Harmonic Oscillator Potential Approximation

thelonious
Messages
15
Reaction score
0

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?
 
Physics news 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/\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.
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top