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Problem integrating complex function

  1. Nov 26, 2014 #1
    1. The problem statement, all variables and given/known data
    Hello, I have been tasked with the next problem, I have to prove that the next two integrals are complex numbers; but I have no idea of how to attack this problem.
    2. Relevant equations
    ∫dx f*(x) x (-ih) (∂/∂x) f(x) integrating between -∞ and ∞
    ∫dx f*(x) (-ih) (∂/∂x) (x f(x)) integrating between -∞ and ∞
    Where h is a constant, i = √-1 and f* the complex conjugate of the function f

    3. The attempt at a solution
    Well, the only thing I can think of for solving is a direct integration by parts, using u =f* x and dv= (∂/∂x) f(x) dx for the first integral, with du=f* and v= ∫(∂/∂x) f(x)dx=f(x). But then, I find that
    -ih∫u dv= uv -∫v du = -ih( [f f* x]-∞∞+∫f f*dx).
    f f*=1, so I find myself with -ih[(∞) (∞-∞)]

    All help is appreciated.
     
  2. jcsd
  3. Nov 26, 2014 #2

    Matterwave

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    You shouldn't have ##f^*(x)f(x)=1## (where did the x dependence go?) instead, the normalization condition is that ##\int_{-\infty}^\infty f^*(x) f(x) dx = 1##.
     
  4. Nov 26, 2014 #3
    That was a bad mistake I made; even worse for the fact that f is actually a density of probability. Thanks for pointing this error.
    Then I get rid of the ∞-∞, but I still have [f f* x]-∞ , which result in a complex but infinite result
     
  5. Nov 26, 2014 #4

    Matterwave

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    Why is it infinite? What conditions must ##f(x)## fulfill at the boundaries if we are to be able to normalize them? Why do you think ##f^*(x)f(x)## will diverge at both limits?
     
  6. Nov 27, 2014 #5

    Ray Vickson

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    No: f is NOT a probability density. In general, f has complex values, so cannot possibly be a probability density. However, ##|f|^2 = f^*f## is a probability density.

    Also, you did not perform integration by parts correctly. You wrote
    [tex] u = f^*(x) x, \;\; dv = (\partial/\partial x) f(x) \, dx \\
    \Longrightarrow du = f^* \; \text{ and }\; v = f.[/tex]
    The ##v = f## part is OK, but ##du = f^*## is wrong, and does not even make sense, because a ##dx## is missing. Anyway, you need to use the product rule, which you have not done, so it would be wrong even if you supplied the missing ##dx##.
     
  7. Nov 27, 2014 #6
    You are right. Posting at late night is a bad idea.
    Today I have remade my calculations and I think I have arrived to coherent results.
    If I make the integration by parts with u = x f* and dv= (∂/∂x)(f(x)) then du = (f* + x (∂f*/∂x)) dx
    Taking into account that [f f* x]-∞ is 0 because f goes way faster to 0 than x to ∞ and that ∫f f* dx is 1 I find that the original inegral is it's own complex conjugate plus -ih, what means that the operator is not hermitian and therefore, the solution is not real.

    I will use the same procedure for the second integral.
     
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