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Particle moving in potential

  • #1

Homework Statement

The wave function for a particle of mass m moving in the potential
V =
{ ∞ for x=0
{ 0 for 0 < x ≤ a
{ V0 for x ≥ a

{ Asin(kx) for 0 < x ≤ a
{ Ce-Kx for x ≥ a


k = √[(2mE/h2)]

K =√[((2m(V0-E)/h2)]

where h is h-bar in both equations, and remains so throughout this thread.

(a) Apply the boundary conditions at x = a and obtain the transcendental equation which determines the bound state energies E.
(b) If
√[(2mV0a2)/h2] = 3pi

determine the allowed bound state energies. Express your answers in the form of a numerical factor multiplying the dimensional factor (h2/(2ma2)).

(c) Given that the normalization factor for the wave function corresponding to the nth energy En is

An = √[2/a]*√[(Kna)/(1+(Kn*a))]

normalize the wave function for the lowest energy state.
(d) What is the probability that a measurement of the position of a particle in the ground state will give a result ≥ a?

Homework Equations

The Attempt at a Solution

I have completed parts (a) and (b) and I believe they are correct, here is a brief outline of what I did.
At x = a: ψ(a)

Asin(ka) = Ce-Ka (1)


kAcos(kx) = -KCe-Ka (2)

Dividing (2) by (1):

kcot(ka) = -K
For the transcendental: z = ka, z0 = a√[(2mV0)/h2]

∴ -cot(z) = √[(z0/z)2-1]

Now setting z0 = 3pi (part b) and graphing -cot(z) and √[(z0/z)2-1] on the same plot, I find that intersections occur just below zn = npi

z = ka = a√[(2mE/h2)] ≈ npi

∴ Solving for En = (n2pi2h2)/(2ma2)

This is where I hit a road block. I believe the lowest energy state is E1 because it is just below pi (or 1pi), but I'm not sure what I'm supposed to do with this. I want to plug this into Kn, but that creates quite the mess. As for the second portion of ψ(x), I can just normalize that and solve for C using Kn, and I think I have a handle on part (d) as well. Any hints on part (c) or just checking my work so far would be much appreciated.

Homework Statement

Homework Equations

The Attempt at a Solution

Homework Statement

Homework Equations

The Attempt at a Solution


Answers and Replies

  • #2
Simon Bridge
Science Advisor
Homework Helper
I believe the lowest energy state is E1 because it is just below pi (or 1pi)
You mean it is the first non-zero energy eigenstate? ##\small E_0=0## and all.

You have the equations for the wavefunctions, you have decided that n=1 is the lowest one, so the lowest energy wavefunction must be ##\psi_1## - which you found an expression for in part a and b.

Part c asks for the normalized wavefunction - which requires the normalization factor A1: which you have an equation for.

What's the problem?

You could always check by normalizing the hard way.

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