 Solution

ID: past-exam-of-the-particle-and-nuclear-physics-course-of-the-university-of-cambridge/2022/section-b/5/c/solution

Label the identical strange quarks as 1 and 2 and the up quark as 3. In the ground spatial state, the

ss

pair has symmetric space and flavour factors but an antisymmetric colour factor. The Pauli exclusion principle therefore requires a symmetric spin factor: the pair has

S_{ss} = 1

S_{ss}^{2} = (S_{1} + S_{2})^{2} = S_{1}^{2} + S_{2}^{2} + 2 S_{1} \cdot S_{2}

gives

⟨ S_{1} \cdot S_{2} ⟩ = \frac{1}{2} [1 (1 + 1) - \frac{3}{4} - \frac{3}{4}] = \frac{1}{4} .

Next couple

S_{ss}

S_{u}

, where

S_{u} = 1/2

. For total

J

⟨ S_{u} \cdot S_{ss} ⟩ = \frac{1}{2} [J (J + 1) - S_{u} (S_{u} + 1) - S_{ss} (S_{ss} + 1)] .

For the spin-

1/2

Xi baryon, this is

(3/4 - 3/4 - 2) /2 = - 1

. By symmetry the two

u s

- 1/2

M_{u ss, J = 1/2} = m_{u} + 2 m_{s} + A^{'} (\frac{1}{4 m _{s}^{2}} - \frac{1}{m _{u} m _{s}}) .

For the spin-

3/2

state, the same coupling gives

(15/4 - 3/4 - 2) /2 = 1/2

. Each

u s

pair correlation is therefore

1/4

, as expected for a fully symmetric spin state. Hence

M_{u ss, J = 3/2} = m_{u} + 2 m_{s} + A^{'} (\frac{1}{4 m _{s}^{2}} + \frac{1}{2 m _{u} m _{s}}) .

The predicted splitting is a useful independent check:

M_{u ss, J = 3/2} - M_{u ss, J = 1/2} = \frac{3 A ^{'}}{2 m _{u} m _{s}} > 0.

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