New answers tagged

0

Use the definition $<v_x> = \int d^3 v v_x (exp(\frac{E}{k_B T})+1)^{-1}/ Z$ with sum of all states $Z$ and $E = \frac{m}{2}(v_x^2+v_y^2+v_z^2)$, introduce spherical coordinates and make the substitution $z = |\vec{v}|/(\sqrt{2 k_B T/m})$. What is if $T=0$?


3

Because we have observed processes where the photon number is not conserved. For example positronium can decay into 2 or 3 (or more) photons. This means that it is not possible to assign a global conserved charge to photons. For neutrinos we can assign lepton number and so far we have not observed a process that would violate total lepton number (lepton ...


0

Kinetic energy in 1D, method 1. Free electrons. Assume no potential energy at the moment. Zero temperature. \begin{equation} n_e=\int_0^{E_F}g(\epsilon)d\epsilon=\int_0^{E_F}\frac{1}{\pi\hbar}\sqrt{\frac{m}{2\epsilon}}d\epsilon=\frac{\sqrt{2mE_F}}{\pi\hbar} \end{equation} \begin{equation} E_F=\frac{\pi^2\hbar^2n_e^2}{2m} \end{equation} Kinetic energy, ...


6

It probably does not mean anything. That paper concerns the quantization of electromagnetic waves in less than three spatial dimensions. In fact, there are a number of decades-old results showing that it is often possible to evade the spin-statistics relationship in lower-dimensional systems. While these kinds of results (including this new one) may be ...


6

Yes, it is not possible to construct a totally antisymmetric spin state with more than two electrons. This is just a statement of Pauli's exclusion principle.


2

In a degenerate gas of fermions, the fermions fully occupy momentum states from zero up to a momentum corresponding to the Fermi energy. It is the momentum of the fermions that leads to degeneracy pressure. As long as the kinetic energy of particles at the Fermi energy is much less than $kT$, then the fermions can be considered completely degenerate, so ...


2

Yes, there is systematic way called the Noether procedure. Simply you write down all possible 2-derivative fermionic terms with arbitrary coefficients and vary the action using the SUSY transformation rules. Then, you fix the coefficients to obtain the invariance up to a total derivative. When you have 4-derivatives, there are two cases: a. Off-Shell ...


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 ...



Top 50 recent answers are included