# Tag Info

## New answers tagged fermions

0

Why don't we call the fermions in the standard model force carriers? Because the Standard Model is what it is. By the way, I think it's far less complete than people make it out to be, and that it comes with some unfortunate baggage. Consider for example an electron and a positron interacting. People say they interact via virtual-photon force carriers: ...

0

If we identify a force as a scattering process, i.e. with a mediator of some interactions, then this need not be a vector boson of course. One can speak of "Higgs" force for instance if the process under consideration is mediated by the the Higgs (which is a scalar). There are also numerous cases where the interaction is mediated by a fermion. Therefore, ...

2

First, let's clear up some terminology: the usual statement "Majorana fermions are their own antiparticles" is correct, but confusing because the words we usually use to describe neutrinos are made for Dirac fermions. If neutrinos had no mass at all, there would be two independent types of neutrino: a left-handed and a right-handed neutrino. These particles ...

4

Well, $1/2\otimes1/2=0\oplus1$, so a system with two fermions has integer spin. But it is still a two fermion system, and therefore its wavefunction must be antisymmetric, as usual. This is not specific to Cooper pairs, but is basic Quantum Mechanics... [what is specific to Cooper pairs is that their size is $\gg a_0$, which means they are highly ...

2

It's because always even number of Fermionic fields appear in the Hamiltonian. for example the Dirac Lagrangian for free electron: $\mathcal{L}=i\hbar\bar \psi(\gamma^\mu\partial_\mu-m)\psi$ has two $\psi$s. Invariance of the theory upon a global gauge transformation requires that each term in the Lagrangian have an even number of Fermionic fields. Since an ...

4

Essentially you are asking why only scalars are allowed to develop a vacuum expectation value (VEV). A scalar (as the name suggests) does not point to any direction -it has spin 0- therefore it can have a VEV without breaking the Lorentz symmetry. On the other hand, a boson with higher spin, e.g. a vector (spin 1) would spontaneously break Lorentz by ...

0

First, observe that the Slater determinant you have written down is the linear combination of the singlet state and the z-spin-0 state of the triplet. Vice versa, you can produce the singlet and triplet states as linear combinations of Slater determinants. Whether you prefer the Slater determinant or the singlet/triplet formalism for writing down your ...

3

It is not true that all superconductors are gapped. For example, d-wave superconductors in cuprates are gapless. The energy gap in the superconductor arises from the fact that breaking the Cooper pair requires finite energy. The low-lying quasi-particle excitations are all pair breaking excitations, so they are gapped from the ground state by the amount of ...

-4

Do Weyl fermions carry electric charge? That depends on whether Weyl fermions exist, and whether they are what people say they are. See this from the article mentioned by John Rennie: 'Whereas electrons and all the other known fermions have mass, in 1929, mathematician and physicist Hermann Weyl theorized that massless fermions that carry electric ...

2

We should probably start by pointing out that no Weyl fermion has ever been observed. The recent observations are of quasiparticles that behave like Weyl fermions. Speaking rather loosely (and at the risk of upsetting the QFT experts hereabouts) a Dirac fermion can be viewed as a sum of two Weyl fermions, and the observations are of paired quasiparticles ...

1

The fermion masses result from Yukawa interactions after EWSB: $$m_f = \frac1{\sqrt 2} y_f v$$ Thus the Yukawa couplings govern lepton and quark masses. Of course the masses should be diagonalized. For the leptons, as there no neutrino masses in the SM, the lepton interactions and masses can be simultaneously diagonlized, whereas differences in up- and ...

0

In spinor space, the fermionic fields $\psi$ are vectors, for example in the 4x4 representation of the $\gamma$-matrices: $\psi^\dagger = \begin{pmatrix} \psi_1^* & \psi_2^* & \psi_3^* & \psi_4^* \end{pmatrix}^{T}$ and \$\psi=\begin{pmatrix} \psi_1 \\ \psi_2 \\ \psi_3\\ \psi_4 ...

4

The situation is not symmetric at all: This diagram describes a force between two fermions, but a diagram such as just doesn't exist (in the Standard Model). Fermions can in fact mediate a force between bosons, like in: Such diagrams are highly suppressed loop diagrams though, and the one above would after renormalization be seen as just one ...

3

I think that this is more about the historical construction of the theory than about the actual interactions. In a lagrangian, two fields A, B interact when there is a product term of both such as AB. So, I see no real fundamental distinction there, even with more complicated expressions. But when one introduces the interaction bosons, it's by the mean of ...

Top 50 recent answers are included