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6

Neutron degenerate matter can undergo a phase transition to a superfluid state. The process is thought to be analogous to Cooper-pairing, but the coupling interaction is of order 1 MeV, so can occur at temperatures below about $10^{9}$ K in neutron star interiors. The neutrons in the deep interior (which dominate the interior 100:1) can form a superfluid; ...


3

At least in the context of ultracold atomic fermions, the answer is no. The creation of a degenerate fermi gas is, unlike a BEC, not a phase transition. One major caveat: if there is an attractive force between the fermions, one can get a BCS-like phase transition to condensation of paired fermions. This is, of course, the case for electrons in metals, as ...


1

Electron and holes are Fermions (particles with spin 1/2). This means that no two particles can share the same microstate. The Fermi-Dirac distribution describes how Fermions fill the available states consistent with this property. Bosons on the other hand (particles with integer spin) can occupy the same state. The Bose-Einstein distribution describes how ...


1

If we're dealing with a finite-dimensional Hilbert space, then the answer appears to be that you can always find a way to label the eigenvalues such that they are all differentiable (at least). See this paper and references therein: A. Parusinski & A. Rainer, "A New Proof of Bronshtein's Theorem". In particular, Theorem 2.4 of the paper states ...


5

Yes, neutrinos should obey Fermi-Dirac statistics and yes, the Pauli Exclusion Principle should operate for neutrinos. But let's examine how dense the neutrino population has to be for this to be important. The Fermi momentum is given by $$ p_F = \left( \frac{3}{8\pi}\right) h n_{\nu}^{1/3} $$ where $n_{\nu}$ is the neutrino number density. In order to be ...


1

The mechanism for "giving mass" to elementary bosons and fermions is different. With bosons, it is related to the gauge symmetry ($SU(3)_c \times SU(2)_L \times U(1)_Y$) which is partially broken (and become $SU(3)_c \times U(1)_{em})$. The unbroken part imposes its associated bosons (gluons and photon) to be massless to respect this symmetry. With ...


-1

I'm wondering if there is any explanation for why bosons(specifically gauge bosons) can be massless (photon and gluon) but we don't see any fundamental massless fermions. It's because a fermion is a "body", and because "the mass of a body is a measure of its energy content". See Einstein's E=mc² paper. He talks about a body and an electron here. "The ...


2

Perhaps the easiest way to see that there should be a Grassmann sign factor $(-1)^{|A| |B|}$ in the definition of time ordering $$\tag{1} {\cal T} \left\{ A(t_A) B(t_B)\right\} ~:=~ \theta(t_A-t_B) A(t_A) B(t_B) + (-1)^{|A| |B|} \theta(t_B-t_A) B(t_B) A(t_A), \qquad $$ is to go to the classical limit $\hbar\to 0$. Here $|A|$ denotes the Grassmann parity, ...


1

Use the property $$ (\sigma^\mu)_{\alpha{\dot\alpha}} ({\bar \sigma}_\mu)^{{\dot \beta}\beta} = - 2 \delta_\alpha^\beta \delta_{\dot\alpha}^{\dot\beta} $$



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