HW4_soln

# HW4_soln - PHY 5667 Problem Set no. 4 solution Problem 1...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: PHY 5667 Problem Set no. 4 solution Problem 1 Amp. = 0|aF 2′ aF 1′ × ˆ ×T 1 ·2· 2! † dtd3 x(−iλ)ψF ψF ψB † † dt′ d3 x′ (−iλ∗ )ψF ′ ψF ′ ψB ′ × × a† 1 a† 2 |0 F F = (−iλ)(−iλ∗ ) dt d3 x dt′ d3 x′ aF 2′ aF 1′ ψ † ψ F ψ B ψ † ′ ψ F ′ ψ † ′ a† 1 a† 2 F F BF F + aF 2′ aF 1′ ψ † ψ F ψ B ψ † ′ ψ F ′ ψ † ′ a† 1 a† 2 F F BF F + aF 2′ aF 1′ ψ † ψ F ψ B ψ † ′ ψ F ′ ψ † ′ a† 1 a† 2 F F BF F = (−iλ)(−iλ∗ ) + aF 2′ aF 1′ ψ † ψ F ψ B ψ † ′ ψ F ′ ψ † ′ a† 1 a† 2 F F BF F dt d3 x dt′ d3 x′ aF 2′ ψ † aF 1′ ψ † ′ ψ B ψ † ′ ψ F a† 2 ψ F ′ a† 1 F F B F F − aF 2′ ψ † aF 1′ ψ † ′ ψ B ψ † ′ ψ F a† 1 ψ F ′ a† 2 F F B F F − aF 2′ ψ † ′ aF 1′ ψ † ψ B ψ † ′ ψ F a† 2 ψ F ′ a† 1 F F B F F + aF 2′ ψ † ′ aF 1′ ψ † ψ B ψ † ′ ψ F a† 1 ψ F ′ a† 2 F F B F F ≡ C1 − C2 − C3 + C4 Let’s start with the ﬁrst contraction C1 . Use that for arbitrary ﬁelds (boson or fermion) † ψ i ψj = ψ i a† = j dω (2π ) i d3 k exp −iω (ti − tj ) + ik · (xi − xj ) k2 (2π )3 ω − 2m + iε 2 kj 1 ti + ikj · xi exp −i 2m (2π )3/2 1 to show that C1 = (−iλ)(−iλ∗ ) dt d3 x dt′ d3 x aF 2′ ψ † aF 1′ ψ † ′ ψ B ψ † ′ ψ F a† 2 ψ F ′ a† 1 F F B F F = −|λ|2 = −|λ|2 1 (2π )6 dt d3 x dt′ d3 x exp i d3 k (2π )3 ω − dω (2π ) × × exp −i = −|λ|2 dt d3 x dt′ d3 x′ aF 2′ aF 1′ ψ † ψ F ψ B ψ † ′ ψ F ′ ψ † ′ a† 1 a† 2 F F BF F 1 (2π )6 i k2 2mB 2 k2 t + ik2 · x 2 mF dω (2π ) + iε ′ k22 ′ t − ik2 · x 2 mF 2 k1 ′ t + ik1 · x′ 2 mF i k2 2mB + iε ′ 2 k2 k2 +2 2 mF 2 mF × dt exp i −ω − × ′ k12 ′ ′ t − ik1 · x′ × 2 mF exp −iω (t − t′ ) + ik · (x − x′ ) × exp −i d3 k (2π )3 ω − exp i ′ dx exp i k + k2 − k2 x ′ dt′ exp i ω + t k2 k12 −1 2 mF 2 mF t′ × ′ dx′ exp i −k + k1 − k1 x′ Where the integral over t and t′ run from ti to tf . Under the assumption we can replace ti → −∞, tf → ∞, i.e. the timespan of the experiment is extremely long compared to the timescale of the scattering process, the integrals over t, t′ , x and x′ evaluate to δ -functions: = −|λ|2 1 (2π )2 dω d3 k i ω− k2 2mB ′ k12 + iε × ′ k22 k2 k2 − 2 −ω δ − 1 +ω 2 mF 2 mF 2 mF 2 mF i 1 × = −|λ|2 ′ 2 ′ (2π )2 k1 −k12 − |k1 −k1 |2 + iε 2mF 2mB ×δ ×δ ′ k12 k ′2 k2 k2 + 2− 1− 2 2 mF 2 mF 2 mF 2 mF ′ δ k + k2 − k2 ′ ′ δ k1 + k2 − k1 − k2 The other contractions can be obtained from C1 by interchanging indices: C2 = C1 under k1 ↔ k2 ′ ′ C3 = C1 under k1 ↔ k2 ′ ′ C4 = C1 under k1 ↔ k2 and k1 ↔ k2 2 ′ δ −k − k1 + k1 So ﬁnally we end up with the result: 1 Amp. = −|λ|2 δ (2π )2 × 2 ′ k1 −k12 2mF − − ′ k ′2 k2 k2 k12 + 2− 1− 2 2 mF 2 mF 2 mF 2 mF i ′ |k1 −k1 |2 2mB − + iε i 2 ′ k2 −k12 2mF i 2 ′ k1 −k22 2mF − ′ |k1 −k2 |2 2mB ′ ′ δ k1 + k2 − k1 − k2 + + iε − ′ |k2 −k1 |2 2mB + iε i 2 ′ k2 −k22 2mF − ′ |k2 −k2 |2 2mB + iε Problem 2 ′ ′ In the process B (kB ) + F (kF ) → B (kB ) + F (kF ), the initial and ﬁnal states are: |i = a† a† |0 FB f | = 0|aB ′ aF ′ So we get: Amp. = 0|aB ′ aF ′ × ˆ ×T 1 ·2· 2! † dtd3 x(−iλ)ψF ψF ψB † † dt′ d3 x′ (−iλ∗ )ψF ′ ψF ′ ψB ′ × × a† a† |0 BF = −|λ|2 = −|λ|2 † † † † dt d3 x dt′ d3 x′ aB ′ aF ′ ψF ψF ψ B ψF ′ ψF ′ ψ † ′ a† a† + aB ′ aF ′ ψF ′ ψF ′ ψ B ′ ψF ψF ψ † a† a† BFB BFB † † † † dt d3 x dt′ d3 x′ ψF ′ ψF ψF a† aF ′ ψF ′ aB ′ ψ † ′ ψ B a† + ψF ′ ψF ψF a† aF ′ ψF ′ aB ′ ψ † ψ B ′ a† F F B B B B ≡ C1 + C2 In the second term in the second line I have interchanged t ↔ t′ and x ↔ x′ . This is allowed since they are dummy variable we integrate over. In this way, we see that eﬀectively C2 diﬀers from C1 only in the way the boson line goes. The corresponding Feynman diagrams are: 3 C2 C1 ′ kB kF ′ kF k = kB + kF ′ kF ω= 2 kF 2 mF kB ′ k = kF − kB kB 2 kB 2 mB + kF ′ kB ω= 2 kF 2 mF − 2 kB ′ 2 mB F ermion line Boson line C1 = −|λ|2 × dt d3 x dt′ d3 x′ exp i = −|λ|2 × d3 k (2π )3 ω − dω (2π ) 1 (2π )6 2 mF exp iω (t − t′ ) − ik · (x − x′ ) × iε ′ t − ikF · x′ − i 3 dk (2π )3 ω − dω (2π ) dt exp i ω − i k2 2mF + ′2 kF ′ i k2 2mF + iε 2 k2 kF − B t× 2 mF 2 mB F × ×δ + 2 kB 2mB |kF +kB |2 2mF 2 2 kF kB − ′2 k ′2 kF + B− − 2 mF 2 mB 2 mF 2 mB 4 B ′ ′ dx′ exp i k − kF − kB · x′ ′ ′ =(2π )3 δ (k−kF −kB ) i 2 kF 2mF ′2 k ′2 kF + B t× 2 mF 2 mB F =(2π )3 δ (−k+kF +kB ) 1 (2π )2 × « „ k′2 k′2 B F =(2π )δ −ω + 2m + 2m B dx exp i −k + kF + kB · x × = −|λ|2 2 kF k ′2 k2 ′ t + ikF · x + i B t′ − ikB · x′ − i B t + ikB · x 2 mF 2 mB 2 mB dt′ exp i −ω + « „ k2 k2 B F =(2π )δ ω − 2m − 2m 1 × (2π )6 × + iε ′ ′ δ kF + kB − kF − kB k2 k2 ′ B Using C2 = C1 under kB ↔ −kB and EB ′ ↔ −EB ′ (i.e. 2B′ ↔ − 2m ) we m end up with: 1 i i Amp. = −|λ|2 +2 2 2 2 2 kB ′ kB kF kF −kB 2 (2π )2 + 2mB − |kF2+kB | + iε − 2mB − |kF2mF ′ | + iε 2mF mF 2mF ×δ ′2 k ′2 k2 k2 kF + B− F− B 2 mF 2 mB 2 mF 2 mB ′ ′ δ kF + kB − kF − kB Problem 3 ′ ′ In the process ψ (k1 ) + ψ (k2 ) → ψ (k1 ) + ψ (k2 ), the initial and ﬁnal states are: |i = a† a† |0 12 f | = 0|a2′ a1′ and the amplitude in ﬁrst order of pertubation theory is given by: 3 ˆ Amp. = 0|a2′ a1′ T −iλ ψ † (∂i ψ † )ψ (∂i ψ ) a† a† |0 12 dtdx i=1 3 = −iλ a2′ a1′ ψ † (∂i ψ † )ψ (∂i ψ )a† a† + a2′ a1′ ψ † (∂i ψ † )ψ (∂i ψ )a† a† 12 12 dtdx i=1 a2′ a1′ ψ † (∂i ψ † )ψ (∂i ψ )a† a† + a2′ a1′ ψ † (∂i ψ † )ψ (∂i ψ )a† a† 12 12 Now use that k2 1 ∂ exp −i A t + ikA · x 3/2 i 2m (2π ) 1 k2 = ikA,i , exp −i A t + ikA · x 3/2 2m (2π ) (∂i ψ )a† = A = ikA,i ψa† A 1 ∂i exp (2π )3/2 1 = −ikA,i , (2π )3/2 aA (∂i ψ † ) = = −ikA,i aA ψ † 5 2 kA t − ikA · x 2m k2 exp i A t − ikA · x 2m i to show 3 (k1′ ,i · k2,i + k2′ ,i · k2,i + k1′ ,i · k1,i + k2′ ,i · k1,i ) Amp. = −iλ i=1 ′ ′ = −iλ(k1 + k2 ) · (k1 + k2 ) dtdx a2′ ψ † a1′ ψ † ψa† ψa† 1 2 From this we read oﬀ the Feynman rule ′ k1 k1 ′ ′ = −iλ(k1 + k2 ) · (k1 + k2 ) ′ k2 k2 6 dtdx a2′ ψ † a1′ ψ † ψa† ψa† 1 2 ...
View Full Document

## This note was uploaded on 12/14/2011 for the course PHY 5667 taught by Professor Okui during the Fall '10 term at FSU.

Ask a homework question - tutors are online