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reviewed Approve suggested edit on Lebesgue integration
2d
comment Understanding Well Defined States
$H$ is the generator of time translations. So the explanation in my comment above applies to the time profile of a wavefunction $\psi(x,t)$ which is an energy eigenstate of the Hamiltonian: $H \psi(x,t) = E \psi(x,t) \implies \partial_t \psi(x,t) = E \psi(x,t)$.
Apr
21
reviewed Approve suggested edit on Stable equilibrium
Apr
21
comment Understanding Well Defined States
Naively speaking, such a wavefunction (evolution) is continuous but not not smooth, so that signals something very weird. More seriously, I think you're imagining an analog with particle in a box, whose Hamiltonian is quadratic $H = \partial_x^2$. So an oscillator over a finite duration continuously connected with "zero" wavefunction is an energy eigenstate with an eigenvalue of $-1$ i.e. a phase shift of $\pi$. On th eother hand, $H \sim \partial_t$ is not quadratic, so it shifts phase of the oscillation only by $\frac{\pi}{2}$ -- and your wavefunction is NOT an eigenstate of this operator.
Apr
20
comment How could there be a truly “pure” state?
Your question can be rephrased to: "How can there be truly isolated systems?" So the answer is the same :-)
Apr
19
reviewed Approve suggested edit on Universe inflation
Apr
19
comment What are threshold corrections?
@whistles: I think it might be exactly related to your infinite set of higher-dimensional operators. Each of these operators is suppressed by a suitable power of the mass scale where you've integrated out something. As you approach that mass scale (from below), you cannot truncate your calculation... but need to "resum" the full set of effective higher-dim operators. I think some such "non-perturbative" correction that happens close to a threshold scale is what the phrase refers to. A "threshold" typically refers to an energy scale where you start producing/sensing new degrees off freedom.
Apr
19
answered Understanding Well Defined States
Apr
19
reviewed Approve suggested edit on Understanding Well Defined States
Apr
18
reviewed Approve suggested edit on Universal gravity at small distance
Apr
17
reviewed Approve suggested edit on $U(1){\times}U(1)$ local gauge invariance derivative
Apr
14
answered Entanglement in single particle state
Apr
14
reviewed Approve suggested edit on What is the notion of a spatial angle in general relativity?
Apr
12
comment Can dimensional regularization solve the fine-tuning problem?
Every regulator carries an opinion about the UV physics. Intuitively, I think dimensional regularization assumes that there is no new UV physics. Further, one consistent way to compute divergent integrals is to just drop the power-law divergences, which is what dim-reg does. So it is unphysical inasmuch as you expect to see new UV physics.
Apr
12
answered why do the electroweak vacuum have to be charge and color neutral?
Apr
12
comment why cannot fermions have non-zero vacuum expectation value?
@Paul: VEV = Vacuum expectation value i.e. a property of the vacuum. So if any object in a non-trivial representation of the Poincare algebra picks up a VEV, then some of the spacetime symmetries will be spontaneously broken by the vacuum (state). One can expect the same to apply to fluctuations around the vacuum. That would mean that the corresponding conserved quantities are not really conserved. And as far as we can see, conservation of energy-momentum and angular momentum apply quite perfectly to our universe.
Apr
12
comment why cannot fermions have non-zero vacuum expectation value?
@innisfree: In any "derivatively-coupled" theory, one cannot have a vev, right? Since there is no special point in field space.
Apr
11
reviewed Approve suggested edit on quadrupole moment and higher for simple current loop
Apr
10
comment Is $\langle k \vert k_1k_2\rangle=0$
In such a case, you'd have a 1-particle state in the initial Fock space and a 2-particle space in the final Fock space and an insertion of the time evolution operator in between them. In the case of an interacing theory with a corresponding trivalent vertex, the evolution operator can indeed cause one particle to decay into two.
Apr
10
comment Is $\langle k \vert k_1k_2\rangle=0$
In other words, a two particle state is orthogonal to a one particle state.