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Quantuam Mechanics: Discussion of unbound states of a combination well and barrier

  1. Dec 8, 2007 #1
    1. The problem statement, all variables and given/known data

    (*I have solved similar problems before and know the method, but my results don't make sense, so I want to make sure I understand this particular potential. Also, I am new to Latex so ignore the equal portion of the less/greather than or equal two signs.*)

    I made a diagram of the potential I am working with.


    I need to find the transmission (T) and reflection (R) coefficients.

    Just to make sure, what have you seen. The potential V(x) is as follows

    [tex]0[/tex] for x[tex]\le[/tex] -a
    [tex]V_{1}[/tex] for -a[tex]\leq[/tex]x[tex]\leq[/tex]0
    [tex]V_{0}[/tex] for 0[tex]\leq[/tex]x[tex]\leq[/tex]+b
    [tex]0[/tex] for x[tex]\ge[/tex] +b

    2. Relevant equations
    3. The attempt at a solution
    I don't need to worry about bound states E < 0. So, there are two remaining cases

    Case 1: [tex] V_{1}[/tex][tex]\le[/tex][tex]E[/tex][tex]\le[/tex][tex]V_{0}[/tex]

    Case 2: [tex] V_{0}[/tex][tex]\le[/tex][tex]E[/tex]

    Now I found R and T for case one and currently working on case two. From what other problems I have done R + T always equals 1, but for case one I actually got an interesting expression. If I assume that [tex] V_{1}[/tex] is small (shallow well) and if [tex] V_{0}[/tex] is small or b is small (the sums depends on the exponent of V0 and b) , then the sum approaches one. I have gone over my algebra a few times. Is it possible they are times that R+T, I have never encountered a problem that has. It's not like I have done that many either. I have done a simple barrier, step potential, drop potential. I have seen the finite well and delta function well examples.

    Without the assumptions, it is implying that there is some absorption, but that doesn't make sense to me... where would it be absorbing... the well? I figure it kinda just bounces back and forth... so R + T still equals one.

    Since I know the method of solving these problems and I don't want to bother you with equations. I am just giving you those little constants (think they are called eigenvalues) that determine the solutions to the Schrödinger's eq. for each separate religion.

    [tex]\frac{\sqrt{-2mE}}{\hbar}[/tex] for x[tex]\le[/tex] -a
    [tex]\frac{\sqrt{2m(E+V_{1})}}{\hbar}}[/tex] for -a[tex]\leq[/tex]x[tex]\leq[/tex]0
    [tex]\frac{\sqrt{2m(V_{0}-E)}}{\hbar}[/tex] for 0[tex]\leq[/tex]x[tex]\leq[/tex]+b
    [tex]\frac{\sqrt{-2mE}}{\hbar}[/tex] for x[tex]\ge[/tex] +b
    Last edited: Dec 8, 2007
  2. jcsd
  3. Dec 8, 2007 #2


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    R+T should always be one.

    For -a<x<0, V(x)=V1, and V1 is negative, right? If so, then you should have E-V1 under the square-root in your expression for the wavenumber. (The eigenvaule of the hamiltonian is E.)
  4. Dec 8, 2007 #3
    When going over [tex]V_{1}[/tex] the wave function should still be the same inside the well, so




    Last edited: Dec 8, 2007
  5. Dec 8, 2007 #4
    Ok, it won't let me change, but for the last eq. it should be negative k sub two squared... my wavenumber for the wave function over the well.
  6. Dec 8, 2007 #5
    I am still struggling with this problem... so I am wondering if it has something to do with my math.

    Because the of Euler's formula, you can write trig. functions into exp. functions and vice versa. Do I need all my wave functions in exp. form. I know if I convert the exp. functions of the barrier into a linear combination of the hyperbolic functions... I get something that is not well behaved since the limit of sinh and cosh go to infinity as x goes to infinity.

    So, I am now under the assumption that sin(x) + cos(x) used in the finite well solution is not sufficient form for the unbound states, since going from exp. commonly used for the free particle to a linear combination of trig. results in a complex constant. However, I am under the impression it doesn't matter. And after working everything in the exp. for calculating of coefficients are very cumbersome.
  7. Dec 9, 2007 #6


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