 Solution

The position operators and momentum operators are self-adjoint operators. Using conjugate linearity of the adjoint operator, $(cA+dB{)}^{\ast }=\overline{c}{A}^{\ast }+\overline{d}{B}^{\ast }$, givesWrite $D=\sqrt{2m\omega \hslash }$. Bilinearity and antisymmetry of the commutator, together with the canonical commutation relations, give $[P_j,X_k]=-i\hslash {\delta }_{jk}I$. HenceThe terms involving two position operators or two momentum operators vanish by the canonical commutation relations. Finally, reversing products under the adjoint operator, $(AB{)}^{\ast }=B^{\ast }{A}^{\ast }$, gives $[a_j,a_k{]}^{\ast }=-[a_j^{\ast },a_k^{\ast }]$. ThereforeHere $I$ 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).