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Show a real, smooth function of Hermitian operator is Hermitian

  1. Apr 9, 2013 #1
    1. The problem statement, all variables and given/known data

    If B is Hermitian, show that BN and the real, smooth function f(B) is as well.

    2. Relevant equations

    The operator B is Hermitian if [itex]\int { { f }^{ * }(x)Bg(x)dx= } { \left[ \int { { g }^{ * }(x)Bf(x) } \right] }^{ * }[/itex]

    3. The attempt at a solution

    Below is my complete solution. I'm hoping someone can review and let me know if it is okay.

    Given the operator B, we can define the operator BN as B[itex]\cdot[/itex]B[itex]\cdot[/itex]B[itex]\cdot[/itex]...BN. Using this, we can show:

    [itex]
    \int { { f }^{ * }(x){ B }^{ N }g(x)dx= } \\ \int { { f }^{ * }(x){ B }BB...{ B }_{ N }g(x)dx= } \\ { \left[ \int { { g }^{ * }(x){ B }BB...{ B }_{ N }f(x) } \right] }^{ * }=\\ \int { { g }(x){ { B }^{ * } }{ B }^{ * }{ B }^{ * }...{ B }^{ * }_{ N }{ f }^{ * }(x) } =\\ \int { { g }(x){ B }^{ { N }^{ * } }{ f }^{ * }(x) } \\
    [/itex]

    Therefore, BN is also Hermitian.

    Given the operator B, we can define the smooth function f(B) by a Taylor series expansion:

    f(B)=[itex]\sum _{ N=0 }^{ \infty }{ \frac { 1 }{ N! } \frac { { \partial }^{ N }f }{ \partial { B }^{ N } } } { B }^{ N }[/itex]

    Since B0 = 1 and [itex] \frac { { \partial }^{ N }f }{ \partial { B }^{ N } } [/itex] can be treated as scalar derivatives of a normal function (a number) and BN was shown to be Hermitian, we're left with a scalar multiplied by a Hermitian, which is itself Hermitian.
     
  2. jcsd
  3. Apr 16, 2013 #2
    I've sought assistance from others regarding the above, and everyone believes it's okay . However, I'd love some feedback in this thread on my approach.
     
  4. Apr 16, 2013 #3

    Dick

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    I would say exactly the same thing. If B is hermitian then B^N is hermitian. If B^N is hermitian then any power series in B is also hermitian. Do you see any flaw in that?
     
  5. Apr 17, 2013 #4
    Hi Dick and thanks for your reply. I don't see any flaw in the reasoning, but I'm hoping my solution above conveys this clearly.
     
  6. Apr 17, 2013 #5

    Dick

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    It's clear enough to me. But I do think the notation is a little ugly. And are you clear on what something like ##B^*## means? Using bra-ket notation would make this a lot nicer.
     
  7. Apr 17, 2013 #6

    micromass

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    What does the notation ##\frac{\partial^N f}{\partial B^N}## mean? I don't think I've ever encountered that...
     
  8. Apr 18, 2013 #7
    Thanks Dick. I should be practicing my Dirac notation more...

    ##\left< { f }|{ \hat { B } }|{ g } \right> ={ \left< { g }|{ \hat { B } }|{ f } \right> }^{ * }##

    As for ##B^*##, it is the Hermitian conjugate. In this case, I probably should be using the dagger notation ##{ B }^{ \dagger }##.

    I may have been a bit lazy with the notation there because my cut and paste from LaTex didn't work. I'll cut and paste the image.

    I've been shown that a function of an operator f(A) can be defined by a Taylor series expansion:

    EnP6p7u.png

    where 9xJtFlM.png is a scalar calculated in the same way as the derivative of a normal function.
     
  9. Apr 18, 2013 #8

    Dick

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    Sure. So Hermitian means ##<f|Bg>=<Bf|g>## for any f and g. That makes it 'almost' obvious that ##<f|B^ng>=<B^nf|g>##. If it's not sufficiently obvious, it's easy to formalize the proof with induction.
     
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