Past exam of the mathematics course of the University of Cambridge / 2010 / ii / Paper 3 / 33C (Principles of Quantum Mechanics)

The canonical commutation relations are $[\hat{x},\hat{p}]=i\hslash I$ (in several dimensions $[x_i,p_j]=i\hslash {\delta }_{ij}I$, with the position-position and momentum-momentum commutators zero). For Hermitian $x,p$, the commutator is anti-Hermitian, just as $i\hslash I$ is; there is no conflict with their self-adjointness. The annihilation operator and its adjoint obeyThus the quantum harmonic oscillator Hamiltonian is $\boxed{H=\hslash (N+1/2)}$. The commutators $[N,a]=-a$, $[N,a^{†}]=a^{†}$ and the normalized ground state constructNormalization follows inductively from $a{a}^{†}=N+1$. Positivity of $N$ and repeated lowering exclude a noninteger eigenvalue: lowering it until it becomes negative would produce a nonzero vector of negative $N$ eigenvalue. Every eigenstate therefore lowers to the unique ground state, so these normalized ladder states exhaust the oscillator eigenstates.

For two independent oscillators, the tensor products $|n_1,n_2⟩$ have $E=\hslash (n_1+n_2+1)$. The level $n=n_1+n_2$ has degeneracy $n+1$. In its degenerate subspace the perturbation $\lambda x_1{x}_2$ reduces tosince terms that raise or lower both numbers leave the subspace. For $n=0$ its matrix is zero. For $n=1$, in $|1,0⟩,|0,1⟩$, it is $(\lambda \hslash /2)\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$. For $n=2$, in $|2,0⟩,|1,1⟩,|0,2⟩$, it is $(\lambda \hslash /2)\left(\begin{array}{ccc}0& \sqrt{2}& 0\\ \sqrt{2}& 0& \sqrt{2}\\ 0& \sqrt{2}& 0\end{array}\right)$. Their eigenvalues giveThe $n=1$ states are the symmetric and antisymmetric combinations. At $n=2$, the shifted states are $(|2,0⟩\pm \sqrt{2}|1,1⟩+|0,2⟩)/2$, and the zero-shift state is $(|2,0⟩-|0,2⟩)/\sqrt{2}$.