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1d
comment How to prove this determinant is positive-II?
The quadratic form is $q(x,y) = x^T J y$ where $J = diag(I_n,-I_n)$ and the semi group is given bei the real S with $q(Sx, Sx) \ge q(x,x)$ for all $x$ . Now $\frac{d}{dt} q(e^{t A_i} x,e^{t A_i} x) = 2 x^T diag (E_i,F_i) x \ge 0$ at $t = 0$, therefore the $e^{A_i}$ are in the semi group.
2d
comment How to prove this determinant is positive-II?
The $e^{A_i}$ increase the quadratic form that the split orthogonal group leaves invariant . So maybe this is the generalization of Part I : The split orthogonal group is replaced by the semi group that increases the quadratic form.
Feb
3
comment How to prove this determinant is positive-II?
to the warm up : $det(I_n +\prod_i e^{-F_i}) \ge 0$ because $\Vert\prod_i e^{-F_i}\Vert \le 1$ and $F_i$ real . Then also $det(I_n +\prod_i e^{E_i}) = det(\prod_i e^{E_i}) det(I_n +(\prod_i e^{E_i})^{-1}) \ge 0$ .
Jan
24
comment Resolvent of the operator
Calculate the eigenfunctions : T is the Hamiltonian of an electron in a constant magnetic field perpendicular to the plane. The eigenfunctions are the elements of the Landau levels.
Jan
23
comment States in C*-algebras and their origin in physics?
Within ZFC there are states that are not represented by a density matrix.
Jan
13
comment Does quantum mechanics ever really quantize classical mechanics?
And what is the "sample space" in quantum mechanics ? For a sample space you need a Hilbert space basis, but what basis should one use ? According to decoherence theory there are "robust" pointer states. But you have to split the Hilbert space into system and environment and integrate out the environment degrees of freedom if you want to see this.
Jan
11
comment Does quantum mechanics ever really quantize classical mechanics?
The problem with this model is that superpositions of states last forever, since quantum mechanics is a linear theory, but in the classical world we don't see superpositions of macroscopic states. The cure for this problem could be decoherence theory (already mentioned in the question).
Nov
25
comment Number theory and physics
see also : physics.stackexchange.com/questions/26856/…
Nov
7
awarded  Nice Question
Nov
7
comment Unital $C^{*}$ algebras which all elements have path connected spectrum
Example of a connected Cāˆ— algebra : See B.E. Blackadar. A simple unital projectionless Cāˆ—-algebra. J. Operator Theory, 5:63ā€“71, 1981. By the comments of Hannes Thiel and Sam Evington this algebra is an example of a connected C* algebra. But is it also path connected (remark by Ali Taghavi) ?
Sep
12
awarded  Nice Answer
Aug
19
awarded  Yearling
Jul
8
comment Quantum Fields and Infinite Tensor Products
'tiny separable subset' = GNS construction ?
Jun
11
accepted Do non-normal states exist in the Solovay model?
May
28
awarded  Enthusiast
May
26
comment Do non-normal states exist in the Solovay model?
This should be a nonnormal state on a nonseparable Hilbert space in the Solovay model : Let $H = l^2([0,1])$ and $\lbrace e_x : x \in [0,1] \rbrace$ an orthonormal basis of H . Then $f$ defined by $f(A) = \int_0^1 <e_x, A e_x> dx$ for $A \in B(H)$ is a nonnormal state .
May
26
comment Do non-normal states exist in the Solovay model?
@Ashutosh : to lo.logic + math.physics : See the comments to Arnold Neumaiers answer in physicsoverflow.org/31006/why-isnt-the-path-integral-rigorous
May
25
comment Do non-normal states exist in the Solovay model?
@Ashutosh : Yes, at least for separable Hilbert spaces.
May
25
comment Do non-normal states exist in the Solovay model?
@Ashutosh : According to Gleasons theorem for separable Hilbert spaces of dimension other than 2 states correspond to finitely additive measures on pairwise orthogonal projections but normal states correspond to sigma-finite measures, see ncatlab.org/nlab/show/Gleason%27s+theorem .
May
25
comment Do non-normal states exist in the Solovay model?
@Asaf Karagila : All states have norm 1, but non-normal states are discontinous in some other topologies (e.g. ultraweak), see ncatlab.org/nlab/show/state+on+an+operator+algebra .