Problem Set 4

Problem 4

Please note:  As stated in the problem, the ratio is zero.  By CPT, a particle and it's anti particle must have the same decays.  The problem should have been the decays of the B0 and the B0bar to the same charged states.

Form the basis vectors for B0 and B0 bar:

MatrixForm[Bzero  =  {1, 0}]MatrixForm[Bzerobar  =  {0, 1}]

Form B1 and B2:

MatrixForm[B1 = 1/Sqrt[2] ( Bzero  +  Bzerobar)]

MatrixForm[B2 = 1/Sqrt[2] ( Bzero  -  Bzerobar)]

For Bl and Bs similar to Kl and Ks:

MatrixForm[
      Simplify[Bl  =  1/Sqrt[1 + e^2] ( B2  +  e Exp[I phi] B1)]]

MatrixForm[
      Simplify[Bs  =  1/Sqrt[1 + e^2] ( B1  +  e Exp[I phi] B2)]]

Find the linear combination of Bl and Bs that make up the B0 (remember that this is unnormalized):

coeffb0  =  
      Solve[Bzero  ==  a Bl  +  b Bs,  {a, b}][[1]]

Check B0 state:

checkb0 = Simplify[a Bl  +  b Bs /. coeffb0]

Make the time dependent B0 wave function:

MatrixForm[
      b[t_]  =  
        Simplify[
          a Bl Exp[-gl t  +  I ml t]  +  
              b Bs Exp[-gs t  +  I ms t] /. coeffb0]]

Bo/B0 bar parameters:

parameters  =  {gl -> g, gs -> g,  ml -> ms + dm}

Make the B0 decay matrix element for mu+ (the square of the top component of the vector):
	Play with result to make it look good (ie. real)

aa = 
      ((b[t][[1]] /. {I -> -I, -I -> I})*
            b[t][[1]] /. parameters)

bb = 
      Simplify[aa[[2]] /. 
          Power[E, 
              Times[-1, Plus[a_, Times[Complex[0, -1], b_]], c_]] -> 
            Exp[-a c] (Cos[b c] + I Sin[b c])]

cc = 
      Simplify[aa[[3]] /. 
          Power[E, Times[-1, Plus[a_, Times[Complex[0, 1], b_]], c_]] -> 
            Exp[-a c] (Cos[b c] - I Sin[b c])]

gwb[t_]  = Simplify[aa[[1]]*bb*cc]

Do likewise for B0bar:

coeffb0bar  =  
      Solve[Bzerobar  ==  a Bl  +  b Bs,  {a, b}][[1]]

MatrixForm[
      bbar[t_]  =  
        Simplify[
          (a Bl Exp[-gl t  +  I ml t]  +  
                b Bs Exp[-gs t  +  I ms t]) /. coeffb0bar]]

The decay of B0bar to mu+ (wrong sign) is the top element squared (again, play with result):

aa = 
      (bbar[t][[1]] /. {I -> -I, -I -> I})*
          bbar[t][[1]] /. parameters

bb = Simplify[ExpToTrig[aa[[2]] aa[[4]]]]

cc = Simplify[ExpToTrig[aa[[3]] aa[[5]]]]

dd = 
      Simplify[aa[[6]] /. 
          Exp[(a_  -  I b_) c_] -> 
            Exp[a c] (Cos[b c] -  I Sin[b c])]

ee = 
      Simplify[aa[[7]] /. 
          Exp[(a_  +  I b_) c_] -> 
            Exp[a c] (Cos[b c] +  I Sin[b c])]

gwbb[t_] = Simplify[aa[[1]]*bb*cc*dd*ee]

Calculate the time dependent ratio (not asked for in this problem):

ratio  = Simplify[(gwb[t] - gwbb[t])/(gwb[t] + gwbb[t])]

Integrate decay rates to get total width:

intb0 = Integrate[gwb[t], {t, 0, Infinity}][[2]]

intb0bar = Integrate[gwbb[t], {t, 0, Infinity}][[2]]

Ratio given in problem doesn't make sense for this problem.  A better ratio is the wrong signs from B0bar/right signs from B0.

ratio = Simplify[(intb0bar)/(intb0)]

Assume e is small so that we can ignore e^2:

ratio1  =  ratio /. {e^2 -> 0,  Cos[phi] ->  reale/e}

Using the parameters given in Problem 4:

para1 = {g -> 1. ,  
        dm -> 3.1*10^(-10)*1.56*10^(-12)/
              (6.58 * 10^(-22))}

ratio2 = (ratio1 /. para1)

Note: If there were no CP violation, ratio would be .0632.  To reduct this, Re[e] must be negative and of the order of 1.

Solve[ratio2 ==  .001,  reale]

We suspect that e will be small and that no deviation from the expected mixing will be seen.  (Note, we can observe this with the kaon system because we can produce a pure Kl state and see its charge asymmetry.  Here we cannot produce the Bl state because gl=gs.)