Representations of Galilei group 
Show that the operator $U(\alpha, \beta) = e^{i(\alpha \hat{x}^2 + \beta \hat{p}_{x}^2)}$ can represent the space reflection of the 1D Galilei group: $x \to -x; t \to t$.

I don't really know anything about group theory and the concept of "representing" something isn't clear. It sort of makes sense in the case where you have rotations in 3D and you have matrices that "represent" the rotations but I don't know what to do in this case. I would think showing that $\hat{x} [U(\alpha, \beta) | x' \rangle] = -x' [U(\alpha, \beta)|x' \rangle]$ would do it, but that doesn't even seem to be true.
 A: I expand my comment into an answer. The idea is to fix $\alpha, \beta \in \mathbb R$ in order that, if  $P:= U(\alpha, \beta)$ (which is automatically unitary), we have 
(i) $PP=e^{ik}I$ for some $k\in \mathbb R$,
(ii)  $P\hat{x}P^\dagger = -\hat{x}$,
(iii) $P\hat{p}P^\dagger = -\hat{p}$
Since $\hat{x}$ and $\hat{p}$ has to be treated symmetrically, we assume $\alpha=\beta = t$. From (i) and (ii), it arises $PP\hat{x}= \hat{x}PP$
and $PP\hat{p}= \hat{p}PP$. Since the representation of the CCR is irreducible (actually one should deal with the associate Weyl algebra, but it is irrelevant here), Schur lemma implies that $PP= cI$ for some complex $c$. Since $PP$ is unitary (i) must hold automatically, if (ii) and (iii) are satisfied. However it is technically more convenient using (iii) in the following to fix $t$ and next check if that choice satisfies (ii) and (iii).
The Hamiltonian $H = \hat{x}^2 +\hat{p}^2$ has spectrum ($\hbar=1$)
$$E_n = 2n+1\:, \quad n=0,1,\ldots$$
So that
$$e^{it H} = e^{it}\sum_{n=0}^{+\infty} e^{i2nt}|n\rangle \langle n|$$
Let us fix $t$ such that (i) is valid. Condition (i) reads here
$$e^{it H}e^{it H} = e^{2it}\sum_{n=0}^{+\infty} e^{i4nt}|n\rangle \langle n|=
e^{ik}\sum_{n=0}^{+\infty} |n\rangle \langle n|$$
This is possible only if $t= t_m= \frac{\pi m}{2}$ for some $m=0,1,\ldots$. However, for $m=0,2,4,\ldots$ one has
$$e^{it_mH}= I\:,$$
which trivially does not satisfies (ii) and (iii). It remains the case $m=1,3,\ldots$ which produces
$$e^{it_m H} = -\sum_{n=0}^{+\infty} e^{in \pi}|n\rangle \langle n|\:.$$
Therefore, our candidate is the one for $t=\pi/2$ (the other choices produce the same result): 
$$P:= U(\pi/2,\pi/2) = e^{\frac{i\pi}{2}(\hat{x}^2+ \hat{p}^2)}= -\sum_{n=0}^{+\infty} e^{in \pi}|n\rangle \langle n| = \sum_{n=0}^{+\infty} (-1)^{n+1}|n\rangle \langle n|\:.$$
With this choice $PP=I$.
It is worth noticing that, with this definition, $P$ would be also self-adjoint that is an observable.
To check if (i) and (ii) are true define
$$\hat{x}(t) = e^{itH}\hat{x}e^{-itH} \quad \mbox{and}\quad  \hat{p}(t) = e^{itH}\hat{p}e^{-itH}$$
Using twice commutation relations you easily see that
$$\frac{d^2\hat{x}}{dt^2} + 4 \hat{x}(t)=0 \quad \mbox{and} \quad \frac{d^2\hat{p}}{dt^2} + 4 \hat{p}(t)=0\:.$$
The solutions are (using the fact that $d\hat{x}/dt = 2\hat{p}(t)$ and 
$d\hat{p}/dt = -2\hat{x}(t)$ again from CCR)
$$\hat{x}(t) = \hat{x} \cos(2t) + \hat{p} \sin(2t) \quad \mbox{and} \quad 
\hat{p}(t) = \hat{p} \cos(2t) - \hat{x} \sin(2t)\:.$$
You see that, indeed, $$\hat{x}\left(\frac{\pi}{2}\right)= -\hat{x} \quad \mbox{and}\quad \hat{p}\left(\frac{\pi}{2}\right)= -\hat{p}$$
which are equivalent to (ii) and (iii) for $P:= U(\pi/2,\pi/2)$.
