Name: Lingxiao Xu Date: 12/10/14-04:05:45 AM Z
Hello, Kasi
The mail gateway seems not work well enough at this moment, my
response via e-mail, which is sent about an hour ago, is still not on
the mailing list. So let me reply here again, sorry for replying again
here.
As far as I concerned, there are basically two ways to evaluate any
process with explicit Fermion spin.
First, you can just write down the amplitude by hand and make it
to matrix elements squared further, when summing over spins, you just
replace the spinor part of explicit spin with
(DiracSlash[p1]-m).(1+GA[5].DiracSlash[sm]) in
your case. Here p1 is the 4-momentum of muon in center of mass frame and
sm is the spin vector of muon.
Second, you can work with amplitude directly by inserting
(1+GA[5].DiracSlash[sm])/2 after
SpinorVBar[p1,mm] and then sum over spins “normally”, you will
get the exact same answer.
Regards,
Lingxiao Xu
Here is my code for muon decay with explicit muon and electron spins
which I’ve done not far before.I post it directly here now.I hope it
will help since it is almost the same case as yours.
(*reduced matrix elemnet calculated by hand;allow polarization of \
muon and electron, since many experiment can measure it*)
{p1sl, p2sl, p3sl, p4sl, smsl, sesl} =
DiracSlash /@ {p1, p2, p3, p4, sm, se};
t13 = 1/2 GA[a].(1 - GA[5]).(p1sl + m).(1 +
GA[5].smsl).GA[
b].(1 - GA[5]).p3sl
t24 = 1/2 GA[a].(1 - GA[5]).p2sl.GA[
b].(1 - GA[5]).(p4sl + me).(1 + GA[5].sesl)
Msq = Subscript[G, F]^2/2 TR[t13] TR[t24] //
Contract // Simplify
% /. me -> 0(*in the case that the electron mass is neglected*)
1/2 \[Gamma]^a.(1-\[Gamma]^5).(m+\[Gamma]\[CenterDot]p1).(\[Gamma]^5.(\[Gamma]\[CenterDot]sm)+1).\[Gamma]^b.(1-\[Gamma]^5).(\[Gamma]\[CenterDot]p3)
1/2 \[Gamma]^a.(1-\[Gamma]^5).(\[Gamma]\[CenterDot]p2).\[Gamma]^b.(1-\[Gamma]^5).(me+\[Gamma]\[CenterDot]p4).(\[Gamma]^5.(\[Gamma]\[CenterDot]se)+1)
32 Subsuperscript[G, F, 2] (p1\[CenterDot]p2-m p2\[CenterDot]sm) (p3\[CenterDot]p4-me p3\[CenterDot]se)
32 Subsuperscript[G, F, 2] p3\[CenterDot]p4 (p1\[CenterDot]p2-m p2\[CenterDot]sm)
(*on the other hand, we can obtain the same answer from amplitude \
directly, we can proof an important identity:
1/2
(GA[5].smsl+1).(m+p1sl).(1+GA[5].smsl)=(m+p1sl).(1+GA[5].smsl),
\
so that Pmus is like some kind of projection operator in order to fix
\
the spin orientation of the fermion.*)
Pmus = (1 + GA[5].smsl)/2;
Pes = (1 + GA[5].sesl)/2;
{ScalarProduct[se, se] = ScalarProduct[sm, sm] = -1,
ScalarProduct[p1, sm] = ScalarProduct[p4, se] = 0};
M = -I 2 Sqrt[2] Subscript[G, F]
SpinorUBar[p3,
0].GA[a].GA[7].Pmus.SpinorU[p1, m]
SpinorUBar[p4,
me].Pes.GA[a].GA[7].SpinorV[p2, 0] //
FCI
Mstar = ComplexConjugate[M] /. a -> b
Msqprime =
FermionSpinSum[M Mstar] /. DiracTrace -> TR // Contract //
Simplify
% /. me -> 0
-2 I Sqrt[2] Subscript[G, F] \CurlyPhi.\[Gamma]^a.\[Gamma]^7.(1/2 (\[Gamma]^5.(\[Gamma]\[CenterDot]sm)+1)).\CurlyPhi \CurlyPhi.(1/2 (\[Gamma]^5.(\[Gamma]\[CenterDot]se)+1)).\[Gamma]^a.\[Gamma]^7.\CurlyPhi
(I Subscript[G, F] \CurlyPhi.(1-(\[Gamma]\[CenterDot]sm).\[Gamma]^5).\[Gamma]^6.\[Gamma]^b.\CurlyPhi \CurlyPhi.\[Gamma]^6.\[Gamma]^b.(1-(\[Gamma]\[CenterDot]se).\[Gamma]^5).\CurlyPhi)/Sqrt[2]
32 Subsuperscript[G, F, 2] (p1\[CenterDot]p2-m p2\[CenterDot]sm) (p3\[CenterDot]p4-me p3\[CenterDot]se)
32 Subsuperscript[G, F, 2] p3\[CenterDot]p4 (p1\[CenterDot]p2-m p2\[CenterDot]sm)