Consider pion states composed of $q \bar q$ pairs where $q \in \left\{u,d \right\}$ transforms under an $SU(2)$ isospin flavour symmetry. These bound states transform in the tensor product $R_1 \otimes R_2$ of two representations $(R_1, R_2)$ of $SU(2)$. Take $R_2$ as the fundamental representation of isospin with generators $I^i = \sigma^i/2$ and $R_1$ is the conjugate fundamental with generators $-(\sigma^{i})^*/2$. If the third component of isospin is $$I_{\pm}^{R_1 \otimes R_2} = \frac{1}{2} \left( \sigma_1^{R_1 \otimes R_2} \pm i \sigma_2^{R_1 \otimes R_2}\right)$$ I can try and form a representation of this operator using the standard Pauli matrices. Take $|\pi^+ \rangle = |u\rangle |\bar d \rangle \equiv |u \rangle \otimes | \bar d \rangle \equiv |u \bar d \rangle$

Then $$I_{+}^{R_1 \otimes R_2} |u \bar d \rangle = \frac{1}{2} \left( \sigma_1^{R_1 \otimes R_2} \pm i \sigma_2^{R_1 \otimes R_2}\right)|u \bar d \rangle = \frac{1}{2}\left( \sigma_1^{R_1} |\bar d\rangle \otimes \text{Id} |u \rangle + \text{Id} |\bar d \rangle \otimes \sigma_1^{R_2} |u \rangle \pm i(1 \leftrightarrow 2)\right)$$

1)My first question is if I take $|u \rangle \rightarrow (1,0), |\bar d \rangle = (0,1)$ then I have a tensor product of the form $(2 \times 1) \otimes (2 \times 1)$ Is such a tensor product even defined?

Alternatively, I could just construct the representations for $I_{\pm}^{R_1 \otimes R_2}$ and I would end up with $4 \times 4$ matrices. But what would the $4 \times 1$ objects that these operators act on represent? Would a generic vector be something like $(u, d, \bar u, \bar d)$ so for example I would write $u = (1,0,0,0)$ and $\bar d = (0,0,0,1)$ for example?


1 Answer 1


It seems to me that there's a confusion about how the flavour is acting. In your example $R_1 \sim \bar{\bf{2}}$ and $R_2 \sim {\bf{2}}$. Therefore, by multiplying these representations we expect a singlet state ($\ell =0$) and a vector ($\ell =1$), i.e.,

$R_1 \otimes R_2 = \bf{1} \oplus \bf{3}$.

Now, if we identify the fundamental doublet as: $R_2 = \begin{pmatrix} u \\ d \end{pmatrix}\,, $ then $R_1 = \begin{pmatrix} \bar{d} \\ -\bar{u} \end{pmatrix}\,, $ and:

$R_1\otimes R_2 = \frac{1}{\sqrt{2}}(\bar{u}{u}+\bar{d}{d}) \oplus \begin{pmatrix} \bar{d}u \\ \frac{1}{\sqrt{2}}(\bar{d}{d}-\bar{u}{u})\\ -\bar{u}d \end{pmatrix} $

The $\pi^{\pm}$ state you are looking for is in the vector of course since it is electrically charged.


Here I give the derivation based on the isospin matrices. We can define the matrix $$I_{++} =I_+\otimes I_+\,.$$ This matrix can act on the tensor product $R=R_1 \otimes R_2$ as $$ I_{++} R =I_+ R_1 \otimes I_+ R_2=\begin{pmatrix} \bar{d} \\ 0 \end{pmatrix}\otimes \begin{pmatrix} u\\ 0 \end{pmatrix} \equiv |\bar{d}>|u> \equiv |\pi^+> $$

Applying $I_{--}$ and $I_{0^+}\equiv\frac{1}{\sqrt 2} (I_{+-}+I_{-+})$ will give you exactly the vector I derived above. You can get the singlet as $I_{0^-}=\frac{1}{\sqrt 2} (I_{-+}-I_{+-})$

You can represent the $I_\pm$ matrix in a reducible $4 \times 4$ form, of course, and $R$ as $4\times 1$ as $$ I_{++}= \begin{pmatrix} I_+ & 0 \\ 0& I_+ \end{pmatrix}\,, R=\begin{pmatrix} R_1 \\ R_2 \end{pmatrix} $$

You see that you get a natural notation with the definition of $R_1$ I used ; $I_{\pm\pm}$ gives $\pi^{\pm}$ and $I_{0^\pm}$ gives you the $\pi^0$. If you want to use instead $R_1=\begin{pmatrix} \bar{u} \\ \bar{d} \end{pmatrix}$, you should change the definitions of the $I$s accordingly, e.g. $I_{++} \to I_{+-}$.

  • $\begingroup$ Hi xi45! Thanks for your response! I am just wondering why you chose $$R_1 = \begin{pmatrix} -\bar{d} \\ \bar{u} \end{pmatrix}$$- why not for example, $$R_1 = \begin{pmatrix} \bar{u} \\ \bar{d} \end{pmatrix}?$$ $\endgroup$
    – CAF
    May 7, 2016 at 13:33
  • $\begingroup$ Hi! Because in $SU(2)$, $\mathbf{\bar 2} \sim \epsilon {\mathbf{2}}$. Where $\epsilon$ is the anti-symmetric $2\times 2$ tensor. $\endgroup$
    – xi45
    May 7, 2016 at 14:05
  • $\begingroup$ There are two equivalent ways of doing the singlet contraction. This causes some confusion. In principle, you write $R_1=\epsilon \begin{pmatrix} \bar u \\ \bar d \end{pmatrix}=\begin{pmatrix} -\bar d \\ \bar u \end{pmatrix}$, and you do the $SU(2)$ product: $R_1 \otimes R_2$. Or you can define $R_1 =\begin{pmatrix} \bar u \\ \bar d \end{pmatrix}$, but then you do the dot product $R_1.R_2$. Both lead to the same result because the $SU(2)$ has an $\epsilon$; for two generic doublets $v_{1,2}$, we have: $v_1 \otimes v_2 = v_1.(\epsilon.v_2)$. $\endgroup$
    – xi45
    May 7, 2016 at 14:12
  • $\begingroup$ Ok thanks! Could you explain why $$\bf \bar 2 \propto \epsilon \bf 2?$$ $\endgroup$
    – CAF
    May 7, 2016 at 14:38
  • $\begingroup$ I used $\sim$ to say : 'transforms as'. It's not a $\propto$ sign. It's a feature of the $\mathbf{2}$ of $SU(2)$; pseudoreality, check out e.g., scipp.ucsc.edu/~dine/ph222/group_theory_primer.pdf $\endgroup$
    – xi45
    May 7, 2016 at 14:45

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