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| − | There are two things which I'm not sure about: whether in lemma 2 we can use Lebesgue (we need a discrete sequence of dominated functions, whereas h is a continuous variable; can we change h to 1/n and let \( n \rightarrow \infty \) instead of \( h \rightarrow 0 \) without any problem?) and how lemma 2 exactly justifies the passage with \( \color{red}{*} \)
| + | I tried transforming to polar coordinates in an attempt to make the solution shorter but I don't think it works. I think you can abbreviate in the exam at some places and also its all just derivatives so not very difficult, you just have to remember to invert the metric :) |
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| − | Some derivative signs should be partial derivatives, but I guess it's not such a problem...
| + | [[User:Carl|Carl]] ([[User talk:Carl|talk]]) 23:41, 11 June 2015 (CEST) |
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| − | Best, Nick
| + | == Suggestion for less writing work == |
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| − | So, should be okay now... I think the induction step at \( \color{red}{*} \) is clear enough.
| + | Maybe it is possible to not explicitly show that C_2 is conserved, by saying that for u->v and v->u C_1->C_2, so C_2 must be conserved if C_1 is conserved. |
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| − | Best, A.
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| − | Yes, we can. If the limit of f(h) exists for h -> 0, then for every discrete sequence hn f(hn) will have the same limit for n -> infinity. So I guess you can just substitute them and use Lebesgue at will.
| + | == b) Poisson bracket? == |
| − | And \( \color{red}{*} \) isn't even an induction step... that's just applying the lemma we just proved. However, I believe in Lemma 2 the 1/sqrt(2*pi) on the right hand side of the equality is superfluous. It's not being used in the proof, as far as I can see, and it comes back to bite you in \( \color{red}{*} \).
| + | Instead of calculating d/dt(C) I think it would be nice to use the fact that d/dt(C) = {C,h}. Of course it's basically the same but the hamiltonian eqs wouldn't have to be substituted in but are kind of built in. However, so far I didn't manage to make this resulting in a shorter solution.. |
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| − | Cheers, Nathalie
| + | Regarding your u<->v symmetry proposal, at least here you can easily argue with {,}: |
| | + | With C2 = C1(x1 <-> x2, p1 <-> p2), it obviously holds that {C1,C2}=0. Thus {C1,h}=0 ==> {C2,h}=0 by Jacobi identity. |
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| − | Ok, for the lemma and for the corollary it's all clear. I still can't see the induction step from the first derivative to the n-th derivative, though: the proof is for a specific function and doesn't work anymore when you have a derivative of this function inside the integral, instead of the function itself. In fact, maybe we could forget all about lemma 2 and its corollary by using something of this kind: http://en.wikipedia.org/wiki/Differentiation_under_the_integral_sign
| + | [[User:Mario|Mario]] ([[User talk:Mario|talk]]) 20:40, 29 June 2015 (CEST) |
| − | But the problem is that we have \( \infty \) as integration bounds...
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| − | I wonder whether we really have to justify all about this passage (apparently no one has done it in the other groups) or whether we can just do it "the physician's way" (with no offense for anyone ;), because we're dealing with continuous functions with continuous derivatives and so on.
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| − | Cheers, Nick
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| − | Thanks a lot for completing it! Now I think there's really everything, and lemma 2 in this more general form could also be useful for some other exercise. As far as I'm concerned (because I have to correct it), now it's proof-read. Shall I make the tick in the main page or is somebody in particular responsible for all the final checks?
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| − | '[[User:Nick|Nick]] ([[User talk:Nick|talk]]) 00:11, 5 January 2015 (CET)'
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| − | Have you guys considered just using induction to show that \( \hat \Phi_n(k) = (-i)^n\Phi_n(k)\) ?
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| − | If not i could write an alternative solution...
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| − | Best,
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| − | "[[User:Carl|Carl]] ([[User talk:Carl|talk]]) 14:47, 8 January 2015 (CET)"
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| − | Yes, I found some solution on the internet similar to the one of your group at the time I had to solve the exercise. At the end I decided to extend the solution in the Felder-Script. Your solution looks right and elegant to me, I'm only not quite sure whether one can just use properties (2) and (3) of your solution without proof. But I'd be glad if you shared your solution here on the wiki.
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| − | Best, A.
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| − | thanks. what properties 2,3 do you mean?
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| − | "[[User:Carl|Carl]] ([[User talk:Carl|talk]]) 14:19, 9 January 2015 (CET)"
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| − | I've heard there should be a really short version of the exercise which makes use of the proposition at page 96 in the script Fourier_ScahwartzAdded which states \( (x \phi) \hat \: (k) = i d_k \hat \phi (k) \) and \( (d_x \phi) \hat \: (k)=ik \hat \phi (k) \) (under some conditions), but I couldn't get hold of it so far... Has anyone seen it?
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| − | '[[User:Nick|Nick]] ([[User talk:Nick|talk]]) 21:58, 15 January 2015 (CET)'
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| − | Good, that you're bringing this up. I actually was just thinking of making the boring technical Lemma 2 easier by just using that property and comparing. I'm putting that in the proof. And I think you could like the solution of group Y in the dropbox (that's by the way the one I was referring to you, Carl). - A.
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I tried transforming to polar coordinates in an attempt to make the solution shorter but I don't think it works. I think you can abbreviate in the exam at some places and also its all just derivatives so not very difficult, you just have to remember to invert the metric :)
Maybe it is possible to not explicitly show that C_2 is conserved, by saying that for u->v and v->u C_1->C_2, so C_2 must be conserved if C_1 is conserved.
Instead of calculating d/dt(C) I think it would be nice to use the fact that d/dt(C) = {C,h}. Of course it's basically the same but the hamiltonian eqs wouldn't have to be substituted in but are kind of built in. However, so far I didn't manage to make this resulting in a shorter solution..
Regarding your u<->v symmetry proposal, at least here you can easily argue with {,}:
With C2 = C1(x1 <-> x2, p1 <-> p2), it obviously holds that {C1,C2}=0. Thus {C1,h}=0 ==> {C2,h}=0 by Jacobi identity.
