# Particle in semi infinite potential well

1. Oct 18, 2011

### leonne

1. The problem statement, all variables and given/known data
particle moving in the semi infinite potentail well set up and solve SE for the system assume E<Vo

2. Relevant equations
(-h2/2m) d2$\psi$/dx2 +v(x)$\psi$=E$\psi$

3. The attempt at a solution
so in reagion one its infinite so $\psi$=0. reagion 2 is what i am confused about. looking throught the book i saw the gen solution for inside the well to be $\psi$=Ae(ikx)+Be-(ikx) but for this problem its Acos(kx)+Bsin(kx)
in the book does not really show how it finds the gen solution. im just a little confused on how you would solve the SE well i know for inside v=0 and i plug that into SE and from that the book just goes to the gen solution
thanks

edit after looking at lecture note i see he uses the cos for the inside , but in they book for the same infinite pot well they use the e^ are they the same?

2. Oct 18, 2011

### vela

Staff Emeritus
The solutions are equivalent. They're just being expressed in different bases.
\begin{align*}
A e^{ikx} + B e^{-ikx} &= A(\cos kx+i\sin kx)+B(\cos kx-i\sin kx) \\
&= (A+B)\cos kx + i(A-B)\sin kx \\
&= C \cos kx + D \sin kx
\end{align*}Since A and B are arbitrary constants, C=A+B and D=i(A-B) are as well.

3. Oct 18, 2011

o ok thxs