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In the position representation in quantum mechanics, $\hat{\mathbf{p}}=-i\hslash \nabla$. Therefore the orbital angular momentum operator givesAt fixed $r,\theta$, the spherical coordinates obey ${\partial }_{\varphi }x=-y$, ${\partial }_{\varphi }y=x$, and ${\partial }_{\varphi }z=0$. The chain rule consequently gives ${\partial }_{\varphi }=-y{\partial }_x+x{\partial }_y$, henceThe derivative holds the other two spherical coordinates fixed; it is not a derivative along an arbitrary spatial curve.

Factorization. Put $\alpha =\sqrt{m\omega /(2\hslash )}$ and use the canonical commutation relation $[x,p]=i\hslash$:It follows thatThe $-1/2$ arises from quantum operator ordering; treating $x$ and $p$ as commuting would lose the zero-point energy.

commutators. Direct use of $[x,p]=i\hslash$ gives $[a,a^{†}]=1$. The product rule for commutators then yields $[a^{†}a,a]=-a$ and $[a^{†}a,a^{†}]=a^{†}$, hence

Raising the energy. If $H|\varphi ⟩=E|\varphi ⟩$, thenThus the creation operator produces an energy eigenstate with energy $E+\hslash \omega$. It is nonzero: for a normalized physical oscillator eigenstate, $\Vert a^{†}\varphi {\Vert }^2=⟨a{a}^{†}⟩=E/(\hslash \omega )+1/2>0$. Normalize by this norm if a normalized state is wanted.

Lowest wavefunctions and orthogonality. In the position representation in quantum mechanics,The annihilation condition therefore gives ${\psi }_0^{\prime }=-(m\omega /\hslash )x{\psi }_0$. Applying $a^{†}$ to its solution givesTheir inner product is proportional to ${\int }_{-\infty }^{\infty }x{e}^{-m\omega x^2/\hslash }dx=0$, since the integrand is odd. These are the even and odd lowest wavefunctions shown in A2, with energies $\hslash \omega /2$ and $3\hslash \omega /2$.