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$\newcommand{\bvec}[1]{\boldsymbol{\mathrm{#1}}}$ The following question is very much about a mathematical detail in a physics derivation, but I would like to hear what the mathematicians have to say, and if there's any recommended resources for me to look into.

In a proof of the Bloch theorem in Ashcroft & Mermin's Solid State Physics, it uses the translational operator $T_\bvec{R}$ with respect to a translation vector $\bvec{R}$ such that for any function $f(\bvec{r})$ $$ T_\bvec{R}f(\bvec{r}) = f(\bvec{r}+\bvec{R}). $$

I am wondering if such an operator always exists for any function $f(\bvec{r}): \mathbb{R}^n\to\mathbb{R}$ we can think of, and if they can be explicitly constructed.

A simple enough example I can think of right now is $$ T_\bvec{R} = e^{i\bvec{k}\cdot\bvec{R}} $$ for the function $$ f(\bvec{r}) = e^{i\bvec{k}\cdot\bvec{r}}. $$

Jono94
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    As your example gets at, Fourier modes diagonalize the translation operator. So you can translate a function by taking its Fourier transform, pointwise multiplying by $e^{ik\cdot R}$, then taking the inverse Fourier transform of the resultant function. This can sometimes be a useful way to think about translation, particularly when considering linear combinations of different translations (since Fourier modes diagonalize them all). This post may also be useful – Nick Alger Dec 28 '23 at 09:44
  • @NickAlger this is insightful, maybe you could expand it into an answer? A question: what do you mean by "Fourier modes diagonalise the translation operator"? – Jono94 Dec 28 '23 at 13:09

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Note that $T_{\bf{R}}$ is an operator which is applied to functions. In this particular case the codomain also consists of functions. And the definition is as you have written: If $T_{\bf R}f=g$ then $g({\bf r})=f(\bf r+\bf R)$ for all $\bf r$ by definition.

Jens Schwaiger
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