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A plane wave vs the bound state of Hydrogen atom: orthogonal?

  1. Mar 30, 2010 #1
    These days I met one problem and asked a professor for help. But I can not understand his answer. Can you help me explain his answer?
    My question is that whether we can assume that a plane wave is orthogonal to the bound state of Hydrogen atom when t->\infty?

    Professor answers: <\psi(t)(t)|\psi_0(t)>|^{t \to \infty}.Then the time dependence of this scalar product is \exp i(E_p+I_p)t where E_p =p^2/2m and I_p is the ionization energy. But exp(iEt) goes to zero when t goes to infinity so long as E is different from zero. Since there is no laser field E_p + I_p is always positive.

    My questions: As far as I know, when t->\infty, exp(iEt) turns around the central point in the complex plane, but never goes to zero.

    Professor answers: Regarding lim_{t -> \pm \infty} \exp{iEt}, you are completely right that |\exp{iEt}|=1 for any t, but that is not the point. Because of the rapid oscillations of both the real and the imaginary part of \exp{iEt} for t -> \pm \infty while E \ne 0, the function goes to zero "for all practical purposes". The latter statement means that \int dE f(E) \exp{iEt} will go to zero for t -> \infty for any sufficiently smooth function f(E) (called a test function). This is a pico summary of what is called theory of "distributions" or "improper functions". The best known example is the delta function \delta(x), which is zero for any value of x except x=0, while \int dx \delta(x) f(x) = f(0). One of many limit relations that can be used to define the delta function is i\pi\delta(x) = \lim_{t ->\infty} \exp{ixt}/x. This implies what I wrote down above.

    I cannot follow him. :redface:
  2. jcsd
  3. Apr 1, 2010 #2
    I now answer the question posed in the topic title.

    Let [itex]\phi(k,r)=e^{ikr \cos\theta}[/itex] be the plane-wave with wavenumber [itex]k[/itex] pointing in the z-direction,
    and let [itex]\psi_{100}(r,\theta,\phi)=R_{10}(r)Y_0^0(\theta,\phi)[/itex] be the Hydrogen atom ground state.

    The overlap may now be computed:

    [tex]\langle\psi|\phi\rangle=\int_0^\infty dr\,r^2 R_{10}(r)\int d\Omega\, e^{ikr \cos\theta}Y^0_0(\theta,\phi)^*[/tex]
    [tex]=\int_0^\infty dr\,r^2 R_{10}(r)\,\, (2\sqrt{\pi})\, j_0(k r)[/tex]
    [tex]=2\sqrt{\pi}\int_0^\infty dr\,r^2 R_{10}(r)\, j_0(k r)[/tex]

    Now, using [itex]R_{10}(r)=2 a_0^{-3/2} e^{-r/a_0}[/itex], I can perform the radial integral, giving

    [tex]\langle\psi|\phi\rangle = \frac{8\sqrt{\pi}a_0^{3/2}}{(k^2a_0^2+1)^2}\neq0\,.[/tex]

    And thus, a plane wave and the Hydrogen atom ground state are not orthogonal.
  4. Apr 1, 2010 #3
    Yes, you are right. But the professor mentioned "for all practical purposes". I don't know his meaning.
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