(adsbygoogle = window.adsbygoogle || []).push({}); 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

(-h^{2}/2m) d^{2}[itex]\psi[/itex]/dx^{2}+v(x)[itex]\psi[/itex]=E[itex]\psi[/itex]

3. The attempt at a solution

so in reagion one its infinite so [itex]\psi[/itex]=0. reagion 2 is what i am confused about. looking throught the book i saw the gen solution for inside the well to be [itex]\psi[/itex]=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?

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# Particle in semi infinite potential well

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