Degenerate Perturbations

Consider the anisotropic harmonic oscillator potential from the tutorial, for which the Hamiltonian is $H= -(\hbar^2/2M)\nabla^2+(1/2)M\omega^2(x^2+y^2)+(\gamma/2)M\omega^2(x+y)^2$, where $\gamma$ is small and positive. In this problem you will solve the potential exactly and compare the energies to those found in the tutorial using the degenerate perturbation theory.
a) Find $u$ and $v$ in terms of $x$ and $y$ using substitution such that
$H=-(\hbar^2/2M)\nabla^2+(M\omega^2/2)[u^2+(1+2\gamma)v^2]$
b) Determine the exact energy of the first 3 states of the perturbed potential Explain.
c) Approximate the exact energies using $\sqrt{(1+2\gamma)} \approx 1+\gamma+…$

a)
Consider:
$H=-ℏ^2/2M*∇^2+1/2*Mω^2 (x^2+y^2 )+1/2*Mω^2 γ(x+y)^2$
And
$H=-ℏ^2/2M*∇^2+1/2*Mω^2 (u^2+(1+2γ) v^2)$
Therefore
$x^2+y^2+γ(x+y)^2=u^2+(1+2γ) v^2$
$x^2+y^2+γ(x+y)^2=u^2+v^2+2γv^2$
$x^2+y^2=u^2+v^2 and 2v^2=(x+y)^2$
The coordinate transformation is
$u^2=1/2*(x-y)^2 and v^2=1/2*(x+y)^2$
b)
For the general Schroedinger equation
$[-ℏ^2/2M ∇^2+V(x,y,z) ]ψ(x,y,z)=Eψ(x,y,z)$
If one can write the potential
$V(x,y,z)=V(x)+V(y)+V(z)$
Then the Sch. equation “decouples” into 2 different equations
$[-ℏ^2/2M*d^2/(dx_i^2 )+V(x_i ) ] ψ_i (x_i )=E_i ψ_i (x_i ) with ψ(x,y,z)=ψ_1 (x) ψ_2 (y) ψ_3 (z)$
And
$E=E_1+E_2+E_3$
In our case the Sch equation “decouples” into 3 different equations (ccordinates u,v and w):
$-ℏ^2/2M*(d^2 ψ_1)/(dw^2 )=E_1 ψ_1$
with $E_1=n^2*ℏω$ (inifinite well) $n=1,2,3…$

$[-ℏ^2/2M d^2/(du^2 )+1/2 Mω^2 u^2 ]*ψ_2=E_2 ψ_2$
with $E_2=ℏω(n+1/2)$, $n=0,1,2…$ (simple harmonic osc.)

$[-1/2M*d^2/(dv^2 )+1/2*Mω^2 (1+2γ) v^2 ]*ψ_3=E_3 ψ_3$
with $E_3=ℏω'(n+1/2)$, $n=0,1,2,…$
The last equation is for perturbed harmonic oscillator and by comparing with the simple harmonic osc. one has:
$ω’2=ω2 (1+2γ)$ or $ω’=ω*\sqrt{(1+2γ)}$

Therefore the exact energies of the perturbed potential will be:
$E=ℏω [(n+1)^2+(n+1/2) ]+ℏω\sqrt{(1+2γ)} (n+1/2)$ where $n=0,1,2,…$
$E_0=ℏω+1/2*ℏω+1/2*ℏω√(1+2γ)=ℏω/2(3+\sqrt{(1+2γ)})$
$E_1=4ℏω+3/2 ℏω+3/2 ℏω√(1+2γ)=ℏω/2(11+3\sqrt{(1+2γ)})$
$E_2=ℏω/2(14+5\sqrt{(1+2γ)})$
c)
If we write $\sqrt{(1+2γ)}≈1+γ$ then
$E_0≈ℏω/2 (4+γ),E_2=ℏω/2*(14+3γ),E_3=ℏω/2*(19+5γ)$

Reference
Degenerate Perturbation Theory


valentin68

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