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One dimensional potential function

  1. Sep 29, 2014 #1
    1. The problem statement, all variables and given/known data
    A particle constrained to move in one dimension (x) is the potential field

    V(x)=[(V_0(x-a)(x-b))/(x-c)^2]

    (0 < a < b < c < infinity)

    (a) Make a sketch of V
    (b) Discuss the possible motions, forbidden domains, and turning points. Specifically, if the particle is known to be at x=-infinity with

    E=3(V_0)/(c-b) * (b-4a+3c)

    at which value of x does it reflect?

    This is problem 1.18 in chapter 1 of Richard L. Liboff's book, "Introductory Quantum Mechanics"

    2. Relevant equations
    Equations given

    3. The attempt at a solution
    I assigned a, b, c, the values 1, 2, 3 respectively then I started plugging in values for x starting with 0 and I went all the way up to 10. This gave me a graph that resembled the positive portion of the function [f(x)=1/x] except that there was a discontinuity at x=3 because at that point the denominator is 0. It also gave me that V(0)=(2/9). So what I realized after changing the values of a, b, c, to 1, 10, 100, respectively did I see that no matter what values are assigned to a,b, and c, as x goes to infinity, the difference between the numerator and the denominator gets smaller and smaller, heading towards 1*(V_0). This allowed me to sketch the graph of V (with x plotted on the +x-axis, and V on the +y-axis) which has a peak that goes up to infinity at the value x=c. I ended up just using a=1, b=10, and c=100 for the rest of the problem. The function goes to 1 as x --> (- infinity). going from (- infinity) towards x=a the function goes from 1 down to 0 on the y-axis, then between x=a and x=c the graph peaks going all the way up to infinity on the y-axis then decreasing down to V(x)=1 as x --> infinity.

    I hope that wasn't too confusing, I'm very tired and kind of pressed for time, I can't think of a better way to describe it. Either way I am sure I got it.

    Answering part (b) is proving a little challenging. Since V_0 does not depend on x, then the energy is irrelevant in calculating at what x position does the particle reflect.

    Forbidden domains would be the interval (x=a, x=b) right? I would say that it is forbidden for V(x) to be negative but there is nowhere on the graph where this is negative.

    I can't see where there would be any turning points. I tried taking the derivative and graphed it but it told me that there would be turning points at x=-100 and x=100, but I didn't see any strange activity on the function V(x) at any other value for x besides x=c=100.

    Any help on whether I am close to answering it correctly or if I am completely wrong would be appreciated.
     
  2. jcsd
  3. Sep 30, 2014 #2

    Simon Bridge

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    That is an inefficient, though often a useful, shortcut: but you have to be smart about it. Try assigning a bigger value to ##V_0##, and not going so close to x=3. Better: use your knowledge of functions and critical points (remember secondary school maths?) to discover the shape of the potential.

    What happens to V(x) when x=a and at x=b?
    Is V_0 the same as V(0)

    Are their any turning points, points of inflection, horizontal or vertical assymptotes?
     
  4. Sep 30, 2014 #3

    vela

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    I don't follow your reasoning here.

    No, this isn't correct.

    I think you need to go back and review energy level diagrams. You seem to have fundamental misconceptions that are leading you far astray.

    You also have to get over your need to use specific numbers for a, b, and c. You need to be able to think more generally.
     
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