Particle in semi infinite potential well

[leonne]

Homework Statement

particle moving in the semi infinite potentail well set up and solve SE for the system assume E<Vo

Homework Equations

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

The Attempt at a Solution

so in reagion one its infinite so $\psi$=0. reagion 2 is what i am confused about. looking through 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. I am 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?

 

[vela]

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.

 

[leonne]

o ok thxs