Answered: or a simple harmonic oscillator… | bartleby

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1
For a simple harmonic oscillator potential, Vo(x) =kx² = mo'x?,
%|
2
ħo
and the ground state
2
the ground state energy eigenvalue is E
eigenfunction is
a²x?
то
exp
where a?
%3D
2
Now suppose that the potential has a small perturbation,
1
kx² →
1
-kx? + λxό.
2
Vo(x):
→ V(x) =
Use perturbation theory to find the (first order) corrected eigenvalue, in terms of @.
[7]
1x 3 x 5 x ...× (2n – 1)
2n+1Bn
You will need:
x2" exp(-ßx²) dx =

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