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 Sep 24 comment How can the unstable particles of the standard model be considered particles in their own right if they immediately decay into stable particles? @Yakk Yes, the idea is more or less like this. But White dwarfs do not provide dense enough envirtonment, see en.m.wikipedia.org/wiki/Fermi_energy#White_dwarfs . The states with energies of the order of muon mass are not occupied. Sep 24 comment How can the unstable particles of the standard model be considered particles in their own right if they immediately decay into stable particles? @Yakk actually, yes, in some sense. I don' feel that realistic environment can make a muon (or other unstable elementary particle) stable, but for unstable composite particles example is easy to find. Neutron is unstable, while iron nucleus is stable (and so are neutrons inside it). Or a more spectacular example would be a neutron star. Nov 22 comment Horrifying electron gas model No. Example-state $|\psi\rangle=(a^\dagger_1+a^\dagger_2)|0\rangle$. Then $\langle\psi|a^\dagger_2 a_1|\psi\rangle=1$ Nov 22 revised Kinetic theory of photon gasses edited body Nov 22 comment Horrifying electron gas model As regarding to $\langle a^\dagger_{k-Q}a_l\rangle$ - I guess you problem is that you mess vacuum of the free Hamiltonian with the vacuum of the full system. The first one is being annihilated by $a_k$, while the vacuum state for the full Hamiltonian is not. The expression you are calcualting is in the vacuum of the full system. Nov 22 comment Why is $\nabla\cdot(\hat{\bf r}/r^2)$ giving 0 as answer? Try to write it explicitly in $x,y,z$ components and calculate. It is very instructive and you'll get 0 everywhere, except for badly divergent quantity for $r=0$. Nov 22 answered Kinetic theory of photon gasses Nov 22 comment Creation and annihilation operators in Hamiltonian $a^\dagger_k|0\rangle$ is the eigenstate of the Hamiltonian $H_0=\sum_k\epsilon_ka^\dagger_ka_k$ with continuum spectrum $\epsilon_k$. Sum is an integral for continuum $k$ Nov 22 awarded Editor Nov 22 revised Why we see more diverging light rays than converging light rays? edited body Nov 22 comment Creation and annihilation operators in Hamiltonian Simple answer-just the free kinetik part. If the operators are labeled by momentum, this is virtually always the case. Nov 22 comment Creation and annihilation operators in Hamiltonian "Usually" means that this is the normal convention, and you'll never find anything else. However, you really find different choices of the free part -- in may or may not contain the mass fo the particle, for example. Nov 22 comment Creation and annihilation operators in Hamiltonian The "sums" in such discussions are always infinite. For the case of continuous "label", like momentum $k$, the sum should be replaced by an integral (not much difference, apart form proper normalization of the states, which is delta-function over $k$ in this case). Example: a one particle wavepacket would look like $\int dk \psi(k) a^\dagger_k|0\rangle$. Two particle states would look like $\int dk_1 dk_2 \psi(k_1,k_2)a^\dagger_{k_1}a^\dagger_{k2}|0\rangle$ (with proper symmetry properties for the function $\psi(k_1,k_2)$). Of course the function $\psi$ here should be properly normalized. Nov 22 answered Creation and annihilation operators in Hamiltonian Nov 22 answered Why we see more diverging light rays than converging light rays? Feb 22 awarded Supporter Sep 20 awarded Teacher Sep 20 answered Can a single particle create a black hole?