mooeypoo

Seeing as we generally delete overly bad quality of posts (like, say trolling), that is an inaccurate statement.

I'm very confused, I have no idea what you are talking about... :\ am I even at the right track, or should I try what you propose... I.. meh. I'm lost.

I am not sure if this should be in Math section or in Computer section, but since it's about math, I'm hoping one of the experts here could help me out. I have an equation related to my research that I want to model in Mathematica. The point at the moment is to try and graph the angle on a circle of a group of particles (the 'n' index counts particles) per time (t index). I managed to produce a lineplot per time if I create individual functions for the particles (defining Phi1[t_], Phi2[t_], etc) but I'm looking for a way to model both indices at once so I can see the relationship between all particles. I looked things up in the Mathematica manual, but, alas, I can't find what to do. How do I represent a series with two indices so I can plot them later? anyone knows? Thanks! ~mooey

That's how the question was phrased. Honestly, I'm very confused, but now I feel like an idiot. Of course I know the exponential of a matrix, we learned it in math two semesters ago. Bah. I think I panicked a bit by the fact I'm looking at something new (commutators) and couldn't see how to start. Exponential of a matrix: [math]A^{b}= \sum \frac{1}{n!} b^{n}[/math] (gah! it's not rendering again.. here's the latex in plaintext: A^{b}= \sum \frac{1}{n!} b^{n} Anyways, I'll work on it that way and post my result. Thanks for the help -- and the patience! ~moo

Yikes! I see it. Okay, I think I have an idea of how to solve it now. I'll try and post my solution. Thanks

Last question for this series. I started answering this one and got stuck. Again, this subject is new to me, so forgive me if I miss obvious stuff.. I'm getting used to the format and the method. I don't want to copy an answer, of course. I would appreciate an explanation and/or a way to continue so I can go on to other similar problems and also understand how to solve these type of questions in the future. Thanks! Problem 3: The Hamiltonian for three-dimensional system is of the form: [math]H=\frac{p_{i}p_{i}}{2m}+V[/math] where V depends only on r. Obtain the commutator [math] [x_{k}p_{k},H][/math] -- Here's what I have so far: [math] [x_{k}p_{k},H] = [x_{k}p_{k},\frac{p_{i}p_{i}}{2m}+V] =[/math] [math] [x_{k}p_{k},\frac{p_{i}p_{i}}{2m}] + [x_{k}p_{k}, V] =[/math] [math]\left( x_{k}p_{k}\frac{p_{i}p_{i}}{2m} - \frac{p_{i}p_{i}}{2m}x_{k}p_{k} \right) + \left( x_{k}p_{k} V - V x_{k}p_{k} \right) = [/math] Now, in the left part, I recognize that I might be able to rebuild the commutator from the p's... I tried: [math]\frac{1}{2m} \left( x_{k} \left( p_{k}p_{i}p_{i} \right) - p_{i}p_{i}x_{k}p_{k} \right) [/math] I can switch between x_k and p_k and get the remainder -ihbar: [math]\frac{1}{2m} \left( x_{k} \left( p_{k}p_{i}p_{i} \right) + i\hbar\left( p_{i}p_{i}p_{k} \right)x_{k} \right) [/math] These two can be switched for the commutator, but I have an extra factor of ihbar in there... meh... where do I go from here, and am I even in the right direction? Also, the next question is exactly the same, only V® is defined so that r carries all three dimensions of x. So I think that this means this current question treats r without much relation to x (otherwise why the distinction in the next question?) and hence, V® has no x, and is commutable with both p and x and can be neglected. Am I right? Help. Thanks! ~moo

Hey again, I'm dealing with QM and the subject is entirely new to me. I'm having troubles figuring out how to start the questions. Here's another in my frustrated series: Problem 2: If U is a unitary matrix, show that the absolute value of every matrix element is less than or equal to 1, ie: [math]|U_{ij}|\leq 1[/math] So in this case, here's how I start my solution (couldn't find the 'cross' symbol, so I'll use * instead): U is a unitary matrix, so [math]U^{*}U=UU^{*}=I_n[/math] and [math]U^{*}=U^{-1}[/math] And so, I started: [math]UX = \lambda X[/math] [math]U^{*}UX=U^{*}\lambda X=\lambda U^{*} X[/math] [math]I_{N}X=\lambda U^{*} X[/math] And I am thinking that since I_{N} is the identity matrix, if I can find a way to show that lambda is smaller or equal to that, then I finish the proof. But... I ... don't... see how... I'm stuck! help! ~moo

Hey guys, I got a homework assignment I'm having a lot of trouble with. This is post #1, I'll post a few more regarding the other problems in the homework. I'm not looking for an answer, I'm looking for assistance in how to proceed. This is new to me and very confusing, and my confidence in QM is very low after last semester's course (we all failed, if you remember... had a makeup test that also didn't go well, but I ended up getting a C+, which I usually consider as failure, but in this case, at least it's *passing*... err) Anyways, problem 1: If A,B are two linear operators on a Hilbert space, show that a^(A) B e^(-A) = B + [A,B] + 1/(2!) [A,[A,B]] + ... + 1/(n!) [A, [A, [A, ... [A,B] ... ]]] + ... where the n-th term involves n successive commutators with A's. (for some reason, this doesn't render in latex, there seems to be a problem with the factorial... if you guys want, I am leaving the tex syntax as-is, in case it helps: e^{A}B e^{-A} = B + \left[ A, B \right] +1/2 \left[ A, \left[ A, B \right] \right] + \ldots + 1/n \left[ A, \left[ A, \ldots \left[ A, B \right] \ldots \right] \right] \right] + \ldots Argh. I don't even know where to START.. I looked up online rules of commutators (and we dealt with commutators last semester, so I know the rules), and I know that the above is defined as an identity, but I can't see how to start with the proof for it. The only thing that popped to my mind is, perhaps, using the identity: [math][A^{n},B] = nA^{n-1}[A,B][/math] But I'm not sure how that helps me any here... Anyone? Help..

Nice! I joined Been a while since I had it running, though. Good to start again.

If my previous reply (written nicely and non-staffy without the big bad red font) wasn't clear, here's the clarification: ! Moderator Note This forum includes netiquette and etiquette standards in its rules. Not only is there no need for you to use such obtuse tone and language, these type of posts will not be accepted here. You know this already. My bad for thinking I could hint this without wearing my mod hat. Next time, I'll wear my mod gloves, and start taking action. We have a debate, people. Please, be nice, and express yourselves like the intelligent people you are. I know you can. ~moo

You're a usually very eloquent person, I would assume you could find another way of putting the above sentence that wouldn't sound so obtuse, let alone when you're finally agreeing. ~moo

I'm going to plug myself shamelessly, but it DOES answer the request, so I forgive myself. My blog, "SmarterThanThat" has a few science experiments (including videos) for you to do at home. Hope you enjoy! http://www.smarterthanthat.com/ (go over the articles, or click on "Experiments" at the top for the list of video experiments). Enjoy! ~moo

I assumed this is about Internet Piracy, and hence it's only about internet people. Mr Skeptic's right, if that wasn't the meaning, you should be more accurate.

This sounds like an interesting survey. Will you update us when there will be results? (I'm aware this takes time )

! Moderator Note What seems to be a trend with you, though, is an affection to flaming the debate. This really isn't necessary, Emilio. People are taking the time to answer and debate with you. Whether you agree with them or not, being obtuse and 'in-your-face' will not help your arguments, and is against forum rules and etiquette. Welcome to the forum.

HTML isn't really programming, it's a markup language. The difference is that HTML (just like "BBCode" that's used in this forum) defines how things look, and not really anything code-related. Programming languages are more complicated, and use things like loops (Do.. while..) and conditions (If/else, etc). If you want to learn HTML, you can start with w3schools, that are usually quite good for beginners: http://www.w3schools.com/html/default.asp Look up "Learning HTML" in google, and you'll find lots. If you want to start learning programming, the first step would be to learn the general principles (like what ARE conditions and loops, etc) and the second would be to identify what language you would like to learn, based mostly on what you would like to do with it later. Good luck, ~moo

I just upgraded my G1 (Android, first edition) phone to the new myTouch 3G Slide (Android again, of course). I'm not the least bit sorry, I *love* Android, and though it still isn't COMPLETELY on the same level as iPhone, it's gettting there, and fast. That said, in the spirit of being a science-oriented forum, I shall supply the reference for ydoaPs claim, and yet point out that the survey (ahem) was about number of sexual partners, and not sexual activity per say. Hence, one can simply deduce that Android users are monogamist, while iPhone users are.. well.. wh*res. Severian breaks the mold on that, by his earlier admission and by the fact he's married. ydoaPs, however, is supposed to have 10 iPhones, 13 iPads and 235 iPods to avoid breaking the statistics. Source: http://www.talkandroid.com/10068-survey-android-users-have-less-sex-than-iphone-users/ (there are more) ~moo

Cap'n brought this to my attention. I am working on an iPad specific icon at least. As for the rest, I'm not sure.. probably javascript problems (CSS problems shouldn't affect the keyboard popping up). Cap'n, if you need help figuring it out, let me know, but I don't have an iPad, so it might be hard testing. Of course, if SFN wants to donate an iPad for testing purposes, I won't say no B)

I didn't know those two mix.

I'm quite sorry, rigney, but with due respect, if you're done with the debate, you're free to leave it. You might have started it, but you're not the only one participating. There are others here who might still care to continue it. Call it quits or not, calling me 'girl' and ydoaPs a rear-end is not quite doing justice to your claim that you're not condescending (and it's not much for politeness either). Thank you for your contribution. Now let others continue if you don't want to.

Are you asking whether this happens or are you making the claim?

*YOU* make a claim here, and YOU need to bring us evidence. A YouTube video is *NOT* evidence. I'm not going to start researching the claims the video makes. You need to bring PROPER evidence - something that is supported, like official numbers from official offices in Europe and the USA. Where's the proof for the growth numbers?

Are you being cryptic on purpose, or do you just enjoy telling everyone they don't understand you because they're too young? Can we keep on topic, on track, and avoid being condescending to one another? ~moo

We are a science forum. If you want to be treated seriously, specifically in a debate of this sort that is entirely non-PC and discussing a racial issue, you will need to come up with proper evidence. Random YouTube video and 'quick research' on wiki - without even giving us the research, no less - is far from being good enough, specially for such a claim. Bring proper evidence or there's nothing to discuss. ~moo

May I ask where you're getting these numbers from?