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Particle in Potential Well

  1. Dec 4, 2006 #1
    1. The problem statement, all variables and given/known data
    Consider a particle of mass m in a vee-shaped potential whose analytic form is

    V(x)= -bx (x<=0)
    V(x)= bx (x>=0)

    Use what is known about the uncertainty principle and the simple harmonic oscillator to show that the lowest state energy is ((hbar)^2(b)^2/m)^(1/3). Show that this funky result has the correct dimensions.

    2. The attempt at a solution

    The width of the well is 2bx. I used delta p = hbar/2bx, and then subbed into E=p^2/2m but the b is in the denominator, and I have no clue where the 1/3 power comes from.
     
    Last edited: Dec 4, 2006
  2. jcsd
  3. Dec 5, 2006 #2

    Kurdt

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    You'll have to minimise the energy of a harmonic oscillator which consists of the kinetic energy and the potential you were given. You substitute for the momentum in the kinetic energy term by using the uncertainty principle and then minimise with respect to x.
     
  4. Dec 8, 2006 #3
    I'm not sure how I can mathematically minimize the energy.

    The energy is [tex]\frac{\hbar^2}{8mb^2x^2}+\frac{1}{2}Cx^2[/tex] when substituting in the uncertainty principle.
     
  5. Dec 8, 2006 #4
    From the problem I figure that the total energy is [tex]E=\frac{p^2}{2m}+bx[/tex], so substituting [tex]p=\frac{\hbar}{2bx}[/tex] into that and minimizing gives me a minimum value of [tex]x=\frac{\hbar^\frac{2}{3}}{2^\frac{2}{3}m^\frac{1}{3}b}[/tex], and when this is substituted into the energy, all the b's cancel out! I think somehow the energy equation is wrong, but I'm at my wit's end.
     
    Last edited: Dec 8, 2006
  6. Dec 9, 2006 #5

    Kurdt

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    Sorry what you need to do is minimise for p. Error on my part.

    [tex]x=\frac{\hbar}{2p}[/tex]

    If we sub that into the energy equation we get:

    [tex]E=\frac{p^2}{2m}+\frac{b\hbar}{2p} [/tex]

    To minimise we take the derivative and set equal to zero.

    [tex]0=\frac{p}{m}-\frac{b\hbar}{2p^2} [/tex]

    Rearrange for p:

    [tex]p=(\frac{b\hbar m}{2})^\frac{1}{3}[/tex]

    Then sub back into the energy equation and tidy it up. The only thing is you'll get an annoying term of 2 to a strange power outside.
     
    Last edited: Dec 9, 2006
  7. Dec 9, 2006 #6
    According to my textbook, when you use the uncertainty principle, you essentially have [tex]p=\frac{\hbar}{L}[/tex] where L is the length of the well, which in this case is 2bx. Therefore, I don't quite understand why there is no b in the first equation.
     
  8. Dec 9, 2006 #7

    Kurdt

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    As I said above I made a mistake in an earlier post where I said you had to minimise for x instead of p. I have already started you off above.
     
  9. Dec 9, 2006 #8

    OlderDan

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    The width is not 2bx. The width of the well as experienced by a particle having some particular energy bx_o is from -x_o to +x_o or 2x_o. The subscripts can be dropped for convenience. I think Kurdt has you going in the right direction.
     
  10. Dec 9, 2006 #9
    Thanks for the help guys, I finally got it. The actual question said to prove only that the energy was of the order of that, so it turns out to be proven.
     
  11. Dec 9, 2006 #10

    Kurdt

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    Sorry about the frightful blunder of saying you have to minimise for x instead of p. I hope you understand why you have to minimise for the momentum.
     
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