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Feb 6, 2012 at 16:43 vote accept Mahmood Alaghmandan
Feb 6, 2012 at 16:43
Feb 6, 2012 at 16:41 comment added Mahmood Alaghmandan Yes, you are right. But let me change my question somehow that do not have all elements in $B(H)$. So above I restrict myself to a von Neumann subalgebra of $B(H)$.
Feb 6, 2012 at 10:13 history answered Jochen Wengenroth CC BY-SA 3.0