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Aug
13
comment Can observables with discrete and continous eigenvalues be commuting?
The identity operator has discrete spectrum and commutes with every operator.
May
13
comment Why are there gapless excitations in the anti-ferromagnetic Heisenberg model while the true ground state is a singlet?
to "we do have anti-ferromagnetic magnons which are gapless excitations" : not always true if Haldanes conjecture is true : physics.stackexchange.com/questions/59986/…
May
8
awarded  Yearling
May
4
comment What is the precise definition of state of a quantum system?
Note that if you multiply the vector by a phase factor it is still the same state.
Apr
25
answered Complex conjugate of momentum operator
Feb
4
comment Is the second law of thermodynamics a fundamental law, or does it emerge from other laws?
@Lubos Motl : The second law of thermodynamics can't be true for all times if Poincare recurrence occurs. And Poincare recurrence occurs according to your answer in physics.stackexchange.com/questions/94122/… .
Jan
17
comment Is Poincare recurrence relevant to our universe?
The cosmic horizon depends on the observer, so does the finite-dimensional Hilbert space depend on the observer ? If yes, how are the Hilbert spaces related ?
Nov
17
comment Physical Interpretation of Relationship Between Hall Conductivity and Berry Curvature?
@Everett You : It's a counter example to the quantization to the integer value and to Laughlin's Argument.
Nov
16
comment Physical Interpretation of Relationship Between Hall Conductivity and Berry Curvature?
@Everett You : The fractional quantum hall effect is a counter example.
Nov
12
comment What exactly is $\hat{\psi}^\dagger(x)$? An operator or a function?
to "I'd need to think about it, but I believe the notion of an unbounded-operator valued distribution doesn't even make sense mathematically" : It makes sense, see e.g. Streater-Wightman axioms. $\hat{\psi}^\dagger(x)$ must be a distribution since $[\hat{\psi}(y),\hat{\psi}^\dagger(x)]=\delta(y-x)$
Nov
11
comment What exactly is $\hat{\psi}^\dagger(x)$? An operator or a function?
$\hat\psi^\dagger$ is not an operator valued function but an operator valued distribution.
Oct
9
comment The state of Indefinite metric in Quantum Electrodynamics
Do you mean this : en.wikipedia.org/wiki/Gupta%E2%80%93Bleuler_formalism ?
Oct
1
comment Is the Lorentz group compact (and if not, is U(1)?)
to "only compact groups have finite-dimensional representations" : No, all unitary irreducible representations of abelian groups are one dimensional no matter whether they are compact or not.
Sep
30
comment Maximum theoretical data density
Is the answer also true shortly after the big bang ?
Sep
29
answered Applications of Algebraic Topology to physics
Sep
28
comment Does the Hilbert space of the universe have to be infinite dimensional to make sense of quantum mechanics?
@Arnold Neumaier : You are on the wrong track, see planetmath.org/directintegralofhilbertspaces
Sep
13
comment Does the Hilbert space of the universe have to be infinite dimensional to make sense of quantum mechanics?
@Arnold Neumaier : Take the poincare group and a Hilbert space of 2 particles : This is a separable Hilbert space and a continuous direct integral over the one-particle spaces corresponding to the center of mass motion. Or look at $L^{2}(\mathbb{R}^{2})$ which is a continuous direct integral of copies of $L^{2}(\mathbb{R})$.
Sep
12
comment Superconducting Wavefunction Phase (Feynman Lectures)
Neumann boundary conditions imply that $\theta$ must be constant, see maths.qmul.ac.uk/~wjs/MTH5102/laplace10.pdf
Sep
12
comment Does the Hilbert space of the universe have to be infinite dimensional to make sense of quantum mechanics?
@Arnold Neumaier : I think, the direct integral of separable Hilbert spaces is again separable. For instance, if you have a unitary representation of a locally compact separable group on a separable Hilbert space, then this Hilbert space can be written as a direct integral over the irreducible representations of the group.
Aug
27
comment Does every hermitian operator represent a measurable quantity?
@joshphysics : each normal (maybe unbounded) operator can be written as an integral over projection operators (which are bounded). So if you know the expectation values of the projection operators you know also the expectation value of the normal operator.