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I need to find the ration of cross sections for the following two reactions:

$$\rm p + d\rightarrow He^3 + \pi^0$$ and $$\rm p+d\rightarrow H^3 + \pi ^+.$$

Since the cross section for a reaction $\sigma \propto |M_{if}|^2$ = $|\langle \psi_f|A_{if}|\psi_i\rangle|^2$, where $A_{if}$ is the isospin operator, I need to find the "isospin wave function" of the initial and final states.

I have a problem computing the isopsin of $\rm He^3 $and $\rm H^3 $.

  • Isospin of the initial state : $I=1/2$ and $I_3=1/2$
  • Isopsin of $\pi^0 : I(\pi^0)=1, I_3(\pi^0)=0$ and $\pi^+ : I(\pi^+) =1, I_3 (\pi^+)=+1$.

If I follow the principle of conservation of isospin and use the formula for the charge of particles $Q=I_3 + \frac{1}{2} Y$ then this must hold : $I_3(\mathrm{He}^3)=1/2$ and $I_3(\mathrm{H}^3)=-1/2$ but I can't figure out the "sizes" of isospins $I(\rm He^3) $ and $I(\rm H^3)$, they can both have $I=1/2$ or $I=3/2$...

rob
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Luka8281
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  • Your two mirror nuclei are members of the isodoublet, I =1/2. – Cosmas Zachos Jun 19 '17 at 20:03
  • So $He^3$ which is composed of 2 protons and 1 neutron has a $T_3$ value of $\frac{1}{2} + \frac{1}{2} -\frac{1}{2}= \frac{1}{2}$ , $T$ can take values $\frac{1}{2} $ and $\frac{3}{2}$ so the $T$ value has to be $\frac{1}{2}$, because it's the smallest allowed value that coresponds to $T_3=\frac{1}{2}$ ??? – Luka8281 Jun 19 '17 at 20:18
  • Yes, it is in an isodoublet with tritium, not in an isoquartet..... – Cosmas Zachos Jun 19 '17 at 20:26
  • Well, For an isoquartet, you'd need a neutronless Li isotope and a 3-neutron state on its ends, states that aren't there... – Cosmas Zachos Jun 19 '17 at 22:41

1 Answers1

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There are a few different ways to think about this, which have different levels usefulness for this problem and for similar problems you may encounter later.

  1. Isospin is a symmetry of the strong interaction --- if your initial state has total isospin $I=\frac12$, then so does your final state.

  2. Strong states with the same total isospin tend to have similar properties, even though their electric charges are different. Generally you talk about this in nuclear physics with mirror nuclei. But one of the arguments that the deuteron has zero isospin is that the other partners in the $I=1$ triplet, the $\rm nn$ and $\rm pp$, are not bound. How many bound $A=3$ states are there?

  3. You can use the heuristic that light nuclei typically have the smallest allowed value for the total isospin.

  4. This point was going to be "you can look them up," with a link to the NNDC, but today I can't trivially find ground-state isospin information.

rob
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  • I was already familiar with the points 1. and 2. ... But you answered my question with points 3. and 4. !!! Thank you very much! – Luka8281 Jun 19 '17 at 20:14
  • So $He^3$ which is composed of 2 protons and 1 neutron has a $T_3$ value of $\frac{1}{2} + \frac{1}{2} -\frac{1}{2}= \frac{1}{2}$ , $T$ can take values $\frac{1}{2} $ and $\frac{3}{2}$ so the $T$ value has to be $\frac{1}{2}$, because it's the smallest allowed value that coresponds to $T_3=\frac{1}{2}$ ??? – Luka8281 Jun 19 '17 at 20:17