Where Does a Particle Reflect in a One-Dimensional Potential Field?

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SUMMARY

The discussion centers on a particle's reflection in a one-dimensional potential field defined by the equation V(x) = [(V_0(x-a)(x-b))/(x-c)^2], where 0 < a < b < c < infinity. Participants analyze the potential's behavior, noting a discontinuity at x=c and identifying forbidden domains between x=a and x=b. The energy expression E=3(V_0)/(c-b) * (b-4a+3c) is discussed, with emphasis on the particle's reflection point and the absence of turning points in the specified range. The conversation highlights the need for a deeper understanding of energy level diagrams and critical points in potential functions.

PREREQUISITES
  • Understanding of one-dimensional potential fields in quantum mechanics
  • Familiarity with the concepts of turning points and forbidden domains
  • Knowledge of energy level diagrams and their significance
  • Ability to analyze functions and their critical points
NEXT STEPS
  • Review the concept of potential energy in quantum mechanics, focusing on V(x) functions
  • Study energy level diagrams to understand particle behavior in potential fields
  • Learn about critical points and their implications in function analysis
  • Explore the implications of discontinuities in potential functions and their physical interpretations
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Students and educators in quantum mechanics, particularly those studying potential fields and particle dynamics, as well as anyone seeking to deepen their understanding of energy levels and function analysis in physics.

castrodisastro
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Homework Statement


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"

Homework Equations


Equations given

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.
 
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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.
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?
 
castrodisastro said:
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.
I don't follow your reasoning here.

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.
No, this isn't correct.

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.
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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