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In Griffiths' Introduction to Elementary Particles, it is mentioned p. 179 that the $\pi^0$ is a singlet under $SU(2)$ isospin. But it is also part of the $\pi^-,\pi^0,\pi^+$ isospin triplet. How can it be both?

Don't particles of a given $SU(2)$ multiplet mix under a corresponding transformation?

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3 Answers 3

up vote 8 down vote accepted

In a travesty of overlapping historical notations, we have both strong isospin, under which the neutron and proton are the two projections of the nucleon and are raised and lowered by the pion, and weak isospin, in which the left-handed parts of the $(u,d)$, $(e,\nu_e)$, $(c,s)$, $(\mu, \nu_\mu)$, $(t,b)$, and $(\tau,\nu_\tau)$ doublets are raised and lowered by the $W^\pm$ bosons. The six right-handed quarks and the six right-handed leptons are weak isospin singlets.

The $\pi^0$ is the neutral member of a strong isospin triplet, but is a singlet under weak isospin.

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Nice answer! How can you show that it is a singlet under weak isospin? –  Melquíades Apr 25 '14 at 23:24
I was afraid you'd ask that, that's where my memory goes squishy. It may be as simple as observing that the $\pi^0$ can't couple to a $W^+$ due to charge conservation. –  rob Apr 26 '14 at 1:27

$\pi^0$ is a singlet under $SU(2)$ isospin because if you interchange the u by d (as a rotation in 2- dimensional flavor space) then you will not be able to mix up $\pi^0$ with any other pion, and hence it is a singlet. It is in this sense you should understand it.

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Hmmm, but strong isospin also rotates u into d, and takes the $\pi^0$ to $\pi^\pm$. I have a vague memory of something subtle involving a negative sign appearing under strong isospin rotations; I don't remember that argument well enough to decide whether it applies to weak isospin, or whether that's the difference. –  rob Apr 27 '14 at 2:13

The pi zero is not a singlet, but part of an Ispin triplet. The eta meson and the omega meson are Ispin SU(2) singlets.

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