 Solution

In the position representation $p=-i\hslash d/dx$, so $a{\psi }_0=0$ becomesIntegrating and using the Gaussian integral for wavefunction normalization givesThe creation operator gives $|1⟩=a^{†}|0⟩$, and thereforeIts norm is one since $\int x^2|{\psi }_0{|}^{2}dx=\hslash /(2m\omega )$. Moreover ${\psi }_0$ is even and ${\psi }_1$ is odd, so the overlap integrand ${\psi }_0^{\ast }{\psi }_1$ is odd and integrable. Integration over the whole real line gives $\boxed{⟨0|1⟩=0}$, establishing orthogonality.