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Potential wells

  1. May 2, 2008 #1
    1. The problem statement, all variables and given/known data
    A potential well with V(x)=0 for x<0 and V(x) = Vo for x>0. The particle has energy E greater than Vo and is incident from the left side. Calculate the reflection coefficient.

    2. Relevant equations

    Relfection coefficient is given by R= (B*B)/(A*A)

    3. The attempt at a solution
    I think that the time independent Schrodinger equation for the two regions are
    x<0 (-hbar/2m)dphisquared/dphi =Ephi(x)
    and
    x>0 ((-hbar/2m)dphisquared/dphi +Vo phi = Ephi(x)

    the solutions to this
    x<0 Aexp(ik1x) +Bexp(-ik1x) = phi(x)
    x>0 Cexp(ik2x)

    is this right
    carrying this through and equating two solutions at x=0
    A+b = C
    and equating differentials at x=0 gives
    ik1(A-B) = -ik2C/k1

    but when I solve for A and B i get
    A=(c/2)(1-ik2/k1)
    B=(c/2)(1+ik2/k1)

    when I find R i get R=1 which must wrong because particle is not always reflected it should pass over the potential barrier?
     
  2. jcsd
  3. May 2, 2008 #2

    Dick

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    Things started to break down when you evaluated the derivatives. Why is there a k1 on the left side? And why didn't the i's cancel out in the final solution?
     
  4. May 3, 2008 #3

    siddharth

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    Apart from the extra k1 on the left side which Dick pointed out, why is there a minus sign on the right side?
     
  5. May 3, 2008 #4
    sorry should the differentials be (A-B)=-ik2C/k1?
    in which case i still get the same solutions for A and B and a coefficient of 1
     
  6. May 3, 2008 #5

    siddharth

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    Why is there a minus sign on the left hand side?
     
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