Lambda baryon decays

  • #1
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I've been asked to find the ratio between the cross sections of the two folowing decais:
lamdec.gif


Using the CKM matrix and the feynman diagrams for both decays, I reach the conclusion that the Ratio is exactly 1. However, consulting this document,

http://pdg.lbl.gov/2012/tables/rpp2012-tab-baryons-Lambda.pdf

We clearly see it is not. I do not understand where the problem lies.

My reasoning is the following:
- The decays are exactly the same, except for the final arrangement of the quarks
- Since the vertices are the same, the probability of each interaction is the same in both cases.
- Since the probabilities are the same, the interaction rates are the same.
- Since the mass of the products is roughly equal in both cases, the density of final states is the same
- The cross section is the same.
 

Answers and Replies

  • #2
Vanadium 50
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Two words: Clebsch-Gordon
 
  • #3
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Two words: Clebsch-Gordon
Are you talking about the fact that the neutral pion is in fact a superposition of two mesons? I thought about that, but I do not understand how I can insert that into Fermi's golden rule and obtain the experimental result.
 
  • #4
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Ok, after searching for a couple hours I think I found it:

We have a final state with I=1/2 and I3=-1/2. This state is a linear combination of |n, pi0)=|1/2,-1/2)x|1,0) and |p, pi-)=|1/2,1/2)x|1,-1). Therefore, to find the ratio between branches we just need to find a ratio between the coefficients of the linear combination which are the Clebsch-Gordan coefficients.

Is that it?
 
  • #6
Vanadium 50
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Got me on that one!
 
  • #7
samalkhaiat
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Is that it?
Yes, provided you know what all that has to do with the weak decay. Once you “choose” the final state, [itex]|N \pi \rangle[/itex], to be [itex]|1/2 , -1/2\rangle[/itex], you are done: The Clebsch-Gordan expansion gives you
[tex]| p \pi^{-} \rangle = \sqrt{1/3} \ | \frac{3}{2} , - \frac{1}{2} \rangle - \sqrt{2/3} \ |\frac{1}{2} , - \frac{1}{2} \rangle ,[/tex]
[tex]|n \pi^{0} \rangle = \sqrt{2/3} \ | \frac{3}{2} , - \frac{1}{2} \rangle + \sqrt{1/3} \ |\frac{1}{2} , - \frac{1}{2} \rangle .[/tex]
Solving these for “the final state”, you get
[tex]|\frac{1}{2} , -\frac{1}{2} \rangle = -\sqrt{2/3} \ |p \pi^{-} \rangle + \sqrt{1/3} \ |n \pi^{0} \rangle .[/tex]
From this you obtain the result
[tex]\frac{\sigma \left( \Lambda \to p\pi^{-}\right)}{\sigma \left( \Lambda \to n\pi^{0}\right)} = \frac{(-\sqrt{2/3})^{2}}{(\sqrt{1/3})^{2}} = 2 ,[/tex] which agrees with the experimental data very well.
Okay, now here are my questions to you:
1) In the final state [itex]|N\pi \rangle[/itex], the nucleon has [itex]I_{N}=\vec{1/2}[/itex] and the pion has [itex]I_{\pi} = \vec{1}[/itex]. Therefore, the final state must have [itex]I_{N\pi} = \vec{3/2}[/itex] or [itex]\vec{1/2}[/itex]. So, what made you “choose” [itex]|\frac{1}{2} , -\frac{1}{2} \rangle[/itex] to be the final state? What is wrong with the state [itex]|\frac{3}{2} , -\frac{1}{2} \rangle[/itex]?
2) Closely related to (1) is the question about the initial state [itex]|\Lambda \rangle[/itex], which we know it is an iso-singlet [itex]|0 , 0\rangle[/itex]: In order to find [itex]\sigma (\Lambda \to N\pi)[/itex], we need to evaluate the matrix elements [tex]T(\Lambda \to N\pi) = \langle N \pi | \mathcal{H}^{|\Delta S| = 1}| \Lambda \rangle ,[/tex] where [itex]\mathcal{H}^{|\Delta S| = 1}[/itex] is the low energy (non-leptonic) weak Hamiltonian. So, where is the initial state in the above mentioned analysis? Why was there no mention of the initial state [itex]|\Lambda \rangle[/itex]?
If you know the answers to those questions, you are okay. Otherwise, you will have to ask better questions.
 
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