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1d

comment 
How to prove this determinant is positiveII?
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 positiveII?
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 positiveII?
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 
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Number theory and physics
see also : physics.stackexchange.com/questions/26856/… 
Nov
7 
awarded  Nice Question 
Nov
7 
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Unital $C^{*}$ algebras which all elements have path connected spectrum
Example of a connected Cā algebra : See B.E. Blackadar. A simple unital projectionless Cāalgebra. J. Operator Theory, 5:63ā71, 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 nonnormal states exist in the Solovay model? 
May
28 
awarded  Enthusiast 
May
26 
comment 
Do nonnormal 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 
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Do nonnormal states exist in the Solovay model?
@Ashutosh : to lo.logic + math.physics : See the comments to Arnold Neumaiers answer in physicsoverflow.org/31006/whyisntthepathintegralrigorous 
May
25 
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Do nonnormal states exist in the Solovay model?
@Ashutosh : Yes, at least for separable Hilbert spaces. 
May
25 
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Do nonnormal 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 sigmafinite measures, see ncatlab.org/nlab/show/Gleason%27s+theorem . 
May
25 
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Do nonnormal states exist in the Solovay model?
@Asaf Karagila : All states have norm 1, but nonnormal states are discontinous in some other topologies (e.g. ultraweak), see ncatlab.org/nlab/show/state+on+an+operator+algebra . 