# I Why are delta resonances close in width

1. Feb 3, 2017

### snorkack

The width of Δ resonances is around 110...120 MeV.
All four of them. With modest differences. The difference in width between Δ0 and Δ++ is estimated from 5 to 9 MeV.
Why?
Δ+ and Δ0 resonances have two options to decay.
Δ+→p+π0
Δ+→n+π+
and correspondingly
Δ0→p+π-
Δ0→n+π0
In contrast, Δ++ and Δ- have only one option each:
Δ++→p+π+
Δ-→n+π-
So, from pure consideration of statistics/phase space, shouldn't Δ++ and Δ- live twice as long as Δ0 and Δ+, with widths of 50...60 MeV range?

2. Feb 3, 2017

### Orodruin

Staff Emeritus
You are implicitly assuming that Δ+→p+π0 should have the same decay rate as Δ++→p+π+. This is not the case.

You can understand this better from a symmetry point of view. In each case you have a state in an isospin 3/2 representation decaying and so this symmetrises the final state isospin. The Δs decay into a particular isospin combination in the final state, it is just that in the case of the ++ and -, this state is a set of physical states while in the + and 0, the state is a linear combination of the physical states and so when you project out the physical states, each channel obtains a factor of 1/2.

3. Feb 3, 2017

### Orodruin

Staff Emeritus
Just to put some more detail into it.

The $\Delta$ baryons form an isospin quadruplet, meaning you can transform the $\Delta$ baryons into each other using isospin transformations. At the same time, protons and neutrons form an isospin doublet $N$ while the pions form an isospin triplet $\pi$. Having a final state containing an $N$ and a $\pi$, you can have the (strong) decay $\Delta \to N + \pi$ only if isospin is conserved. The representations in the final state are of the form $2\otimes 3 = 4 \oplus 2$, where the 4-representation occurs when the isospins are parallel and the 2-represenation occurs when the isospins are anti-parallel. Due to isospin conservation, only the decay into the 4-state is allowed. The (normalised) state in the 4-representation with a third component of 1/2 is given by $(\sqrt 2 (1/2)(0) + (-1/2)(1))/\sqrt 3$. As you can see, this is a state that is a linear combination of $p+\pi^0$ ((1/2)(0)) and $n+\pi^0$ ((-1/2)(1)).

Let me also correct myself, the branching ratios of both channels are not the same. The branching ratio of $\Delta^+$ into $p+\pi^0$ should be twice as large as that into $n+\pi^+$ based on these symmetry arguments. The $p+\pi^0$ partial width is therefore 2/3 of the $\Delta^{++} \to p + \pi^+$ width and the $n+\pi^+$ is 1/3.