A2 - 1D simple harmonic oscillator perturbed by a harmonic potential

consider a harmonic potential that perturbs a 1D SHO:

V = \frac{1}{2} m α^{2} x^{2}

the Hamiltonian:

H = - \frac{h^{2}}{2 m} \frac{d^{2}}{d x^{2}} + \frac{1}{2} m ω^{2} x^{2} + \frac{1}{2} m α^{2} x^{2}

the solution for the unperturbed system is known:

E_{n} = (n + \frac{1}{2}) ℏ ω

the first eigenstate:

ϕ_{0}^{0} = {(\frac{m ω}{π ℏ})}^{\frac{1}{4}} \exp (- \frac{m ω x^{2}}{2 ℏ})

therefore, the first-order energy correction:

\begin{aligned} E_{0}^{1} & = \int_{- \infty}^{\infty} {(\frac{m ω}{π ℏ})}^{\frac{1}{4}} \exp (- \frac{m ω x^{2}}{2 ℏ}) \frac{1}{2} m α^{2} x^{2} {(\frac{m ω}{π ℏ})}^{\frac{1}{4}} \exp (- \frac{m ω x^{2}}{2 ℏ}) d x \\ = {(\frac{m ω}{π ℏ})}^{\frac{1}{2}} \frac{1}{2} m α^{2} \int_{- \infty}^{\infty} x^{2} \exp (- a x^{2}) d x \end{aligned}

where, $a = m ω / h$

using the standard integral:

E_{0}^{1} = \frac{α^{2} ℏ}{4 ω}

E = \frac{1}{2} ℏ ω + \frac{α^{2} ℏ}{4 ω}

to check this, find the exact solution:

\begin{matrix} H = - \frac{ℏ^{2}}{2 m} \frac{d^{2}}{d x^{2}} + \frac{1}{2} m (ω^{2} + α^{2}) x^{2} \\ E_{0} = \frac{1}{2} ℏ (ω^{2} + α^{2})^{\frac{1}{2}} = \frac{1}{2} ℏ ω {(1 + \frac{α^{2}}{ω^{2}})}^{\frac{1}{2}} \end{matrix}

if $α ≪ ω :$

\frac{1}{2} ℏ ω {(1 + \frac{α^{2}}{ω^{2}})}^{\frac{1}{2}} = \frac{1}{2} ℏ ω + \frac{ℏ α^{2}}{4 ω} + O (α^{4}) + \dots