The Eightfold Way — Question 5

Section 44-6 · Physics pages 7400–7401

Question 5 The missing corner of a decuplet predicts three additive numbers A spin-3/2 baryon decuplet has rows of isospin 3/2, 1, 1/2 and 0 at Y=1,0,-1,-2. All states have B=1. Suppose only the last row is unobserved. Y I3 ∆− ∆0 ∆+ ∆++ Σ∗− Σ∗0 Σ∗+ Ξ∗− Ξ∗0 Ω− Light-flavor weight diagram: horizontal coordinate I3, vertical coordinate Y = B + S; axis origins are offset for clarity. (a) Count the states and infer S and Q of the missing member. (b) Distinguish the missing state total isospin from its spatial spin and give each multiplicity.
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Question 5 - Solution The missing corner of a decuplet predicts three additive numbers Y I3 ∆− ∆0 ∆+ ∆++ Σ∗− Σ∗0 Σ∗+ Ξ∗− Ξ∗0 Ω− (a) Method and calculation N = 4 + 3 + 2 + 1 = 10 . S = Y −B = −3, Q/e = 0 −2/2 = −1 . (b) Interpretation and further calculation I = 0 : 2I + 1 = 1 flavor state . J = 3/2 : 2J + 1 = 4 spin projections . Physical check The predicted corner has the quantum numbers of the Omega minus. Four spin projections do not create four separate weights in this diagram.

Question 5

The missing corner of a decuplet predicts three additive numbers

A spin-3/2 baryon decuplet has rows of isospin 3/2, 1, 1/2 and 0 at Y=1,0,-1,-2. All states have B=1. Suppose only the last row is unobserved.

Diagram shown in the original SVG worksheet.

  1. Count the states and infer S and Q of the missing member.

  2. Distinguish the missing state total isospin from its spatial spin and give each multiplicity.

Question 5 - Solution

The missing corner of a decuplet predicts three additive numbers

Diagram shown in the original SVG worksheet.

(a) Method and calculation

N=4+3+2+1=10.\boxed{N=4+3+2+1=10}.S=Y−B=−3,Q/e=0−2/2=−1.\boxed{S=Y-B=-3,\quad Q/e=0-2/2=-1}.

(b) Interpretation and further calculation

I=0:2I+1=1flavor state.\boxed{I=0:\ 2I+1=1\ \text{flavor state}}.J=3/2:2J+1=4spin projections.\boxed{J=3/2:\ 2J+1=4\ \text{spin projections}}.

Physical check

The predicted corner has the quantum numbers of the Omega minus. Four spin projections do not create four separate weights in this diagram.