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For a quantum harmonic oscillator perturbed by $\lambda X^4$, write $X=\sqrt{\hslash /(2m\omega )}(A+A^{†})$. Sincefirst-order nondegenerate perturbation theory gives

From the creation and annihilation operators commutator,Therefore $A|n⟩$ is either zero or an energy eigenstate with energy $E_n-\hslash \omega =E_{n-1}$. Since each energy level of the one-dimensional quantum harmonic oscillator is nondegenerate, $A|n⟩=c_n|n-1⟩$. Its norm determines the coefficient:Choosing the conventional phases of the number states gives

Solving the given relation for the position operator givesPut $Q=A+A^{†}$. Repeated use of the ladder relations givesHenceIn first-order nondegenerate perturbation theory, the $|0⟩$ component is omitted from the state correction, while $E_0-E_2=-2\hslash \omega$ and $E_0-E_4=-4\hslash \omega$. It follows thatwhich is the first-order ground state of the quartic oscillator.

For a normalized eigenstate $|n⟩$ with a nondegenerate energy eigenvalue, first-order nondegenerate perturbation theory and second-order nondegenerate perturbation theory give

Let $\Delta =E_{+}-E_{-}>0$. Hermiticity givesso in the ordered basis $(|-⟩,|+⟩)$,For a nondegenerate unperturbed level, the first-order nondegenerate perturbation theory formulas areThe diagonal matrix elements vanish, so both linear energy corrections are zero. The state corrections areThus, through linear order,Normalization changes only at quadratic order.

The unperturbed ground state is $|0,0⟩$ with energy $\hslash \omega$. Sincethe first-order energy correction vanishes and first-order nondegenerate perturbation theory gives