Prove that a translation operator times a reflection operator is unitary and Hermitian I am trying to prove some properties of the product of the (unitary) translation operator $\hat{T}(a)\psi(x) = \psi(x-a)$ and the (Hermitian) reflection operator $\hat{R} \psi(x) = \psi(-x)$. In particular, I want to show that $\left(\hat{R}\hat{T}(a)\right)^\dagger$ is both Hermitian and unitary. 
Is it correct just to say: $\left(\hat{R}\hat{T}(a)\right)^\dagger \psi(x) = \hat{T}(-a) \hat{R} \psi(x) = \hat{T}(-a) \psi(-x) = \psi(-x+a)$ while $\hat{R} \hat{T}(a) \psi(x) = \hat{R}\psi(x-a) = \psi(-x+a)$? Because then I show that $\left(\hat{R}\hat{T}(a)\right)^\dagger = \hat{R}\hat{T}(a)$... I'm not sure how to show the next part-- or if it's even true!
 A: Intuitively, shifting then reflecting is not the same as reflecting then shifting. Consider the case of first shifting 1 unit to the right from 0, then reflecting: you end up at $x=-1$. If you reflect first, it does nothing, and then shifting to the right by $1$ means you end up at $x=1$.
The problem is that
$$\hat{T}(-a)\hat{R}\psi(x) =\hat{T}(-a)\psi(-x) = \psi(-(x+a)) = \psi(-x-a),$$
whereas
$$\hat{R}\hat{T}(-a)\psi(x) = \hat{R}\psi(x+a) = \psi((-x)+a) = \psi(-x+a).$$
So these aren't the same. However,
$$\hat{R}\hat{T}(a)\psi(x) = \hat{R}\psi(x-a) = \psi((-x)-a) = \psi(-x-a),$$
which indicates that (since $\psi$ was arbitrary),
$$(\hat{R}\hat{T}(a))^{\dagger}=(\hat{T}(a))^{\dagger}\hat{R}=\hat{T}(-a)\hat{R}=\hat{R}\hat{T}(a).$$
This shows that it's Hermitian. The thing that we show above can also be used to show that it's unitary, but I'll leave that to you.

Here's how I like to understand things, because perhaps it's not clear why some of the above calculations are true. I prefer to work in Dirac notation and consider explicitly the action of these operators on the position basis vectors:
$$\hat{R}|x\rangle = \left|-x\right\rangle,$$
and
$$\hat{T}(a)|x\rangle = |x+a\rangle.$$
For the bra-vectors,
$$\langle x|\hat{R} = \langle-x|,$$
and
$$\langle x|\hat{T}(a) = \langle x-a|.$$
(That is, $\hat{T}(a)$ acts like $\hat{T}(-a)$ if it acts to the left.)
This allows us to compute what happens to the amplitudes by taking inner products. For instance, if we define the transformed vector $|\psi'\rangle=\hat{T}(a)|\psi\rangle$, then
$$\psi'(x) = \langle x|\psi'\rangle = \langle x|\hat{T}(a)|\psi\rangle = \langle x-a|\psi\rangle = \psi(x-a),$$
which indicates that $\hat{T}(a)\psi(x) = \psi(x-a)$.
A: Background
You already seem to know this stuff but it's worth going over again.
So, the adjoint of an operator is the equivalent effect of the operator on the other side of the wavefunction inner product: $$\langle \Phi | \hat A | \Psi\rangle = \int_{-\infty}^{\infty} dx~\Phi^*(x) ~ A[\Psi](x) = \int_{-\infty}^{\infty} dx~{A^\dagger[\Phi]}^*(x) ~ \Psi(x) $$ where $A[\bullet]$ is the linear functional which does this position-space transformation which performs the same operation as this operator $\hat A;$ for example for the $x$-component of momentum $\hat p_x$ the functional is $p[\bullet] = -i \hbar~\frac{\partial~\bullet}{\partial x}.$
For $T_a[\psi](x) = \psi(x - a)$ it's clear that we can $u$-substitute $u = x - a$ to find out that $$\hat T_a^\dagger = \hat T_{-a} = \hat T_a^{-1},$$ which proves that $\hat T_a$ is unitary ($U^\dagger = U^{-1}$) but not Hermitian ($H^\dagger = H$). Similarly the $x$-reflection $\hat R$ takes the form $R[\psi](x) = \psi(-x),$ so a $u$-substitution
gives us $$\langle \Phi|\hat A|\Psi\rangle = -\int_{\infty}^{-\infty} du~\Phi(-u)~\Psi(u)$$ which proves ultimately that $\hat R$ is Hermitian. Since we already know $\hat R$ is self-inverse, this means it's also unitary.
Is $\hat R~\hat T_a$ unitary?
Notice that both the transpose and the inverse operations reverse their constituents when they distribute over the multiplication; both $(A B)^{-1} = B^{-1} A^{-1}$ and $(A B)^\dagger = B^\dagger A^\dagger.$ This is important because it means that the product of two unitary transformations is always unitary; you just start off with $$(A~B)^\dagger = B^\dagger~A^\dagger = B^{-1}~A^\dagger = B^{-1}~A^{-1} = (A B)^{-1},$$ and there you have it.
Is $\hat R~\hat T_a$ Hermitian?
Now that we know it's unitary, probably the easiest way to do this is to check if it's self-inverse, which it is: $\hat R~\hat T_a~\hat R~\hat T_a$ maps $\psi(x)\mapsto \psi(x - a)\mapsto \psi(-x + a) \mapsto \psi(-x) \mapsto \psi(x).$ Since it's unitary and self-inverse, it's Hermitian.
