Quantum Field Theory: Problem Solutions Mark Srednicki University of California, Santa Barbara
[email protected] c 2006 by M. Srednicki All rights reserved. 1 Mark Srednicki 1 Quantum Field Theory: Problem Solutions 2 Attempts at relativistic quantum mechanics 1.1) β 2 = 1 ⇒ eigenvalue-squared = 1 ⇒ eigenvalue = ±1. α21 = 1 ⇒ Tr β = Tr α21 β. Cyclic property of the trace ⇒ Tr α21 β = Tr α1 βα1 . Then {α1 , β} = 0 ⇒ Tr α1 βα1 = −Tr α21 β = −Tr β. Thus Tr β equals minus itself, and so must be zero. Tr αi = 0 follows from this analysis by taking β → αi and α1 → β. 1.2) For notational simplicity, switch to a discrete notation: Z = Z d3x1 . . . d3xn , δxy = δ3 (x − y) , a1 = a(x1 ) , ψ = ψ(x1 , . . . , xn ; t) . (1.40) Using [X, AB . . . C] = [X, A]B . . . C + A[X, B] . . . C + . . . + AB . . . [X, C] , (1.41) which follows from writing out the terms on both sides, we have [a†x ay , a†1 . . . a†n ] = [a†x ay , a†1 ]a†2 . . . a†n + a†1 [a†x ay , a†2 ]a†3 . . . a†n + . . . + a†1 . . . a†n−1 [a†x ay , a†n ] . (1.42) We have [a†x ay , a†i ] = a†x [ay , a†i ]∓ ± [a†x , a†i ]∓ ay , = δiy a†x (1.43) where [A, B]∓ = AB ∓ BA. Using this and ay |0i = 0, we find (a†x ay )a†1 . . . a†n |0i = n X i=1 Similarly, we have (a†x a†y ay ax )a†1 . . . a†n |0i = (a†1 . . . a†n )i→x δiy |0i . n X (a†1 . . . a†n ) i→x |0i i,j=1 (1.44) (1.45) j→y for both bosons and fermions. (Extra minus signs with fermions cancel when we move a†x and a†y into the positions formerly occupied by a†i and a†j .) Now consider Z d3x a† (x