How Does the Uncertainty Principle Relate to Energy in a V-Shaped Potential?

In summary, the conversation discussed a particle in a vee-shaped potential and how to use the uncertainty principle and simple harmonic oscillator to find the lowest state energy. There was some confusion about the width of the well, but it was resolved by correctly minimizing for momentum rather than position. The final result showed that the energy was of the order of ((hbar)^2(b)^2/m)^(1/3), with the correct dimensions.
  • #1
ultimateguy
125
1

Homework Statement


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.
 
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  • #2
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.
 
  • #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.
 
  • #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.
 
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  • #5
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.
 
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  • #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.
 
  • #7
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.
 
  • #8
ultimateguy said:
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.

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.
 
  • #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.
 
  • #10
ultimateguy said:
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.

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.
 

Related to How Does the Uncertainty Principle Relate to Energy in a V-Shaped Potential?

1. What is a particle in a potential well?

A particle in a potential well is a concept in quantum mechanics that describes the behavior of a particle (such as an electron) confined by a potential energy barrier. The particle is able to move freely within the well, but its energy is limited by the potential barrier on either side.

2. How is the behavior of a particle in a potential well different from that of a free particle?

A free particle has no potential energy barrier restricting its movement, so it is able to move with constant velocity. However, a particle in a potential well is confined by the potential barrier and its energy is limited, causing its movement to be more complex and dependent on its energy state.

3. What is the significance of the potential well in quantum mechanics?

The potential well is a fundamental concept in quantum mechanics that helps explain the behavior and properties of particles at the atomic and subatomic level. It is used to describe the confinement and behavior of particles in a variety of systems, such as atoms, molecules, and solid-state materials.

4. How does the depth and width of a potential well affect the behavior of a particle?

The depth and width of a potential well determine the energy states and behavior of a particle within it. A deeper and narrower well will result in a particle with higher energy levels and more confined movement, while a shallower and wider well will result in lower energy levels and more freedom of movement for the particle.

5. Can a particle in a potential well escape from the well?

Yes, a particle in a potential well can escape if it gains enough energy to overcome the potential barrier. This can happen through various processes such as absorption of a photon or collision with another particle. However, the probability of escape decreases as the depth and width of the potential well increase.

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