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1、Solutions to Peskin , = 1,3. (30) Then, we defi ne M = 13, and it can be easily shown that M1M = ()T, and M1M = 1. Then we have: tr(12n) = tr(M11MM12M M1nM) = tr ( 1)T(2)T (n)T = tr ( n21)T = tr(n21).(31) With this formula in hand, we can derive the equality uL(p1)uL(p2) = uR(p2)uR(p1)(32) as follow
2、s LHS = C uR0/p1/p2uR0= C tr(/p1/p2) = C tr(/p2/p1) = C uL0/p2/p1uL0= RHS, in which C (2(p 1 k0)(p2 k0) )1. (c)The way of proving the Fierz identity uL(p1)uL(p2)ab= 2uL(p2) uL(p1) + uR(p1) uR(p2)ab(33) has been indicated in the book. The right hand side of this identity, as a Dirac matrix, which we
3、denoted by M, can be written as a linear combination of 16 matrices listed in Problem 3.6. In addition, it is easy to check directly that M = M5. Thus M must have the form M = ( 1 5 2 ) V + ( 1 + 5 2 ) W. Each of the coeffi cients V and Wcan be determined by projecting out the other one with the aid
4、 of trace technology, that is, V = 1 2 tr (1 5 2 ) M= uL(p1)uL(p2),(34) W= 1 2 tr (1 + 5 2 ) M= uR(p2)uR(p1) = uL(p1)uL(p2).(35) The last equality follows from (32). Substituting V and W back, we fi nally get the left hand side of the Fierz identity, which fi nishes the proof. 5 Notes by Zhong-Zhi X
5、ianyuSolution to P&S, Chapter 5 (draft version) (d)The amplitude for the process at leading order in is given by iM = (ie2) uR(k2)uR(k1) i s vR(p1)vR(p2).(36) To make use of the Fierz identity, we multiply (33), with the momenta variables changed to p1 k1and p2 k2, by v R(p1) a and v R(p2) b, and al
6、so take account of (32), which leads to uR(k2)uR(k1) vR(p1)vR(p2) = 2 vR(p1)uL(k2) uL(k1)vR(p2) + vR(p1)uR(k1) uR(k2)vR(p2) = 2s(p1,k2)t(k1,p2).(37) Then, |iM|2= 4e4 s2 |s(p1,k2)|2|t(k1,p2)|2= 16e4 s2 (p1 k2)(k1 p2) = e4(1 + cos)2,(38) and d d (e+ Le R + L R) = |iM|2 642Ecm = 2 4Ecm (1 + cos)2.(39)
7、It is straightforward to work out the diff erential cross section for other polarized pro- cesses in similar ways. For instance, d d (e+ Le R + R L) = e4|t(p1,k1)|2|s(k2,p2)|2 642Ecm = 2 4Ecm (1 cos)2.(40) (e)Now we recalculate the Bhabha scattering studied in Problem 5.2, by evaluating all the pola
8、rized amplitudes. For instance, iM(e+ Le R e+ Le R) = (ie)2 uR(k2)uR(k1) i s vR(p1)vR(p2) uR(p1)uR(k1) i t vR(k2)vR(p2) = 2ie2 s(p1,k2)t(k1,p2) s s(k2,p1)t(k1,p2) t .(41) Similarly, iM(e+ Le R e+ Re L) = 2ie 2t(p1,k1)s(k2,p2) s ,(42) iM(e+ Re L e+ Le R) = 2ie 2s(p1,k1)t(k2,p2) s ,(43) iM(e+ Re L e+
9、Re L) = 2ie 2 t(p1,k2)s(k1,p2) s t(k2,p1)s(k1,p2) t ,(44) iM(e+ Re R e+ Re R) = 2ie 2t(k2,k1)s(p1,p2) t ,(45) iM(e+ Le L e+ Le L) = 2ie 2s(k2,k1)t(p1,p2) t .(46) Squaring the amplitudes and including the kinematic factors, we fi nd the polarized diff erential cross sections as d d (e+ Le R e+ Le R)
10、= d d (e+ Re L e+ Re L) = 2u2 2s ( 1 s + 1 t )2 ,(47) 6 Notes by Zhong-Zhi XianyuSolution to P&S, Chapter 5 (draft version) d d (e+ Le R e+ Re L) = d d (e+ Re L e+ Le R) = 2 2s t2 s2 ,(48) d d (e+ Re R e+ Re R) = d d (e+ Le L e+ Le L) = 2 2s s2 t2 .(49) Therefore we recover the result obtained in Pr
11、oblem 5.2: d d (e+e e+e) = 2 2s t2 s2 + s2 t2 + u2 ( 1 s + 1 t )2 .(50) 4Positronium lifetimes In this problem we study the decay of positronium (Ps) in its S and P states. To begin with, we recall the formalism developed in the Peskin & Schroeder that treats the problem of bound states with nonrela
12、tivistic quantum mechanics. The positronium state |Ps, as a bound state of an electron-positron pair, can be represented in terms of electron and positrons state vectors, as |Ps = 2MP d3k (2)3 (k)Cab 1 2m|e a(k) 1 2m|e+ b (k),(51) where m is the electrons mass, MPis the mass of the positronium, whic
13、h can be taken to be 2m as a good approximation, a and b are spin labels, the coeffi cient Cabdepends on the spin confi guration of |Ps, and (k) is the momentum space wave function for the positronium in nonrelativistic quantum mechanics. In real space, we have 100(r) = (mr)3 exp(mrr),(52) 21i(r) =
14、(mr/2)5 xiexp(mrr/2).(53) where mr= m/2 is the reduced mass. Then the amplitude of the decay process Ps 2 can be represented in terms of the amplitude for the process e+e 2 as M(Ps 2) = 1 m d3k (2)3 (k)CabcM(e a(k)e + b (k) 2).(54) We put a hat on the amplitude of e+e 2. In the following, a hatted a
15、mplitude always refer to this process. (a)In this part we study the decay of the S-state positronium. As stated above, we have to know the amplitude of the process e+e 2, which is illustrated in Figure 3 with the B replaced with , and is given by icM = (ie)2 (p1) (p2) v(k2) i(/ k1/p1+ m) (k1 p1)2 m2
16、 + i(/ k1/p2+ m) (k1 p2)2 m2 u(k1),(55) where the spinors can be written in terms of two-component spinors and in the chiral representation as u(k1) = (k 1 k 1 ) ,v(k2) = ( k 2 k2 ) .(56) 7 Notes by Zhong-Zhi XianyuSolution to P&S, Chapter 5 (draft version) We also write as: = ( 0 0 ) , where = (1,i) and = (1,i) with ithe three Pauli matrices.Then the amplitude can be brought into the f
