By Peskin and Schroeder

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**Extra resources for An Introduction to Quantum Field Theory**

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If the ground state j0i is to be invariant under translations, we must have j0i = eiP x j0i. Furthermore, since asqy creates momentum q, we can use Eq. 48) to compute h0j arp asqy j0i = h0j arp asqy eiP x j0i = ei(p;q) x h0j eiP x arpasqy j0i = ei(p;q) x h0j arp asqy j0i : This says that if h0j apr asqy j0i is to be nonzero, p must equal q. Similarly, it can be shown that rotational invariance of j0i implies r = s. ) From these considerations we conclude that 56 Chapter 3 The Dirac Field the matrix element can be written h0j arp asqy j0i = (2 )3 (3) (p ; q) rs A(p) where A(p) is so far undetermined.

The only nonzero0term in this latter quantity has the structure arpy arp as0y ] = (2 )3 (3) (p)ar0y r s the other three terms in the commutator either vanish or annihilate the vacuum. 47) of u(0) to obtain the last expression. The sum over r is accomplished most easily by; choosing the spinors r to be eigenstates of 3 . We then nd that for s = 10 , the ;one-particle state is an eigenstate of Jz with eigenvalue +1=2, while for s = 01 , it is an eigenstate of Jz with eigenvalue ;1=2. This result is exactly what we expect for electrons.

43). In practice it is often convenient to work with speci c spinors . A useful ; choice here would be eigenstates of 3 . For example, if = 10 (spin up along the 3-axis), we get u(p) = pE ; p3;1 pE + p3;01 0 p ;0 ;! 3 Free-Particle Solutions of the Dirac Equation while for = ;0 1 (spin down along the 3-axis) we have pE + p3;0 pE ; p3;10 ;! p 47 ;0 2E 01 : (3:53) 1 In the limit ! 1 the states degenerate into the two-component p spinors of a massless particle. 53) are eigenstates of the helicity operator, i (3:54) h p^ S = 21 p^i 0 0i : A particle with h = +1=2 is called right-handed, while one with h = ;1=2 is called left-handed.