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Is it always possible that for two operators $\hat A$ and $\hat B$, which have a commutator of $[\hat A, \hat B] = i \hbar$, we can write the action of these two operators to a function $f(a)$, which depends on the eigenvalues $a$ of $\hat A$, in the following way

$$ \hat A f(a) = a f(a) \\ \hat B f(a) = -i \hbar \frac{\partial}{\partial a}f(a). $$

Or stated differently: Does the form of the coordinate representation of the position and the momentum operator hold also for other operators with the commutator of $i \hbar$ or is this representation unique for position and momentum operator?

I remember vaguely that there is a theorem which exactly makes a statement about this.

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Well, the position and momentum operators are defined as operators which satisfy the relation $[\hat{A},\hat{B}]=i\hbar$. We call them $\hat{X}$ and $\hat{P}$ (respectively) out of convention. So yes, any operators which satisfy that relation will have the same coordinate forms as the position and momentum operators, because if those operators satisfy that relation, they are position and momentum operators by definition. However, do note that you have not written down the most general form of the position and momentum operators; they are $$ \begin{align} \langle a| \hat{A} |f\rangle &= a\langle a|f\rangle \\ \langle a| \hat{B} |f\rangle &= -i\hbar\frac{\partial}{\partial a} + \langle a|f\rangle \end{align} $$ These satisfy the communication relation (try it yourself!). Any pair of operators that satisfy the relation will be of these forms in the coordinate base.

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  • $\begingroup$ +1 for "communication relation"! $\endgroup$ – yngabl Nov 23 '19 at 20:40

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