xcthulhu

@hobz: In all politeness I suspect you don't really know nonstandard analysis. You don't have to study it for very long to understand that it's nothing like physics at all. Just think about infinity in this formalism. For instance, if you have an infinitesimal [math]dx[/math] then [math]\frac{1}{dx}[/math] is an "unlimited" or infinitary number. Let's call [math]\frac{1}{dx}=\omega[/math]. Evidently [math]\omega + 1[/math] is infinitary, too (and unlike cardinal arithmetic and ordinal arithmetic, [math]\omega + 1 = 1 + \omega \neq \omega[/math]). In fact all of the points [math]\{\omega + r\ |\ r \in \mathbb{R}\}[/math] are going to be infinitary, too. Nonstandard analysts like to call all of the points a finite distance away from some particular infinitary number a galaxies. A tedious question arises: how many galaxies are there? With difficulty, we can prove that [math]\omega / 2[/math] is less than every number in the galaxy around [math]\omega[/math]. Also, apparently [math]e^\omega[/math] is in a separate galaxy. So... with a little work you can show there must be at least uncountably many galaxies. But how many really are there? Where's physics based intuition to help us like in physics problems? (btw, the answer to this is undecidable) In nonstandard analysis there are "integers" that are greater than every finitary integer. One can prove bizarro "infinitary" prime factorization theorems. Just like for the real unlimited numbers, there is no smallest galaxy around an unlimited integer. Another tedious question: Can you always "round" an unlimited number to an infinitary integer? Again, physics, analysis, really nothing helps you answer this. (my answer: Yes in some models but I'm not sure about all of them) In physics, you don't bother with actual nonstandard analysis, because it's rediculous. You wave your hands and pretend, or you use functionals on the category of Hilbert spaces or manifolds or whatever (if you want to make proper use of your PhD).

No question really, just registering a complaint. I'll get off my soap-box now I suppose.

It occurs to me that because I already showed that if [math]\psi[/math] is negative on [b,a] then [math]\psi'©<\psi'(d)[/math] for all [math]c,d\in[b,a][/math]. But since I also showed that for all [math]x < a[/math] it must be taht [math]\psi(x) < \psi(a)[/math], we know that [math]\psi[/math] is negative on the interval [math](-\infty,a][/math], whence we know that [math]\psi(x) << \psi(a)x[/math] on that interval. Hence it must go to negative infinity. That wasn't so hard I just ran out of time last time. Anyway, so either a function is constant or else it is not, in which case one of the four lemmas apply. In each case the function evidently blows up somehow, which is prohibited by assumption.

I suppose I'll tackle (d) today. We first take [math]\psi[/math] to be real based on yesterday's results. Today's problem is just a bit of analysis. We can rewrite the equation for the stationary solution as: [math](V(x) - E) \psi(x) = \frac{\hbar^2}{2m}\psi''(x)[/math] If [math]E < V_{min}[/math], then it must be that [math]\psi(x)[/math] and [math]\psi''(x)[/math] have the same sign. From this we can deduce the following 4 lemmas: (1) If [math]\psi(a) > 0[/math] and [math]\psi'(a) > 0[/math], then [math]\psi[/math] is increasing on [math][a,\infty)[/math] and [math]\lim_{x \to \infty} \psi(x) = \infty[/math] (2) If [math]\psi(a) > 0[/math] and [math]\psi'(a) < 0[/math], then [math]\psi[/math] is decreasing on [math](-\infty,a][/math] and [math]\lim_{x \to -\infty} \psi(x) = \infty[/math] (3) If [math]\psi(a) < 0[/math] and [math]\psi'(a) < 0[/math], then [math]\psi[/math] is decreasing on [math][a,\infty)[/math] and [math]\lim_{x \to \infty} \psi(x) = -\infty[/math] (4) If [math]\psi(a) < 0[/math] and [math]\psi'(a) > 0[/math], then [math]\psi[/math] is increasing on [math](-\infty,a][/math] and [math]\lim_{x \to -\infty} \psi(x) = -\infty[/math] All of the proofs of these are rather similar. I'll do (4). Assume [math]\psi(a) < 0[/math] and [math]\psi'(a) < 0[/math] and consider any interval [b,a] for which [math]\psi[/math] is negative. I argue that [math]\psi[/math] must be decreasing on the entire interval. For consider any two points [math]c,d\in [b,a][/math] where c < d, then we know by [math]L^2[/math] continuity and the mean value theorem that there's some point e such that c < e < d and [math]\psi''(e) = \frac{\psi'© - \psi'(d)}{c - d}[/math] But evidently [math]\psi(e) < 0[/math] hence [math]\psi''(e) < 0[/math], hence [math]0 > \psi'© - \psi'(d)[/math]. But then that just means that [math]\psi'© < \psi'(d)[/math]. Since by assumption [math]\psi'(a) < 0[/math] then it must be that [math]\psi'(x) < 0 \forall x \in [b,a][/math], which is to say the stationary solution is decreasing on the whole interval [b,a]. We can see from the above that it must be that [math]\psi(x) < \psi(a)[/math] for all [math]x < a[/math]. For if there was some [math]\psi© > 0[/math] for [math]c < a[/math] then we find some [math]c < b < a[/math] where [math]\psi(b) = 0[/math] (by continuity and the intermediate value theorem) and for all [math]\epsilon > 0[/math] we have [math]\psi[/math] is negative on [math][b+\epsilon,a][/math]. Then that means that [math]\psi(b) = \lim_{x\to b} \psi(x) < \psi(a)[/math] (by continuity), which is a contradiction. The proof of the blowup is tricky and I will have to deal with it later. ~XCT

There was a post a while back on "1/0", and I was disappointed to see a post of mine deleted. So I'll be brief and repeat my previous remark. While 1/0 is generally undefined for fields, it is defined on the Riemann Sphere for the so-called extended complex numbers. On this complex manifold, 1/0 = [tex]\infty[/tex], the "point at infinity". Wolfram|Alpha and Mathematica implement this sort of complex arithmetic, albeit a bit inconsistently. Here's what Wolfram|Alpha thinks 1/0 is: http://www.wolframalpha.com/input/?i=1%2F0

You and Plato. How much time have you spent in industry? Anyway, I think we can both agree that nonstandard analysis is practically useless and of limited real world utility.

I suppose my favorite language is Haskell, but that's a poor choice for a beginner language. I started with C. I think C or Python are good beginner languages.

a) Schrodinger's equation (in one dimension) asserts: [math]i \hbar \frac{\partial \Psi}{\partial t} = -\frac{\hbar^2}{2 m} \frac{\partial^2 \Psi}{\partial x^2} + V\Psi[/math] If we assume seperability, then [math]\Psi(x,t) = \psi(x)\phi(t)[/math]. We can then rewrite the above differential equation as follows: [math]i \hbar \frac{\phi'(t)}{\phi(t)} = -\frac{\hbar^2 \psi''(x)}{2m \psi(x)} + V(x)[/math] This implies both sides are constant, which is the allowable energy, which we will label [math]E[/math]. Now suppose that E were complex; such that [math]E = K + i J[/math]. Then [math]\phi(t) = A e^{-J t} e^{K i t}[/math] according to the differential equation. From normality we have that [math]\int_{-\infty}^{\infty} |\Psi|^2 d x= 1[/math], but since [math]\Psi(x,t) = \psi(x)\phi(t) = A \psi(x) e^{-J t} e^{K i t}[/math], then [math]\int_{-\infty}^{\infty} |\Psi|^2 d x = \int_{-\infty}^{\infty} \Psi^\ast\Psi d x = A^2 e^{-2 J t} \int_{-\infty}^{\infty} \psi^\ast(x)\psi(x) dx[/math] Hence: [math]A^2 e^{-2 J t} = \int_{-\infty}^{\infty} \psi^\ast(x)\psi(x) dx[/math] (a constant) Thus it must be that [math]J[/math] is zero. b) We have the following rules for conjugation: [math](f + g)^\ast = f^ast + g^ast[/math] [math](f \cdot g)^\ast = f^\ast \cdot g^\ast[/math] [math](a f)^\ast = a f^\ast[/math] if [math]a[/math] is a real scalar [math](1/f)^\ast = 1/f^\ast[/math] [math](f')^\ast = (f^\ast)'[/math] Now add in the observation that from separability we have that [math]E = -\frac{\hbar^2 \psi''(t)}{2m \psi(t)} + V[/math]. Hence the conjugates of both sides of this equation must be equal. But E is its own conjugate since it is real, as well as V (since the potential energy is also defined to be real). With this and using the rules we can push conjugation through to arrive at: [math]E = -\frac{\hbar^2 (\psi^\ast)''(t)}{2m \psi^\ast(t)} + V[/math] (a "stationary solution") Which means that if [math]\psi[/math] indeed satisfies the above so too does [math]\psi^ast[/math]. Now it's easy to see that we can rewrite the above as: [math]E\psi = \hat{H} \psi[/math] and [math]E\psi^\ast = \hat{H} \psi^\ast[/math] Note that [math]\hat{H}[/math] is a linear operator. We add these two equations or subtract them from one another and multiply by i, giving us: [math]E(\psi+\psi^\ast) = \hat{H} (\psi+\psi^\ast)[/math] and [math]E i (\psi-\psi^\ast) = \hat{H} i (\psi-\psi^\ast)[/math] (we know that [math]\hat{H}[/math] distributes because it's a linear operator) Either way, we can see that we have obtained a new stationary solution [math]\psi[/math] which is completely real. c) We can rewrite the stationary solution equation as: [math]V(x) - E = \frac{\hbar^2 \psi''(x)}{m^2 \psi(x)}[/math] Assuming that [math]V(x)[/math] is even then we have that [math]V(x) - E = V(-x) - E = \frac{\hbar^2 \psi''(-x)}{m^2 \psi(-x)}[/math] Now note that [math](\psi \circ (\lambda x. -x))''(x) = \psi''(-x)[/math]. From this we can deduce that it must be that if [math]\psi(x)[/math] is a solution then [math]\psi(-x)[/math] is a solution, too. Thus we have [math]\psi(x) + \psi(-x)[/math] (an even function) and [math]\psi(x) - \psi(-x)[/math] (an odd function) are both stationary solutions. I ran out of time on (d), but I'll tackle it later....

The easiest way is to install linux (which will make your computer much harder to hack), use Chrome in incognito mode all the time, block all cookies and javascript, and use open wireless networks in cafes whenever possible while avoiding entering any passwords. ~XCT

Hi DrRocket, You appear to be an expert. I have some really pedantic caveats with what you say here. (1) Specifically, Abraham Robinson uses a non-principle ultrafilter to obtain an elementary extension of the real numbers called an ultra-product. While the existence of non-principle ultrafilters is indeed a consequence of the axiom of choice, it's not logically equivalent. I don't mean to get all reverse mathematics up in here but you can obtain them in models of set theory without choice, namely they are guaranteed to exist on the real numbers as a consequence of the axiom of determinacy. I forget if you can obtain them from various forms of Konig's lemma... Reading Abraham Robinson will turn your mind into mush, btw. If you can just accept the existence of a non-principle ultra-filter as an article of faith then the non-standard reals aren't any harder to obtain than the reals via Cauchy sequences. I like Goldblatt's Lectures on the Hyper-reals which takes this approach. Applying the transfer theorems is as difficult as you say, however. (2) There are other approaches to the construction of infinitesimals. One approach is smooth infinitesimal analysis (here's Bell's introduction to the subject). Most mathematicians despise this approach completely because it's purely intuitionistic, which is hard if you aren't a disciple of Brouwer. Well... do I really need to mention that the mathematical community is vast? A modern application of nonstandard analysis is in the semi-automated proof assistant Isabelle/HOL. Typically a automated theorem prover like VAMPIRE (invoked by Isabelle) will stumble on quantifier manipulation, requiring more interaction from the mathematician/programmer. Isabelle/HOL has a mechanism "proof by transfer theorem" that turns ordinary analysis problems into non-standard analysis problems, performing quantifier elmination in the process. It's considered the preferred way to tackle analysis once you get advanced enough. And I don't mean to straw man you, but I suspect you are thinking something to the extent "But computer proof isn't real mathematics, it's just nerds with computers." I would suggest reading Donald MacKenzie's Computers and the Sociology of Mathematical Proof. TL;DR: Apparently the British government is of the opinion that computer assisted proof constitutes mathematical proof.

Here is a little problem I stumbled across from Bracewell's The Fourier Transform and Its Applications. I don't know the answer, so if anyone can help me out I would be grateful (this is just for self-study). I have a theory what the answer is like but I'm not altogether sure.

Hi, I am an amateur trying to teach myself physics by trying to do at least one problem every day. While these are textbook exercises, nobody has assigned them to me... so helping me isn't cheating unless you think I should be able to handle these problems myself. These are some beginner problems from Griffiths' Introduction to Quantum Mechanics I will post my answers later today. I love feedback and further reading if anybody has any suggestions.

Hi, I'm xcthulhu. I recently decided to start teaching myself physics by doing one problem every day. I'm happy that there are so many experts active on this forum, because it will be nice to have some feedback as I try to teach myself physics this way. ~XCT

Hi, I'm a student at Universiteit van Amsterdam, and for a final project for a cognitive science class me and some other students have devised an online survey. We'd greatly appreciate it if you could help us out in collecting data! Here is the URL of the questionnaire: http://www.w-d.org/questionnaire/ Cheers, ~xcthulhu Merged post follows: Consecutive posts mergedBump. I know I'm new to this forum, but I'd greatly appreciate any participation and feedback! Thanks! ~XCT