= Solution
From the <creation and annihilation operators> commutator,
$$
[H,A]=\hbar\omega[A^\dagger A,A]=-\hbar\omega A.
$$
Therefore $A|n\rangle$ is either zero or an <energy eigenstate> with energy $E_n-\hbar\omega=E_{n-1}$. Since each energy level of the one-dimensional <quantum harmonic oscillator> is <nondegenerate>, $A|n\rangle=c_n|n-1\rangle$. Its <norm> determines the coefficient:
$$
|c_n|^2=\langle n|A^\dagger A|n\rangle=n.
$$
Choosing the conventional phases of the <number states> gives
$$
\boxed{A|n\rangle=\sqrt n\,|n-1\rangle.}
$$
Solving the given relation for the <position operator> gives
$$
X=\sqrt{\frac{\hbar}{2m\omega}}(A+A^\dagger).
$$
Put $Q=A+A^\dagger$. Repeated use of the ladder relations gives
$$
Q|0\rangle=|1\rangle,
\qquad Q^2|0\rangle=|0\rangle+\sqrt2|2\rangle,
$$
$$
Q^3|0\rangle=3|1\rangle+\sqrt6|3\rangle,
\qquad
Q^4|0\rangle=3|0\rangle+6\sqrt2|2\rangle+2\sqrt6|4\rangle.
$$
Hence
$$
X^4|0\rangle=\left(\frac{\hbar}{2m\omega}\right)^2
\left(3|0\rangle+6\sqrt2|2\rangle+2\sqrt6|4\rangle\right).
$$
In <first-order nondegenerate perturbation theory>, the $|0\rangle$ component is omitted from the state correction, while $E_0-E_2=-2\hbar\omega$ and $E_0-E_4=-4\hbar\omega$. It follows that
$$
\boxed{|0_\lambda\rangle
=|0\rangle-\frac{\hbar\lambda}{4m^2\omega^3}
\left(3\sqrt2|2\rangle+\sqrt{\frac32}|4\rangle\right)+O(\lambda^2),}
$$
which is the <first-order ground state of the quartic oscillator>.
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