Simple explanation for cross section ratio

[kelly0303]

Hello! Is there any simple (i.e. using some physics arguments, without actually doing the math) explanation for why $$\sigma(pp \to \pi^+d)/\sigma(np\to\pi^0d)=2$$ where d is the deuteron? Thank you!

 

[Gaussian97]

Well, what do you mean with "no math" there is a way to deduce using isospin symmetry and little assumptions. If you compare the amount of math of this method compared with the actual computation of a cross-section, this is almost without maths, but you still need some mathematical knowledge of the representations of SU(2) group. Is that what you want?

 

[kelly0303]

Well, what do you mean with "no math" there is a way to deduce using isospin symmetry and little assumptions. If you compare the amount of math of this method compared with the actual computation of a cross-section, this is almost without maths, but you still need some mathematical knowledge of the representations of SU(2) group. Is that what you want?

Thank you! yes, i would like something like that, based on some symmetries, rather than actually blindly calculating some feynman diagrams.

 

[Gaussian97]

Well, then the idea is:
Because the two scatterings are almost identical (they only differ in the exchange of $p\leftrightarrow n$ and $\pi^0 \leftrightarrow \pi^+$), then we expect that the cross-sections will be like
$$\sigma(pp\rightarrow \pi^+ d) = \sigma_0 |\mathscr{M_{pp}}|^2$$
$$\sigma(pn\rightarrow \pi^0 d) = \sigma_0 |\mathscr{M_{pn}}|^2$$
with $\sigma_0$ the same constant for both processes. Also, the amplitude will be $\mathscr{M}_{i\rightarrow f}= \left<f\right|T\left|i\right>$. Again, if isospin is a symmetry we expect
$$\mathscr{M}_{pp\rightarrow \pi^+ d} = \mathscr{M}_0 \left<\pi^+ d\right.\left|pp\right>$$
$$\mathscr{M}_{pn\rightarrow \pi^0 d} = \mathscr{M}_0 \left<\pi^0 d\right.\left|pn\right>$$
So, the ratio between the two cross-sections will be simply
$$\frac{\sigma(pp\rightarrow \pi^+ d)}{\sigma(pn\rightarrow \pi^0 d)}=\left|\frac{\left<\pi^+ d\right.\left|pp\right>}{\left<\pi^0 d\right.\left|pn\right>}\right|^2$$.

Then, using that
$$\left|p\right>=\left|I=\frac{1}{2}, I_3 =\frac{1}{2} \right>$$
$$\left|n\right>=\left|I=\frac{1}{2}, I_3 =-\frac{1}{2} \right>$$
$$\left|\pi^+\right>=\left|I=1, I_3 =1 \right>$$
$$\left|\pi^0\right>=\left|I=1, I_3 =0 \right>$$
$$\left|d\right>=\left|I=0, I_3 =0 \right>$$
I let you compute the $2$