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Inequality involving positive definite operator

  1. Feb 20, 2009 #1
    1.
    Given that [tex]|\langle f|g\rangle|^2 \leqslant \langle f|f\rangle\langle g|g\rangle[/tex]
    prove that [tex]|\langle f|H|g\rangle|^2 \leqslant \langle f|H|f\rangle \langle g|H|g\rangle[/tex]
    where [tex]H[/tex] is a Hermitian and positive definite operator.




    3. I tried to identify [tex]H|g\rangle[/tex] as a state and put it into the given inequality, but not so help.

    Is there any ideas to prove this inequality? Thanks in advance.
     
  2. jcsd
  3. Feb 20, 2009 #2
    How about trying sqrt(H)?

    You will also need to think about the conditions H should meet for its square root to exist.
     
  4. Feb 20, 2009 #3
    Thank you so much!!
    I got it, and in this way, Hermitian and positive definite properties are both used.
    Thanks!
     
  5. Feb 20, 2009 #4

    Ben Niehoff

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    Science Advisor
    Gold Member

    The problem is kind of silly to begin with. If you can show that

    [tex]\langle \cdot | H | \cdot \rangle[/tex]

    satisfies the defining properties of an inner product:

    1. Positive definite,

    2. Linear on second slot,

    3. Conjugate-linear on first slot,

    then the expression automatically obeys the Schwartz inequality, because the Schwartz inequality is true of inner products in general.
     
  6. Feb 20, 2009 #5

    Well, you are right. I keep forget that an inner product doesn't always have to be what we usually consider as 'the inner product'.
     
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