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The neutral pion, π0, should have an isospin of 0. To be certain, I am referring to the third component of isospin, $I_3$.

The reason for this is very simple: I3 is an additive quantum number. The valence quarks of the neutral pion are up and antiup, or down and antidown. Up and antiup have the opposite (additive inverse) $I_3$ values to each other, as do down and antidown (and just like any particle-antiparticle pair). The sum of opposite numbers is 0 and so, of course, the $I_3$ of the neutral pion is 0.

However, browsing through the Particle Data Group (PDG) particle listings, I noticed something peculiar: the $I^G$ value of the neutral pion is listed as 1-. Here, $I$ is isospin. This is the PDF file from PDG containing this information on the neutral pion.

The same situation occurs with other mesons having the same valence quarks as the neutral pion, such as the neutral rho meson.

So, what is going on here? Are PDG referring to something other than the value which should be 0, as described above? Is there any quantum number notated $I$ that can have a value of 1 while $I_3$ equals 0? (By the way, no, in this case the answer is not "weak isospin".) If so, what is the explanation for this divergence being possible?

Qmechanic
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R. Paul
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2 Answers2

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Isospin is a made-up concept to explain the interactions between the neutron and proton via pions. Effectively there is a certain symmetry that is explained by the SU(2) group nicely. You can use it to predict branching ratios and hence rates of decays.

We assign isospin of +1/2 to u quark and -1/2 to d quark. You can look into Young's tableaux and derive 2 ⊗ 2¯ = 1 ⊕ 3. That means that if you combine a quark and an antiquark you get 1 symmetric and 3 antisymmetric states (much like the singlet and triplet that you get in the S and P shells of Helium. You can have both electrons pointing up,down or somewhere in the x,y plane in the same direction, which would always give S=1, or they can be pointing in different directions whichever axis you pick S=0)

Now the triplet has I=1 and Iz=-1,0,1 and can easily be identified as the 3 pions. The singlet is a bit more tricky as you need the s-quark to make a nonet of mesons to identify a singlet.

Luthelin
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Total Isospin $I$ is different from its third component $I_3$. What you did was add the third components of the valence quarks, which is somehow ok as long as you are certain of their values. On the other hand, as with angular momentum, total isospin is not so straightforward to establish.

If two quarks carry $I=1/2$ total isospin each, a bound system made from these can either have $I=1$ or $I=0$. The $I=1$ configuration is a triplet, and therefore there are three states associated with it. These are just $\pi^+$ ($I_3=1$), $\pi^0$ ($I_3=0$) and $\pi^-$ ($I_3=-1$).

Johann Liebert
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