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.


2 Answers 2


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.


One can use the special unitary (2) SU(2) group representations whenever one has a classification of two states to find representations of states consisting of the original two assignments..

Originally isospin was used to classify nuclear states assigning to the proton isospin 1/2 and to the neutron isospin -1/2. So yes the representation of |1/2 1/2> |1/2>|1/2> describes the isospin state of two protons. Isospin had been used to successfully classify nuclear states and nuclear resonances into representations before quarks were discovered to be the underlying components.

When one goes to the quark framework, it is the quark and antiquark that are the two states represented in the SU(2)combinations that give mesons that have been observed. These support the hypothesis of the existence of quarks and antiquarks by classifying observed mesons into representations of SU(2) that have the same quantum numbers and are differing in isospin only.

Correspondingly the representations joining three quarks organize the observed baryon resonances.

Now your question about the two quarks |u>|u> is valid, it is called a diquark, and there are researchers looking for the effects of the existence of diquarks, but they have not been confirmed.


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