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| bio | website | |
|---|---|---|
| location | ||
| age | 41 | |
| visits | member for | 10 months |
| seen | Oct 26 '12 at 6:50 | |
| stats | profile views | 260 |
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Feb 19 |
awarded | Nice Question |
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Nov 20 |
awarded | Popular Question |
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Oct 15 |
awarded | Nice Question |
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Oct 6 |
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What are imaginary numbers? @MakotoKato: That depends very much on what you mean with "the Schrödinger equation". If you mean the equation exactly as normally formulated, then yes, because there's an explicit i in it. But that doesn't tell us much because nature doesn't care about which of many equivalent ways to formulate a theory we use. |
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Oct 5 |
answered | How to solve this conditional probability problem? |
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Oct 5 |
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An element within a set within a set from a power set @alexthebake: Exactly. That's what I wrote in the second paragraph and proved in the third. |
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Oct 5 |
answered | Does a Lattice with min and max operator reduce to a total order? |
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Oct 5 |
revised |
An element within a set within a set from a power set added 329 characters in body |
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Oct 5 |
answered | An element within a set within a set from a power set |
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Oct 5 |
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How to draw metaballs? In Mathematica, I think RegionPlot3D would be well suited for this. The colouring of the balls would then be done with a suitable ColorFunction. |
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Oct 5 |
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What are imaginary numbers? @MakotoKato: I can't tell for sure (for that, I'd have to know much more), but I definitely wouldn't exclude the possibility. |
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Oct 5 |
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Help explaining the sign of a permutation @mixedmath: Exactly if permutations act on the right, I get $(ab)(ac)$ to mean $a\mapsto c$, $b\mapsto a$, $c\mapsto b$. Example: $(ab)(ac)a = (ab)c = c$. |
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Oct 5 |
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What are imaginary numbers? @MakotoKato: Just another argument which just came to me: You know that the Schrödinger equation has complex solutions. However if you look at the gauge transformations of the electromagnetic field and how they affect the wave function, you'll notice that you can completely gauge away the complex phases, and then get a real wave function and a corresponding electromagnetic potential and vector potential (if there were already such potentials to begin with, they are modified accordingly). That way, you get a completely real (but more complicated) form of quantum mechanics. |
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Oct 4 |
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Is there a function like this? (ii) is clearly wrong because $-C\not\subset A$. |
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Oct 4 |
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What are imaginary numbers? @MakotoKato: It might for example be an artefact of the fact that Lie groups are most easily worked with in their $GL(d,\mathbb C)$ representation. Also, if you look at the space of density matrices, they form a convex set in a real vector space (that's despite the density matrices are denoted by complex matrices). Unitary transformations translate into orthogonal transformations which keep that set invariant. And Hermitian operators are associated with linear function in that space. Who says that this real vector space is not more fundamental than the complex space of Hilbert space vectors? |
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Oct 2 |
answered | Help explaining the sign of a permutation |
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Oct 2 |
awarded | Good Question |
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Sep 30 |
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What are imaginary numbers? could be described without complex numbers. However I know for sure that Hilbert spaces are not the only way to describe quantum mechanics (although I don't know enough about those alternative descriptions which have been developed to say whether they use complex numbers; but then, there are almost certainly others which we haven't yet found, and I see no reason to assume that all of them use complex numbers). |
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Sep 30 |
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What are imaginary numbers? First: We don't know whether it is the fundamental law of the universe. It's the best we currently have. But then, classical mechanics was considered the fundamental laws for centuries, and today we know it isn't. Second: We use complex numbers for it because it turns out that they are useful in formulating the concepts which we use to model the universe. Note that we do not model the universe with complex numbers. We model the universe with operators and rays in Hilbert spaces. Now those Hilbert spaces are described with complex numbers. I don't know if the very same Hilbert spaces … |
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Sep 30 |
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Prime numbers and their products There are more than $2^{120}$ primes below $2^{128}$ (using the formula from this question). Given that to store one number, you need $128$ bits = $16$ bytes, you'd need more than $2^{124}$ bytes to store all those numbers. With 1 TB = $2^{40}$ bytes, that means you'd need $2^{64}\approx 1.8\cdot 10^{19}$ Terabyte disks to store those numbers. So no, the list definitely would not be small enough to fit in memory. |
