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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?

Qmechanic
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Whelp
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  • It is on p 170. It is an alarming (and shameful!) "typo". He means η, (5.92), and *not* π0, (5.91). It is hard to catch, but, frankly, inexcusable. – Cosmas Zachos Mar 06 '22 at 20:47

3 Answers3

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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.

It is possible in principle (I think) for a particle like the $\pi^0$ to be the neutral member of a strong isospin triplet, but a singlet under weak isospin.

However, that doesn't seem to be the case for the $\pi^0$. The "lowering operator" vertex $\pi^\pm \leftrightarrow W^\pm\pi^0$ actually does exist, as evidenced by the existence of the decay mode $\pi^+\to \pi^0 e^+ \nu_e$, which has branching ratio $10^{-8}$. You have a fresh comment suggesting that your textbook makes this claim incorrectly.

rob
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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
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    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 01:27
  • Excessively academic nitpick: I think I can see a survining $ZZ\pi^0\pi^0$ vertex, but I could be wrong there... – Cosmas Zachos Mar 07 '22 at 18:00
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    @CosmasZachos Non-academic non-nitpick: the $W^\pm\pi^0$ vertex is actually allowed, and the isospin-lowering decays actually do happen (though they are kinematically and energetically suppressed). This very old answer (v1) was wrong; I've edited. There is no need to hunt for a double-weak vertex. Single-weak $\pi Z$ vertices contribute to parity violation in purely hadronic weak interactions, but when I left that field the theory situation was kind of an inscrutable mess. – rob Mar 07 '22 at 18:54
  • How embarrassing I skipped it... $g W^+_\mu \pi^- \partial^\mu \pi^0$, and the likes, alright... – Cosmas Zachos Mar 07 '22 at 19:48
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$\pi^0$ 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.

user44895
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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 02:13
  • The neutral pion *does couple* to he other pions, as manifest in hundreds of terms in the chiral lagrangian. This is a thoroughgoingly specious argument. – Cosmas Zachos Mar 07 '22 at 19:55
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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.

Jerrold Franklin
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