Answers and Comments on the Exercises in Java

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Answers and Comments on the Exercises
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If(TIu + dd)if, If (llU - dd)1f (j!
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DO -+ K - '17+ due to c -+ s(du), DO -+ '17- '17+ due to c -+ d(du), with, in each case, II as a spectator. In Section 12.11, we see that the c -+ d transmutation is "Cabibbo-suppressed" in comparison to c -+ s.
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2.23 The hadronic state must have I = 0 and C = + 1. The dominant hadronic decay modes are observed to be 1/1' -+ 1/1'17+'17-,1/1'17 '17 , 1/1T/ with branching ratios 33, 17, and 3%, respectively. 2.24 Radiative transitions occur between states of opposite C parity; the (lil = 1) transitions are E l or M l according to whether the relative parity of the levels
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is odd or even. The approximate equality of the 1/1' -+ Xy branching ratios is due to the balancing compensation of the 21 + 1 and phase space (k 3 ) factors which occur in the E l transition probability formula.
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f(qq -+ e+e-) is proportio~al to Ip) and Iw) are (Ull =+= dd)/V2, respectively. Since the Ull and dd annihilations are indistinguishable, we must add amplitudes; and so
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2.26 2.27
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See, for example, Close (1979). See, for example, Close (1979).
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2.28 The strong ~:+-+ A:w+ decay, followed by the weak decays A: -+ Aw+w+w- and A -+ pw-. 2.29 The eigenvalues of
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0 1 '02
are - 3 and + 1 for S
m m Q
0 and 1, respectively.
~(81'82 +
(8 1 + 8 2).83),
Center-of-Mass Frame: PI
(1Ecm ' -p)
Laboratory Frame: PI = (E 1ab 'Plab)'P2 = (M,O) . .. s _ ( E 1ab + M )2 - (2 - M 2) E 1ab Advantages of Fixed Target: higher flux because of the density of the target;
Answers and Comments on the Exercises
choice of reaction 'lTp, Kp, 'lTd, and so on; need detectors over less solid angle about the interaction point and hence a less expensive experiment.
3.5 (a) Let E and E' be the energies of the outgoing e- and e+, respectively; and so the matrix element contains
CHAPTER 4 4.1 To satisfy the periodic boundary conditions, the allowed values of Px are such that Lpx = 2'ITn, where n is an integer. Hence, the number of allowed states in the range Px to Px + dpx is L dPx/2'IT. 4.2 From (4.30), we have
3 3 _ 1 d Pe d PD (4)( ) dQ - - 4 2E ~E~ 2 PA + PB - Pc - PD 'IT e D
1 d3Pe1 4'IT 2 2E 2E ~(EA + EB - Ee - ED)' e D
In the center-of-mass frame
== W = E A + E B ),
= -2
pJ dPl drl 4E E ~ (w - E e - ED)' 4 'IT e D
2 ( me 2 + PI2)1/2 + (m D -t PI2)1/2 '
we have
E e + ED
W dd = PI( + 'PI e
Substituting for dPI' we find
dQ dQ
1 -4 ( E 1 E ) dWdrl~(W - E - ED) PI e 4 'IT e + D
_1_ PI drl.
4'IT 2 41S
Answers and Comments on the Exercises
From (4.32),
Neglecting masses, we may write in the center-of-mass frame
PA = (p,p), Pc
PB = (p, PD = (p,
q = PD - PB'
where P = Ipl = and use (4.35). 4.5
IP'I. Substitute these values into (4.18), with
Using (4.43) and p
Pc - PD) =
Lm; + 2pl + 2PA '(PB PB = (E, -k;),
If e-e--+ e-e- is the s channel process A
PA = (E,k;),
+ B -+ C + D, then
PD = (E, -k f )
Pc = (E,k f ),
where E
m 2 )1/2. So, for example,
since k;.k f = we have t ~ 0 and u
t = (PA - PC)2 = -(k; - k f )2 = -2k 2(1 - cosO) k2 cos 0. As k 2 ~ 0, we have s ~ 4m 2; and since -1 ~ cos 0 ~ 1,
4.7 If we keep PA'" ,PD defined as for the s channel AB -+ CD process, then for AD -+ CB in the center-of-mass frame,
PA = (E,k;), -PD = (E, -k;),pc = (E,k f ), -PB = (E, -k f ),
where k;, -k;,k f , -k f are the three-momenta of A, D, C, and B, respectively. The required results now follow upon using (4.43). 4.8
See, for example, Martin and Spearman (1970), 4, Section 3.
+ Pc)
+ PB)
+ PB
- PD) '(PD
+ PB)
-2PA '(PD
+ PB)
u - s.
See, for example, Aitchison (1972), 8, Section 1.
0= yVJv(iyl'JI' - m)I/J
+ yl'yv)avJI'I/J - myvJA
+ im 1/J
+ m )1/J.
Answers and Comments on the Exercises
5.3 The momentum p' = (p sin 0,0, p cos 0) is obtained from p = (0,0, p) by a rotation through 0 around the y axis. The required helicity eigenspinor u(p') may therefore be obtained from u(1)(p) of (5.27) using (2.26): cos "2
o - I02 sm "2 : . . 0:
O : 0 : cos "2