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?(adsbygoogle = window.adsbygoogle || []).push({});

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.

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

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