Harmonic Oscillator Potential Approximation

AI Thread Summary
The discussion centers on approximating the ground state energy of a particle in a potential defined by V(x) = κ(x² - l²)². The stable equilibrium point is identified at x = l, and the potential is expanded around this point to resemble a harmonic oscillator. The expansion results in V approximating to (1/2)(8κl)(x - l)², which aligns with the form of a harmonic oscillator potential. The ground state energy is derived using the formula E₀ = (ħ/2)√(8κl²/m). The key takeaway is that the approximation leads to a clear understanding of the ground state energy in terms of the harmonic oscillator model.
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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?
 
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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/\hbardx from l-\epsilon to l+\epsilon
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=\hbarω/2, and V=1/2*kx2=1/2*8κl2(x-l)2.
ω=√(8kl2/m), so E0=\hbar/2*√(8kl2/m)
 
Right.
 
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