 Z m 3 + m 4 - E CM dxδ ( x ) p f E CM = d Ω 16 π 2 p E E CM θ ( E CM - m 3 - m 4 ) . So we get d Ω = 1 2 E 1 2 E 2 | ~v 1 - ~v 2 | 1 16 π 2 p f E CM | M | 2 = 1 64 π 2 E 2 CM p f p i | M | 2 .
Physics 253a Notes 17 in the center of mass frame. (This d Ω is the spherical angle dφd cos θ , so that d 3 p 3 = p 2 3 dp 3 d Ω.) Let us look at the non-relativistic limit. Consider the Born approximation d Ω = m 2 e 4 π 2 | ˜ V ( k ) | 2 .