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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

2 Answers 2

up vote 3 down vote accepted

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
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@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

You can justify thus:

$\psi=\ Ae^{i\frac{(px-E t)}{h}}$ 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.

Therefore uncertainty in position tends to infinity and in momentum it is zero for the given case.

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