# One dimensional potential function

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1. Sep 29, 2014

### castrodisastro

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. Sep 30, 2014

### Simon Bridge

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?

3. Sep 30, 2014

### vela

Staff Emeritus