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I am following the Griffiths Book on Quantum Mechanics, and am following the derivation for the wave function for Delta-Function Potentials.

$$V(x) = -\alpha \delta(x)$$

In the scattering states, where $E > 0$, when solving the Schrodinger Equation in the range $x < 0$, we are left with

$$ \psi(x) = Ae^{i\kappa x} + Be^{-i\kappa x}, \text{ where } \kappa \equiv \frac{\sqrt{2mE}}{\hbar} $$

Whereas in the boundstate, we had the $Ae^{-\kappa x}$ term having to be zero due to the blow-up at -infinity, why do we not have the same problem here? The book states: "this time we cannot rule out either term, since neither of them blows up." Isn't the -infinity still blowing up for the $B$ term?

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  • $\begingroup$ Hello, and welcome to Stack Exchange Physics! Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better. $\endgroup$
    – Daniel Griscom
    Commented Feb 4, 2016 at 17:23
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    $\begingroup$ Notice the factor of $i$ in the exponent: those are complex exponentials, which always have modulus 1. $\endgroup$
    – march
    Commented Feb 4, 2016 at 17:39

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Neither one of the terms blow up because of the complex exponential. This exponential is real for the bound state case.

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