# What is the isospin of the neutral pion?

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?

• $I^2$ vs $I_3$, i.e. the standard theory of angular momentum… – user154997 Oct 28 '17 at 1:47
• They give the value of the "length" I of the isospin vector describing the multiplets. – anna v Oct 28 '17 at 3:46

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 $\pi^-$ ($I_3=-1$).