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Math Help - Hermiticity Question

  1. #1
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    Hermiticity Question

    Please help, I have no idea!

    Consider the set of functions {f(x)} of the real variable x defined on the interval −inf < x < inf
    that go to zero faster than 1/x for x --> ħinf, i.e.
    lim(x-->ħinf) xf(x) = 0
    For unit weight function, determine which of the following linear operators is Hermitian when acting upton {f(x)}:
    (a) d/dx + x
    (b) −i d/dx + x^2
    (c) ix d/dx
    (d) i d3/dx3


    I tried using definition of a hermitian operator as <u, Hv> = (<v, Hu>)* but I don't get what u and v are, unless they are just general members of the set {f(x)} in which case I'm still confused!

    Thanks!
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  2. #2
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    Quote Originally Posted by joker_900 View Post
    Please help, I have no idea!

    Consider the set of functions {f(x)} of the real variable x defined on the interval −inf < x < inf
    that go to zero faster than 1/x for x --> ħinf, i.e.
    lim(x-->ħinf) xf(x) = 0
    For unit weight function, determine which of the following linear operators is Hermitian when acting upon {f(x)}:
    (a) d/dx + x
    (b) −i d/dx + x^2
    (c) ix d/dx
    (d) i d3/dx3


    I tried using definition of a hermitian operator as <u, Hv> = (<v, Hu>)* but I don't get what u and v are, unless they are just general members of the set {f(x)} in which case I'm still confused!
    In these example, the space V on which H acts is the set of functions f(x) on the real line for which \textstyle\lim_{|x|\to\infty} xf(x) = 0. (They had better be differentiable functions for the question to make sense, though this is not specified.) The definition of H being hermitian is: \langle Hf,g\rangle = \langle f,Hg\rangle, for all f, g in V. The inner product \langle f,g\rangle means the integral (with respect to the unit weight function) of the product of f(x) and the complex conjugate of g(x).

    In part (a), for example, the operator H is differentiation plus multiplication by x, so that Hf(x) = f'(x) + xf(x). You have to check whether the integrals of \bigl(f'(x) + xf(x)\bigr)g(x)^* and f(x)\bigl(g'(x) + xg(x)\bigr)^* (where the asterisk denotes the complex conjugate) are equal. It looks as though integration by parts will figure somewhere in the calculation.
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