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Direct product of operators

  1. Aug 7, 2013 #1
    Is this correct in infinite dimensional Hilbert spaces?
    ## (\hat{A}_1 \otimes \hat{A}_2)^{-1}=\hat{A}^{-1}_1 \otimes \hat{A}^{-1}_2 ##
    ## (\hat{A}_1 \otimes \hat{A}_2)^{\dagger}=\hat{A}^{\dagger}_1 \otimes \hat{A}^{\dagger}_2 ##
    ## (\hat{A}_1 +\hat{A}_2) \otimes \hat{A}_3=(\hat{A}_1 \otimes \hat{A}_2)+(\hat{A}_2 \otimes \hat{A}_3) ##
    ## \hat{A}_1 \otimes (\hat{A}_2+\hat{A}_3)=(\hat{A}_1 \otimes \hat{A}_2)+(\hat{A}_1 \otimes \hat{A}_3) ##
    ## \hat{1} \otimes \hat{1}=\hat{1} ##
    ## (\hat{A}_1 \otimes 0)=(0 \otimes \hat{A}_2)=0 ##
    Can you tell me a book where I can see this properties. I found this only for operators which acts in finite dimensional Hilbert spaces.
     
  2. jcsd
  3. Aug 14, 2013 #2
    Your question belongs to the part of mathematics usually discussed within "Operators theory". But here is essentially the positive answer to your questions (with somewhat different notation) - provided appropriate care is being taken:

    stratila58.jpg

    Taken from "Lectures on von Neumann Algebras" by Serban Stratila and Laszlo Zsido, Abacus Press 1975. You can also find it in online Bratteli and Robinson book "Operator algebras and quantum statistical mechanics", Vol. 1, but there it is more complicated as tensor product of not just two but a family of Hilbert spaces is being considered.
     
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