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Leo32

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About Leo32

  • Birthday November 9

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  • Location
    Near Ghent, Belgium
  • Interests
    Reading
  • College Major/Degree
    Electronics Engineer / Master Japanology
  • Favorite Area of Science
    Quantum Mechanics (for the time being)
  • Occupation
    Project Manager

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  1. Hi y'all, Working myself at walking pace through Shankar's "Principles of Quantum Mechanics", there is some intreguing jump towards Hamiltons canonical equations that I'm missing. It could be something obvious, since Wikipedia doesn't provide any additional info either. So far, the canonical momentum in Lagrangian formalism was defined as p= dL/dq(dotted) Terribly sorry, but I can't get the LaTex working... and I know that should be partial derivates, and also imagine the p and q's being indexed. A certain deduction, which I still understand results in dH/dq = -dL/dq The strange thing, which I don't understand at all, is that dL/dq is equalled to p(dotted). No idea where this pops out of, or what the reasoning is behind it. I see it provides a nice symmetry in Hamilton's canonical equations, but other than that, I see no reason why. Given that p, q and q(dotted) are defined yet, it's not just a question of defining a new thing called p(dotted) I suppose, so where oh where is the reason behind this ? Cheers ! Leo
  2. Vending Menace, Atheist, Thank you for the in depth replies. Atheist, reason why I didn't include the complex conjugation was that the text I'm trying to read at this point didn't introduce it yet... I promise not to ask any questions anymore trying to remember the formula's I read before going to sleep the night before LOL. The minus should have been a multiplication... Boy that Dirack Delta function is something strange and new! Just to make sure that I kind of get it, is following expression correct ? [math]\int_{-\infty}^{+\infty}dx\delta(x) = 1[/math] Greetz, Leo
  3. Back at home, I see that the second formula indicates that the eigenfunctinos form a complete set... Anybody any idea ?
  4. Sigh, started a lot of languages only speak 5 more or less in an acceptable way... Ever started, no speak: latin, greek (classical), russian, classical Japanese, classical chinese, chines, korean Speak more or less: dutch, english, french, german (to be prepared ;-), Japanese And yes, I majored in Japanese, and I loooooved the engrish site! Most funny thing I saw was a slightly naive and looking young Japanese girl with a T-Shirt saying "Take me I'm yours" over the chest...
  5. First, let's check whether I understand the formula you mean by: Uin-Uuit((1-(1+jwrc))jw2rc+1=0 Using the fantastic tool implemented here (check for details how to use it in the Quick Latex tutorial thread), you can compose this in real looking mathematics. I'll just asume I understood correctly and start from: [math]U_{in}-U_{out} \frac{j\omega2RC+1}{1-(1+j\omega RC)}=0[/math] If you want to express H(w) = Uin/Uout, that would be [math] \frac{j\omega2RC+1}{1-(1+j\omega RC)}[/math] [math]= \frac{j\omega2RC+1}{-j\omega RC}[/math] [math]= \frac{j\omega2RC}{-j\omega RC}+\frac{1}{-j\omega RC}[/math] [math]= -2 + \frac {j}{\omega RC}[/math] Of course, the end result will only be as perfect as the formula which I started with... Give me a buzz if I got it all wrong in the first place. Next to the general rule for multiplication given by Mandrakeroot, one for dividing a complex number: [math]\frac{1}{a+bj}=\frac{a}{a^2-b^2}-\frac{b}{a^2-b^2}j[/math] just multiply top & bottom with [math]a-bj[/math] Greetz, Leo
  6. First, let's check whether I understand the formula you mean by: Uin-Uuit((1-(1+jwrc))jw2rc+1=0 Using the fantastic tool implemented here (check for details how to use it in the Quick Latex tutorial thread), you can compose this in real looking mathematics. [math]U_{in}-U_{out} \frac{j\omega2RC+1}{1-(1+j\omega RC)}=0[/math] If you want to express H(w) Uin/Uout, that would be [math] \frac{j\omega2RC+1}{1-(1+j\omega RC)}[/math] [math]= \frac{j\omega2RC+1}{-j\omega RC}[/math] [math]= \frac{j\omega2RC}{-j\omega RC}+\frac{1}{-j\omega RC}[/math] [math]= -2 + \frac {j}{\omega RC}[/math] Of course, the end result will only be as perfect as the formula which I started with...
  7. While trying to teach myself a little quantum mechanics in the evenings, I came across the following sequence of formula's and text. Although the subject is quantum mechanics, the question I have is a mathematical one. [math]\int_{-\infty}^{+\infty}dx\psi_m(x)\psi_n(x)=\delta_{mn}[/math] these functions form a complete set [math]\sum_{n=1}^{\infty}\psi_n(x)-\psi_n(x')=\delta(x-x')[/math] The question I have is: are both statements required for the set of [math]\psi_n(x)[/math] to be complete ? Or which of both is sufficient ? Greetz, Leo
  8. Suicide attacks have indeed everything to do with desperation. Look at the Israeli-Palestinian conflict. Most of the Palestinian people comitting suicide attacks are wel trained young people. Imagine yourself being completely at the mercy of the Israëli people (I'm intentionally leaving any reference to religion out of this argument). You want to use your hard-earned college degree ? Well, let's see day by day if you are allowed through the border, if you are harrassed at random for a few hours, or just have to stay home and try again next day. You're sick and need urgent medical help ? Well, you'll most probably depend on Israëli made equipment and medicine. The list goes on and on and on... So there you are, young, well educated and nothing which you could call a normal life in front of you. Then one day, some member of your family is shot, and you decide you should do somethig about it. If you decide on armored resistance, there is not a lot else than suicide bombing. Once you've decided this, of course you become more religious active, that is a normal reaction. Look at what most people go through when they hear they're terminally ill. What I'm trying to put foreward here is that first, there is a decision out of desperation to become a suicide bomber, and only after that religion comes in. The organisations supporting the suicide bombers of course tell another story. Especially in Islamic countries, Islam provides the right set of arguments. Note that the Islam has always been closely bound with life in general, be it political, social, ... The prophet himself remained politically very active, even after receiving the Words, in Medina and other places to defend his religion... I'm sure most of you have wondered at times why the religious reasoning is so "irrational" at times, and even contradics at times earlier arguments based on the same principle. Desperation is the word. Another point is that there have been a number of conflicts using suicide bombers where Islam played no role at all: the Kurds, Tamil Tigers, ... Hope this all makes a bit sense... Leo
  9. Happy to see that my cat story triggers responses which point out the essential questions to understand the entanglement issue. Vending Menace: Indeed, the question that needs to be asked is which one of both possibilities is the correct interpretation: 1: does the collapse of state of cat 1 initiate an information transport faster than light, which causes the state of cat 2 to collapse OR 2: is the info on the end result of the colapse of cat 2 already contained within it's quantum state, and as such taken "hand carried". Jakiri: Indeed, you can only check with another information transfer, but this one does not require to be faster than light, it can be a simple phone call. Greetz, Leo
  10. Quantum entanglement is quite a challenging subject, and I would like to see if we can make it any easier to think about when using the Schrödinger cat analogy. For this, we can make abstraction of the atom which decays and triggers a poison to kill the cat. Let's just say that the cat itself is in a quantum state of being dead/alive as long as we don't look inside the box. Once we do, the quantum state collapses and we see the cat as being either dead or alive. For starters, I'll refrain from inserting an analogy to tackle the Heisenberg uncertainty. I'll leave that for another thread. Now, suppose we take 2 boxes with cats and manipulate them in such a way that they become entangled. After this, one of the boxes is taken to the other side of the world. A first thing we can do is open one of the boxes and look. This causes the quantum state of this cat to collapse. Entanglement would mean that we can now 100% sure what state the other cat is in, even without taking a peek. It might look as if the information which obtained by the state to collapse, the cat is dead, or the cat is alive, can be taken over large distances at a speed greater than light. Remember however that the cat, and thus it's quantum state, had to be taken "hand carried" to the other side of the world, significantly slower than speed of light. A second step might be to influence the quantum state of the cat on one side such that the chances of it being found alive are much higher than they are in the original experiment, after the boxes have been separated. Some of the texts on entanglement make me think, and I'm not sure about this, that this increased chance of the cat being found alive is "transported" at a speed higher than that of light. In other wordt, the quantum state of the remote cat would be changed immediately together with the manipulated quantum state of the local cat. Checking this might be done by doing the experiment over and over again, while collapsing the quantum state each time to get an idea of probability shifts, but that's a practicality outside the scope of this thread. The big question now is of course, did I remain on the path of quantum mechanics, or, at what point did I leave it ? Greetz, Leo
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