III. The adjoint of an operator $\hat Q$ is the operator such $\hat Q ^ {\dagger}$

$\begin{align}\langle f | \hat Q g \rangle = \langle \hat Q ^ {\dagger} f | g\rangle\end{align}$
for all f and g.

Using the well-known fact that the position operator and the momentum operator are Hermitian (they must be, since they are observables!), we have that

$\setcounter{equation}{1}\begin{align} \langle f| \sqrt{\frac{m \omega_0}{2 \hbar}} \left(\hat x + i \frac{\hat p}{m \omega_0...$

Thus, the operator $\hat a$ and its adjoint are different, so statement II is CORRECT.

II. First of all, the fact that an operator is Hermitian does NOT imply that it represents observable. (The converse, however, is true: the fact that an operator represents an observable DOES imply that the operator is hermitian.)
An operator $\hat Q$ is Hermitian if and only if

$\setcounter{equation}{5}\begin{align}\langle f | \hat Q g \rangle = \langle \hat Q f | g\rangle\end{align}$
for all f and g.

If $\hat Q$ were Hermitian, then

$\setcounter{equation}{6}\begin{align}\hat Q = \sqrt{\frac{m \omega_0}{2 \hbar}} \left(\hat x - i \frac{\hat p}{m \omega_0} \r...$
because of the work above and the fact that the Hermitian adjoint of an operator is unique. But equality is clearly NOT true, therefore, the assumption that $\hat Q$ is Hermitian is false.

$\setcounter{equation}{7}$

Solution to 1996 Problem 100