Past exam of the quantum mechanics quantum theory and quantum computers course of the University of Oxford / 2009 / A / 3

With the specified ladder-operator normalization, $A_{+}^{†}=A_{-}$ andTherefore $\Vert A_{-}\psi {\Vert }^2=2m(E-\hslash \omega /2)\Vert \psi {\Vert }^2\ge 0$. This proves $\boxed{E\ge \hslash \omega /2}$, with equality exactly when $A_{-}\psi =0$. In position representation this equation is $\hslash {\psi }^{\prime }+m\omega x\psi =0$, whose general solution isFor a unit vector $|C|=(m\omega /\pi \hslash {)}^{1/4}$; its phase is arbitrary. The commutators $[H,A_{\pm }]=\pm \hslash \omega A_{\pm }$ make $A_{+}$ raise and $A_{-}$ lower energies by $\hslash \omega$. Repeated raising of the Gaussian produces all levels $(n+1/2)\hslash \omega$. Conversely lowering any nonzero energy eigenvector must eventually terminate, since otherwise it would produce an energy below the lower bound. A terminating vector is annihilated by $A_{-}$ and hence has energy $\hslash \omega /2$. Thus no other eigenvalues occur.

Bilinearity of the two-dimensional commutators giveswhile all momentum-momentum and position-position commutators remain zero. Equal angles give a canonical position-momentum pair. Perpendicular angles give commuting components, which belong to independent coordinate directions.

The orthogonal coordinates $X_{+}=(X_1+X_2)/\sqrt{2}$, $X_{-}=(X_1-X_2)/\sqrt{2}$ preserve the kinetic sum, andHence $\boxed{{\omega }_{+}=\omega ,{\omega }_{-}=3\omega }$ and $H=H_{+}+H_{-}$. Every cross-mode position/momentum pair commutes, so $[H_{+},H_{-}]=0$. On the tensor-product oscillator basis,For each $N\ge 0$, allowed pairs have $n_{-}=0,1,\dots ,\lfloor N/3\rfloor$ and $n_{+}=N-3n_{-}$. ThusEach pair contributes one product eigenvector; counting them accounts for the energy-level degeneracy.