= Solution
Use the <canonical commutation relations>
$$
[X_j,P_k]=i\hbar\delta_{jk}I,\qquad [X_j,X_k]=[P_j,P_k]=0.
$$
Writing $c_j=\sqrt{m\omega_j/(2\hbar)}$, $d_j=1/\sqrt{2m\hbar\omega_j}$, direct expansion of the <commutator> gives
$$
[a_{j,-},a_{k,+}]=\hbar(c_jd_k+d_jc_k)\delta_{jk}I
=\boxed{\delta_{jk}I}.
$$
The <Fock vacuum> $\psi_{0,0}$ is annihilated by both $a_{j,-}$. Successive <creation operators> generate
$$
\psi_{n_1,n_2}\propto(a_{1,+})^{n_1}(a_{2,+})^{n_2}\psi_{0,0}.
$$
Hence, in the order of part (a), the six <energy eigenstates> are proportional to
$$
\boxed{\psi_{0,0},\quad a_{1,+}\psi_{0,0},\quad a_{2,+}\psi_{0,0},\quad
(a_{1,+})^2\psi_{0,0},\quad a_{1,+}a_{2,+}\psi_{0,0},\quad(a_{2,+})^2\psi_{0,0}.}
$$
If the <Fock vacuum> has unit <norm>, dividing the general expression by $\sqrt{n_1!n_2!}$ gives normalized <number states>. Normalization is not required here, but using this convention makes the later change of basis unambiguous.
Back to article page
