Quantum Field Theory: Problem Solutions Mark Srednicki University of California, Santa Barbara
[email protected]
c
2006 by M. Srednicki All rights reserved.
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Mark Srednicki
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Quantum Field Theory: Problem Solutions
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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