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For example in this paper (also on arxiv), the author writes:

However, for an $n \bar n$ state, $\gamma^2(K \bar K) = \frac{1}{3} \gamma^2(\pi \pi)$, ...

Where $n \bar n = \frac{1}{\sqrt2} (u\bar u + d\bar d)$ and $\gamma$ is the partial width of the decay.

Is this a straightforward calculation I don't see or does it require heavy theory?

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  • $\begingroup$ What exactly is the question? $\endgroup$ Dec 22, 2017 at 14:09

1 Answer 1

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This result follows from assuming perfect $SU(3)_{flavour}$ symmetry and the OZI rule. The theory is not very heavy but there are multiple steps.

We start with $SU(3)_{flavour}$ symmetry, i.e the assumption that the strong interaction is indifferent to the flavour of light $u$, $d$, and $s$ quarks. This symmetry is actually broken by the different masses and charges of the quarks, but it is usually not a bad approximation. (This is why the eightfold way worked so well in explaining meson states observed in the 1960s and which led to the development of the quark model.)

Following the "ideally mixed" Equation 11 of Curtis Meyer's lecture on Light and Exotic Mesons, we start by writing the $n \bar n$ state as $$n \bar n =\frac{1}{\sqrt{2}} (u\bar u + d\bar d) = \sqrt{\frac{1}{3}}f_8 + \sqrt{\frac{2}{3}}f_1 $$ in terms of the two isoscalar $SU(3)_{flavour}$ states

$$f_8=\frac{1}{\sqrt{6}}\left(u\bar u + d \bar d - 2 s \bar s\right) \textrm{ and } f_1=\frac{1}{\sqrt{3}}\left(u\bar u + d \bar d + s \bar s\right)$$

The decay amplitudes into pions and kaons are $$\gamma(n \bar n \rightarrow \pi\pi) = \sqrt{\frac{1}{3}}\gamma (f_8\rightarrow \pi\pi) + \sqrt{\frac{2}{3}}\gamma(f_1 \rightarrow \pi\pi) = \sqrt{\frac{1}{3}} C_{f_8\rightarrow \pi\pi} g_T + \sqrt{\frac{2}{3}}C_{f_1\rightarrow \pi\pi})g_1$$ $$\gamma(n \bar n \rightarrow K\bar{K}) = \sqrt{\frac{1}{3}} C_{f_8\rightarrow K\bar{K}} g_T + \sqrt{\frac{2}{3}}C_{f_1\rightarrow K\bar{K}}g_1$$ where $g_T$ and $g_1$ are coupling constants and the $C$ are $SU(3)$ Clebsch–Gordan coefficients determined by the hypercharge $Y$ and isospin $I$ of the initial and final states.

Since kaons and pions are members of an $SU(3)$ octet and the $n \bar n$ state has $(Y,I)=(0,0)$, for $n \bar n$ decays into $\pi\pi$ or $K\bar{K}$ we must look at the $(Y,I)=(0,0)$ listings of $\mathbf{8}\bigoplus\mathbf{8}$ Clebsch–Gordan coefficients in a source such as Table 8.4 of Unitary Symmetry and Elementary Particles by D. B. Lichtenberg (or we could just look at the bottom of page 9 of Meyer). Since pions, kaons, and anti-kaons have $(Y,I)=(0,1)$, $(1,\frac{1}{2})$ and $(-1,\frac{1}{2})$, we find that $$C_{f_8\rightarrow \pi\pi} = -\frac{\sqrt{15}}{5} = -\sqrt{\frac{12}{20}} \textrm{ and } C_{f_1\rightarrow \pi\pi} = \frac{\sqrt{6}}{4} = \sqrt{\frac{3}{8}}$$ $$C_{f_8\rightarrow K\bar{K}} = \frac{\sqrt{10}}{10} = \sqrt{\frac{2}{20}} \textrm{ and } C_{f_1\rightarrow K\bar{K}} = \frac{1}{2} = \sqrt{\frac{2}{8}}$$ $$C_{f_8\rightarrow \bar K K} = -\frac{\sqrt{10}}{10} = -\sqrt{\frac{2}{20}} \textrm{ and } C_{f_1\rightarrow \bar K K} = -\frac{1}{2} = -\sqrt{\frac{2}{8}}$$ (Note that all $\pi\pi$ states are included in the $(I_1,Y_1;I_2,Y_2) = (1,0;1,0)$ row in the Clebsch-Gordan table, but $K\bar K$ and $\bar K K$ are in different $(\frac{1}{2},1; \frac{1}{2},-1)$ and $(\frac{1}{2},-1; \frac{1}{2},1)$ rows, so we must calculate them separately. This is not obvious from Meyer's derivation, where the $K\bar{K}$ amplitudes at the bottom of page 12 are a factor of 2 inconsistent with the $K\bar{K}$ rates shown in his Fig. 4.)

Inserting these values for the coefficients in the $n \bar n$ decay amplitudes we have: $$\gamma(n \bar n \rightarrow \pi\pi) = -\sqrt{\frac{1}{3}} \sqrt{\frac{12}{20}} g_T + \sqrt{\frac{2}{3}}\sqrt{\frac{3}{8}}g_1 = -\sqrt{\frac{1}{5}} g_T + \frac{1}{2}g_1$$ $$\gamma(n \bar n \rightarrow K\bar{K}) = \sqrt{\frac{1}{3}} \sqrt{\frac{2}{20}} g_T + \sqrt{\frac{2}{3}}\frac{1}{2}g_1 = \sqrt{\frac{1}{30}} g_T + \sqrt{\frac{1}{6}}g_1$$ $$\gamma(n \bar n \rightarrow \bar K K) = -\sqrt{\frac{1}{30}} g_T - \sqrt{\frac{1}{6}}g_1$$ (The first two of these are just Meyer's Eqs. 21 & 22 for ideal mixing angle.)

In order to figure out the relation between $g_T$ and $g_1$, we consider the amplitude for the decay of the $$s\bar{s} = \sqrt{\frac{2}{3}}f_8 - \sqrt{\frac{1}{3}}f_1 $$ state into pions, i.e. $$\gamma(s\bar{s}\rightarrow \pi\pi) = \sqrt{\frac{2}{3}} C_{f_8\rightarrow \pi\pi} g_T - \sqrt{\frac{1}{3}}C_{f_1\rightarrow \pi\pi}g_1 =-\sqrt{\frac{2}{5}} g_T - \sqrt{\frac{1}{8}}g_1.$$ This decay can only happen if the $s\bar{s}$ quarks annihilate each other, which is suppressed according to the OZI rule. The rate is zero under the assumption of perfect OZI suppression, in which case $$\sqrt{\frac{2}{5}} g_T = -\sqrt{\frac{1}{8}}g_1 \qquad \Longrightarrow \qquad g_1=-\sqrt{\frac{16}{5}} g_T$$

From this we can calculate the $ n\bar n$ decay rates: $$\gamma^2(n \bar n \rightarrow \pi\pi) = \left(-\sqrt{\frac{1}{5}} g_T - \frac{1}{2}\sqrt{\frac{16}{5}} g_T\right)^2 = \frac{9}{5} g_T^2$$ $$\gamma^2(n \bar n \rightarrow K\bar{K}) = \left(\sqrt{\frac{1}{30}} g_T - \sqrt{\frac{1}{6}}\sqrt{\frac{16}{5}} g_T\right)^2 = \frac{9}{30} g_T^2$$ $$\gamma^2(n \bar n \rightarrow \bar K K ) = \frac{9}{30} g_T^2$$

Finally, the ratio of $n \bar{n}$ decays into two pions or kaons is $$\frac{\gamma^2(n \bar n \rightarrow \pi\pi)}{\gamma^2(n \bar n \rightarrow K \bar{K} )+\gamma^2(n \bar n \rightarrow \bar K K )} = \frac{\frac{9}{5} g_T^2}{2\frac{9}{30} g_T^2}=3$$ i.e. the value stated by Amsler and Close.

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