- #1
Logan Rudd
- 15
- 0
I'm working on a problem out of Griffith's Intro to QM 2nd Ed. and it's asking to find the bound states for for the potential ##V(x)=-\alpha[\delta(x+a)+\delta(x-a)]## This is what I'm doing so far:
$$
\mbox{for $x\lt-a$:}\hspace{1cm}\psi=Ae^{\kappa a}\\
\mbox{for $-a\lt x\lt a$}\hspace{1cm}\psi=Be^{-\kappa x}+Ce^{\kappa x}\\
\mbox{and for $x\gt a$:}\hspace{1cm}\psi=De^{-\kappa a}
$$
However, this is what the solution reads:
$$
\mbox{for $x\lt a$:}\hspace{1cm}\psi=Ae^{-\kappa a}\\
\mbox{for $-a\lt x\lt a$}\hspace{1cm}\psi=B(e^{\kappa x}+e^{-\kappa x})\\
\mbox{and for $x\lt-a$:}\hspace{1cm}\psi=Ae^{\kappa a}
$$
Can someone explain what I'm doing wrong - why they are getting the coefficients they are - and what I should be doing? Maybe go over the general way of approaching these kind of problems? I'm also wondering why they don't have the case for $x>a$?
Thanks
$$
\mbox{for $x\lt-a$:}\hspace{1cm}\psi=Ae^{\kappa a}\\
\mbox{for $-a\lt x\lt a$}\hspace{1cm}\psi=Be^{-\kappa x}+Ce^{\kappa x}\\
\mbox{and for $x\gt a$:}\hspace{1cm}\psi=De^{-\kappa a}
$$
However, this is what the solution reads:
$$
\mbox{for $x\lt a$:}\hspace{1cm}\psi=Ae^{-\kappa a}\\
\mbox{for $-a\lt x\lt a$}\hspace{1cm}\psi=B(e^{\kappa x}+e^{-\kappa x})\\
\mbox{and for $x\lt-a$:}\hspace{1cm}\psi=Ae^{\kappa a}
$$
Can someone explain what I'm doing wrong - why they are getting the coefficients they are - and what I should be doing? Maybe go over the general way of approaching these kind of problems? I'm also wondering why they don't have the case for $x>a$?
Thanks