Meson Compostions: Difference Between π0 and η

[Shen712]

The composition of the π⁰ is (uu_bar - dd_bar/√2; while the composition of the η meson is (uu_bar + dd_bar)/√2. Why is there a - sign in π⁰ while is there a + sign in η? How are the signs determined?
By the way, can I type Latex symbols on this site? I tried to type the anti up quark by typing \overline{u}, but it did not show what I wanted. How should I type it?

 

[Mark44]

By the way, can I type Latex symbols on this site? I tried to type the anti up quark by typing \overline{u}, but it did not show what I wanted. How should I type it?

Yes, LaTeX is supported here. See our tutorial here -- https://www.physicsforums.com/help/latexhelp/.

 

[Orodruin]

Because $\pi$ and $\eta$ are different linear combinations of states in flavour space.

 

[Shen712]

Because $\pi$ and $\eta$ are different linear combinations of states in flavour space.

Can you explain more clearly? How are the linear compositions different? And how does one lead to a - sign while the other lead to a + sign?

 

[Orodruin]

They are in different flavour states. It is that simple. If you have two vectors x and y, x+y is a different linear combination than x-y. In this case, one ($\pi_0$) is part of an isospin triplet and the other ($\eta$) is an isospin singlet.

 

[ChrisVer]

In the flavor symmetric model, aka the down and up quarks are in an SU(2) doublet (aka 2-dimensional representation), which is a good approximate symmetry as long as the quarks are massless $m_u , m_d \approx 0$, when you go to build mesons by combining the doublet with the anti-doublet you have:
$\textbf{2} \otimes \bar{\textbf{2}} = \textbf{3} \oplus \textbf{1}$
The (iso)triplet you get contains the pions (well with some combinations for the charged pions*): $\pi^{\pm} , \pi^0$ since it's built from the combination : $\bar{q} \sigma^i q$ with sigma the pauli matrices, while the (iso)singlet is the $\eta$ meson and it's constructed from the identity matrix (not pauli-matrices): $\bar{q} \textbf{1} q$. The isotriplet is symmetric under the exchange $u\leftrightarrow d$ while the isosinglet is antisymmetric (similar to spin-1/2 from quantum mechanincs).
This is a simplified explanation (yet it works for your case). The actual (physical) $\eta$ states are produced as a mixture of states. These states are predicted by flavor SU(3) symmetry (adding strange quarks in, since they are also quite light), they are in an iso-octet and iso-singlet: $\textbf{3} \otimes \bar{\textbf{3}} = \textbf{8} \oplus \textbf{1}$. Because the u,d,s quarks are not exactly massless, the SU(3) symmetry is not exact and mixings happen.

*In fact I wrote the charged pions immediately, but the quark composition you get out of this procedure is for $\textbf{3}$:
$\phi^1 = \frac{1}{\sqrt{2}} (u\bar{d} + \bar{u} d) ,~~\phi^2= \frac{i}{\sqrt{2}} (u\bar{d} - \bar{u}d) ,~~\phi^3= \frac{1}{\sqrt{2}} (u\bar{u} - d \bar{d} )$
The charged pions come from combining $\phi^{1,2}$ ... in a similar manner as you built the ladder operators in spin-1/2 case... $\pi^\pm = \frac{1}{\sqrt{2}} ( \phi^1 \mp i \phi^2)$

 

[samalkhaiat]

The composition of the π⁰ is (uu_bar - dd_bar/√2; while the composition of the η meson is (uu_bar + dd_bar)/√2. Why is there a - sign in π⁰ while is there a + sign in η? How are the signs determined?
By the way, can I type Latex symbols on this site? I tried to type the anti up quark by typing \overline{u}, but it did not show what I wanted. How should I type it?

Where is the $s \bar{s}$ component of your $\eta$? Scalar mesons (i.e., the $J^{P} = 0^{-}$ representation of the Lorentz group) contain 2 $\eta$-particles: the $T = S = 0$ member of the $SU(3)$ octet, usually denoted by $\eta_{8}$ and having the quarks content $$\eta_{8} = \frac{1}{\sqrt{6}} (u \bar{u} + d \bar{d} - 2 s \bar{s}) ,$$ and there is the $SU(3)$ singlet (invariant) state $$\eta_{1} = \frac{1}{\sqrt{3}} \sum_{i = u,d,s} q^{i} \ \bar{q}_{i} = \frac{1}{\sqrt{3}} (u \bar{u} + d \bar{d} + s \bar{s}) \ .$$ The exact quark content follows from the $SU(3)$ decomposition $3 \otimes \bar{3} = 8 \oplus 1$, which you can represent as follow $$q^{i} \bar{q}_{j} = \left( q^{i} \bar{q}_{j} - \frac{1}{3} \delta^{i}_{j} \sum_{k = u,d,s} q^{k} \bar{q}_{k}\right) + \frac{1}{3} \delta^{i}_{j} \sum_{k} q^{k}\bar{q}_{k} \ .$$ The first term on the right hand side is the $3 \times 3$ traceless hermitian mesons matrix $$\{8\}^{i}_{j} = q^{i} \bar{q}_{j} - \frac{1}{3} \delta^{i}_{j} \sum_{k = u,d,s} q^{k} \bar{q}_{k} \ .$$ For the $0^{-}$ (i.e., scalar) mesons, the off-diagonal matrix elements are easily recognised, for example $\{8\}^{2}_{3} = d \bar{s} = K^{0} \ , \{8\}^{3}_{1} = s \bar{u} = K^{-}$ and the two charged pions $\{8\}^{1}_{2} = u \bar{d} = \pi^{+} \ , \{8\}^{2}_{1} = d \bar{u} = \pi^{-}$. Now, before considering the diagonal elements of the $SU(3)$ meson matrix $\{8\}^{i}_{j}$, and in order to settle the sign issue for $\pi^{0}$, let us repeat the same thing for the iso-spin group $SU(2)$. That is decomposing the tensor product $2 \otimes 2 = 3 \oplus 1$ by subtracting the $SU(2)$-invariant trace. So, in terms of quarks $$q^{m} \bar{q}_{n} = \left( q^{m} \bar{q}_{n} - \frac{1}{2} \delta^{m}_{n} \sum_{l = u,d} q^{l}\bar{q}_{l} \right) + \frac{1}{2} \delta^{m}_{n} \sum_{l = u,d} q^{l} \bar{q}_{l} \ .$$ From this we identify the triplet (i.e., iso-spin one) pion states which are contained in the matrix $\{3\}^{m}_{n}$:$$\pi^{+} = |1 , +1\rangle = \{3\}^{1}_{2} = u \bar{d} \ ,$$$$\pi^{0} = |1 , 0 \rangle = \sqrt{2} \{3\}^{1}_{1} = \frac{1}{\sqrt{2}} (u \bar{u} - d \bar{d}) \ ,$$$$\pi^{-} = |1 , -1 \rangle = \{3\}^{2}_{1} = d \bar{u} \ .$$ Now, you know the exact form of $\pi^{0}$ state, go back to consider the diagonal elements of the $SU(3)$ matrix $\{8\}^{i}_{j}$: $$\{8\}^{1}_{1} = u \bar{u} - \frac{1}{3} \sum_{j = u,d,s} q^{j}\bar{q}_{j} = \frac{1}{\sqrt{2}} \pi^{0} + \frac{1}{6} \left( u \bar{u} + d \bar{d} - 2 s \bar{s} \right) \ ,$$ $$\{8\}^{2}_{2} = - \frac{1}{\sqrt{2}} \pi^{0} + \frac{1}{6} \left( u \bar{u} + d \bar{d} - 2 s \bar{s} \right) \ ,$$$$\{8\}^{3}_{3} = - \frac{2}{6} \left( u \bar{u} + d \bar{d} - 2 s \bar{s}\right) \ .$$ This led to the identification of $\eta_{8}$ with the state $$\eta_{8} = \frac{1}{\sqrt{6}} \left( u \bar{u} + d \bar{d} - 2 s \bar{s}\right) \ .$$