Lecture-5 Perturbation Theory and Feynman Diagrams U. Robkob, Physics-MUSC SCPY639/428 September 3, 218
From the previous lecture We end up at an expression of the 2-to-2 particle scattering S-matrix S = out; q 1, q 2 p 1, p 2 ; int = d 4 x 1 d 4 x 2 d 4 y 1 d 4 y 2 e ip 1 x 1 +ip 2 x 2 iq 1 y 1 iq 2 y 2 ( x 2 1 m1) 2 ( 2 x2 m2) 2 ( 2 y1 m3) 2 ( 2 y2 m4 2 ) out; T [ϕ(x 1 )ϕ(x 2 )ϕ(y 1 )ϕ(y 2 )], in (1) We have to evaluate the time-ordered correlation function out; T [ϕ(x 1 )ϕ(x 2 )ϕ(y 1 )ϕ(y 2 )], in (2)
Interacting picture We have three pictures of quantum mechanics formulations: a) Schrodinger picture Let ψ(t) S be Schrodinger state vector, indiced by S, satisfy Schrodinger equation i t ψ(t) S = H ψ(t) S (3) where H = H + V is the system Hamiltonian and normally H H(t). Let A be any physical property of the system, normally A A(t), and we will denote it as the Schrodinger operators A S. Its quantum expectation value is A (t) = S ψ(t) A S ψ(t) S (4) b) Heisenberg picture We can construct time evolution operator U(t, ) such that ψ(t) S = U(t, ) ψ() S (5)
Interacting picture Note that U(t, ) itself satisfy Schrodinger equation (3); i t U(t, ) = HU(t, ) U(t, ) = e iht (6) when U(, ) = 1. From (4)., we can rewrite A (t) = S ψ() U (t, )A S U(t, ) ψ() S (7) Let us define Heisenberg state and operator; ψ H = ψ() S, and A H (t) = U (t, )A S U(t, ) (8) Then we can determine ) i t A H (t) = i t (U (t, )A S U(t, ) = [A H (t), H] (9) It is called Heisenberg equation.
Interacting picture c) Interacting picture Let us define a new time evolution operator Or U (t, ) = e ih t ψ(t) S = U (t, ) ψ(t) I (1) ψ(t) I = U (t, ) ψ(t) S = U (t, )U(t, ) ψ H (11) i t ψ(t) I = U (t, ) ( H + H) U(t, ) ψ H i t ψ(t) I = V I (t) ψ(t) I (12) From (11), w can construct the interacting time evolution operator U I (t, ) = U (t, )U(t, ) ψ(t) I = U I (t, ) ψ H (13) A I (t) = U I (t, )A H(t)U I (t, ) (14) i t A I (t) = [A I (t), H ] (15)
Interacting picture We can see from (12) that U I (t, ) itself satisfy equation i t U I (t, ) = V I (t)u I (t, ) (16) By direct integration, we get U I (t, ) = 1 i Using the identity t = 1 + ( i) t t t1 dt 1 dt 2 V I (t 1 )V I (t 2 ) = 1 2! dt V I (t )U I (t, ) dt 1 V I (t 1 ) t t1 +( i) 2 dt 1 dt 2 V I (t 1 )V I (t 2 ) +... (17) t t dt 1 dt2t [V I (t 1 )V I (t 2 )](18)
Interacting picture And its generalization t = 1 n! t1 dt 1 dt 2... t tn 1 t dt 1 dt 2... t dt n V I (t 1 )V I (t 2 )...V I (t n ) dt n T [V I (t 1 )V I (t 2 )...V I (t n )] (19)
Interacting picture From (17), we will have U I (t, ) = 1 + n=1 i n n! t t dt 1... dt n T [V I (t 1 )...V I (t n )] (2) U I (t, ) = T (e i ) t dt V I (t ) (21) Back to equation (2), all field operators are Heisenberg operators: From (2), we can write ϕ(x) ϕ H ( x, t) = U I (t)ϕ I ( x, t)u I (t) (22) out, T [ϕ(x 1 )ϕ(x 2 )ϕ(x 3 )ϕ(x 4 )], in = out, T [U I (T, t 1 )ϕ I (1)U I (t 1, t 2 )ϕ I (t 2 )U I (t 2, t 3 )...ϕ I (3)U I (t 3, t 4 )ϕ I (4)U I (t 4, T )], in (23)
Interacting picture In the limit T, we can use non-interacting ground state instead of interacting ground state, in/out. From (23), we will have out, T [ϕ(x 1 )ϕ(x 2 )ϕ(x 3 )ϕ(x 4 )], in = lim T T [U I (T, t 1 )ϕ I (1)... ϕ I (4)U I (t 4, T )] U I (T, T ) = T [U I (, )ϕ I (1)ϕ I (2)ϕ I (3)ϕ I (4)] U I (, ) (24) See more discussion from Peskin & Schroder, page 86-87. Note that U I (, ) = T (e i ) dt V I (t ) (25)
Perturbation theory To calculate (24), we will do the expansion of (25) into infinite series. We first rewrite (24) in the form out, T [ϕ(x 1 )ϕ(x 2 )ϕ(x 3 )ϕ(x 4 )], in = N (26) D N = N (n) and D = D (n) (27) n= We first determine D-series: n= D () = 1 (28) D (1) = i D (2) = = ( i)2 2! dt T [V I (t )] (29) dt 1 dt 2 T [V I (t 1 )V I (t 2 )] (3)
Perturbation theory In ϕ 3 -interaction model, we have V I (t) = g 3! d 3 xϕ 3 I (x) (31) Then we have D (1) = i D (2) = ( i)2 g 2!3! d 4 x 1 T [ϕ 3 I (x 1)] (32) dx2 4 T [ϕ 3 I (x 1)ϕ 3 I (x 2)] (33) dx 4 1
Wick s theorem We have to define: a) Normal ordering: : ϕ(x 1 )ϕ(x 2 ) : : ϕ(x 1 )ϕ(x 2 ) : = {}}{ b) Contraction: ϕ(x 1 )ϕ(x 2 ) = T [ϕ(x 1 )ϕ(x 2 )] : ϕ(x 1 )ϕ(x 2 ) :, so that {}}{ ϕ(x 1 )ϕ(x 2 ) = T [ϕ(x 1 )ϕ(x 2 )] = D F (1, 2) (34) The theorem say that: for any 2N-point correlation function, it can be written in term summation of all possible normal orderings and contractions. We can see from (32) that D (1) =. And from (33) we have D (2) = ( i)2 g 2!3! dx 4 1 dx 4 2 {D(1, 1)D(1, 2)D(2, 2) +D(1, 2)D(1, 2)D(1, 2)} (35)
Feynman diagrams We assign the diagrams: The we have diagrams for D (2) as Note that D (3) =, you can try for D (4) diagrams.
Feynman diagrams Do the same analysis for N-series. We have N () = T [ϕ(x 1 )ϕ(x 2 )ϕ(y 1 )ϕ(y 2 )] (36) = D(1, 2)D(1, 2 ) + D(1, 1 )D(2, 2 ) + D(1, 2 )D(2, 1 ) (37) when 1 = x 1, 2 = x 2.1 = y 1, 2 = y 2. Diagrams for N () are They are all trivial diagrams. next we have N (1) = (38) N (2) = ( ig)2 d 4 z 1 d 4 z 2 T [ϕ 3 (z 1 )ϕ 3 (z 2 ) 2!3!...ϕ(x 1 )ϕ(x 2 )ϕ(y 1 )ϕ(y 2 )] (39)
Feynman diagrams With the help from Wick s theorem, we have N (2) = ( ig)2 d 4 z 1 d 4 z 2 2!3! { D(1, 1 )D(2, 1 )D(1, 2 )d(2, 1 )D(2, 2 ) +D(1, 1 )D(1, 1 )D(1, 2 )D(2, 2 )D(2, 2 ) + D(1, 1 )D(1, 2 )D(1, 2 )D(2, 2 )D(2, D1 ) } (4)
Feynman diagrams Note that N (2) disc contains trivial diagrams together with disconnected diagrams. They will cancelled with disconnected diagrams from D-series, left only the connected diagrams of the N-series, which are all non-trivial.
Amplitudes Back to S-matrix in (1), we will do the same expansion as S = M (n) (41) n= where M(n) is related to n-order of non-zero N (n ) con diagrams.let us assume that m 1 = m 2 = m 3 = m 4 = m for simplicity. First we determine M (), we have M () = d 4 x 1 d 4 x 2 d 4 y 1 d 4 y 2 e ip 1 x 1 +ip 2 x 2 iq 1 y 1 iq 2 y 2 ( 2 x 1 m 2) ( 2 x 2 m 2) ( 2 y 1 m 2) ( 2 y 2 m 2) { D(1, 2)D(1, 2 ) + D(1, 1 )D(2, 2 ) + D(1, 2 )D(2, 1 ) } (42) M () a + M () b + M () c (43)
Amplitudes Let us determine M a () first M a () = d 4 x 1 d 4 x 2 d 4 y 1 d 4 y 2 e ip 1 x 1 +ip 2 x 2 iq 1 y 1 iq 2 y 2 ( 2 x 1 m 2) ( 2 x 2 m 2) ( 2 y 1 m 2) ( 2 y 2 m 2) D(1, 2)D(1, 2 )(44) Since ( 2 x 1 m 2) D(1, 2) = δ (4) (x 1 x 2 ) and ( 2 y1 m 2) D(1, 2 ) = δ (4) (y 1 y 2 ), we will have from (44) M a () = d 4 x 2 d 4 y 2 e i(p 1+p 2 ) x 2 i(q 1 +q 2 ) y 2 ( 2 x2 m 2) ( 2 y 2 m 2) (45) Integration by part two times on x 2 and y 2, we have M () a = ((p 1 + p 2 ) 2 + m 2 )((q 1 + q 2 ) 2 + m 2 ) d 4 x 2 d 4 y 2 e i(p 1+p 2 ) x 2 i(q 1 +q 2 ) y 2 δ (4) (p 1 + p 2 )δ (4) (q 1 q 2 ) (46)
Amplitudes It shows nothing but energy-momentum conservation. Similar results wwill be derived for M () b and M c (). Next we do evaluate M (1), which is related to connected diagrams from N (2) ; M (1) = ( ig)2 d 4 z 1 d 4 z 2 2!3! d 4 x 1 d 4 x 2 d 4 y 1 d 4 y 2 e ip 1 x 1 +ip 2 x 2 iq 1 y 1 iq 2 y 2 ( 2 x 1 m 2) ( 2 x 2 m 2) ( 2 y 1 m 2) ( 2 y 2 m 2) { D(1, 1 )D(2, 1 )D(1, 2 )d(2, 1 )D(2, 2 ) +...... + D(1, 1 )D(1, 1 )D(1, 2 )D(2, 2 )D(2, 2 ) +...... + D(1, 1 )D(1, 2 )D(1, 2 )D(2, 2 )D(2, D1 ) } (47) = M (1) a + M (1) b + M (1) c (48)
Amplitudes Determine d 4 x 1 d 4 x 2 M (1) a = ( ig)2 2!3! d 4 z 1 d 4 z 2 d 4 y 1 d 4 y 2 e ip 1 x 1 +ip 2 x 2 iq 1 y 1 iq 2 y 2 ( 2 x 1 m 2) ( 2 x 2 m 2) ( 2 y 1 m 2) ( 2 y 2 m 2) D(1, 1 )D(2, 1 )D(1, 2 )d(2, 1 )D(2, 2 ) = ( ig)2 d 4 z 1 d 4 z 2 e i(p 1+p 2 ) z 1 i(q 1 +q 2 ) z 2 D(z 1, z 2 ) 2!3! = ( ig)2 d 4 z 1 e i(p 1+p 2 q 1 q 2 ) z 1 dz 2 e i(q 1+q 2 ) (z 1 z 2 ) D(z 1, z 2 ) 2!3! = ( ig)2 D(k)(2π) 4 δ (4) (p 1 + p 2 q 1 q 2 )(49) 2!3! where k = q 1 + q 2 = p 1 + p 2 and i D(k) = k 2 m 2 (5) + iɛ
Amplitudes In general the non-zero amplitudes can be written in the form ( M (n) = im (n) S n (2π) 4 δ (4) p i,in ) q i,out i i }{{} energy mentum conservation (51) where S n is symmetry factor and, for example, We will have im (1) a = ( ig) 2 D(k), k + p 1 + p 2 = q 1 + q 2 (52) im (1) b = ( ig) 2 D(k ), k = p 1 q 1 = p 2 q 2 (53) im (1) c = ( ig) 2 D(k ), k = p 1 q 2 = p 2 q 1 (54)
Feynman rules at tree level diagrams Field Lagrangian will tell us about field propagator and interaction vertex as L = 1 2 µϕ µ ϕ 1 2 m2 ϕ 2 We assign } {{ } field propagator a) field propagator: D(k) = i k 2 m 2 +iɛ b) vertex factor: ig g 3! ϕ3 }{{} vertex (55) c) count symmetry factor S n, for each type of Feynman graph d) the transition amplitude M will be written as in (51)