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To: "M. Jennifer Markus"
cc: murayama@lbl.gov
Subject: Re: Me, again
Reply-to: murayama@lbl.gov
In-reply-to: Your message of "Sun, 23 Nov 1997 10:25:42 PST."
<2.2.16.19971123102609.24474ea4@uclink4.Berkeley.edu>
Date: Sun, 23 Nov 1997 15:56:35 -0800
From: Hitoshi Murayama
> -- using template mhl.format --
> Thanks for the message. It did help.
Good!
> I just wanted to run a couple of things by you (I am going to spell out the
> Greek letters, so I hope this is clear):
>
> On number 2: In the typed notes you gave us Gamma(i ->f) is equal to the
> integral of (1/2M)d(sigma)(Feynman amplitude)squared. So when I solve for
> d(sigma)(Feynman amplitude)squared, I just multiply the Gamma(i->f) I solved
> for in number one by twice the mass of the kaon, right? But if this is
> true, my units don't seem to come out to (GeV)squared.
Gamma has a dimension of energy. The mass of the kaon has also a
dimension of energy once you set c=1. Then their product has a
dimension of energy squared.
> On number 5: I divided my answer for number 4 by my answer for number 2
> because I believe that that ratio is approximately epsilon (I realize it has
> a factor of epsilon and (epsilon)squared, but it seemed a reasonable thing
> to do because for number 6 I got an answer that was within 5% of the value
> in PDG) Was this the correct approach?
You need to think yourself on this. A hint is that epsilon is defined
as a mixture of K1 state in KL. Using the fact that it is only K1
that decays into two-pion states, write down the decay amplitude of KL
into two-pion states in terms of K1 and K2.
> And finally for number 6: I multiplied Gamma(Ks->pi(0)pi(0)) by my epsilon
> value and I took that product for the KL->pi(0)pi(0). Is that correct?
This folows whatever you do on problem 5. It is true though that you
are supposed to get the prediction consistent with the observed
branching fraction.
Hitoshi