Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Particle in semi infinite potential well

  1. Oct 18, 2011 #1
    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[itex]\psi[/itex]/dx2 +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

    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. jcsd
  3. Oct 18, 2011 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    The solutions are equivalent. They're just being expressed in different bases.
    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.
  4. Oct 18, 2011 #3
    o ok thxs
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook