f p i cos \u03b8 32 Let us expand on equation 29 Q E i E f 2 E 2 i m 2 e E 2 f m 2 e

F p i cos θ 32 let us expand on equation 29 q e i e

This preview shows page 5 - 8 out of 23 pages.

f || p i | cos θ . (32) Let us expand on equation 29: Q = ( E i + E f ) 2 + ( E 2 i - m 2 ˜ e ) + ( E 2 f - m 2 ˜ e ) + 2 q ( E 2 f - m 2 ˜ e )( E 2 i - m 2 ˜ e ) cos θ (33) = 2 E 2 i + 2 E 2 f - 2 m 2 ˜ e + 2 E i E f + 2 q ( E 2 f - m 2 ˜ e )( E 2 i - m 2 ˜ e ) cos θ , (34) we can now identify s = 4 E 2 i = 4 E 2 f (the square of the total energy available), such that Q = 2 n s 4 + s 4 - m 2 ˜ e + E i E f + q ( E 2 f - m 2 ˜ e )( E 2 i - m 2 ˜ e ) cos θ o . (35) Note we are dealing with identical particles in the center-of-mass frame, and as such E i = E f : Q = 2 n s 4 + s 4 - m 2 ˜ e + s 4 + ( E 2 i - m 2 ˜ e ) cos θ o (36) = 2 3 s 4 - m 2 ˜ e + s 4 - m 2 ˜ e cos θ (37) = 1 2 s (3 + cos θ ) - 2 m 2 ˜ e (1 + cos θ ) . (38) Therefore, the first term in the matrix element is i M t - 1 = ie 2 t Q = ( ie 2 ) s (3 + cos θ ) - 4 m 2 ˜ e (1 + cos θ ) 2 t . (39) We can argue the second term: i M t - 2 = - ie 2 (1 - ξ ) ( p 1 - p 3 ) 2 + i ( p 1 + p 3 ) μ ( p 1 - p 3 ) μ ( p 1 - p 3 ) ν p 2 ( p 2 + p 4 ) ν , (40) vanishes because ( p 1 + p 3 ) μ ( p 1 - p 3 ) μ = ( E i + E f , p i + p f ) · ( E i - E f , p i - p f ) (41) = (2 E i )(0) - ( p i + p f ) · ( p i - p f ) (42) = ( p i + p f ) · ( p f - p i ) = | p f | 2 - | p i | 2 (43) = ( E 2 f - m 2 ˜ e ) - ( E 2 i - m 2 ˜ e ) = 0 , (44) and thus the whole term vanishes. Page 5 of 23
Image of page 5
Dylan J. Temples Quantum Field Theory II : Solution Set One 2.1.2 u -Channel Matrix Element The matrix element for this channel is i M u = ( - ie ) 2 ( p 1 + p 4 ) μ - i p 2 + i g μν - (1 - ξ ) p μ p ν p 2 ( p 2 + p 3 ) ν , (45) with p = p 1 - p 4 . First, let us argue the second term in the photon propagator vanishes here as well. This term is proportional to the Lorentz invariant quantity ( p 1 + p 4 ) μ ( p 4 - p 1 ) μ = ( E 1 + E 4 , p 1 + p 4 ) · ( E 4 - E 1 , p 4 - p 1 ) , (46) we are free to evaluate this in any frame, so we select the center of mass frame: E 1 E i = E 4 E f , so ( p 1 + p 4 ) μ ( p 4 - p 1 ) μ = (2 E i , p i + p f ) · (0 , p f - p i ) = - | p f | 2 - | p i | 2 (47) = ( E 2 f - m 2 ˜ e ) - ( E 2 i - m 2 ˜ e ) = 0 , (48) and we can see that as long as the incoming and outgoing scalar has the same mass, this term will always vanish - we will use this result in the next section. The matrix element is then i M u = ( - ie ) 2 ( p 1 + p 4 ) μ - i ( p 1 - p 4 ) 2 + i ( p 2 + p 3 ) μ , (49) which is proportional to the dot-product: ( p 1 + p 4 ) μ ( p 2 + p 3 ) μ = (2 E, p i - p f ) · (2 E, - p i + p f ) , (50) in the center-of-momentum frame, using the same definitions as before, with E = E i = E f . Carrying out the product yields ( p 1 + p 4 ) μ ( p 2 + p 3 ) μ = 4 E 2 - -| p f | 2 - | p f | 2 + 2 | p i || p f | cos θ (51) = s + ( E 2 - m 2 ˜ e ) + ( E 2 - m 2 ˜ e ) - 2 q ( E 2 - m 2 ˜ e )( E 2 - m 2 ˜ e ) cos θ (52) = s + s 2 - 2 m 2 ˜ e - 2 E 2 cos θ - 2 m 2 ˜ e cos θ (53) = s 2 (3 - cos θ ) - 2 m 2 ˜ e (1 - cos θ ) . (54) Inserting this into the matrix element, and performing some simplification yields i M u = ie 2 u s 2 (3 - cos θ ) - 2 m 2 ˜ e (1 - cos θ ) . (55) 2.1.3 Squaring the Amplitude We will collect our results here: i M t = ie 2 ( p 1 + p 3 ) · ( p 2 + p 4 ) t = ( ie 2 ) s (3 + cos θ ) - 4 m 2 ˜ e (1 + cos θ ) 2 t (56) i M u = ie 2 ( p 1 + p 4 ) · ( p 2 + p 3 ) u = ( ie 2 ) s (3 - cos θ ) - 4 m 2 ˜ e (1 - cos θ ) 2 u . (57) Page 6 of 23
Image of page 6
Dylan J. Temples Quantum Field Theory II : Solution Set One We can examine the dot products: ( p 1 + p 3 ) · ( p 2 + p 4 ) = p 1 · p 2 + p 1 · p 4 + p 3 · p 2 + p 3 · p 4 (58) We must note that since the final state particles are identical, the matrix elements have a relative sign difference, so the total matrix element for this process is i M = i M t - i M u , (59) so the modulus squared is |M| 2 = |M t | 2 + |M u | 2
Image of page 7
Image of page 8

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture