= Quadratic excitation of an oscillator ground state
{title2=$X^2|0\rangle\propto |0\rangle+\sqrt2\,|2\rangle$}
Multiplying the <quantum harmonic oscillator> <ground state> by $x^2$ produces a <quantum superposition> of only its $n=0$ and $n=2$ <energy eigenstates>. In the conventional phase choice for the <creation and annihilation operators>, $X=\sqrt{\hbar/(2m\omega)}(b+b^\dagger)$ and $b|0\rangle=0$. Therefore
$$
X^2|0\rangle=\frac{\hbar}{2m\omega}(|0\rangle+\sqrt2\,|2\rangle).
$$
After <wavefunction normalization>, the <energy measurement> probabilities are $1/3$ at $\hbar\omega/2$ and $2/3$ at $5\hbar\omega/2$. Rephasing the second basis state can change the displayed relative coefficient's sign but cannot change these <probabilities>. For example, defining the unscaled <creation operator> as $a_+=P+im\omega X=i\sqrt{2m\hbar\omega}\,b^\dagger$ reverses the sign of the $n=2$ basis function generated from $(a_+)^2|0\rangle$.
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