One-dimensional linear harmonic oscillator perturbation

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Firben
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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



ψn(x) = [itex]\frac{α}{√π*2<sup>n</sup>*n!}[/itex]1/2 * e2x2[itex]\frac{1}{2}[/itex]

E1n = <ψ0n|H'|ψ0n>

The Attempt at a Solution




E10 = <ψ00|H'|ψ00> =

∫[itex]\frac{α}{√π*2<sup>n</sup>*n!}[/itex]1/2 * e2x2[itex]\frac{1}{2}[/itex]*[itex]\frac{α}{√π*2<sup>n</sup>*n!}[/itex]1/2 * e2x2[itex]\frac{1}{2}[/itex]*H'* dx

Is this right ? What is α in this case ?
 
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Firben said:

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



ψn(x) = [itex]\frac{α}{√π*2<sup>n</sup>*n!}[/itex]1/2 * e2x2[itex]\frac{1}{2}[/itex]

E1n = <ψ0n|H'|ψ0n>

The Attempt at a Solution

E10 = <ψ00|H'|ψ00> =

∫[itex]\frac{α}{√π*2<sup>n</sup>*n!}[/itex]1/2 * e2x2[itex]\frac{1}{2}[/itex]*[itex]\frac{α}{√π*2<sup>n</sup>*n!}[/itex]1/2 * e2x2[itex]\frac{1}{2}[/itex]*H'* dx

Is this right ? What is α in this case ?
Your expression for the n-th wavefunction is missing the Hermite polynomial Hn(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.