My book says that the solution to a singular potential at zero problem has the form(adsbygoogle = window.adsbygoogle || []).push({});

[tex]

\begin{array}{ccc}

\psi(x) &=& A exp(Kx) for x< 0 \\

& =& Aexp(-Kx) for x > 0 \end{array}

[/tex]

How do you get that from the general solution of the Schrodinger equation

[tex]\hbar^2/2m d^2 \psi(x)/dx^2 = E \psi(x) = -|E|\psi(x) [/tex]

which is

[tex]

\begin{eqnarray*}

psi(x) &=& A exp(Kx) + B exp(-Kx) for x< 0 \\

& &= Cexp(-Kx) + Dexp(Kx)for x > 0

\end(eqnarray*}

[/tex]

by "imposing the condition that the wavefunction be square integrable and continuous at x = 0".

Obviously the second condition gives you A + B = C + D but I do not see how that helps reduce to a single coefficient.

And why are my equation arrays not working? :(

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# Homework Help: Singular potential problem

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