Neutral Xi Decay: Why \Xi^0--> \Lambda^0+\pi^0?

[maxverywell]

Is there any explanation why the $$\Xi^0$$ decays to $$\Lambda^0$$ + $$\pi^0$$ and not to

$$\Lambda^0$$ + $$\pi^-$$+$$\pi^+$$ ?

 

[Meir Achuz]

The mass of the Lambda and two pions is greater than that of the Xi.

 

[maxverywell]

But in each case we have four quarks: u, d, $$\bar{u}$$ and $$\bar{d}$$.

Actually, two W bosons decay to $$\pi^-$$ ($$\bar{d}u$$) and $$\pi^+$$ ($$\bar{u}d$$) as you can see in the following Feynman diagram (the first one):

Why these four quarks combine to produce a neutral pion? And what is the meaning of factor $$\sqrt{2}$$? The number of quarks in mesons is supposed to be integer.

 

[Vanadium 50]

Where are you getting the left diagram from?

 

[tom.stoer]

You have two diagrams, the first one with one, the second one with two W's exchanged. But the decay rate is nearly identical. I can't believe that!

 

[astrak]

I not understand very well what meaning sqrt(2) in the left picture. Because I think some particle is composite from quarks, or leptons. But we can take only part of particle? :-)

 

[tom.stoer]

it's nothing else but a normalization

$$\vec{e}_x = (1,0)$$
$$\vec{e}_y = (0,1)$$
$$\vec{e}_{xy} = (\vec{e}_x + \vec{e}_y)/\sqrt{2} = (1,1)/\sqrt{2}$$

with

$$|\vec{e}_x| = |\vec{e}_y| = |\vec{e}_{xy}| = 1$$

 

[maxverywell]

Can you explain these please? Do we have four or two quarks?
Quarks are real particles, not vectors, I don't understand the meaning of superposition in the case of real entities like quarks.

 

[tom.stoer]

You have no idea regarding quantum mechanics?

QM and QFT uses state vectors in infinite dimensional vector spaces (these spaces have nothing to do with ordinary space time!). In these vector spaces you define (infinitly many) basis vectors. The square root is nothing else but the normalization in such a space.

But that's irrelevant for our discussion.

So where's this diagram from, including the lifetimes?

 

[maxverywell]

So it's superposition of two quantum states and If we make "measurement" we will get $$u\bar{u}$$ or $$d\bar{d}$$ ?

The diagram is from here http://hyperphysics.phy-astr.gsu.edu/hbase/particles/imgpar/xidec2.gif

I think the following diagram is more correct:

http://teachers.web.cern.ch/teachers/archiv/HST2002/feynman/exampl10.gif

 

[tom.stoer]

According to the particle data group this is correct: http://pdg.lbl.gov/2010/download/rpp-2010-booklet.pdf , p. 145

(still I do not understand the lifetimes)

Regardig the neutral pion: you always detect a neutral pion as an isospin singulet; that's why it reads

$$|\pi^0\rangle = (|u\bar{u}\rangle - |d\bar{d}\rangle)/\sqrt{2}$$

The "+" is incorrect in your diagram. The reason for the "-" and for this superposition is that the pions sit in the 3-dim. rep. of the SU(2) isospin symmetry.

 

[Vanadium 50]

OK, so now you have the right diagram. Now you can make progress.

As for the lifetimes, there is a factor of 2 because the udbar state has 100% overlap with the pi+, but the uubar state has only 50% overlap with the pi0.

 

[astrak]

QM and QFT uses state vectors in infinite dimensional vector spaces (these spaces have nothing to do with ordinary space time!). In these vector spaces you define (infinitly many) basis vectors. The square root is nothing else but the normalization in such a space.

Are you talking about Hilbert space? So if I understand your replies, square root is only probability measurement of pí 0, or pí - (probability decays will be this particles)?

 

[maxverywell]

As far as I understand it, we have two-quark system of four different quarks $$u,\bar{u},d,\bar{d}$$ with six basis (isospin) states $$|u\bar{u}\rangle$$, $$|d\bar{d}\rangle$$,$$|u\bar{d}\rangle$$, $$|d\bar{u}\rangle$$, $$|uu\rangle$$ and $$|dd\rangle$$ that form an abstract 6-dimensional (Hilbert) space. But the states with total isospin 1 are:
$$|\pi^+\rangle=|u\bar{d}\rangle=|1,1\rangle$$
$$|\pi^0\rangle=(|u\bar{u}\rangle+|d\bar{d}\rangle)/\sqrt{2}=|1,0\rangle$$
$$|\pi^-\rangle=|d\bar{u}\rangle=|1,-1\rangle$$
So these new three basis vectors (triplet states) form 3-dimensional isospin space. Any rotation in this space (any action of SO(3) Lie group) leaves invariant the Hamiltonian of the strong interactions.

Is it correct?

 

[tom.stoer]

nearly correct

you do not have six but three basis vectors, as the coupling of two 1/2 isosopin states only gives you a triplet (it's like the spin case; don't confuse this with the symbols for u and d)

there is a sign error; the neutral pion looks like

$$|\pi^0\rangle=(|u\bar{u}\rangle-|d\bar{d}\rangle)/\sqrt{2}=|1,0\rangle$$

The Lie group is not SO(3) but SU(2) b/c SO(3) does not have 1/2 representations. The isospin=1 triplet is OK for the SO(3) (but it would be the fundamental rep. for the SO(3)) whereas for the SU(2) the fundamental rep. is the isopsin 1/2 dublet with the quarks itself (whereas the triplet is the adjoint rep. of SU(2)).

The Hamiltonian is nearly invariant under SU(2) isospin; the symmetry is broken due to the slightly different u- and d-quark masses. But for most purposes isospin symmery is exact or nearly exact (there are even cases where the quarks are treated as massless).

There is also a singulet state which (for a reason I do not understand) has never been discussed for isospin but only (after discovery of the third flavour strangeness) in the context of the SU(3)-flavor singulet / octet.