A γ 5 g 2 v g 2 a tr 6 p 2 γ μ 6 p 1 γ ν 2 g v g

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A γ 5 )] = ( g 2 V + g 2 A )Tr[ 6 p 2 γ μ 6 p 1 γ ν ] + 2 g V g A Tr[ 6 p 2 γ μ γ 5 6 p 1 γ ν ] = 4( g 2 V + g 2 A )( p μ 2 p ν 1 + p μ 1 p ν 2 - η μν p 1 · p 2 ) - 8 ig V g A μνρσ p 1 ρ p 2 σ Multiplying together two of these traces isn’t the end of the world. The first bit is symmetric in μν , while the second is antisymmetric, so there are no cross terms. Using the relation μνρσ μναβ = 2( δ ρ α δ σ β - δ ρ β δ σ α ) , we find X spins |M| 2 = 32 ( s - m 2 Z ) 2 (( g 2 A - g 2 V ) 2 p 13 p 24 + ( g 4 A + 6 g 2 V g 2 A + g 4 V ) p 14 p 23 ) = 8 ( s - m 2 Z ) 2 (( g 2 A - g 2 V ) 2 t 2 + ( g 4 A + 6 g 2 V g 2 A + g 4 V ) u 2 ) = 16 E 2 ( s - m 2 Z ) 2 (( g 2 A + g 2 V ) 2 (1 + cos 2 θ ) + 8 g 2 A g 2 V cos θ )
Phys 253a 4 (c) Figure 2 says that the cross section is zero at cos θ = 1. We can’t achieve this, but we can make the cross section vanish at cos θ = - 1 if we choose g A = ± g V . This makes physical sense. If g A = ± g V , then the current that Z couples to is either 2 g V ψγ μ P L ψ = 2 g V ψ L σ μ ψ L , or 2 g V ψγ μ P R ψ = 2 g V ψ R σ μ ψ R In other words, it involves only one chiral half of the neutrino spinor. Suppose Z interacts only with the left-handed half. We know from the last problem set that helicity is the same as chirality for external states of massless particles. Thus, we have that Z interacts with left helicity neutrinos and right helicity antineutrinos. The scattering process looks like this in the COM frame: a left helicity neutrino, moving in the +ˆ z direction, with spin in the - ˆ z direction, collides with a right helicity antineutrino, moving in the - ˆ z direction, with spin in the - ˆ z direction. Together, they create a polarization vector in the - ˆ z direction: 2 g V v 2 γ μ P L u 1 (0 , 1 , - i, 0) Suppose the scattering angle is zero, so that the decay products are headed in the same direction as the initial particles. Then we would have 2 g V u 3 γ μ P L v 4 (0 , 1 , - i, 0) * = (0 , 1 , i, 0) To see what happens when θ = π , we should rotate this 180 degrees around the x or y axes. We get (0 , 1 , - i, 0), which has zero inner product with the initial polarization (0 , 1 , - i, 0). Thus, the backscattering cross section is zero when Z couples to only left-handed neutrinos. The same argument works if Z couples only to right-handed neutrinos. Parity reflects x → - x , so in particular takes cos θ → - cos θ for distributions that are symmetric around the beam axis. Figure 2 is not invariant under this transformation. (d) Since the weak force couples to left handed particles and right handed antipar- ticles, the electron will be left handed and the positron right handed. (e),(f) The crystal comes in two forms, one which has a higher index of refraction for left circularly polarized light and one which has a higher index of refraction for
Phys 253a 5 right circularly polarized light. These are connected to the two possibilities for the relation between how the parallelogram of the top face points and which way the top of the crystal leans relative to the bottom. Let’s label one of these geometric configurations “left” and the other “right.” Then regardless of which crystal the aliens have, they’ll associate a unique “left” or “right” to the two kinds of circularly polarized light they have. Now tell the aliens to take their charge carrier (which will be the electron if they’re made of matter and the positron if they’re made of antimatter) and

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