Why does the singular potential solution reduce to a single coefficient?

[ehrenfest]

My book says that the solution to a singular potential at zero problem has the form
<br /> \begin{array}{ccc}<br /> \psi(x) &amp;=&amp; A exp(Kx) for x&lt; 0 \\<br /> &amp; =&amp; Aexp(-Kx) for x &gt; 0 \end{array}<br />

How do you get that from the general solution of the Schrodinger equation
\hbar^2/2m d^2 \psi(x)/dx^2 = E \psi(x) = -|E|\psi(x)
which is
<br /> \begin{eqnarray*}<br /> psi(x) &amp;=&amp; A exp(Kx) + B exp(-Kx) for x&lt; 0 \\<br /> &amp; &amp;= Cexp(-Kx) + Dexp(Kx)for x &gt; 0<br /> \end(eqnarray*}<br />

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? :(

 

[olgranpappy]

the condition that it be square integrable means the we must set
B=D=0.

 

[ehrenfest]

How does square integability imply that?

 

[Gokul43201]

As a quick test, you want to evaluate each of those (modulus squared) integrals over their respective domains. Ask yourself which ones will converge and which will diverge (for non-zero coefficients)?

 

[ehrenfest]

I see. Thanks.