# Question concerning Isospin symmetry

I'm currently taking an introductory course to particle physics and I'm now trying to understand the concept of isospin. However I do have some trouble.

So let's write the up- and down Quark as a Flavour-Doublet (u,d). We then obtain $\mid u \rangle = \mid \frac{1}{2} \frac{1}{2}\rangle$ and $\mid d \rangle = \mid \frac{1}{2} \frac{-1}{2}\rangle$. Now I want to know what particles I can get by combining these two quarks. So let's write $\frac{1}{2} \otimes \frac{1}{2} = 0 \oplus 1$. So I have a singlett and a triplet. Most books now tell me that the triplet corresponds to the pion-triplet and the singlet to the $\eta$-particle.

Now I can for example write by using Clebsch Gordon coefficients that

$\mid 1,1 \rangle = \mid \frac{1}{2} \frac{1}{2}\rangle \mid \frac{1}{2} \frac{1}{2}\rangle = \mid u u \rangle$.

But this doens't correspond to any pion (They are mostly products of ups and anti-downs etc). However some books interpret this as the product of two proton wave function.

This is quite confusing? Can someone help me?

I'd me more than happy. Thanks in advance.

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I'll just write $\left| u \right >$ for the up state and $\left |d \right >$ for the down state. Now, when dealing with $SU(2)$ group, the representations for particles and antiparticles are equivalent (i.e. $\mathbf 2 = \mathbf{ \bar 2}$) but this doesn't mean they are completely same! They differ by an unitary transformation that swaps up and down (and can also introduce phase factors). In particular, we can have the following identification $\left| \bar u \right > = \left |d \right>$ for the anti-up state and $\left |\bar d \right > = -\left | u \right>$ for anti-down.

Now, all the possible combinations in $\mathbf 2 \otimes \mathbf{\bar 2}$ are $$\left |d \right >\left |\bar d \right > ,\quad \left |d \right >\left |\bar u \right >, \quad \left |u \right >\left |\bar d \right >, \quad \left |u \right >\left |\bar u \right >$$ with the 3rd component of isospin being respectively $0$, $-1$, $1$, $0$. Now, $\mathbf 2 \otimes \mathbf{\bar 2} = \mathbf 1 \oplus \mathbf 3$. The triplet representation is characterized by being symmetric and this means that its generator corresponding to the 3rd component of the isospin $0$ must be ${1 \over \sqrt 2} (\left |d \right >\left |\bar d \right > - \left |u \right >\left |\bar u \right >)$ because when we swap first and second factor we get $${1 \over \sqrt 2} (\left |\bar d \right >\left | d \right > - \left |\bar u \right >\left | u \right >) = {1 \over \sqrt 2} (-\left |u \right >\left | \bar u \right > + \left |d \right >\left | \bar d \right >).$$ This is $\pi^0$ we wanted. For the same reason the antisymmetric state must be ${1 \over \sqrt 2} (\left |d \right >\left |\bar d \right > + \left |u \right >\left |\bar u \right >)$ and this is $\eta$.

Excercise: check that the remaining generators of the triplet (representing $\pi^{\pm}$) are symmetric.

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