= Solution
Use the <canonical commutation relations> $[\hat x,\hat p_x]=[\hat y,\hat p_y]=i\hbar$, with cross-coordinate <commutators> zero. For either coordinate,
$$
[\hat a,\hat a^\dagger]=2\alpha\beta\hbar,
\qquad
\hat a^\dagger\hat a=\alpha^2\hat x^2+\beta^2\hat p_x^2-\alpha\beta\hbar.
$$
Thus $2\alpha\beta\hbar=1$. To obtain the required <Hamiltonian>, the ratio of its <position operator> and <momentum operator> coefficients must be preserved:
$$
\frac{\alpha^2}{\beta^2}=m^2\omega^2,
\qquad \alpha=m\omega\beta,
$$
where the positive root follows from the requested signs. Consequently
$$
\boxed{\alpha_x=\alpha_y=\sqrt{\frac{m\omega}{2\hbar}},
\qquad \beta_x=\beta_y=\frac1{\sqrt{2m\hbar\omega}}.}
$$
The <annihilation operators> are dimensionless; $\alpha$ has units of inverse length and $\beta$ has units of inverse <momentum>. Operators belonging to different coordinates commute automatically. Writing the <number operators> as $\hat N_j=\hat a_j^\dagger\hat a_j$, substitution gives
$$
\boxed{\hat H=\hbar\omega(\hat N_x+\hat N_y+1).}
$$
The constant is the sum of the two <zero-point energies> of this <two-dimensional isotropic harmonic oscillator>.
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