# Can a neutral pion decay to a neutrino and an anti-neutrino?

Is it allowed? such as a electron neutrino and an electron anti-neutrino.
And why?
Now, I am confused.....

Thanks.

This decay is forbidden by angular momentum conservation if neutrinos are purely massless.

Staff Emeritus
This decay is forbidden by angular momentum conservation if neutrinos are purely massless.

But they are not. So the decay occurs with some rate. (It's GIM suppressed, though).

malawi_glenn
Homework Helper
Strong- and EM-decay would dominate altough right?

There is no strong decay of the pi0.
It is not a strong interaction.

malawi_glenn
Homework Helper
Ah yeah thats right! I forgot :-)

Hello,

But they are not. So the decay occurs with some rate. (It's GIM suppressed, though).

what do you mean by GIM suppressed ?

From PGD, existing limits are :
Br(pi0 -> nunu) < 2.7e-7
Br(pi0 -> nunugamma) < 6e-4
I have no idea, how much we should expect from SM theory, do you know ?

Hi
I have no idea, how much we should expect from SM theory, do you know ?
I wanted to quote the same values as you just did. You can find the original references for those upper bounds, they describe the various existing predictions.

See the original reference for the value quoted Upper Limit on the Branching Ratio for the Decay $$\pi^0 \to \nu \bar\nu$$
They provide references for :
$$Br(\pi^0 \to \nu \bar\nu) = 3\times 10^{-8} \left(\frac{m_{\nu}}{m_{\pi^{0}}}\right)^{2} \sqrt{1-4\left(\frac{m_{\nu}}{m_{\pi^{0}}}\right)^{2}}$$

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Staff Emeritus
what do you mean by GIM suppressed

Suppressed by the mechanism of Glashow, Illiopoulos and Maiani. This is a cancellation that occurs in flavor changing neutral currents: decays like K0L -> mu+ mu- and pi0 -> nu nubar. If the neutrinos were exactly degenerate the cancellation would be exact and this decay wouldn't occur. Because they have slightly different masses, the decays are suppressed.

Suppressed by the mechanism of Glashow, Illiopoulos and Maiani. This is a cancellation that occurs in flavor changing neutral currents: decays like K0L -> mu+ mu- and pi0 -> nu nubar. If the neutrinos were exactly degenerate the cancellation would be exact and this decay wouldn't occur. Because they have slightly different masses, the decays are suppressed.

GIM mechanism start to be old for me last time I studied it.
I understand that for K0 we have FCNC, but why can't we have :
u ubar + ... -> Z0 -> nu nubar ?
This diagram is suppressed by mnu/MZ, but why do we need CKM and stuff like that ?
Can you explain a bit more ?

Staff Emeritus
The pi0 carries no weak charge, so it cannot couple directly to a Z, just as it contains no electric charge and cannot couple directly to a photon.

arivero
Gold Member
But suppose that the up quark were massless. Could the neutral pion decay into a pair of up, anti-up quarks?

See the original reference for the value quoted Upper Limit on the Branching Ratio for the Decay $$\pi^0 \to \nu \bar\nu$$
They provide references for :
$$Br(\pi^0 \to \nu \bar\nu) = 3\times 10^{-8} \left(\frac{m_{\nu}}{m_{\pi^{0}}}\right)^{2} \sqrt{1-4\left(\frac{m_{\nu}}{m_{\pi^{0}}}\right)^{2}}$$

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Staff Emeritus
But suppose that the up quark were massless. Could the neutral pion decay into a pair of up, anti-up quarks?

The pion is already composed of u-ubar quarks, so I don't know what a decay to u-ubar would mean. Since quarks are confined, you'll end up with a pi0 anyway, so that's not much of a decay.

There is an additional subtlety - a pi0 composed of massless quarks would itself be massless (it becomes a Goldstone) and massless particles do not decay.

Hello,

The pi0 carries no weak charge, so it cannot couple directly to a Z

Your argument looks sensible (why no weak charge ? I would say only T_3 = 0) but I'm not sure to understand it completly. Looking at http://doc.cern.ch//archive/electronic/hep-ph/0501/0501117.pdf ,
it seems that pi0 -> Z -> nu nubar vanish only for massless neutrinos.

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Staff Emeritus
That very paper points out that at tree level pi0 -> Z -> nu nubar is zero. For massive neutrinos you can have a box diagram involving two W's, a diagram that is both GIM and helicity suppressed, but there is no symmetry forcing it to be exactly zero.

As for only T_3 being zero, that alone is enough. The SU(2) nature of the weak interaction tells you that this coupling is zero. (The (1,0)+(1,0)=(1,0) Clebsch-Gordan coefficient is zero) Now that I think about it, I think the other SU(2) symmetry, isospin, also blocks this for the same reason.

arivero
Gold Member
There is an additional subtlety - a pi0 composed of massless quarks would itself be massless (it becomes a Goldstone) and massless particles do not decay.

Well, it is a subtlety but it is not an additional subtlety (to the original question). It roots on chiral aspects too, doesn't it?

Staff Emeritus
If the quarks were massless, the pion (unlike the proton) would also be massless. That would prevent it decaying by any mechanism whatsoever - not just the weak interaction.

arivero
Gold Member
Yes, but why does the mass of the pion depends of the mass of the quarks? It is because of chiral symmetry breaking, isn't it?

Staff Emeritus
Yes, but chiral symmetry breaking is essentially a QCD phenomenon. The fact that the weak interaction treats states of different chirality differently isn't necessary to break chiral symmetry: the masses of the u and d quarks do it by themselves.

It's broken by terms in the Lagrangian like u_L(m)u*_R.