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Apr
24 |
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A curious eigenvalue inequality
With the notation I used in my previous comment the conjectured inequality $tr((U^*AU+B)^2) \le tr((A+B)^2)$ is then equivalent to $tr((R B^2 R)^{1/2} B) \le tr((B R^2 B)^{1/2} B)$ . |
Apr
24 |
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A curious eigenvalue inequality
If we set $R=B^{-1} A U$ then $R$ is Hermitian and we have to show : $\rho((R B^2 R)^{1/2} + B)\le \rho((B R^2 B)^{1/2} + B)$ . |
Apr
23 |
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A curious eigenvalue inequality
Could there exist a completely positive map that maps $A+B$ to $U^* A U + B$ and the identity to itself ? If yes this would be the solution. |
Mar
23 |
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A log inequality for positive definite trace-one matrices
Can you also prove the last inequality of the question ? |
Mar
2 |
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How to prove this determinant is positive-II?
What one needs is $A_i^T J + J A_i \ge 0$ but I haven't checked whether this is a generalization. |
Feb
24 |
answered | How to prove this determinant is positive-II? |
Feb
5 |
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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. |
Feb
4 |
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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 |
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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 |
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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 |
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States in C*-algebras and their origin in physics?
Within ZFC there are states that are not represented by a density matrix. |
Jan
13 |
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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 |
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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 Cā-algebra. 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 |
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Quantum Fields and Infinite Tensor Products
'tiny separable subset' = GNS construction ? |
Jun
11 |
accepted | Do non-normal states exist in the Solovay model? |