= Solution
{c}
The <position operators> and <momentum operators> are <self-adjoint operators>. Using conjugate linearity of the <adjoint operator>, $(cA+dB)^*=\overline cA^*+\overline dB^*$, gives
$$
\boxed{a_j^*=\frac{m\omega X_j-iP_j}{\sqrt{2m\omega\hbar}}.}
$$
Write $D=\sqrt{2m\omega\hbar}$. Bilinearity and antisymmetry of the <commutator>, together with the <canonical commutation relations>, give $[P_j,X_k]=-i\hbar\delta_{jk}I$. Hence
$$
\begin{aligned}
D^2[a_j,a_k]&=im\omega[X_j,P_k]+im\omega[P_j,X_k]\\
&=-m\omega\hbar\delta_{jk}I+m\omega\hbar\delta_{jk}I=0,\\
D^2[a_j,a_k^*]&=-im\omega[X_j,P_k]+im\omega[P_j,X_k]\\
&=2m\omega\hbar\delta_{jk}I.
\end{aligned}
$$
The terms involving two <position operators> or two <momentum operators> vanish by the <canonical commutation relations>. Finally, reversing products under the <adjoint operator>, $(AB)^*=B^*A^*$, gives $[a_j,a_k]^*=-[a_j^*,a_k^*]$. Therefore
$$
\boxed{[a_j,a_k]=[a_j^*,a_k^*]=0,\qquad [a_j,a_k^*]=\delta_{jk}I.}
$$
Here $I$ is the <identity operator>. These are the relations for two independent pairs of <creation and annihilation operators>. The <operator algebra> in this question is understood on a common invariant <operator domain>, such as finite <linear combinations> of the <number states> constructed in part (c).
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