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Infinite potential well with a step

  1. Apr 7, 2013 #1
    1. The problem statement, all variables and given/known data
    Hey guys:) Maybe you will be able to help me with this problem i got as an assignment for my quantum mechanics course, it goes as follows:
    a particle of mass m moves in the potential

    for x<0 infinity,
    for 0<x<a -U,
    for a<x<b 0,
    for b<x infinity.
    a) Sketch the potential.
    b) We are looking for a solution with energy E=0. Set up the Schrödinger equation (in different areas), and determine the general solutions. Put E=0 from the beginning of Schrödinger equation.
    c) Use the continuity conditions for reducing the number of constants. Note that continuity condition always applies to ψ(x) at all points, but it only applies to dψ(x)/dx at points where the potential is finite. In the case above, we can not require continuity of dψ(x)/dx in points x=0 or x=b but in the point x=a.
    d) Determine the condition for U to be valid.

    2. Relevant equations



    3. The attempt at a solution
    Of course i attemted to solve this by solving the schrodinger's equations for indicated ranges, however i faced the problem by having to present E as 0 from the very beginning.
     
  2. jcsd
  3. Apr 7, 2013 #2

    mfb

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    Just work with E, and replace every occurence of "E" with "0" afterwards if you like. Setting E to 0 just makes the equations easier, not more complicated.
     
  4. Apr 8, 2013 #3
    I tried to calculate the schrodinger equations for the regions-ok, i have two wavefunctions psi1=Asin(k1x) with k1=root(2Um)/hbarred, and psi2=Csin(k2x) with k2=root(2Em)/hbarred, where E=0. Those two wavefunctions equalize in point a, that's when I have this equation Asin(k1x)=Csin(k2x), so it is Asin(k1x)=C*0. I am not sure if I got it right. Where am I making a mistake?
     
  5. Apr 8, 2013 #4

    mfb

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    How did you get psi2?
     
  6. Apr 8, 2013 #5
    I just divided the range into regions, psi2 is the wavefunction for region a<x<b. I am not sure if I am doing it right?
     
  7. Apr 8, 2013 #6

    mfb

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    "psi2=Csin(k2x)" <- how did you get this?
    If you solve the Schrödinger equation, you should get a different result.
     
  8. Apr 8, 2013 #7
    for this region V(x)=0, and energy is also equal 0. I put this to the schrodinger equation and it looked something like this:
    -(hbarred)^2/2m*psi''=E*psi(x)
    then I derived this: k=root(2mE)/ℏ
    from border conditions I got this: psi=Csin(kx), where k is equal to 0. In the end I could show this wavefunction as 0, but decided to show it like this and maybe get something from it later.
     
  9. Apr 8, 2013 #8

    mfb

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    This cannot be derived from the equation above.
    Did you see any particular reason to use a sin here? I am sure it was not the result of your calculation.

    Fine. Now we can use E=0:
    $$-\frac{\hbar^2}{2m}\psi''=0$$
    Can you simplify and solve this? Please solve it yourself, and don't try to use solutions to other problems.
     
  10. Apr 8, 2013 #9
    it states that second derivative of psi is equal to 0, but how am I supposed to use this information?
     
  11. Apr 8, 2013 #10

    mfb

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    Can you give a general description of all functions with zero second derivative?
     
  12. Apr 8, 2013 #11
    Sorry, forgot about linear function:P Now I have two equations, first one for 0<x<a (psi=Asin(kx)), and second one for a<x<b (psi=Cx+d). Now the only thing that comes to my head is to equalize them in point a, is this right?
     
  13. Apr 8, 2013 #12

    mfb

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    Right.
    You get an additional constraint on psi2 for x=b. This gives you two conditions with 3 free parameters, but the global normalization is irrelevant.
     
  14. Apr 10, 2013 #13
    I think I got lost about this point. I have psi for 0<x<a (Asin(kx)), and I have psi for a<x<b (Cx+d), and I know that they equalize in point a, and their derivatives should equalize too, but I still don't know how to get the constants A, C and d. It is probably something simple, but I am missing it.
     
  15. Apr 10, 2013 #14

    mfb

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    Can you write that as an equation?
    And that one, too.

    You have an additional constraint at x=b.
     
  16. Apr 10, 2013 #15
    I get a set of three equations:
    Asin(ka)=Ca+d
    Akcos(ka)=C
    Cb+d=0

    I am probably missing some ridiculously simple step, but only thing I obtain is equation
    Asin(ka)=Akcos(ka)(a-b) and this is where I no longer know what to do. What am I missing?
     
  17. Apr 10, 2013 #16

    mfb

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    Asin(ka)=Akcos(ka)(a-b)
    sin(ka)=kcos(ka)(a-b)
    tan(ka)=k(a-b)
    a and b are constants, you can solve the equation for k numerically or graphically.
     
  18. Apr 10, 2013 #17

    vela

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    You can divide A out and rearrange slightly to get
    $$\tan ka = ka\left(1-\frac{b}{a}\right).$$ This is the condition you're looking for in part (d). For given values of ##a## and ##b##, this relationship is satisfied for only certain values of ##ka##, which corresponds to certain values of U. You can see this by plotting the two sides of the equation and seeing where the curves intersect.
     
  19. Apr 15, 2013 #18
    I'm having a little problem with plotting this, mostly due to the fact that I don't exactly know the values of a and b, only bit of information is that b>a. Please help me uderstand this?
     
  20. Apr 15, 2013 #19

    mfb

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    Plot it for some different values of b/a? The general pattern of the solutions is the same for all ratios, just the numerical values are different.
     
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