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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
    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. jcsd
  3. Oct 18, 2011 #2

    vela

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    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.
     
  4. Oct 18, 2011 #3
    o ok thxs
     
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