= Solution
For part (a), the <adjoint operator> is
$$
A^\dagger=\frac{m\omega X-iP}{\sqrt{2m\hbar\omega}}.
$$
Using the <canonical commutation relation> $[X,P]=i\hbar I$,
$$
[A,A^\dagger]
=\frac1{2m\hbar\omega}[m\omega X+iP,m\omega X-iP]
=\boxed{I}.
$$
Thus $A$ and $A^\dagger$ are the <creation and annihilation operators>[annihilation and creation operators] of the <quantum harmonic oscillator>.
For part (b), write
$$
K=\frac12(A^{\dagger2}-A^2),
\qquad S(\gamma)=e^{-\gamma K}.
$$
Since $[A,K]=A^\dagger$ and $[A^\dagger,K]=A$, differentiation of the <unitary conjugation> gives
$$
\frac d{d\gamma}A(\gamma)
=-S^\dagger[A,K]S=-A^\dagger(\gamma),
\qquad
\frac d{d\gamma}A^\dagger(\gamma)=-A(\gamma).
$$
The initial conditions are $A(0)=A$ and $A^\dagger(0)=A^\dagger$. Solving this pair of <linear ordinary differential equation>[linear ordinary differential equations] gives the <Bogoliubov transformation>
$$
\boxed{A(\gamma)=A\cosh\gamma-A^\dagger\sinh\gamma.}
$$
This is the transformation implemented by the <single-mode squeeze operator>.
For part (c), invert the definition of $A$:
$$
X=\sqrt{\frac{\hbar}{2m\omega}}(A+A^\dagger),
\qquad
P=-i\sqrt{\frac{m\hbar\omega}{2}}(A-A^\dagger).
$$
The transformation above implies
$$
S^\dagger XS=e^{-\gamma}X,
\qquad
S^\dagger PS=e^\gamma P.
$$
The <ground state> has zero position and momentum means and variances $\hbar/(2m\omega)$ and $m\hbar\omega/2$. Hence the <squeezed vacuum state> $|\gamma\rangle=S(\gamma)|0\rangle$ has
$$
(\Delta X)^2_\gamma=e^{-2\gamma}\frac{\hbar}{2m\omega},
\qquad
(\Delta P)^2_\gamma=e^{2\gamma}\frac{m\hbar\omega}{2}.
$$
The reciprocal squeezing factors cancel, so
$$
\boxed{\Delta X\,\Delta P=\frac\hbar2.}
$$
The state therefore continues to saturate the <Robertson uncertainty principle>[Heisenberg uncertainty relation].
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