Harmonic Oscillator Potential Approximation

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thelonious
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Homework Statement



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?

Homework Equations



ground state harmonic oscillator ~ AeC*x2


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.
Next, I expanded V(x) about this point and found:
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?
 
on Phys.org
thelonious said:

Homework Statement



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?

Homework Equations



ground state harmonic oscillator ~ AeC*x2


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.
Next, I expanded V(x) about this point and found:
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?
You don't want to do anything. You want the potential in terms of powers of (x-l), which is exactly what you have.
$$V \cong \frac{1}{2}(8\kappa l)(x-l)^2$$
 
So, in calculating <0|H|0>, I will have to integrate ∫(x-l)2e-mωx2/[itex]\hbar[/itex]dx from l-[itex]\epsilon[/itex] to l+[itex]\epsilon[/itex]
Right?
 
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?
 
So E=[itex]\hbar[/itex]ω/2, and V=1/2*kx2=1/2*8κl2(x-l)2.
ω=√(8kl2/m), so E0=[itex]\hbar[/itex]/2*√(8kl2/m)