= Solution
Put $g=k\hbar/(2m\omega)$, which has units of <energy>. In the original oscillator basis,
$$
H=\hbar\omega(a^\dagger a+b^\dagger b+1)
+g(ab+a^\dagger b^\dagger+a^\dagger b+ab^\dagger).
$$
Write $A=\widetilde a$, $B=\widetilde b$, $C_2=\cosh2\theta$ and $S_2=\sinh2\theta$. The transformed number and pair terms obey
$$
a^\dagger a+b^\dagger b+1
=C_2(A^\dagger A+B^\dagger B+1)-S_2(AB+A^\dagger B^\dagger),
$$
$$
ab+a^\dagger b^\dagger
=C_2(AB+A^\dagger B^\dagger)-S_2(A^\dagger A+B^\dagger B+1).
$$
Hence the coefficient of $AB+A^\dagger B^\dagger$ is $gC_2-\hbar\omega S_2$, which vanishes for
$$
\boxed{\tanh2\theta=\frac g{\hbar\omega}=\frac{k}{2m\omega^2},
\qquad\theta=\frac12\operatorname{artanh}\frac{k}{2m\omega^2}.}
$$
\b[This cancels the requested pair terms but does not diagonalize the printed <Hamiltonian>.] Its number-conserving coupling also transforms:
$$
a^\dagger b+ab^\dagger
=C_2(A^\dagger B+AB^\dagger)
-\frac{S_2}{2}(A^2+B^2+A^{\dagger2}+B^{\dagger2}).
$$
After the pair cancellation, the last creation terms have coefficient $-gS_2/2$, nonzero for $k\ne0$. They create two quanta from the $A,B$ vacuum, so that vacuum is not an <eigenstate> of the literal <Hamiltonian>. This is an inconsistency in the ground-state assertion used by the next part, present in the printed PDF.
For the actual coordinate-coupled system, the <normal coordinates> $q_\pm=(x\pm y)/\sqrt2$ and their conjugate momenta diagonalize $H$ into independent oscillators with
$$
\boxed{\omega_\pm=\sqrt{\omega^2\pm k/m}.}
$$
A normalizable <ground state> requires $|k|<m\omega^2$. Eliminating just the two indicated cross-pair terms is not sufficient to establish that <ground state>.
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