What Quantum Conditions Govern a Particle in a Step Potential?

[derravaragh]

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

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

with

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 n^th energy E_n is

A_n = √[2/a]*√[(K_na)/(1+(K_n*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)

dψ/dx

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 z_n = npi

z = ka = a√[(2mE/h²)] ≈ npi

∴ Solving for E_n = (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 K_n, but that creates quite the mess. As for the second portion of ψ(x), I can just normalize that and solve for C using K_n, 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.

 

[Simon Bridge]

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.