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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 $\psi(x)$?

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up vote 5 down vote accepted

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 $x$).

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I appreciate your answer, and thanks for including the link for translational operators... that is very helpful! – curiousGeorge119 Oct 31 '13 at 16:59

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 $$ f(x, x', a) = \delta (x' - (x - a)) + \delta (x' - (x + a)). $$

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Thanks for your post, Andrew McAdams. Is the kernel always related to the null space, or does it have different meanings depending on the application? – curiousGeorge119 Nov 1 '13 at 13:09
@curiousGeorge119 . The first statement is right. – Andrew McAddams Nov 2 '13 at 0:01

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.

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