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Hermitian Operator

  1. Apr 26, 2005 #1
    Is the second derivative with respect to position a hermitian operator? (i.e. d^2/dx^2)? Can anyone prove it? I don't think it is. Thanks
     
  2. jcsd
  3. Apr 26, 2005 #2

    Galileo

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    That operator is basically [itex]\hat p_x^2[/itex], which we know to be Hermitian.
     
  4. Apr 26, 2005 #3

    dextercioby

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    It is essentially selfadjoint under certain conditions.([tex] \hat{A}=\hat{A}^{\dagger}=\bar{\hat{A}} [/tex])The first to use it in QM was John von Neumann.Check out his book:"Mathematische Grundlagen der Quantenmechanik".

    Other sources:Ahiezer & Glazman:"The theory of Linear Operators in Hilbert Space".
    J.Prugoveçki:"Quantum Mechanics in Hilbert Space".
    Birman & Solomjak:"Spectral Theory of Self-Adjoint Operators in Hilbert Space".



    Daniel.
     
  5. Apr 27, 2005 #4
    Hello PhysKid24,

    Galileo already gave you a clue. Use the definition of the hermitian operator:
    [itex](\phi, \hat{A} \psi) = (\hat{A}^{\dagger} \phi, \psi)[/itex].

    So what you have to show is:
    [itex]\hat p_x^2 = (\hat p_x^2)^{\dagger} [/itex] ,
    that is
    [itex](\phi,\hat p_x^2 \psi) = (\hat p_x^2 \phi, \psi) [/itex] .

    Hint: You already know that [itex]\hat p_x = \hat {p_x}^{\dagger} [/itex],
    that is
    [itex](\phi, \hat{p_x} \psi) = (\hat{p_x}^{\dagger} \phi, \psi) = (\hat{p_x} \phi, \psi)[/itex].
     
  6. Apr 27, 2005 #5

    dextercioby

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    Useless,u'd still have to prove the differential operator is at least hermitean...


    Daniel.
     
  7. Apr 27, 2005 #6
    The subset of functions where p is hermitian is not the same as the one where p^2 is hermitian . Therefore, p hermitian on a subset of functions =/=> p^2 hermitan on this subset. We just have a common subset of functions where p and p^2 are both hermitian.

    Seratend.
     
  8. Apr 27, 2005 #7

    Galileo

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    What do you mean? [itex]\langle \psi |\hat p^{\dagger}|\varphi \rangle = \langle \psi | \hat p|\varphi \rangle[/itex] for any vector in the state space and it does imply that the same holds for [itex]p^2[/itex].
     
  9. Apr 27, 2005 #8
    No. P is an unbounded operator (i.e. a non conitnuous operator). In other words, p is not defined on most of the functions of L2(|R,dx) (or a vector of a given separable abstract hilbert space).

    For example, if f(x) is a continuous derivable and integrable function (subset of L2) , that has a limit 0 as x --> +00, we have (in hbar units):

    <f|p|f>= int_{R} if*(x)f'(x)dx= [f*(x)f(x)]_{-oO, +oO} -int_{R} if'*(x)f(x)dx = -int_{R} if'*(x)f(x)dx = <f|-id/dx|f>=<f|p+|f>

    where p+ is the hermitian conjugate of p.

    now, <f|p^2|f>= -int_{R} f*(x)f''(x)dx=-[f*(x)f'(x)]_{-oO, +oO} +int_{R} if'*(x)f'(x)dx= -[f*(x)f'(x)]_{-oO, +oO} + [f*(x)f(x)]_{-oO, +oO} - int_{R} if''*(x)f(x)dx

    = -[f*(x)f'(x)]_{-oO, +oO} - int_{R} if''*(x)f(x)dx

    therefore if lim_{x-->+/-oO} f*(x)f'(x)=/=0, p2 is not hermitian ((<f|p^2+)|f>=/= <f|(p^2|f>).

    Seratend.
     
  10. Apr 27, 2005 #9

    dextercioby

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    I think you're diverging from the initial question.


    Daniel.
     
  11. Apr 27, 2005 #10
    prego?

    Seratend.
     
  12. Apr 27, 2005 #11

    dextercioby

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    You haven't shown the operator is at least hermitean on its domain.

    Daniel.
     
  13. Apr 27, 2005 #12
    I do not understand what you mean.
    I think I have demonstrated that p is an hermitan operator for the vector space of continuous derivable and L2 functions (i.e. the set where p=d/idx is the classical derivation). And that this vector space is not the L2 separable hilbert space.
    And In addition, to get a hermitian operator on this subspace (the set where p^2=-d^2/dx^2), we need to restict these functions to the subspace where (lim_{x--> +/-oO}=f(x)f'(x)=a where a is any constant).
    However, we may enlarge this domain (extending the classical derivation) but this is not the subject of this thread, I think.

    Seratend.

    Simply, not understanding what you mean.
     
  14. Apr 28, 2005 #13

    Galileo

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    Seratend. I read your post. When I mean vectors in the state space, I mean physically realizable states. At the very least they go to zero at infinity and are uniformly continuous. In that case, clearly:

    [tex]\lim_{x \to \infty} f^*(x)f'(x)=0[/tex]
     
  15. Apr 29, 2005 #14
    Ok, I understand you now (and for all practicall purposes it is ok).
    However, for the ones who are interrested, you cannot strictly say that all the physically realizable states (in the sense <psi|psi> <+oO) are uniformly continuous and vanishes to 0 at the infinity vicinity (even if it is sufficient for all practical purposes and with some density results of continuous functions) as this only a subclass of possible states.
    We must not forget that all the limits we get are based mainly on the L2 norm and not the usual |x| norm of the limit symbol. In other words, there are lot of points in |R where "[itex]\lim_{x \to \infty} f^*(x)f'(x)\neq 0[/itex]".

    Seratend.
     
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