28

Two particles forming an $SU(2)$ doublet means that they transform into each other under an $SU(2) $ transformation. For example a proton and neutron (which form such a doublet) transform as, \begin{equation} \left( \begin{array}{c} p \\ n \end{array} \right) \xrightarrow{SU(2)} \exp \left( - \frac{ i }{ 2} \theta_a \sigma_a \right) \left( \begin{array}{...


12

You probably know that the electrons in atoms occupy a series of energy levels, the $1s$, $2s$, $2p$, etc orbitals. Although the structure of nuclei is complicated, basically the same idea applies to nuclei as well as atoms. This happens because you can't put more than one fermion into the same quantum state. The electrons in atoms are fermions, and so are ...


11

The isospin is different. $I=0$ for the $\Lambda^0$ and $I=1$ for the $\Sigma^{0}$. This makes the $\Lambda^0$ an isospin singlet state but the $\Sigma^0$ is part of an isospin triplet. There are quite few other examples e.g. compare a proton (uud with $I=1/2$) with a $\Delta^{+}$ (uud with $I=3/2$).


10

Symmetry and statistics. The quarks being fermions dictate a fully antisymmetric wavefunction of the three constituents of the baryon. The color wavefunction is antisymmetric, so the combined spin&flavor wavefunction must be symmetric. The spin 1/2 combination (SU(3) octet) is of mixed symmetry and so is the flavor symmetry of the baryon octet you ...


10

Both SU(3) flavor and SU(2) isospin are approximate symmetries of the Standard Model at low energies. Consider physics below the proton mass, where we can talk about the pions and kaons that are the avatars of these symmetries. At energies this low, it doesn't make sense to talk about the heavy quarks (charm, bottom, top), so we're left with the light quarks:...


10

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9

[...] $\Delta^+ \rightarrow p + \pi^0$, [...] $\Delta^+ \rightarrow n + \pi^+$, which process is favored: the proton and neutral pion or neutron and charged pion [?] Since the kinematics (and corresponding "phase space" factors) for the two final states are presumably as good as equal, the evaluation of the branching ratio $$\text{BR} := \frac{\Gamma[ ...


9

You didn't understand any of these questions right. Antiquarks and their bound states, including the antineutrons, are produced and observed as easily as bread and butter. Lots of details experiments with e.g. antineutrons have been performed, e.g. Scattering of antineutrons with hydrogen http://www.sciencedirect.com/science/article/pii/...


9

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 ...


8

An anti-particle transforms in the conjugate transformation $\bar{\mathbf r}$ than the particle ${\mathbf r}$ does under the given symmetry group, $SU(2)$ in your case. No surprise thus in general that the generators $T^i_{\mathbf r}$ and $T^i_\bar{\mathbf r}=-T^{i\,*}_{\mathbf r}$ act differently on ${\mathbf r}$ and $\bar{\mathbf r}$ respectively. This is ...


8

Isospin was "invented" to model experimental properties. Isospin was introduced by Werner Heisenberg in 1932[3] to explain symmetries of the then newly discovered neutron: The mass of the neutron and the proton are almost identical: they are nearly degenerate, and both are thus often called nucleons. Although the proton has a positive charge, and ...


8

First, the neutron $n$ and the proton $p$ don't have "excited counterparts" in the decuplet, either, do they? Now, the two multiplets are completely different. One has eight $SU(3)_{\rm flavor}$ components, the other has ten. So it's clearly invalid to call one group "excitations" of the other. Because the quark content (charge and strangeness) is the same,...


7

There is an easy partial answer to the question. The proton is (uud) and the neutron is (udd). The up quark is 2.2MeV and the down quark is 4.7MeV. So there is a 2.5MeV mass increase with the neutron. The proton is 938.272MeV and the neutron is 939.565MeV which is then 1.293MeV heavier. That is odd! The neutron has a smaller mass difference than just based ...


6

First at all - if I understood right - the existence of antiquarks is hypothetical. If one not agree with this please refer to experimental data which shows their observation. Everything we observe can be considered hypothetical for each of us. It is a hypothesis that you have a screen and are reading this. Maybe it is all a hypotheis in my mind , or your ...


6

They do. It's the third component of isospin, which came about before Murray Gell-Mann's quark model. \begin{equation} I_3 = \frac{(N_u-N_\bar{u})-(N_d-N_\bar{d})}{2} \end{equation}


6

A similar question is the following. How can $\pi^0$ and $\eta$ in the $SU(3)_F$ meson octet both have the same $SU(3)_F$ flavor content? One could answer that this is because $\pi^0$ is part of an isospin triplet of pions with $I=1$, while $\eta$ is an isospin singlet with $I=0$. Or one may point out that their explicit ket linear combinations of quark ...


6

Usually when one says a function $f$ respects a group action, they mean that $f$ and the group action commute. To be specific, let $g\in G$, $f: X \rightarrow Y$, and the group action of $G$ on $X$ and $G$ on $Y$ is defined and denoted by $g\cdot x$ and $g \cdot y$ respectively. Then $f$ is respects the group action if $\forall x\in X, g\in G$ \begin{...


5

I don't know what background you bring to the question. So at the risk of sounding patronizing, let me give a down-to-earth answer. I wonder if this helps. Think of rotations on the (Real) 2-dimensional plane $\mathbb{R}^2$. You can rotate the X-axis into the Y-axis, and the Y-axis into the negative X-axis. This group of 2d rotations is called $SO(2)$. ...


5

Because of $u\leftrightarrow d$ isospin symmetry. For a more detailed explanation, see e.g. chapter 8 of 't Hooft's lecture notes. The pdf file is available here.


5

The reason the signs are flipped from what you expect has to do with the fact that the antiquark transforms in the opposite way under isospin rotations. If the ordinary quark doublet is a column vector $$q=(u, d)^T$$ and transforms under rotations as $$q\rightarrow U(R) q$$ the antiquark doublet is a row vector $$\bar{q}=(\bar{u}, \bar{d})\rightarrow \bar{q} ...


5

To start 3 quarks u,d,s form an irreducible representation for SU(3) group. Now $3\times 3\times 3 = **10+8+8+1**$ for the SU(3) group. The first irreducible representation 10 has 10 purely symmetric states. The next representation 8 has 8 mixed symmetric states. The next representation 8 has 8 mixed anti-symmetric states. The last 1 is purely anti-symmetric ...


5

OK, I'll get to the answer with insouciance that would make Paul Gordan turn in his grave, and only then, separately, I'll dismiss your dilemma by summary reassurance via the Racah recouping coefficients. This is not the most complete answer, but it is by far the fastest one, exploiting a non-universal piece of luck, here. I'll be pretending isospin is spin ...


5

Except for small electromagnetic decay branching ratios like $\Delta\to N\gamma$, essentially 100% of the branching ratio of the $\Delta$ is to $N\pi$ final states. This is just kinematics. The only other hadronic final state with baryon number one that is energetically allowed is $N\pi\pi$, which has an extremely small phase (it is just barely allowed). ...


5

A phenomenological model can be described in terms of the liquid drop model and semi-empirical mass formula. This ascribes a nucleus a binding energy of $$B(A,Z) = a_v A - a_s A^{2/3} - a_c Z^2A^{-1/3} - a_a \frac{(A-2Z)^2}{A},$$ together with a smaller "pairing term", which I am going to ignore. Here, $A$ is the total number of nucleons and $Z$ is the ...


5

You are contrasting the Weak interactions to the Strong interactions, where the breaking is spontaneous (and large) as contrasted to explicit (and small), respectively. ($e^{-}, \nu_{e}$) are a doublet under weak isospin. This means the the corresponding SM lagrangian is invariant under an SU(2) gauge group -- you could rotate the fields under such a ...


4

It's a (known) linear combination of 0 and 1. The electromagnetic field operator can be written as something like $$A_\mu = B_\mu\cos\theta_W + W_\mu^3\sin\theta_W$$ where $B_\mu$ is the $U(1)$ hypercharge operator and $W_\mu^3$ is the measured component of the weak isospin operator. $B_\mu$ is not associated with any weak isospin, so that's where the 0 ...


4

By isospin I assume you mean weak isospin which is an exact gauge symmetry of the standard model. There is another thing called flavour isospin which is an approximate global symmetry of the strong interaction. Also, this is a long reply because I've included background and example that you may or may not need. If I am not explaining things well enough for ...


4

First at all - if I understood right - the existence of antiquarks is hypothetical. Your understanding is entirely incorrect. Anti-quarks are a work-a-day reality in the particle physics world. The annihilation of quarks and anti-quarks to form lepton pairs (i.e. Drell-Yan scattering) is not merely regularly observed, it is used a physics tool to probe the ...


4

The 3 pions can be considered as 3 states of the same particle, the isospin being used to label the 3 states. Since pions are bosons, the total wave function must be symmetric (Pauli principle). The total wave function is the (tensorial) product of space-wave function, spin wave-function and isospin wave-function. Spin wave-function is symmetric since pions ...


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