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Let H be a Hilbert space and A, B two non-commuting bounded linear operators. Let Com(A,B) be the set of bounded linear operators C which commute both with A and B.

Question 1 : What is known about Com(A,B) ?

And in the quantum mechanical context :

QUestion 2 : What is known about Com(A,B) when H is a complex Hilbert space, and all the operators are Hermitian ?

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2  
Your title and tagging are confusing, since have nothing to do with the question. –  Wadim Zudilin Jul 4 '10 at 13:55
    
I've re-tagged the problem as 'linear algebra'. (The problem title is appropriate, but vague; quantum observables are hermitian operators on a Hilbert space.) –  Niel de Beaudrap Jul 4 '10 at 17:59

2 Answers 2

In both cases, Com(A,B) is a weakly closed algebra (i.e., closed in the weak operator topology). In the second case, it is a *-algebra too, so it is a von Neumann algebra.

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If I understood you right:

If we have two-commuting bounded linear operators (if they Hermitian), in quantum physics it's mean, if I'm right, that we can't make such observation and determine the state of both parameters of the system. But the suggestion, that we have such operator that commute to both can possibly make both states — A and B determined simultaneously.

So, for my small opinion, Com(A, B) didn't exist or equal to zero.

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2  
Com(A,B) includes the identity operator, at the very least. –  userN Jul 4 '10 at 16:24
    
C(A,B) is not void since it contains all scalar multiples of the identity operator. –  Elemer E Rosinger Jul 4 '10 at 17:15
    
Your comment that in the Hermitian case Both observables may end up being simultaneously measurable is precisely the motivation for my question. –  Elemer E Rosinger Jul 4 '10 at 17:17

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