# Tag Info

25

Mesons are not elementary, they are composed of quarks. So take a look at their quark content. The charmed eta meson consists of a charm and an anti-charm quark, denoted $c\overline{c}$. An anti charmed eta meson would therefore be an anti-charm and an anti-anti-charm (which is just a charm) quark, i.e. $\overline{c}c$, which is obviously the same as $c\... 21 No, a three-quark baryon can not be be made out of two quarks and one anti-quark (and vice versa) as this would necessarily give the particle color. Each quark carries one of three colors (red, blue, green) and each anti-quark respectively carries anti-color. Color is an additive quantity when constructing particle and the result must be color-neutral, i.e. ... 19 Pentaquarks contain three quarks and a quark-antiquark pair, and they are baryons, since baryons are defined as having an odd number of valence quarks. 11 Briefly, a hadron has to be color singlet${\bf 1}$under the$SU(3)_C$color gauge group, due to color confinement. Examples: A single quark$q$transforms in the fundamental representation${\bf 3}$of$SU(3)_C$, and is hence not allowed. See also related Phys.SE post here. A diquark$qq$belongs to the tensor representation${\bf 3}^{\otimes 2}:={\bf 3}\...

10

As far as I understand, due to conservation of angular momentum, the resulting system of neutral pions would need to have angular momentum 1, therefore, the identical neutral pions would be in an anti-symmetric state, which does not seem possible as they are bosons. Note that a neutral rho meson can decay into two neutral pions and a $\gamma$, although this ...

8

The total angular momentum of a meson is the sum of the spins of the two quarks and their orbital angular momentum. Excited states can have $L>0$ and therefore $J>1$.

8

This is a good question, and the basic answer is "there is no rigorous difference" (which is the main problem with experimental efforts to identify glueballs). 1) Electrically charged states, or neutral particles that are part of an isospin multiplet are clearly mesons. This means your question can be narrowed down to distinguishing isospin singlet mesons ...

8

That is more or less what we do, except... Annihilation through a photon is electromagnetic rather than strong, so this is a suppression. Annihilation through a single gluon can't happen because the gluon has a colour+anticolour charge, whereas the initial meson is colourless. The gluon has to have a charge like red-antiblue and you can't make that from a ...

7

To toot my own horn a little1 may I suggest the series of papers published by the $f_\pi$ collaboration from Hall C at Jefferson Lab: Determination of the pion charge form factor for $Q^2=0.60-1.60 \,\mathrm{GeV^2}$, 2007 DOI:10.1103/PhysRevC.75.055205 and arXiv:nucl-ex/0607007 Charged pion form factor between $Q^2=0.60$ and $2.45 \,\mathrm{GeV^2}$. I. ...

7

They have 2 main differences. The first of them is very straightforward: They have different spins: As you pointed out, both are bound states of 2 spin 1/2 particles, therefore you can find the possible spins of such a bound state using the usual rules of angular momentum addition in Quantum Mechanics. 1/2 + 1/2 can give you either 3 spin 1 states (the ...

5

As anna mentioned, there are non quark models which clarify exotic hadrons. In principle, they are allowed in Quantum Chromodynamics (QCD). Non-quark models predict 1.hybrid mesons: Include quark anti-quark pair and gluon. 2.Glueballs: Gluons are their own bound states. 3.Exotic hadrons as in figure below which exchange pion at low energies (couple of ...

5

The fact of the matter is that there are no stable mesons, that might conceivably form states bound by the strong force, as the nucleus is bound. Within the nucleus there exist virtual mesons, i.e. described as pions etc but not on mass shell To a large extent, the nuclear force can be understood in terms of the exchange of virtual light mesons, such as ...

5

The color charges are paired (color with anticolor), but there's no gauge invariant meaning to the identification of the color (RGB). And due to QM, the quark states are a superposition over all the colors (and antiquarks over the anticolors). The wikipedia page is pretty clear: http://en.wikipedia.org/wiki/Color_charge. As it notes, you also can't ...

5

According to the Particle Data Group [1], Mesons are [strongly interacting particles that] have baryon number $\mathcal{B}=0$. In the quark model, they are $q\bar{q}'$ bound states of quark $q$ and antiquark $\bar{q}'$. This definition has two sentences, which in fact are somewhat different. The second one defines the quark-model mesons, that you're ...

5

K$^+$ and K$^-$ are antiparticles - they contain the same quarks, but in a different configuration. K$^+$ is $u \overline{s}$, while K$^-$ is $\overline{u} s$. $\Sigma^+$ and $\Sigma^-$ are not antiparticles, they contain different sets of quarks. $\Sigma^+$ is $uus$ and $\Sigma^-$ is $dds$. The $\Sigma \pm$ have different masses, as the $u$ and $d$ quarks ...

4

The constraint on the singlet state $\eta' = \frac{1}{\sqrt{3}}(u\bar u+d\bar d+s\bar s)$ is that it be flavorless, just as the constraint on the singlet state for SU(2) is that it have no angular momentum. To decompose $3\times 3^* = 8 + 1$ you first need to know what are the irreps. You can build these with raising and lowering operators as is done ...

4

1) The Lie group $G$ behind the meson octet is the $SU(3)_{\rm Flavor}$ Lie group over the three lightest quark flavors $u$, $d$, and $s$. More precisely, the fundamental $SU(3)$ representation $$V~=~\text{Fund} ~=~{\bf 3}$$ is a linear span of the $u$, $d$, and $s$ quarks. The three complex coefficients are collected in a $3\times 1$ column vector $v\in V$...

4

I think this is so that the B-meson has a positive 'bottomness' by definition. Bottomness is a quantum number which is the (signed) difference between the number of bottom antiquarks and bottom quarks. The bottom quark has an electric charge of -1/3, so the bottomness is given the same sign as the charge, which results in the bottom quark having a ...

4

Assume for the sake of argument that this decay channel happens. Now, ask yourself how you are going to prove it? How about we compute the rate for each channel and see if the real rate is the sum? This works when the added channel is a reasonable fraction of the dominate channel, but you're talking about comparing a strong decay to a weak one. This is ...

4

How can we measure meson decay constants? I am not an experimental physicst, but I think that the best way to obtain the decay constant is to study processes like $\pi^+\to \mu^+ \nu$ and extract them from the branching ratio: $$\rm{Br}(\pi^+\to \mu^+\nu)=\dfrac{G_F^2 m_{\pi^+} m_\mu^2}{8 \pi}\left(1-\dfrac{m_\mu^2}{m_{\pi^+}^2} \right)^2 f_{\pi^+}^2 |V_{... 4 The Yukawa meson theory was a theory developed to describe the force between protons and neutrons in the nucleus. It works very well at low energies but fails completely at energies high enough that the internal structure of the hadrons becomes important. The strong force, or more precisely quantum chromodynamics, describes the interactions between quarks, ... 4 This is a good question that puzzled theorists for a while, until the modern understanding of chiral symmetry breaking in QCD clarified itself. The crucial thing to note is that the quadratic formula you are quoting is valid, and necessary, for pseudoscalar mesons only---the abnormally light pseudogoldstone bosons of spontaneously broken chiral symmetry. By ... 4 The antiparticle of eta meson is the eta meson itself.$$ \eta = \dfrac{1}{\sqrt{6}}(u\bar{u} + d\bar{d} -2s\bar{s}) As you can see, since the eta is composed of the terms $q\bar{q}$, therefore taking the charge conjugate of $\eta$ will take it back to $\eta$.

4

I think you may be asking a few related questions, so here's my attempt to answer some of them: Are there examples of scalar Goldstone bosons? Yes, at least at a hypothetical level. One simple example in particle physics is the dilaton, the Goldstone boson of spontaneous breaking of scale symmetry. Remarks: When scale symmetry is broken by strong ...

4

I'm not an expert in HEP, but I guess what you mean by asymmetry in the decay of K meson is the $\textbf{CP violation}$ process and Cronin together with Fitch demonstrated this in their experiment back in 1964. CP invariance is one of the conservation laws which govern the general behavior of the physical particles (C for charge conjugation and P for parity)...

4

For two particles to interfere, you need them to have similar quantum numbers. For example, a $\rho^+$ could not interfere with the neutral $\phi$ or $\omega$, since the respective transition would violate charge. What about the $\rho^0 =(u\bar{u}-d\bar{d})/\sqrt{2}$? It sure has a similar mass to the $\omega=(u\bar{u}+d\bar{d})/\sqrt{2}$, since these ...

4

Note that in the 3-dimensional complex space spanned by basis $\boldsymbol{\lbrace}\boldsymbol{u}\overline{\boldsymbol{u}},\boldsymbol{d}\overline{\boldsymbol{d}},\boldsymbol{s}\overline{\boldsymbol{s}}\boldsymbol{\rbrace}$, this basis is replaced by $\boldsymbol{\lbrace}\boldsymbol{\pi^{0},\boldsymbol{\eta},\boldsymbol{\eta}^{\prime}}\boldsymbol{\rbrace}$ ...

4

Yes, there are some oddball baryons that contain four quarks and one anti-quark. They're called pentaquarks. These could be thought of as a baryon with three quarks plus a meson with a quark anti-quark pair, but sticking to each other more than they should. There are few hadrons that survive more than a tiny fraction of a second, and only the proton and ...

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