# Uncertainty principle when in position (or momentum) eigenstate [duplicate]

Suppose we want to calculate the expectation value $$\langle x|[\hat{x},\hat{p}]|x\rangle,$$ where $$|x\rangle$$ is an position eigenstate, so that $$\hat{x}|x\rangle=x|x\rangle$$ and $$\langle x|\hat{x} = x\langle x|.$$ If we 'forget' for a moment the canonical commutation relation, we get $$\langle x|\hat{x}\hat{p}|x\rangle-\langle x|\hat{p}\hat{x}|x\rangle = x\langle x|\hat{p}|x\rangle- x\langle x|\hat{p}|x\rangle = 0,$$ which, according to the generalized uncertainty principle $$\Delta x\Delta p \geq\left|\frac {\langle-i[\hat{x},\hat{p}]\rangle} {2}\right|,$$ means that when we are on a position (or momentum) eigenvector we can measure simultaneously (as accurately as we want) position and momentum (!?).

What's wrong with my reasoning?

• Position eigenvector is not a physical state
– hft
Commented Jun 27, 2023 at 17:02
• $\langle x'|[\hat x,\hat p]|x\rangle = i\hbar\delta(x-x')$. You are evaluating this at $x=x'$.
– hft
Commented Jun 27, 2023 at 17:04
• Expanding it out doesn't really help: $\langle x'|[\hat x, \hat p]|x\rangle = -i(x' - x)\frac{d}{dx'}\delta(x - x')$. The $\delta'(x-x')$ is going to infinity as the $x-x'$ is going to zero...
– hft
Commented Jun 27, 2023 at 17:09
• Possible duplicates: physics.stackexchange.com/q/14116/2451 and links therein. Commented Jul 14, 2023 at 8:20

Your teacher has presumably taught you the way out of all these pseudoparadoxes: Consider Gaussian states of vanishing width. Instead of doing that, effectively hijacking your question to something more interesting, I will remind you of the erroneous way you are multiplying 0 to infinities of different orders!

The short answer is that, technically (physics aside: there is little physics in $$|x\rangle$$ unnormalizable states!), $$\Delta x=0$$, indeed, but $$\Delta p=\infty$$, to satisfy Robertson's formal inequality. You are effectively pleading for a formal way to bypass your technical mistake ("you get"; no you don't!). Take $$\hbar=1$$ w.l.o.g.

For starters, remind yourself of the identity $$\epsilon \delta(\epsilon)=0, ~~~~\leadsto \\ \epsilon \partial_\epsilon \delta(\epsilon)= -\delta(\epsilon).$$

You are effectively considering matrix elements implicitly under the final step operation $$\lim_{\epsilon \to 0}$$. Specifically, evaluating the canonical commutator, $$\langle x+\epsilon|[\hat x, \hat p]|x\rangle = i\delta (\epsilon),$$ divergent.

The same follows from your formal pathway, $$\langle x+\epsilon|[\hat x, \hat p]|x\rangle =(x+\epsilon)\langle x+\epsilon| \hat p |x\rangle - x \langle x+\epsilon| \hat p |x\rangle=\epsilon \langle x+\epsilon| \hat p |x\rangle\\ =-i\epsilon \int \!\! dy~~ \langle x+\epsilon|y\rangle \partial _y \langle y|x\rangle= -i\epsilon \partial_\epsilon \delta (\epsilon)= i\delta(\epsilon),$$ by dint of $$\hat p=-i \int \!\! dy~~ |y\rangle \partial _y \langle y|$$.

So, taking the limit, you indeed find a divergent $$\Delta x \Delta p \geq \delta (0)/2~ .$$

(You can take a further derivative in the leading identity, $$\partial_\epsilon^2\delta (\epsilon) = {2\over \epsilon^2} \delta (\epsilon)$$.)