Neutral Pion to Electron Positron Pair
geo
Why is the decay of a neutral pion via a photon to an e+ e- pair not
allowed?
Thanks,
-geo
Uncle Al
>
> Hi,
>
> Why is the decay of a neutral pion via a photon to an e+ e- pair not
> allowed?
<http://en.wikipedia.org/wiki/Pion#Neutral_pion_decays>
<http://superweak.wordpress.com/2006/07/31/dalitz-plots/>
<http://www.nikhef.nl/pub/experiments/zeus/theses/wouter_verkerke/latex2html/node60.html>
A neutral particle obviously cannot radiate photons directly. 1.198%
of neutral pion decay is Dalitz decay into a photon and an
electron-positron pair. The electromagnetic decay is a three-point
interaction: it decays into two virtual and charged kaons or protons,
one emits a photon, then the virtual antiparticles annihalate to
produce a second photon.
Decay of the pion into two electrons and two positrons is "double
Dalitz" decay". One pion in 30,000 decays in this way.
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/lajos.htm#a2
nma...@gmail.com
http://en.wikipedia.org/wiki/Richard_Dalitz
Patrick Nolan
> geo wrote:
>>
>> Hi,
>>
>> Why is the decay of a neutral pion via a photon to an e+ e- pair not
>> allowed?
>
><http://en.wikipedia.org/wiki/Pion#Neutral_pion_decays>
><http://superweak.wordpress.com/2006/07/31/dalitz-plots/>
><http://www.nikhef.nl/pub/experiments/zeus/theses/wouter_verkerke/latex2html/node60.html>
>
> A neutral particle obviously cannot radiate photons directly. 1.198%
> of neutral pion decay is Dalitz decay into a photon and an
> electron-positron pair. The electromagnetic decay is a three-point
> interaction: it decays into two virtual and charged kaons or protons,
> one emits a photon, then the virtual antiparticles annihalate to
> produce a second photon.
>
> Decay of the pion into two electrons and two positrons is "double
> Dalitz" decay". One pion in 30,000 decays in this way.
>
(http://pdg.lbl.gov/2008/tables/rpp2008-tab-mesons-light.pdf),
a neutral pion decays into an electron-positron pair once in
15,500,000 times.
If you work out the Feynman diagrams, you can see that the usual
2-photon decay is the process with the fewest vertices. To get
an e+e- pair, the dominant process would be annihilation of the
two photons. That adds two more electromagnetic vertices,
which accounts for the low branching ratio.
There are plenty of other decay modes, some smaller branching
ratios.
geo
Why is the following Feynman diagram not allowed?
u u-bar -> virtual photon -> e+ e-
or
d d-bar -> virtual photon -> e+ e-
I don't want to hear about the other decay channels and the virtual
photon loop channel that produces e+e- with alpha^2 suppression. I can
read about that in the PDG too. I'm asking about the simplest decay
channel I can draw and why that is not allowed.
Thanks,
-geo
On Apr 4, 9:29 am, Patrick Nolan <p...@glast2.Stanford.EDU> wrote:
> On 2009-04-03, Uncle Al <Uncle...@hate.spam.net> wrote:
>
> > geo wrote:
>
> >> Hi,
>
> >> Why is the decay of a neutral pion via a photon to an e+ e- pair not
> >> allowed?
>
> ><http://en.wikipedia.org/wiki/Pion#Neutral_pion_decays>
> ><http://superweak.wordpress.com/2006/07/31/dalitz-plots/>
> ><http://www.nikhef.nl/pub/experiments/zeus/theses/wouter_verkerke/late...>
Patrick Nolan
> Let me rephrase my question:
>
> Why is the following Feynman diagram not allowed?
>
> u u-bar -> virtual photon -> e+ e-
> or
> d d-bar -> virtual photon -> e+ e-
>
> I don't want to hear about the other decay channels and the virtual
> photon loop channel that produces e+e- with alpha^2 suppression. I can
> read about that in the PDG too. I'm asking about the simplest decay
> channel I can draw and why that is not allowed.
>
I have a conjecture which really should be checked by someone who knows
what they are doing. Suppose that angular momentum must be conserved
at each vertex. We know that linear momentum/energy doesn't need to
be conserved, but maybe it's different for angular momentum. Since
the pion has zero spin, the quark-antiquark pair must be in a state
of zero total angular momentum. They can't combine to produce a
massless photon, which must have a spin of +1 or -1. A similar diagram
would work with a massive intermediary, such as a Z, but the rate
would be greatly suppressed.
Maybe angular momentum isn't exactly conserved at each vertex.
It seems plausible that the propagator for a virtual photon
with zero spin would be much suppressed.
This doesn't apply to free particle annihilation, such as
mu+ mu- -> e+ e- because the angular momentum of the mu pair is
not constrained to be zero.
Igor Khavkine
> On 2009-04-04, geo <souvik1...@gmail.com> wrote:> Let me rephrase my question:
>
> > Why is the following Feynman diagram not allowed?
>
> > u u-bar -> virtual photon -> e+ e-
> > or
> > d d-bar -> virtual photon -> e+ e-
> I have a conjecture which really should be checked by someone who knows
> what they are doing. Suppose that angular momentum must be conserved
> at each vertex. We know that linear momentum/energy doesn't need to
> be conserved, but maybe it's different for angular momentum. Since
> the pion has zero spin, the quark-antiquark pair must be in a state
> of zero total angular momentum. They can't combine to produce a
> massless photon, which must have a spin of +1 or -1. A similar diagram
> would work with a massive intermediary, such as a Z, but the rate
> would be greatly suppressed.
I think that's basically correct, except for the part about mass. The
photon does have spin 1 and so does the Z boson. So, if the quark-
antiquark-photon vertex is forbidden by angular momentum conservation,
then so is the quark-antiquark-Z vertex.
On the other hand, a quark-antiquark or an electron-positron pair
cannot annihilate into a single photon, but could annihilate into a
single Z. The reason is that the time-like massive fermion 4-momenta
cannot be added up to a null photon 4-momentum. The Z 4-momentum is
not null, hence it is not excluded.
> Maybe angular momentum isn't exactly conserved at each vertex.
> It seems plausible that the propagator for a virtual photon
> with zero spin would be much suppressed.
It is. It is enforced by the contraction of the spin and vector
indices at the vertex. The quark and antiquark spinor indices must be
contracted with the photon vector index using gamma matrices. However,
following usual spin addition rules, such a contraction must
necessarily be zero if the quark-antiquark pair is in the zero spin
singlet state.
> This doesn't apply to free particle annihilation, such as
> mu+ mu- -> e+ e- because the angular momentum of the mu pair is
> not constrained to be zero.
Right. For the up and down quarks, which make up the pion, the
equivalent situation is the decay of the rho^0 meson into an e+ e-
pair through a single intermediate photon. The rho is a vector meson,
which is just an excited state of the pion, where the u and d quarks
are combined into a spin 1 triplet.
Hope this helps.
Igor
Patrick Nolan
> On Apr 6, 9:05 pm, Patrick Nolan <p...@glast2.Stanford.EDU> wrote:
>> On 2009-04-04, geo <souvik1...@gmail.com> wrote:> Let me rephrase my question:
>>
>> > Why is the following Feynman diagram not allowed?
>>
>> > u u-bar -> virtual photon -> e+ e-
>> > or
>> > d d-bar -> virtual photon -> e+ e-
>
>> I have a conjecture which really should be checked by someone who knows
>> what they are doing. Suppose that angular momentum must be conserved
>> at each vertex. We know that linear momentum/energy doesn't need to
>> be conserved, but maybe it's different for angular momentum. Since
>> the pion has zero spin, the quark-antiquark pair must be in a state
>> of zero total angular momentum. They can't combine to produce a
>> massless photon, which must have a spin of +1 or -1. A similar diagram
>> would work with a massive intermediary, such as a Z, but the rate
>> would be greatly suppressed.
>
> I think that's basically correct, except for the part about mass. The
> photon does have spin 1 and so does the Z boson. So, if the quark-
> antiquark-photon vertex is forbidden by angular momentum conservation,
> then so is the quark-antiquark-Z vertex.
What I meant to say is that the photon, being massless, can't have
a state of helicity 0. Any massive boson can have helicity 0.
>
> On the other hand, a quark-antiquark or an electron-positron pair
> cannot annihilate into a single photon, but could annihilate into a
> single Z. The reason is that the time-like massive fermion 4-momenta
> cannot be added up to a null photon 4-momentum. The Z 4-momentum is
> not null, hence it is not excluded.
But consider Bhaba scattering. That's e+e- -> e+e-. One of the
two Feynman diagrams has the two particles annihilating into a
single photon, which then turns back into two particles. The
conservation of 4-momentum must not apply strictly to internal,
virtual photons. I've heard this described as an application
of Heisenberg's uncertainty relationship, since the photon
"exists" for a very short time.