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12

The deuterium nucleus is a boson, with spin $\hbar$ (and positive parity). Unlike other stable nuclei, deuterium doesn't have any bound excited states; however if it did they would also have integer spin. The deuterium atom is a fermion, which may have spin $\frac12\hbar$ or $\frac32\hbar$, to be combined with the orbital angular momentum (which is zero in ...


-3

Protons, electrons, and neutrons too. Hydrogen is a boson, deuterium is a fermion.


0

I've clarified my doubt. The expression $ (\bar{\psi} \chi )^{T }=\chi^{T }\bar{\psi} ^{T } $ is simply the definition of trasposition (the same argument for the adjoint), independently on the nature of the objects in the parenthesis. What is subtle is that, the following is true: $ \bar{\psi} \chi =-\chi^{T }\bar{\psi} ^{T } =-(\bar{\psi} \chi )^{T }$ ...


0

I think the short answer is that they mix, because nothing prevents them from not doing it. The fact is that states with equal quantum numbers in general do mix. The experiment will tell you how much they mix. If you find a very or almost zero mixing you can start thinking to add a symmetry that prevents some mixing to happen, which was the case of a lot ...


1

A fermion is just a particle of half-integer spin. Being a lepton for a particle is a matter of definition of global symmetries of the theory. This means that a lepton can in principle be both a fermion or a boson, although all known leptons are fermions (electron, muon, tau and their neutrinos). One example of bosonic lepton is the weak triplet Higgs ...


15

A fermion is any particle, elementary or composite, that obeys Fermi-Dirac (as opposed to Bose-Einstein) statistics relating to how identical particles behave when you swap two of them. Due to an important but complicated result, this is taken to amount to having half-integer spin. A lepton is one type of elementary particle with spin 1/2. The only leptons ...


9

A fermion is any particle characterized by Fermi–Dirac statistics and obeying the Pauli exclusion principle. So for example quarks are fermions, as are Helium-3 atoms. A fermion does not have to be an elementary particle. I'm not even sure that it has to be spin $\tfrac{1}{2}$, though I can't think of any fermions that aren't. A lepton is a spin ...


4

The Standard Model includes 12 elementary known as fermions that respect the Pauli exclusion principle. They include six quarks (up, down, charm, strange, top, bottom), and six leptons (electron, electron neutrino, muon, muon neutrino, tau, tau neutrino) (ref) All leptons are fermions, but not all fermions are leptons.


2

Note that Sommerfeld's model simply generalizes Drude's theory of metals by taking into account the fact that electrons are fermions, so Pauli exclusion becomes a very important factor. In Sommerfeld's model, there's no effective mass to talk about, as one basically ignores the atoms(nuclei) in the system and considers free moving fermions. So there, your ...


0

I think you can consider the anti-commutator, then use the linearity of the trace, as follows: $$ \{\overline{\psi},\chi\} = 0$$ $$ \{\overline{\psi},\chi\}^T = 0$$ $$ (\overline{\psi}\chi)^T + (\chi\overline{\psi})^T = 0$$ $$ (\overline{\psi}\chi)^T + \overline{\psi}^T\chi^T = 0$$ $$ (\overline{\psi}\chi)^T - \chi^T\overline{\psi}^T = 0$$ $$ ...


0

We are interested in computing $$ \Pi^{\mu\nu,ab}(x) = \langle V^{\mu,a}(x) V^{\nu,b}(0) \rangle = \langle : {\bar \psi}(x) \gamma^\mu M^a \psi(x) : : {\bar \psi}(0) \gamma^\nu M^b \psi(0) : \rangle $$ Several structures of this quantity can be directly computed. Firstly, let us study the index structure. Firstly, due to translational invariance we have ...


1

In my understanding $:\hat{f}_2\hat{f}_1^\dagger\hat{f}_2^\dagger:$ is defined to be equal to $\hat{f}_1^\dagger\hat{f}_2^\dagger\hat{f}_2$. Otherwise there would be no need to write the two "$:$" and give this operation a special name. What your are doing is applying the ordinary commutator relations. Therefore $$ ...


2

What you are missing is that you are calculating the Jacobian, not simply multiplying $d\psi$ by $d\bar\psi$. The determinant also goes downstairs instead of upstairs, because that's how Grassmann numbers roll. See http://en.m.wikipedia.org/wiki/Berezin_integral for details.


4

It's pretty much what Javier Badia said in the comments: Grassmann numbers anticommute. $$\chi_1 \chi_2 = -\chi_2 \chi_1\tag{1}$$ or in this case, $\chi\bar\chi = -\bar\chi\chi$. Note that this implies the square of any Grassman number is zero, if you set $\chi_1 = \chi_2 = \chi$ in equation (1). Using these properties and some very careful algebra, you ...


1

If you assume that whatever generates the mixing patterns of quarks and leptons (beyond the SM) has no underlying symmetry and that nature chose $V^{CKM}$ and $V^{PMNS}$ randomly within the set of $3\times3$ unitary matrices, then it is natural to expect mixing between families because the probability of randomly selecting $V^{CKM}=V^{PMNS}={\mathbb 1}$ is ...



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