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ft1ls07p_08

# ft1ls07p_08 - MIT OpenCourseWare http/ocw.mit.edu 8.323...

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MIT OpenCourseWare http://ocw.mit.edu 8.323 Relativistic Quantum Field Theory I Spring 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms .

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Alan Guth, 8.323 Lecture, May 13, 2008 , p. 1. MASSACHUSETTS INSTITUTE OF TECHNOLOGY Physics Department λφ 4 1 1 λ = ( µ φ ) 2 m 2 φ 2 φ 4 . 2 2 4! Then H = H 0 + H int , where H int = d 3 x λ φ 4 ( x ) . 4! Goal: to perturbatively calculate matrix elements of the Heisenberg field φ ( x, t ) = e iHt φ ( x, 0) e iHt in the state | , the true ground state of the interacting theory. –1–
Alan Guth, 8.323 Lecture, May 13, 2008 , p. 2. Choose any reference time t 0 , at which the interaction picture operators and Heisenberg operators will coincide. Define φ I ( x, t ) e iH 0 ( t t 0 ) φ ( x, t 0 ) e iH 0 ( t t 0 ) . At t = t 0 , can expand Heisenberg φ and π in creation and annihilation operators: φ ( x, t 0 ) = (2 d π 3 p ) 2 2 1 E p a p e ip · x + a p e ip · x . π ( x, t 0 ) = φ ˙ ( x, t 0 ) = (2 d π 3 p ) 2 2 1 E p iE p a p e ip · x + iE p a p e ip · x . a p creates state of momentum p , but not energy E p not single particle. But [ φ ( x, t 0 ) , π ( y, t 0 )] = 3 ( x y ) = a p , a q = (2 π ) 3 δ (3) ( p q ) . –2– Can write φ I ( x, t ) for all t : φ I ( x, t ) = d 3 p (2 π ) 2 1 2 E p a p e ip · x + a p e ip · x x 0 = t t 0 . How to express φ ( x, t ) : φ ( x, t ) = e iH ( t t 0 ) e iH 0 ( t t 0 ) φ I ( x, t ) e iH 0 ( t t 0 ) e iH ( t t 0 ) U ( t, t 0 ) φ I ( x, t ) U ( t, t 0 ) , where U ( t, t 0 ) = e iH 0 ( t t 0 ) e iH ( t t 0 ) . Differential equation for U : i ∂t U ( t, t 0 ) = e iH 0 ( t t 0 ) ( H H 0 ) e iH ( t t 0 ) = e iH 0 ( t t 0 ) H int e iH ( t t 0 ) = e iH 0 ( t t 0 ) H int e iH 0 ( t t 0 ) e iH 0 ( t t 0 ) e iH ( t t 0 ) = H I ( t ) U ( t, t 0 ) , where H I ( t ) = e iH 0 ( t t 0 ) H int e iH 0 ( t t 0 ) = d 3 x λ 4! φ 4 I ( x, t ) . –3–

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Alan Guth, 8.323 Lecture, May 13, 2008 , p. 3. i U ( t, t 0 ) = H I ( t ) U ( t, t 0 ) with U ( t 0 , t 0 ) = I ∂t implies the integral equation t U ( t, t 0 ) = I i d t H I ( t ) U ( t , t 0 ) . t 0 To first order in H I , t U ( t, t 0 ) = I i d t 1 H I ( t 1 ) . t 0 To second order in H I , t t t 1 U ( t, t 0 ) = I i d t 1 H I ( t 1 ) + ( i ) 2 d t 1 d t 2 H I ( t 1 ) H I ( t 2 ) . t 0 t 0 t 0 To third order, t t 1 t 2 U ( t, t 0 ) = . . . + ( i ) 3 d t 1 d t 2 d t 3 H I ( t 1 ) H I ( t 2 ) H I ( t 3 ) . t 0 t 0 t 0 –4– Note that t 1 t 2 t 3 . Can rewrite 3rd order term as U ( t, t 0 ) = . . . + ( i ) 3 t t 0 d t 1 t 1 t 0 d t 2 t 2 t 0 d t 3 H I ( t 1 ) H I ( t 2 ) H I ( t 3 ) = . . . + ( i ) 3 3! t t 0 d t 1 t t 0 d t 2 t t 0 d t 3 T { H I ( t 1 ) H I ( t 2 ) H I ( t 3 ) } , where T {} is time-ordered product (earliest time to right). Finally, U ( t, t 0 ) = I + ( i ) t t 0 d t 1 H I ( t 1 ) + ( i ) 2 2! t t 0 d t 1 t t 0 d t 2 T { H I ( t 1 ) H I ( t 2 ) } + . . . T exp i t t 0 d t H I ( t ) . –5–
Alan Guth, 8.323 Lecture, May 13, 2008 , p. 4. Generalize to arbitrary t 0 : U ( t 2 , t 1 ) T exp i t 2 t 1 d t H I ( t ) , where (for t 1 < t 0 < t 2 ) U ( t 2 , t 1 ) = T exp i t 2 t 0 d t H I ( t ) T exp i t 0 t 1 d t H I ( t ) = U ( t 2 , t 0 ) U 1 ( t 1 , t 0 ) . Given U ( t, t 0 ) = e iH 0 ( t t 0 ) e iH ( t t 0 ) , have U ( t 2 , t 1 ) = e iH 0 ( t 2 t 0 ) e iH ( t 2 t 0 ) e iH ( t 1 t 0 ) e iH 0 ( t 1 t 0 ) = e iH 0 ( t 2 t 0 ) e iH ( t 2 t 1 ) e iH 0 ( t 1 t 0 ) .

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