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I Why are delta resonances close in width

  1. Feb 3, 2017 #1
    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. jcsd
  3. Feb 3, 2017 #2

    Orodruin

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    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.
     
  4. Feb 3, 2017 #3

    Orodruin

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    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.
     
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