= Solution
The <number operator> <commutators> established above imply that $\hat a_x^\dagger|n\rangle$ is proportional to $|n+1\rangle$ and $\hat a_x|n\rangle$ to $|n-1\rangle$. Their <norms> determine the magnitudes of the proportionality constants:
$$
\|\hat a_x^\dagger|n\rangle\|^2
=\langle n|\hat a_x\hat a_x^\dagger|n\rangle=n+1,
\qquad
\|\hat a_x|n\rangle\|^2
=\langle n|\hat a_x^\dagger\hat a_x|n\rangle=n.
$$
Choose the <phase> of the normalized ground <eigenstate> once, then define recursively
$$
|n+1\rangle=\frac{\hat a_x^\dagger|n\rangle}{\sqrt{n+1}},
\qquad
|n\rangle=\frac{(\hat a_x^\dagger)^n}{\sqrt{n!}}|0\rangle.
$$
This removes the otherwise arbitrary relative <phases> of the <eigenstates>. The <Hermitian adjoint> relation then fixes the lowering coefficient with the same positive sign, giving
$$
\boxed{\hat a_x^\dagger|n\rangle=\sqrt{n+1}|n+1\rangle,
\qquad \hat a_x|n\rangle=\sqrt n|n-1\rangle.}
$$
At $n=0$, the second expression means the <zero vector>, $\hat a_x|0\rangle=0$; it does not introduce a physical state $|-1\rangle$.
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