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The term interaction is sometimes used interchangeably with force, and other times interchangeably with potential energy. Although force and potential energy certainly aren't the same thing, what justification is there for using the same term to cover both?

Short Answer

Expert verified

Because force and potential energy often represent the same thing that is causing the particle to move.

Step by step solution

01

Given data

The force and the potential energy of the particle are both related if the force is a conservative force.

02

Concept of vector gradient of potential energy 

If the force is conservative it can be defined as negative of the vector gradient of the potential energy:

F=-U

03

Conservative Force

If the force is conservative (energy of the particle is conserved under the application of the force if the particle moves in a closed loop or, the change in energy of the particle is zero when it returns to its original position under the application of the force), it can be defined as negative of the vector gradient of the potential energy:

F=-U

So, the force and the potential energy of the particle are both related if the force is a conservative force. If we know the potential energy, we know the force and if we know the force, we know the potential energy (only upto a constant but it does not matter because zero of the potential energy can be defined anywhere). And in physics, we often work with conservative forces, for example: gravitational force, electromagnetic force, spring force etc.

04

Gravitational potential energy

So, the two are analogous and therefore the term interaction is used to cover both of them. For example, while working with gravity, whether we say "the gravitational force acting on a particle at a particular point in space is this much" or "the gravitational potential energy of the particle at that particular point in space is that much", both the statements describe the same phenomenon.

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Most popular questions from this chapter

There are mathematical solutions to the Schrödinger equation for the finite well for any energy, and in fact. They can be made smooth everywhere. Guided by A Closer Look: Solving the Finite Well. Show this as follows:

(a) Don't throw out any mathematical solutions. That is in region Il (x<0), assume that (Ce+ax+De-ax), and in region III (x>L), assume thatψ(x)=Fe+ax+Ge-ax. Write the smoothness conditions.

(b) In Section 5.6. the smoothness conditions were combined to eliminate A,Band Gin favor of C. In the remaining equation. Ccanceled. leaving an equation involving only kand α, solvable for only certain values of E. Why can't this be done here?

(c) Our solution is smooth. What is still wrong with it physically?

(d) Show that

localid="1660137122940" D=12(B-kαA)andF=12e-αL[(A-Bkα)sin(kL)+(Akα+B)cos(kL)]

and that setting these offending coefficients to 0 reproduces quantization condition (5-22).

a) Taking the particle’s total energy to be 0, find the potential energy.

(b) On the same axes, sketch the wave function and the potential energy.

(c) To what region would the particle be restricted classically?

To describe the matter wave, does the function Asin(kx)cosωthave well-defined energy? Explain

Under what circumstance does the integral x0xbdxdiverge? Use this to argue that a physically acceptable wave function must fall to 0 faster than|x|1/2 does as xgets large.

What is the probability that a particle in the first excited (n=2) state of an infinite well would be found in the nuddle third of the well? How does this compare with the classical expectation? Why?

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