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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    $\begingroup$ Notice the factor of $i$ in the exponent: those are complex exponentials, which always have modulus 1. $\endgroup$
    – march
    Feb 4, 2016 at 17:39

1 Answer 1


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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