Slide 2                      S.Olsen 
                                                    CUSeeMe  4/12/96 
                 psi'-->k1(1270)^+ k^- +cc  
                            and
                        k1(1400)^+ k^- + cc

Here
       k1(1400) =  K_a cos\theta  - K_b sin\theta    <--theta near 45 deg
       k1(1270) =  K_a sin\theta  + K_b cos\theta       (mixing near maximal)
and
       k_a =   k1(1400) cos\theta + k1(1270) sin\theta
       k_b =  -k1(1400) sin\theta + k1(1270) cos\theta

Br(k1(1270)-->krho) = 42%  -->k*pi 44% ;    Br(k1(1400)-->k*pi) = 92 %

psi'-->K_b k is SU(3)-allowed   psi'-->K_a k violates SU(3)

If only K_b k,   expect roughly equal k1(1270) and k1(1400)

  ANALYSIS

     -Use psi'-->k+k-pi+pi- selected events

           chisq<12.  m_pipi<2.4    1.05>m_kk<3.0
           veto events where k's are unambiguously identified as pions
                 i/e  pion_wt/(pi_wt + k_wt) <0.1

           rho cut:  |m_pipi - 0.77|<0.15
           kstr cut: |m_kpi - 0.896|<0.05

           form       k rho mass distribution        Fig 4c
                      kstr pi "      "               Fig 4f

     -Fit k rho mass distribution two ways:

      i)  WMD-inspired line shape for k1(1270)-->k rho + f_bkg (as before)
                 
          Nevts = 65.7 +/- 12.7  <--background overestimated?   Fig 5

      ii)              "                   + BW["k*(1680)"]<--Mass around 1.7
                                                              Width " 0.3
          Nevts = 86.1 +/- 12.1  <---background underestimated?  Fig 6

          use avg of i) & ii) and include difference in syst error

       -MC acceptance for k1-->k rho = 0.127 +/- 0.005
        "      "          k1-->k* pi = 0.088 +/- 0.006 <--in spite of rho cut

        *******************************************************************
        * Br(psi'-->k1(1270)+ k- +cc) = (0.64 +/- 0.10 +/- 0.13) x 10^{-3}*
        *******************************************************************
 
         PDG has no entry for J/psi