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## Homework Statement

I am trying to derive the mass of [itex]\Xi[/itex] using the formula:

[tex]M\left(baryon\right)=m_1 + m_2 + m_3 + A' \left[\frac{S_1 \cdot S_2}{m_1 m_2} +\frac{S_1 \cdot S_3}{m_1 m_3} + \frac{S_2 \cdot S_3}{m_2 m_2\3}\right][/tex]

## Homework Equations

We have:

[tex] S_1 \cdot S_2 + S_1 \cdot S_3 + S_2 \cdot S_3 = \frac{\hbar^2}{2}\left[j\left(j+1\right)-2/4\right] = -3/4 \hbar[/tex] for octet

and also:

[tex]\left(S_u+S_d\right)^2 = S_u^2 + S_d^2 + 2S_u \cdot S_d [/tex]

## The Attempt at a Solution

What I don't get it the last equation. In the case of [itex] \Sigma [/itex], is equal to [itex]2\hbar^2[/itex] because the isospin is 1 (and therefore [itex]S_u \cdot S_d = \hbar^2 /4[/itex]. Following the pattern, since the isospin of [itex]\Xi[/itex] is 1/2, I tried to figure out [itex]S_s \cdot S_s[/itex] which is needed, since the quark content for [itex]\Xi[/itex] is uss. I got [itex] S_s \cdot S_s = -3/8\hbar^2[/itex] which doesn't give the right answer.

The right answer should be:

[tex]M_\Xi = 2*m_s + m_u + \frac{\hbar^2}{4}A'\left(\frac{1}{m_s^2}-\frac{4}{m_u m_s}\right) [/tex]

Can someone help me? thanks!

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