Help with understanding an Operator definition The operator $\hat{F}$ is defined by $F\psi(x)=\psi(x+a)+\psi(x-a)$
Does this mean $\hat{F}=(x+a)+(x-a)$ and that $\hat{F}$ is operating on $\psi(x)$?
 A: No, you cannot say that $\hat{F}=(x+a)+(x-a)$ (which would simplify to $2x$), you specified its definition in the original problem:
$$
\hat{F}\psi(x)=\psi(x+a)+\psi(x-a)
$$
 With words, you could say that $\hat{F}$ is a (bi-directional?) translational operator. But symbolically, there is no other way to state what $\hat{F}$ means, except through how it operates on $\psi(x)$ (or any general function of $x$).
A: My answer is only the addition for Kyle Kanos answer. You may use the integral representation of the operator terms of the kernel of the operator:
$$
\hat {F}\Psi (x) = \int \limits_{-\infty}^{\infty} f(x, x', a)\Psi(x')dx' = \psi (x + a) + \psi (x - a),
$$
where $f(x, x', a)$ is called kernel of operator.
So
$$
f(x, x', a) = \delta (x' - (x - a)) + \delta (x' - (x + a)).
$$
A: Just to add to the other answers here, you can solve this equation for $\hat F$ by noting that for a smooth function $\psi(x)$
\begin{equation}\cosh(aD)\psi(x)=\psi(x+a)+\psi(x-a)=\hat F\psi(x)\end{equation}
Where $D\equiv\frac{d}{dx}$.
You can prove this by writing $\cosh(aD)=\frac{1}{2}\left( e^{aD}+e^{-aD}\right)$ and noting what an exponential of a derivative operator does to a function. 
