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Self-adjoint operator problem

  1. Sep 21, 2009 #1
    1. The problem statement, all variables and given/known data

    Show that if the operator Q satifies

    [tex](\phi,Q\phi)=(Q\phi,\phi)[/tex] for all [tex]\phi[/tex], then Q is self-adjoint , that is

    [tex](\varphi,QX)=(Q\varphi,X).[/tex] Consider the functions

    [tex]\phi_1=\varphi+X[/tex],[tex] \phi_2=\phi+i*X[/tex]

    Note: X is NOT a matrix. Could not find the latex code for the curvy X so i just typed X

    2. Relevant equations



    3. The attempt at a solution

    [tex](\phi,Q\phi)=(Q\phi,\phi)[/tex] =[tex](\phi_1,Q\phi_1)=(Q\phi_1,\phi_1)=(\varphi+X),Q(\varphi+X), [/tex] (Q\phi,\phi)=[tex](\phi_2,Q\phi_2)=(Q\phi_1,\phi)=(\varphi+X)i,Q(\varphi+X)i[/tex] ? Am I off in the right direction?
     
  2. jcsd
  3. Sep 22, 2009 #2
    is my question to esoteric
     
  4. Sep 22, 2009 #3

    kuruman

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    It is hard to sort out what you are doing because your Latex code is mangled up. I would start with what is known, namely

    [tex]((\varphi+\chi),Q(\varphi+\chi))=(Q(\varphi+\chi),(\varphi+\chi))[/tex]

    and develop each side separately, i.e.

    Left side =
    Right side =

    Set the two sides equal and use that

    [tex](\chi,Q\chi)=(Q\chi,\chi);\ (\varphi,Q\varphi)=(Q\varphi,\varphi)[/tex]

    to get some terms to cancel. After that you should be able to see how to finish. BTW "curvy X" is \chi.
     
  5. Sep 22, 2009 #4

    shouldn't I plug in [tex]\phi_2[/tex] as well ?
     
  6. Sep 22, 2009 #5

    kuruman

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    Yes, that is part 2. Do this one first to see how it works, then do the other one following the same procedure.
     
  7. Sep 22, 2009 #6
    [tex]
    ((\varphi+\chi),Q(\varphi+\chi))=q(\varphi+\chi), (\varphi+\chi))=q*(\varphi+\chi), (\varphi+\chi))[/tex], q is some number.
    So now what?
     
  8. Sep 23, 2009 #7

    kuruman

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    Are you given that Qφ = qφ where q is some number? It does not appear so. You need to distribute Q and get four terms on each side.
     
  9. Sep 23, 2009 #8
    no, it is given in a section of my textbook but not in the problem.[tex](\varphi+\chi),Q(\varphi+\chi)=(\varphi+\chi),Q(\varphi)+Q(\chi)=(\varphi,Q\varphi)+(\chi,Q(\chi)[/tex] given that
    [tex]
    (\phi,Q\phi)=(Q\phi,\phi)
    [/tex],then [tex](\varphi,Q\varphi)+(\chi,Q(\chi)=Q(\varphi,\varphi)+Q(\chi,\chi)?[/tex]
     
  10. Sep 23, 2009 #9

    kuruman

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    What does your textbook say the definition of the notation

    [tex](\varphi,Q\varphi)[/tex]

    is? Is there any reason to believe that

    [tex](\varphi,Q\varphi)=Q(\varphi,\varphi) ?[/tex]

    The parentheses with the comma in between them mean something. What is it?
     
  11. Sep 23, 2009 #10
    it is the notation for the inner product. I think that I can only move Q out side when its a constant, but Q is a matrix. my book says: [tex](\varphi,Q\varphi)=(Q\varphi,\varphi)[/tex]
     
  12. Sep 23, 2009 #11

    kuruman

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    Right. So you cannot move Q outside the parentheses. It is a matrix and it needs a vector to operate on. This means that Q needs to be inside the parentheses either on the left or on the right side of the comma. What you wrote in posting #8 is incorrect. Try again.
     
  13. Sep 23, 2009 #12
    [tex]
    (\varphi+\chi),Q(\varphi+\chi)=(\varphi+\chi),Q(\varphi)+Q(\chi)=(\varphi,Q\varphi)+(\chi,Q(\chi)= (\varphi,Q\varphi)+(\chi,Q(\chi)=(Q\varphi,\varphi )+(Q\chi,\chi)=(Q\varphi+Q\chi,\varphi+\chi)=(Q(\varphi+\chi),(\varphi+\chi)?
    [/tex]

    therefore[tex] (\varphi+\chi),Q(\varphi+\chi))=(Q(\varphi+\chi),(\varphi+\chi))[/tex] and is self-adjointed?
     
    Last edited: Sep 23, 2009
  14. Sep 23, 2009 #13

    kuruman

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    One more time. Start with the left side and split it into two terms

    [tex]((\varphi+\chi),Q(\varphi+\chi))=((\varphi+\chi),Q\varphi)+((\varphi+\chi),Q\chi) [/tex]

    the last result can be split into four terms

    [tex]=(\varphi,Q\varphi)+(\chi,Q\varphi)+(\varphi,Q\chi)+(\chi,Q\chi) [/tex]

    Now do the same with the right side which is

    [tex](Q(\varphi+\chi),(\varphi+\chi))[/tex]

    then set the two sides equal.
     
  15. Sep 23, 2009 #14
    [tex]
    (Q(\varphi+\chi),(\varphi+\chi))=((Q\varphi,(\varphi+\chi))+(Q\chi,(\varphi+\chi)=
    ((Q\varphi,(\varphi+\chi))+(Q\chi,(\varphi+\chi)=(Q\varphi,\varphi)+((Q\varphi,\chi)+(Q\chi, \varphi)+(Q\chi,\chi)=(\varphi,Q\varphi)+((\varphi,Q\chi)+(\chi, Q\varphi)+(\chi,Q\chi) ?[/tex]
     
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