So I'm having problems with the double infinite potential well given by
$$V(x)= \left\{\begin{array}{ll}
\infty & -\infty < x < -a-b \\
0 & -a-b< x < -a \\
V_0 & -a < x < a \\
0 & a < x < a + b \\
\infty & a+b < x < \infty \\
\end{array}\right.$$
I have to use the fact that the potential well is symmetric about x=0. I have solved the Schrödinger equation in all the regions and end up with
$$ψ1=A\sin(k_1x)+B\cos(k_1x)$$
$$ψ2=Ce^{k_2x}+De^{−k_2x}$$
$$ψ3=E\sin(k_1x)+F\cos(k_1x)$$
where $k_1=\sqrt{\frac{2mE}{\hbar^2}}$ and $k_2=\sqrt{\frac{2m(V_0-E)}{\hbar^2}}$. I can use the symmetry argument and get C=D which means cosh for even and sinh for odd wave functions. I know that I'm meant to get $A\sin(k_1(a+b−x))$ between the $-a-b$ and $-a$ and between $a+b$ and $a$ but I'm not sure how. If I know how to do that it may help me normalise the wave function which is key at the moment. Many thanks in advance!
Best Answer
One of the important condition to get the relationship between $A$ and $B$ in the left region (or in other words to get that the wave function there is proportional to $\sin(x-(-a-b))$ is to realize that the wave function must be zero for $x<-a-b$, because the potential is infinite there. Thus we have that $\psi_1|_{x=-a-b}=0$. This should give you the condition you need. There is a similar condition on the right side. I am not sure I understood your question. If not, leave a comment.