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I wonder why the neutral pi meson is

$$ | \pi^0\rangle = \frac{1}{\sqrt{2}}\left(\vert u\overline {u}\rangle - \vert d \overline{d} \rangle \right) $$

and not

$$ | \pi^0\rangle = \frac{1}{\sqrt{2}}\left(\vert u\overline {u}\rangle + \vert d \overline{d} \rangle \right) .$$

Pions are a pair of quark and antiquark which both are isospin doublets. Since pions are isospin triplets $\pi^0$ should have plus in the quarks decomposition (?). I would expect minus in the isospin singlet state which, unfortunately, does not seem to be realized in the nature.

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2 Answers 2

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The reason the signs are flipped from what you expect has to do with the fact that the antiquark transforms in the opposite way under isospin rotations. If the ordinary quark doublet is a column vector $$q=(u, d)^T$$ and transforms under rotations as $$q\rightarrow U(R) q$$ the antiquark doublet is a row vector $$\bar{q}=(\bar{u}, \bar{d})\rightarrow \bar{q} U(R)^\dagger.$$

But $SU(2)$ has a special property called being "pseudoreal" so we can write the antiquarks as a column vector that transforms normally like $$(-\bar{d}, \bar{u})^T\rightarrow U(R)(-\bar{d}, \bar{u})^T$$ This is related to the Pauli matrix $\sigma_2$ being like a charge conjugation operator if you are familiar with that.

To do the addition of isospin in the ordinary way we need both quark and antiquark in the same representation, so the singlet $|\uparrow\downarrow\rangle-|\downarrow\uparrow\rangle$ is in this case $$u\bar{u}-d(-\bar{d})$$ so we pick up a plus sign.

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  • $\begingroup$ I first thought the explanation should come from Clebsch-Gordan coefficients but that method suggests the one with the plus sign. How do the two ideas connect? $\endgroup$ Commented Sep 6, 2022 at 17:11
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    $\begingroup$ @KaanGüven, it does come from Clebsch-Gordan coefficients. But you have to put both isospin 1/2 representations in the fundamental representation first (rather than anti-fundamental). That introduces a sign flip coming from $-\bar{d}$ $\endgroup$
    – octonion
    Commented Sep 6, 2022 at 17:37
  • $\begingroup$ Thank you for your comment. However I have difficulty in understanding why I need to use both in fundamental representations. I mean, why can’t I write them in mixed representations? Is Clebsch-Gordan defined only in fundamental representations? My professor told me that I was missing the fundamentals in quantum field theory, but I don’t know any lecture notes that cover these gaps. I know that in a Lagrangian $\psi$ is in fundamnetal, $\bar{\psi}$ is in anti-fundamental and generators of the symmetry are in adjoint representations, but noone explains why.Could you recommend a way of learning? $\endgroup$ Commented Sep 6, 2022 at 18:51
  • $\begingroup$ @KaanGüven, You might like Georgi's textbook Lie algebras in particle physics. $\endgroup$
    – octonion
    Commented Sep 6, 2022 at 20:03
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Let's understand this from the point of view that pions are pseudogoldstone bosons. Suppose for case of $SU_{L}(2)\times SU_{R}(2)$ we have bilinear form $$ \tag 0 L_{qq} = \bar{q}_{i}q_{i}, \quad i = u, d $$ As we know, below $\Lambda_{QCD}$ scale spontaneously breaking symmetry group down to diagonal group $SU_{V}(2)$ arises. By using the usual technique we may extract goldstone degress of freedom from quark fields, $$ q_{i} \equiv (U\tilde{q})_{i}, \quad U \equiv \text{exp}\left[ i\gamma_{5}\frac{\epsilon_{a}t_{a}}{f_{\pi}}\right], $$ where $t_{a}$ are Pauli matrices and $\epsilon_{a}$ are real-valued coordinate dependent parameters, ans then replace bilinear forms to VEVs: $$ \tag 1 \bar{\tilde{q}}_{i}\tilde{q}_{j} \to V\delta_{ij}, \quad \bar{\tilde{q}}_{i}\gamma_{5}\tilde{q}_{j} \to 0 $$ Note that $\epsilon_{a}t_{a}$ may be parametrized, by using explicit form of Pauli matrices, in a form $$ \tag 2 \epsilon_{a}t_{a} = \begin{pmatrix} \frac{\pi^{0}}{\sqrt{2}} & \pi^{-} \\ \pi^{+} & -\frac{\pi^{0}}{\sqrt{2}}\end{pmatrix}, $$ where $\pi^{\pm} \equiv \epsilon_{1} \pm i\epsilon_{2}$. As we see, we explicitly obtain one neutral degree of freedom, $\pi^{0}$, and two charged, $\pi^{\pm}$. Now let's calculate one-pion transition amplitude from $(0)$, $$ \langle 0 | \bar{q}_{i}q_{i}|\pi^{0}\rangle, $$ by using $(1)$ and $(2)$. We immediately obtain that $$ \langle 0 | \bar{q}_{i}q_{i}|\pi^{0}\rangle \equiv \frac{i}{\sqrt{2}f_{\pi}}\langle 0|\bar{\tilde{q}}_{i}t_{3}^{ij}\tilde{q}_{j}\pi^{0}|\pi^{0}\rangle \simeq \frac{i}{\sqrt{2}f_{\pi}}\langle 0|\bar{\tilde{u}}\tilde{u}-\bar{\tilde{d}}d|0\rangle, $$ which immediately gives statement that the pion is combination of $uu, dd$ with the minus sign because of it is the parametrization of goldstone degree of freedom in case of broken $SU_{L}(2)\times SU_{R}(2)$ group, not $U_{L}(2)\times U_{R}(2)$. The combination with the "plus" sign corresponds to parametrization of $U(1)$ group. As you know, this is $\eta$ meson for $SU_{L}(2)\times SU_{R}(2)$ group and $\eta{'}$ meson for $SU_{L}(3)\times SU_{R}(3)$ group. Of course, these combinations appear in nature.

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  • $\begingroup$ Thank you for sharing this explanation! Can you explain further in eq (1): how do we know that scalar bilinear picks up a vev while the pseudoscalar bilinear does not? $\endgroup$ Commented Aug 22, 2017 at 15:09
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    $\begingroup$ @HenryDeith : the underlying reason is that phenomenologically all of the QCD's staff (the isospin symmetry, vector current conservation, PCAC, Goldberg-Treiman relation) at low energies can be explained by the spontaneous breaking of the $G\simeq SU_{L}(3)\times SU_{R}(3)$ approximate global symmetry down to $SU_{V}(3)$. In terms of the fundamental quark fields, this means nothing but appearance of non-zero VEV of the quark bilinear $\bar{q}q$: it breaks the axial part of $G$, leaving the vectorial part unbroken. $\endgroup$
    – Name YYY
    Commented Aug 22, 2017 at 19:47

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