Recent content by aleazk

Thank you for your argumets... as usual in this topic. Sigh.

Really? Most relativists will tell you it actually is. In fact, the only people I have seen casting doubt on that are precisely string theorists, with very little convincing argumentation. Time to stop giving these people so much credit and start to challenge their BS. Decades have passed, none...

aleazk submitted a new PF Insights post Tools to Enrich our Quantum Mechanics Interpretations Discourse Continue reading the Original PF Insights Post.

Because of results like the Stone-von Neumann theorem. In non-relativistic QM, this theorem asserts that any irreducible unitary realization of the (integrated form of the) canonical commutation relations on a complex Hilbert space ##H## (in principle, not necessarily separable) is unitarily...

What @Shyan says is correct. But I think that, for practical calculations, the following point of view is more convenient. The tensor product Hilbert space ##L^{2}(\mathbb{R}^{3})\otimes\mathbb{C}^{2}## of the two spaces ##L^{2}(\mathbb{R}^{3})## and ##\mathbb{C}^{2}## is isomorphic to the...

Is not the area operator used in the derivation of the black hole entropy formula? At least I remember reading that in some papers (they had some sum over the area eigenvalues and that's how the area enters into the entropy formula; the area operator seemed to be the key point in the...

Hi, I submitted my personal details, when I will be able to write the actual article? (PS: I messed up the email, but now I updated my profile so the two emails are the same now)

False. The metric is enough, as I already told you before: https://www.physicsforums.com/threads/where-do-ctcs-come-from-how-do-i-interpret-these-tensors.811708/#post-5095361 (it's not a general method, but works for most of the well known examples) It depends on the metric, If your metric is...

Check "Introduction to Smooth Manifolds" by J.Lee. From page 361: "...explicitly finding integral manifolds boils down to solving a system of PDEs, we can interpret the Frobenius theorem as an existence and uniqueness result for such equations." And then it goes to show this connection.

8/12, no guesses. My awesome Newton knowledge comes from... ...reading the wikipedia article some weeks ago when I was looking for some specific detail about his life. :woot:

"Philosophy: Beacon or Trap?" An extract from the book "Philosophy of Physics" by M.Bunge, my favorite epistemologist (also a physicist and hardcore realist). I strongly recommend this book to anyone interested in physics (yes, I say physics, rather than "anyone just interested in the philosophy...

But it's precisely in the cases in which the GNSs are all equivalent when the algebraic approach is less powerful! For example, for a system in which there are only a finite number of degrees of freedom (the usual non-relativistic particle), all of those (irreducible) GNS constructions (and, in...

Have you tried the discussion in the Stanford Encyclopedia of Philosophy?: http://plato.stanford.edu/entries/kochen-specker/ It discusses in detail many of the relevant points (I will need to give it a re-read before making further comments, though)

It's true that the operators representing observables (if they are unbounded, their exponentials) can be shown to form C*-algebras. It's also true that quantum states are actually positive trace-class operators of trace 1. But to finish the argument that it's reasonable to use only C*-algebras...

I'm not sure what you are trying to say here. I will sketch the standard argument (which can be found in, e.g., Varadarajan's "Geometry of Quantum Theory") so you can point out which are the points you find problematic. In classical mechanics, linear momentum is indeed the generator of...