# Harmonic Oscillator Potential Approximation

1. Dec 31, 2011

### thelonious

1. The problem statement, all variables and given/known data

A particle is in a region with the potential
V(x) = κ(x2-l2)2
What is the approximate ground state energy approximation for small oscillations about the location of the potential's stable equilibrium?

2. Relevant equations

ground state harmonic oscillator ~ AeC*x2

3. The attempt at a solution

My plan is to find the value of x for which V(x) is at stable equilibrium, then expand about that point, and the expansion should look something like the harmonic oscillator potential (1/2*mωx2). Once the potential is known, I know H = p2/2m + V, and I can use <0|H|0> to find the ground state energy, where |0> is the GS energy eigenstate for the harmonic oscillator.

I have found that the stable equilibrium of V(x) is at x = l.
V = 4κl2(x2-2lx+l2) + O(x-l)3

But what can I do about the 2lx term? I need it to get lost in order to get the HO potential, right?

2. Dec 31, 2011

### vela

Staff Emeritus
You don't want to do anything. You want the potential in terms of powers of (x-l), which is exactly what you have.
$$V \cong \frac{1}{2}(8\kappa l)(x-l)^2$$

3. Dec 31, 2011

### thelonious

So, in calculating <0|H|0>, I will have to integrate ∫(x-l)2e-mωx2/$\hbar$dx from l-$\epsilon$ to l+$\epsilon$
Right?

4. Dec 31, 2011

### vela

Staff Emeritus
You don't need to calculate that expectation value, and no, that isn't how you'd calculate the expectation value either. The idea is you can use the approximation that you have a simple harmonic oscillator. What's the ground-state energy of a harmonic oscillator?

5. Dec 31, 2011

### thelonious

So E=$\hbar$ω/2, and V=1/2*kx2=1/2*8κl2(x-l)2.
ω=√(8kl2/m), so E0=$\hbar$/2*√(8kl2/m)

6. Dec 31, 2011

### vela

Staff Emeritus
Right.