One-dimensional linear harmonic oscillator perturbation

[Firben]

Homework Statement

Consider a one-dimensional linear harmonic oscillator perturbed by a Gaussian perturbation H' = λe^-ax2. Calculate the first-order correction to the groundstate energy and to the energy of the first excited state

Homework Equations

ψ_n(x) = \frac{α}{√π*2ⁿ*n!}^1/2 * e^{-α2x2\frac{1}{2}}

E¹_n = <ψ⁰_n|H'|ψ⁰_n>

The Attempt at a Solution

E¹₀ = <ψ⁰₀|H'|ψ⁰₀> =

∫\frac{α}{√π*2ⁿ*n!}^1/2 * e^{-α2x2\frac{1}{2}}*\frac{α}{√π*2ⁿ*n!}^1/2 * e^{-α2x2\frac{1}{2}}*H'* dx

Is this right ? What is α in this case ?

 

[vela]

Homework Statement

Consider a one-dimensional linear harmonic oscillator perturbed by a Gaussian perturbation H' = λe^-ax2. Calculate the first-order correction to the groundstate energy and to the energy of the first excited state

Homework Equations

ψ_n(x) = \frac{α}{√π*2ⁿ*n!}^1/2 * e^{-α2x2\frac{1}{2}}

E¹_n = <ψ⁰_n|H'|ψ⁰_n>

The Attempt at a Solution

E¹₀ = <ψ⁰₀|H'|ψ⁰₀> =

∫\frac{α}{√π*2ⁿ*n!}^1/2 * e^{-α2x2\frac{1}{2}}*\frac{α}{√π*2ⁿ*n!}^1/2 * e^{-α2x2\frac{1}{2}}*H'* dx

Is this right ? What is α in this case ?

Your expression for the n-th wavefunction is missing the Hermite polynomial H_n(x). It should be
$$\psi_n(x) = \left(\frac{a}{\sqrt{\pi}2^n n!}\right)^{1/2} e^{-a^2x^2/2} H_n(ax).$$ Your expression for the first-order energy correction for the ground state turns out to be fine because H₀(x)=1. The quantity $a$ should be defined in your notes or textbook. It's the characteristic length scale for the harmonic oscillator.