jackeslenoa@yahoo.com
Dec22-04, 05:57 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Hi, I was trying to understand the modal interpretation of\nVermaas and Dieks, and I found some probles frome the very beginning.\nAs far I could see, this interpretation is based in the spectral\ndescomposition of the density matrix, and the porperty assignment are\nmade to the eigenprojectors of that matrix, with probability related to\nde enigenvalues of each eingenprojectors.\nNow, my question is related with the uniqueness of the\nspectral descompostion. It is well know that the there are many basis\nin which de density matrix is diagonal. In the other hand, the spectral\ntheorem asserts that "Every normal operator has a unique set of\ncomplete orthonormal, eigenvectors, the so called eigenbasis with the\ncorresponding set of eigenvalues, and conversely every basis with\neigenvalues determines uniquely a normal operator."\nMaybe i need some mathematics (in fact, i need it), but I see\na contradiction between the to fact listed above. =BFIt is not the same\nthe spectral descomposition and the diagonal form of the density\nmatrix? I always thought it was, but semms not.\n\nJack\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Hi, I was trying to understand the modal interpretation of
Vermaas and Dieks, and I found some probles frome the very beginning.
As far I could see, this interpretation is based in the spectral
descomposition of the density matrix, and the porperty assignment are
made to the eigenprojectors of that matrix, with probability related to
de enigenvalues of each eingenprojectors.
Now, my question is related with the uniqueness of the
spectral descompostion. It is well know that the there are many basis
in which de density matrix is diagonal. In the other hand, the spectral
theorem asserts that "Every normal operator has a unique set of
complete orthonormal, eigenvectors, the so called eigenbasis with the
corresponding set of eigenvalues, and conversely every basis with
eigenvalues determines uniquely a normal operator."
Maybe i need some mathematics (in fact, i need it), but I see
a contradiction between the to fact listed above. =BFIt is not the same
the spectral descomposition and the diagonal form of the density
matrix? I always thought it was, but semms not.
Jack
Vermaas and Dieks, and I found some probles frome the very beginning.
As far I could see, this interpretation is based in the spectral
descomposition of the density matrix, and the porperty assignment are
made to the eigenprojectors of that matrix, with probability related to
de enigenvalues of each eingenprojectors.
Now, my question is related with the uniqueness of the
spectral descompostion. It is well know that the there are many basis
in which de density matrix is diagonal. In the other hand, the spectral
theorem asserts that "Every normal operator has a unique set of
complete orthonormal, eigenvectors, the so called eigenbasis with the
corresponding set of eigenvalues, and conversely every basis with
eigenvalues determines uniquely a normal operator."
Maybe i need some mathematics (in fact, i need it), but I see
a contradiction between the to fact listed above. =BFIt is not the same
the spectral descomposition and the diagonal form of the density
matrix? I always thought it was, but semms not.
Jack