# Quantum Mechanics, Uncertainty Principle— help understanding notes

There is a section of my notes which I do not understand, hopefully someone here will be able to explain this to me. The notes read (after introducing the uncertainty operator):

If the state $\chi_A$ is an eigenstate of $\hat O_A$ then the uncertainty is zero and we measure it with probability 1. However, if $\hat O_B$ is another observable which does not commute with $\hat O_A$, then the uncertainty in any simultaneous measurement of the two observables will be infinite.

I understand the first sentence, but I can't see how to justify/prove the second one. Can someone tell me how the second sentence is justified, please?

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look at here:Uncertainty Principle; Robertson –  Ali May 28 '13 at 12:49

Assuming we have already proved the uncertainty principle(which can be found here), we know:

$$\sigma_A \sigma_B \geq \sqrt{\Big(\frac{1}{2}\langle\{\hat{O}_A,\hat{O}_B\}\rangle - \langle \hat{O}_A \rangle\langle \hat{O}_B\rangle\Big)^{2}+ \Big(\frac{1}{2i}\langle[\hat{O}_A,\hat{O}_B]\rangle\Big)^{2}}=C$$ Where C is a constant.

Since the state we are looking at is an eigenstate of $\hat{O}_A$, we know $\sigma_A=0$; also since $\hat{O}_A$ and $\hat{O}_B$ do not commute, the right hand side($C$) is greater than zero. Ergo:

$$\sigma_B > \frac{C}{\sigma_A}\rightarrow \infty$$

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so if the uncertainty with respect to $\hat O_A$ is 0, then the uncertainty with respect to $\hat O_B$ is undefined? –  user27182 May 28 '13 at 13:15
@user27182 Yes(realistically that is impossible). However, in real cases the uncertainty with respect to $\hat{O}_A$ can only get close to zero, which makes the uncertainty of the other operator increase dramatically. –  Ali May 28 '13 at 13:19
$\psi=\ Ae^{i\frac{(px-E t)}{h}}$ is an eigenstate of a free particle. The momentum $\ p$ is well defined and its in the eignestate of momentum operator (as $O_p\psi=\ -ih\frac{d\psi}{dt}=p\psi$ ) This means the probability of finding the particle with momentum $\ p$ is $\ 1$.
Operators $\ x$ and $\ p$ don't commute. Now look at $\ x$ .The particle has equal probability of being found anywhere on the x axis since $\psi^*\psi=constant$, all over the axis. This means you have no idea where the particle is. Uncertainty is as high as it could be.