How Does an Electron Impact the Walls of an Infinite Potential Well?

[laxatives]

Homework Statement

An electron is confined to an infinite potential well of width L. Find the force it exerts on the walls of the well in the lowest energy state:
a) Estimate the force using uncertainty principle
b) Calculate the force exactly for the ground-state wavefunction

Homework Equations

[tex]\Delta[/tex]x = L
[tex]\Delta[/tex]x * [tex]\Delta[/tex]p [tex]\geq[/tex] [tex]\frac{hbar}{2}[/tex]
[tex]\Delta[/tex]E * [tex]\Delta[/tex]t = [tex]\frac{hbar}{2}[/tex]

The Attempt at a Solution

I used the uncertainty principle to solve for [tex]\Delta[/tex]p and [tex]\Delta[/tex]t and divided to get F=dp/dt = (mc^2)/L. I'm not sure if this is correct. I actually didn't really have much of a start until I started typing this up just now.

For part b: I'm not sure what the ground-state wavefunction is. Does this mean k = 1?
I have Schrodinger's equation for 1D, the Hamiltonian, and Hermite polynomials, but don't really know where to begin to relate it all back to force.Another question:

Homework Statement

Normalize the wavefunction u(x) proportional to sin(pi*x/L) + sin(2pi*x/L) for a particle of mass m bound in an infinitely deep one-dimensional potential well extending from x = 0 to x = L.

Homework Equations

1 = A^2 Integral u(x)^2 dx

The Attempt at a Solution

So I want to solve for A by integrating the u(x)^2 dx from 0 to L since the probability of finding the particle outside of the well is zero. I found a solution online that states A = (2/L)^(1/2) but I keep calculating the integral to be L resulting in A = (1/L)^(1/2). What am I missing?

 

[vela]

Homework Statement

An electron is confined to an infinite potential well of width L. Find the force it exerts on the walls of the well in the lowest energy state:
a) Estimate the force using uncertainty principle
b) Calculate the force exactly for the ground-state wavefunction

Homework Equations

[tex]\Delta[/tex]x = L
[tex]\Delta[/tex]x * [tex]\Delta[/tex]p [tex]\geq[/tex] [tex]\frac{\hbar}{2}[/tex]
[tex]\Delta[/tex]E * [tex]\Delta[/tex]t = [tex]\frac{\hbar}{2}[/tex]

The Attempt at a Solution

I used the uncertainty principle to solve for [tex]\Delta[/tex]p and [tex]\Delta[/tex]t and divided to get F=dp/dt = (mc^2)/L. I'm not sure if this is correct. I actually didn't really have much of a start until I started typing this up just now.

The uncertainty principle relating energy and time doesn't apply here. You want to estimate the time from the width of the well and the momentum of the electron.

For part b: I'm not sure what the ground-state wavefunction is. Does this mean k = 1?
I have Schrodinger's equation for 1D, the Hamiltonian, and Hermite polynomials, but don't really know where to begin to relate it all back to force.

The ground state is the lowest-energy state. I'm not sure what conventions you're using, but usually, the ground state corresponds to n=1; k is typically the spatial frequency 2π/λ.

The Hermite polynomials have nothing to do with this problem. They appear in the solutions for the quantum mechanical simple harmonic oscillator.

From the wording of the question, I'm not sure what they had in mind. I would calculate the force by considering what happens to the energy if the width of the well were increased by an infinitesimal amount.

Another question:

Homework Statement

Normalize the wavefunction u(x) proportional to sin(pi*x/L) + sin(2pi*x/L) for a particle of mass m bound in an infinitely deep one-dimensional potential well extending from x = 0 to x = L.

Homework Equations

1 = A^2 Integral u(x)^2 dx

The Attempt at a Solution

So I want to solve for A by integrating the u(x)^2 dx from 0 to L since the probability of finding the particle outside of the well is zero. I found a solution online that states A = (2/L)^(1/2) but I keep calculating the integral to be L resulting in A = (1/L)^(1/2). What am I missing?

Your answer is correct. The factor sqrt(2/L) is for a single eigenstate. You have a superposition of two eigenstates, so the factor is reduced by 1/sqrt(2).