I'm reading about the uncertainty principle formulated in the language of $ C^*$ algebras. It states that for any state $\omega: A \rightarrow \mathbb{C}$ on a $C^*$ algebra and for any self-adjoint $a, b \in A$, we have $$ \sigma_{\omega}(a)^{2} \sigma_{\omega}(b)^{2} \geq \frac{1}{4} \omega(i[a, b])^{2} $$ So, what about the particle on a ring (one can read about it on wiki)? LHS vanishes here. In contrast to this formulation, in usual QM we have condition for domains of operators. Should this principle be reformulated because of this case?
We know that for a position variable $x$ and momentum $p$, the uncertainties of the two quantities are bounded by  $$\Delta x \Delta p \gtrsim \hbar$$  Now, this is usually first explained with $x$ being a simple linearly measured position and $p$ being linear momentum.  But it should apply to any good coordinate and its conjugate momentum.  It should, for instance, apply to angle $\phi$ about the $z$ axis, and angular momentum $L_z$:  $$\Delta \phi \Delta L_z \gtrsim \hbar$$  The thing is, $\Delta \phi$ can never be greater than $2\pi$.  I mean, you have to have some value of $\phi$ and $\phi$ only runs from 0 to $2\pi$.  Therefore  $$\Delta L_z \gtrsim \hbar/\Delta \phi \geq \hbar/2\pi$$  But, uh-oh!  This means it is impossible for $\Delta L_z$ to be zero, and we should never be able to have angular momentum states with definite $L_z$ values.  Of course, it doesn't mean that.  But I have never figured out how this is not in contradiction with the Schroedinger eqn. calculations that give us states with definite values of $L_z$.  Can anyone help me out?  One answer I anticipate is that $\phi$ is sort of "abstract" in that if you chose your origin at some other point you will get completely different values of $\phi$ and $L_z$, and ipso facto, usual considerations don't apply.  I don't think this will work, though.  Consider a "quantum bead" sliding around on a rigid circular ring and you get the exact same problem with no ambiguity in $\phi$ or $L_z$.  (Well, there will be some limited ambiguity in $\phi$, but still, there won't be in $L_z$.)
How to simulate data that satisfy specific constraints such as having specific mean and standard deviation? This question is motivated by . But I imagine that it would also be useful in teaching contexts where you want to create a dataset that exactly mirrors an existing published dataset.  I know how to generate random data from a given distribution. So for example, if I read about the results of a study that had:   a mean of 102, a standard deviation of 5.2 , and  a sample size of 72.   I could generate similar data using rnorm in R. For example,   set.seed(1234) x &lt;- rnorm(n=72, mean=102, sd=5.2)   Of course the mean and SD would not be exactly equal to 102 and 5.2 respectively:  round(c(n=length(x), mean=mean(x), sd=sd(x)), 2) ##     n   mean     sd  ## 72.00 100.58   5.25    In general I'm interested in how to simulate data that satisfies a set of constraints. In the above case, the constaints are sample size, mean, and standard deviation. In other cases, there might be additional constraints. For example,    a minimum and a maximum in either the data or the underlying variable might be known. the variable might be known to take on only integer values or only non-negative values. the data might include multiple variables with known inter-correlations.   Questions   In general, how can I simulate data that exactly satisfies a set of constraints? Are there articles written about this? Are there any programs in R that do this? For the sake of example, how could and should I simulate a variable so that it has a specific mean and sd? Deleting multiple elements from a list Order of nested random effects in lme4 I just fit a model in lme4, and I'm wondering what the heck I fit...  I have individuals id, and each is measured pass/fail on items that can be described using two factors, f1 and f2. My theory says that id and f2 can interact. So I want to compare these models:  # f2 doesn't matter m1 &lt;- lmer(pass ~ f1 + (1|id), family="binomial")  # f2 is a fixed effect, doesn't interact with individuals m2 &lt;- lmer(pass ~ f1 + f2 + (1|id), family="binomial")  # f2 is a nested random effect, interacting with individuals m3 &lt;- lmer(pass ~ f1 + (1|f2/id), family="binomial")   First, am I right that there's a random effect of each level of f2, and each individual's level of f2 is partially pooled, depending on the number of observations for that individual?  Second, does it make sense to fit f2 as a fixed effect with random individuals nested within each level of f2? How does one write that?  Third, what's this, with f2 and id reversed in the formula? I typed it by accident, but now I want to understand it.  m3weird &lt;- lmer(pass ~ f1 + (1|id/f2), family="binomial") Is my expansion of the state $| x \rangle$ correct? Why shouldn't I mix tabs and spaces? I am planning a trip to Italy this coming January. My plan is to land by plane in Roma and then go to France by train/bus stopping in some Northern Italian cities on the way, enjoying the architecture, the atmosphere, the art and and the food.   I want to do it at my own pace, without heavy scheduling. Ideally, I would just decide which cities to visit, and take a train whenever I feel it. I often do this here in South Korea where I am living, taking advantage of the "standing seats" as it is called here, that is travelling (for a cheap price and no reservation) standing on the romm between two cars. From my experience, it is also doable in Taiwan (at least in some trains).  In France, regional trains usually does not need any reservation, so you can buy a ticket and take the train whenever you want.  Can I expect a similar system in Italy, that is a way to make short travels between close cities without reservation?   Example of "short travel": Pisa - Firenze. Was Harry Potter a direct descendant of Godric Gryffindor? Will I lose previous purchases when I change country in iTunes I lived in country A for a few years, and made a number of purchases on the iTunes store for that country. I have now moved to country B, and want to start using the iTunes store from country B. If I change the country in my account settings, will I lose all my previous purchases, or will I still be able to re-download my purchases from country A from iCloud? How can I find the version of Ubuntu that is installed? Retrieve active locale in Grails application Concurrent updates to GDAL (to 2.1.3) and QGIS (to 2.18.5) on Linux Mint 18.1 from the Ubuntugis repos breaks QGIS, with a noted dependency for the virtual package gdal-abi-2-1-2 being uninstallable.    The stable Ubuntugis ppa currently supplies GDAL 2.1.0 (gdal-abi-2-1-0) while the Ubuntugis-unstable ppa has been updated to GDAL 2.1.3 (gdal-abi-2-1-3).  Has anyone found a way around this?    Update:  QGIS bug report for those interested is here:     UPDATE 2  The ticket has been closed as fixed. Everything should be back to normal now. Check tag classes with watir? How does a wizard or bard learn new spells? Find the highest order bit in C Do fixtures trigger model callbacks? Appending the same string to a list of strings in Python Here's what is happening; everytime I go to Stack Overflow (stackoverflow.com, NOT meta) with Firefox, it gives me the "Welcome back" tab that pops up from the top. Whenever I click in the "click here to refresh the page" link, all it does is refresh the page without logging me back in.  Even if I try to log in using the "log in" button at the top, all it does is pop up the tab that says "Welcome name here, you're being redirected." The problem is I'm not being logged in, not to mention that it keeps refreshing the page over and over again without giving me a choice to log in with Facebook.  For some reason, I was capable of logging in to Meta Stack Overflow using Firefox. I am still unable to log in to Stack Overflow. Any ideas? I am trying to solve the following question which is Ex3 from Arthur Engel, Problem Solving strategies. Here is the question: $\sum_{k=1}^n k^3 {n \choose k}$ and asks to find the sum. I am sincerely having a hard time trying to figure out how to solve it and get $n^2 (n+3)(2)^{n-3}$. I would really appreciate if someone could show the step-by-step process. Thank you! How should an educational institute woo good faculty? For an odd prime $p$, prove that any primitive root $r$ of $p^n$ is also a primitive root of  $p$   So I have assumed $r$ have order $k$ modulo $p$ , So $k|p-1$.Then if I am able to show that $p-1|k$ then I am done .But I haven't been able to show that.Can anybody help me this method?Any other type of prove is also welcomed.