# Limit of the position and momentum commutator [closed]

The commutator of position and momentum operator, $\hat{p}$ and $\hat{x}$, respectively is derived as $[\hat{x},\hat{p}]=i\hbar$. Let $\lim_{x\rightarrow x_{o}} [\hat{x},\hat{p}]=\lim_{x\rightarrow x_{o}}(\hat{x}(-i\hbar \partial/\partial x) -(-i\hbar\partial/\partial x)\hat{x})$. Then we have $$\lim_{x\rightarrow x_{o}}((\hat{x})(-i\hbar\partial/\partial x))-\lim_{x\rightarrow x_{o}}((-i\hbar\partial/\partial x)\hat{x})$$ evaluating the limits we obtain the following:$$(x_{o}p_{o})I-(p_{o}x_{o}I)$$ since $\{x_{o},p_{o}\}\subset \mathbb{R}$ we have that the commutator vanishes for every $x_{o}\in \mathbb{R}$. Can someone explain why the limit of a commutator at any point $x_{o}\in\mathbb{R}$ vanishes but the commutator does not.

## closed as unclear what you're asking by ACuriousMind♦, user36790, Gert, Kyle Kanos, Norbert SchuchDec 30 '15 at 23:27

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• $\hat{x}$ does not refer to a real number. Rather, it is a multiplication operation $\hat{x}: \psi(x) \to x\psi(x)$. Hence $\lim_{x\to x_{0}}$ doesn't make sense. – higgsss Dec 29 '15 at 6:18
• but $\hat{x}:\mathbb{R}\times H\rightarrow H$ where $H$ is the Hilbert state space. Hence, $\hat{x}$ depends on the set of reals. – Ara Vartomian Dec 29 '15 at 6:24
• $\hat{x}: \psi(x) \to x\psi(x)$ simultaneously acts on all $x \in \mathbb{R}$ rather than on a single value $x$. That is, it shouldn't be thought as a map of $R \times H$ onto $H$. – higgsss Dec 29 '15 at 6:36

The simple point is that you cannot take the limit of the $\hat x$ operator, because it is an operator. That's about it.
The position operator $\hat x$ is not a map $\mathbb{R} \times \mathbb{H} \to \mathbb{H}$, but rather $\mathbb{H} \to \mathbb{H}$, just like any other hermitian observable in quantum mechanics.