How Are Energy Levels Determined in a Power Law Potential?

[Dipoleman]

Homework Statement

A particle of mass m is bound in a one-dimensional power law potential V(x)=K*x^B, where B is an even positive integer. Show that the allowed energy levels are proportional to m^[-B/(2+B)].

Homework Equations

Using Schrodinger Eq (one-dimensional)

The Attempt at a Solution

I seem to get stuck, there is an example in the book, however that case is for the Harmonic potential V(x)=(K*x^2)/2

Any hints would be greatly appreciated.

Thanks,
Dipoleman

 

[mark726]

Hi Dipoleman,

To solve this problem, we can use the one-dimensional Schrodinger equation:

-Ĥ^2/2m d^2ψ/dx^2 + V(x)ψ = Eψ

First, let's rewrite the potential as V(x) = Kx^B. Then, we can plug this into the Schrodinger equation and simplify:

-Ĥ^2/2m d^2ψ/dx^2 + Kx^Bψ = Eψ

Next, let's make a substitution: y = x^(B/2). Then, we can rewrite the equation as:

-Ĥ^2/2m (d^2ψ/dy^2)(dy/dx)^2 + Ky^2ψ = Eψ

Substituting in the value for dy/dx, we get:

-Ĥ^2/2m (d^2ψ/dy^2)(B^2/4)y^(B-2) + Ky^2ψ = Eψ

Now, let's make one more substitution: z = y^(B-2)/2. This will simplify the equation further:

-Ĥ^2/2m d^2ψ/dz^2 + (K/4)(B^2 - 4)z^2ψ = Eψ

We can further simplify this by defining a new constant, C = (K/4)(B^2 - 4), giving us:

-Ĥ^2/2m d^2ψ/dz^2 + Cz^2ψ = Eψ

This is now in the form of the Schrodinger equation for the harmonic oscillator potential, which has known solutions for the energy levels. We can use these solutions and substitute back in our original variables to get the energy levels for our power law potential.

The energy levels for the harmonic oscillator potential are given by:

E_n = (n + 1/2)ℏω

Where n is a non-negative integer and ω = √(C/m).

Substituting back in our original variables, we get:

E_n = (n + 1/2)ℏ√[(K/4)(B^2 - 4)/m]

Simplifying further, we get:

E_n = (n +