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Particle in a triangular potential well

  1. Feb 9, 2012 #1
    1. The problem statement, all variables and given/known data
    Particle is in a potential well in the shape of a isosceles right-angled triangle. Need to find the wave function and allowed energies.


    2. Relevant equations
    How to determinate a boundaries for the potential when that line is in a form of some linear function, there is no strict point between potential equal to zero and infinity potential.


    3. The attempt at a solution
    I tried potential as [itex]kx[/itex] for [itex]0<x<\frac{a}{2}[/itex], and [itex]-kx[/itex] for [itex]\frac{a}{2}<x<a[/itex], but seems that is not right.
     
  2. jcsd
  3. Feb 12, 2012 #2

    ehild

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    Plot your potential function. Is it a well? Write up a potential function which has minimum at x=0.


    ehild
     
    Last edited: Feb 12, 2012
  4. Feb 13, 2012 #3
    This plot of potential function
     

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  5. Feb 13, 2012 #4

    ehild

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    Your function is not the same as the plot. -kx is negative in the range a/2<x<a

    By the way, the drawing shows a hill instead of a well.

    ehild
     
  6. Feb 13, 2012 #5
    Storry, like this
     

    Attached Files:

  7. Feb 14, 2012 #6

    ehild

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    That is better. What is the potential function?

    ehild
     
  8. Feb 14, 2012 #7
    That is my main problem, I don't know how to write it, I know that potential is zero inside and ∞ outside, that green area is unreachable for the particle.
     
  9. Feb 14, 2012 #8

    ehild

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    You know the equation of a straight line? You have a straight line for 0<x<a/2 which intersects the x axis at x=a/2, and makes -45° angle with it. And you have an other equation for a/2<x<a, that encloses a positive angle of 45° with the x axis.

    ehild
     
  10. Feb 14, 2012 #9

    vela

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    What is this supposed to mean?

    The potential isn't zero inside. It varies with x.

    If you want to make things a bit simpler, take ehild's suggestion to center the well at x=0.
     
  11. Feb 15, 2012 #10
    Yes, exactly.

    If I center the well at x=0 then I have a function

    [itex]V(x)=
    \begin{cases}
    -x & \text{for } -\frac{a}{2}<x < 0 \\
    x & \text{for } 0 < x < \frac{a}{2} \\
    0 & \text{for } x=0\\
    \infty & \text{elsewhere}
    \end{cases}[/itex]
     
    Last edited: Feb 15, 2012
  12. Feb 15, 2012 #11

    vela

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    Looks good. Now what?
     
  13. Feb 16, 2012 #12
    I need to find ψ and allowed energies inside the well.
     
  14. Feb 16, 2012 #13

    vela

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    Well, we're not here to do your homework for you. Show us your work. What do you have so far?
     
  15. Feb 16, 2012 #14
    I've just realized that the right potential is [itex]V(x)=\left | x \right | \text{for } x\in \left \{ -\infty, \infty \right \}[/itex]
     
    Last edited: Feb 16, 2012
  16. Feb 16, 2012 #15
    So far I have this:

    [itex]-\frac{\hbar^2}{2m}\frac{\mathrm{d^2}\psi }{\mathrm{d} x^2} + |x|\psi=E\psi[/itex]

    [itex]c=\frac{2m}{\hbar^2}[/itex] and [itex]k=\frac{\sqrt{2mE}}{k}[/itex]

    solution is [itex]\psi(x)=C_1\text{Ai}(\frac{cx-k^2}{c^{\frac{2}{3}}})+C_2\text{Bi}(\frac{cx-k^2}{c^{\frac{2}{3}}})[/itex] because x is always > 0, Bi goes to ∞ so C_2 have to be 0. And final solution is [itex]\psi(x)=C_1\text{Ai}((\frac{2m}{\hbar^2})^{\frac{1}{3}}(x-E))[/itex]. I've determined the energy from conditions [itex]\psi^{(1)}_n=\psi^{(2)}_n[/itex] and [itex]\psi'^{(1)}_n=\psi'^{(2)}_n[/itex]. And energies are: for even [itex]E_n=-a'_{n+1}[/itex] and for the odd [itex]E_n=-a_{n+1}[/itex]. Normalization is a bit tricky [itex]C_1^2\int_{-\infty}^{\infty} \text{Ai}^2(x-E_n)dx=1[/itex], I don't know how to solve this integral.
     
    Last edited: Feb 16, 2012
  17. Feb 16, 2012 #16

    vela

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    Your solutions don't work out unit-wise. For example, you can't subtract E from x. They don't have the same units.
     
  18. Feb 16, 2012 #17
    Yes they do, potential at x have the same value as the x coordinate, V(x)=|x|, unit of that x is the same as the unit of E.
     
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