in this state and what does it become in the ultra relativistic ï¿½ limit A boost

# In this state and what does it become in the ultra

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in this state, and what does it become in the ultra-relativistic ( η → ∞ ) limit? A boost along the x axis by η of the wavefunction yields ψ ( x ) exp - 1 2 η σ 1 0 0 σ 1 m ξ + ξ + , (108) see Peskin 3.49. Following Peskin, this can be expressed ψ 0 ( x ) = cosh( η/ 2) 1 0 0 1 - sinh( η/ 2) σ 1 0 0 σ 1 m ξ + ξ + (109) = m cosh( η/ 2) ξ + ξ + - sinh( η/ 2) σ 1 ξ + σ 1 ξ + , (110) let us note σ 1 ξ + = 0 1 1 0 1 0 = 0 1 = ξ - , (111) so ψ 0 ( x ) = m cosh( η/ 2) ξ + ξ + - sinh( η/ 2) ξ - ξ - . (112) Let us act the S z operator on the new wavefunction: S 3 | ψ 0 + i = m 2 cosh( η/ 2) σ 3 ξ + σ 3 ξ + - sinh( η/ 2) σ 3 ξ - σ 3 ξ - (113) = m 2 cosh( η/ 2) ξ + ξ + + sinh( η/ 2) ξ - ξ - . (114) Page 13 of 23
Dylan J. Temples Quantum Field Theory II : Solution Set One Now when we take the expectation value, the cross-terms from the multiplication cancel exactly: h ψ 0 + | S 3 | ψ 0 + i = m 4 E n cosh 2 ( η/ 2)(2 ξ + ξ + ) - sinh 2 ( η/ 2)(2 ξ - ξ - ) o , (115) where we’ve introduced the factor of 2 E for the relativstic normalization. The normalization of the spinors ξ is one, so h ψ 0 + | S 3 | ψ 0 + i = 2 m 4 m cosh η cosh 2 ( η/ 2) - sinh 2 ( η/ 2) = 1 2 cosh η , (116) because after the boost by η , the particle has E = m cosh η . In the ultrarelativistic limit ( η 0) this vanishes. (c) Take the ultra-relativistic limit of the wavefunction. Show that the result is an eigenstate of the helicity operator, ˆ p · S . Interpret the result of part (b) using this result. The helicity operator is ˆ p · S = ˆ p · 1 2 σ 0 0 σ = 1 2 ˆ p · σ 0 0 ˆ p · σ , (117) where ˆ p is the particle’s momentum. Since we have boosted in the x direction from rest: ˆ p = (1 , 0 , 0) , (118) so h = 1 2 σ 1 0 0 σ 1 . (119) The boosted wavefunction ψ 0 ( x ) = m cosh( η/ 2) ξ + ξ + - sinh( η/ 2) ξ - ξ - . (120) in the ultra-relativistic limit becomes ψ 0 ( x ) = me η/ 2 ξ + ξ + - ξ - ξ - = me η/ 2 1 - 1 1 - 1 , (121) because both hyperbolic functions diverge (the negative exponential term vanishes). We can act the helicity operator on the boosted wavefunction: 0 = m 2 e η/ 2 σ 1 0 0 σ 1 1 - 1 1 - 1 = m 2 e η/ 2 0 1 0 0 1 0 0 0 0 0 0 1 0 0 1 0 1 - 1 1 - 1 (122) = m 2 e η/ 2 - 1 1 - 1 1 = - 1 2 me η/ 2 1 - 1 1 - 1 = - 1 2 ψ 0 , (123) which is an eigenstate of helicity n with eigenvalue - 1 / 2. Page 14 of 23
Dylan J. Temples Quantum Field Theory II : Solution Set One 4 Schwartz 9.1. Compton scattering in QED. (a) Calculate the tree-level matrix elements for γφ γφ . Show that the Ward identity is satisfied. Figure 3: The tree-level Feynman diagrams describing the process γφ γφ . (Left) vertex only. (Center) s -channel. (Right) t -channel. Let us label the process as follows: φ ( p 1 ) + γ ( p 2 , ε 2 ) γ ( p 3 , ε * 3 ) + φ ( p 4 ) , (124) where the ε are photon polarization vectors. The tree-level diagrams for this scattering process are shown in figure 3, the corresponding matrix elements will be M 1 , M s , and M t , respectively. The vertex-only diagram is trivial: i M 1 = 2 ie 2 ε 2 · ε * 3 . (125) Since the remaining two processes are mediated through a scalar propagator, we need the appropriate Feynman rule: - i p 2 - m 2 φ + i . (126) Using the Feynman rule for the two scalar-one photon vertex yields, for the s -channel: i M s = ( - ie ) 2 ( p 1 + p ) μ ε