 Solution

The ladder operators obey $[a,a^{†}]=1$ and $N=a^{†}a$ has $[N,a]=-a$, $[N,a^{†}]=a^{†}$. Since $⟨\psi ,N\psi ⟩=\Vert a\psi {\Vert }^2\ge 0$, its eigenvalues are nonnegative. If $N|\nu ⟩=\nu |\nu ⟩$, lowering gives eigenvalue $\nu -1$ and squared norm $\nu$. Repeated lowering cannot produce a negative nonzero eigenstate; the norm product $\nu (\nu -1)\cdots$ forces $\nu$ to be an integer. The final lowered state has $a|0⟩=0$. With a unique ground state all normalized states are

For time-independent perturbation theory, write $H^{\prime }=H+\lambda W$ with $W=\hslash \omega (a^2+a^{†2})$. The standard first correction is $\lambda ⟨n|W|n⟩$ and the second is ${\lambda }^2{\sum }_{j\ne n}|⟨j|W|n⟩{|}^2/(E_n-E_j)$. The first vanishes. Only $j=n-2,n+2$ occur, with ladder coefficients $\sqrt{n(n-1)}$ and $\sqrt{(n+1)(n+2)}$, respectively. ThereforeThe downward term is zero for $n<2$.

Inverting the given position/momentum expressions givesThus $\boxed{\alpha =1-2\lambda ,\beta =1+2\lambda }$. For $|\lambda |<1/2$ both are positive, and this is a quantum harmonic oscillator of effective mass $m/\alpha$ and frequency $\omega \sqrt{\alpha \beta }$. Its exact energies areExpansion gives $E_n^{\prime }=\hslash \omega (n+1/2)[1-2{\lambda }^2+O({\lambda }^4)]$, agreeing with both perturbative corrections.