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seen Dec 3 at 11:18

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?