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A multisymplectic form is not a symplectic form and cannot be canonically quantized. Thus your main question doesn't make sense as made ''more precise''. (By the way, I have never seen a way to quantize a multisymplectic form. You would have to quantize the symplectic form underlying the Peierls bracket constructed from the multisymplectic framework. This ...

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There was an FQXi essay competition on this subject in Spring 2015: "Trick or Truth: the Mysterious Connection Between Physics and Mathematics" Here is the home page with links to winners and other entries: http://fqxi.org/community/essay/winners/2015.1 The competition is meant to encourage an informal style, readable by non-experts, but with some ...

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Often $X$ is a coadjoint orbit of a Lie group. These have a natural symplectic structure; see https://en.wikipedia.org/wiki/Symplectic_reduction

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Greiner's Thermodynamics and Statistical Mechanics is pretty good from a few short readings I did. Also, it has better reviews from almost all of the other popular textbooks on the subject in goodreads.com

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Recall that the (global) conformal group is given by $${\rm Conf}(p,q)~\cong~O(p\!+\!1,q\!+\!1)/\{\pm {\bf 1} \},\tag{1}$$ cf. e.g. this Phys.SE post. Using the embedding $\imath: \mathbb{R}^{p,q}\hookrightarrow \overline{\mathbb{R}^{p,q}}$ into the conformal compactifification $\overline{\mathbb{R}^{p,q}}$, one may show after a short calculation that the ...

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I have communicated with both of these fellows. The mathematics is based on Donaldson's theorem that in four dimensions there exists an infinite number of atlases of charts on a manifold that are homeomorphic but not diffeomorphic. I am not able to go into the mathematics, for it is pretty deep. It centers around the moduli space of self-dual connections $SD/... 0 Conceptually, the idea is that planes and spheres are equivalent from the point of conformal geometry. Conformal transformations map {planes, spheres} to {planes, spheres}, and in fact do this transitively -- any object in the set {planes, spheres} can be obtained from any other by a conformal transformation. Inversion is a reflection against a sphere, and ... 2 UPDATED ANSWER Sorry, I interpreted your question too narrowly. Couette flow occurs without a pressure gradient, due to viscous drag from a boundary surface, and is laminar. If the drag force is increased the flow can become turbulent. If a transient inertial flow begins laminar I think it must remain laminar as it dies out, because the speed of flow ... 0 From a physicists point of view, I would start with the following notes, which are Chapter 9 in John Preskill's Quantum Computing lecture notes: http://www.theory.caltech.edu/~preskill/ph219/topological.pdf, as well as the references within. I would also mention Kitaev's paper https://arxiv.org/abs/cond-mat/0506438 as a specially influential reference. ... 2 I think the particular statement of Gleason's theorem is really important in elucidating this perceived ambiguity. Wikipedia states it as: Theorem. Suppose H is a separable Hilbert space of complex dimension at least 3. Then for any quantum probability measure on the lattice Q of self-adjoint projection operators on H there exists a unique trace class ... 0 Thinking of the sphere$S^n$as the one-point compactification of$\mathbb{R}^n$, we can consider the stereographic projection from the plane defined by$x^0 = 0$to the unit sphere$\{x\in \mathbb{R}^n\,:\,|x| = 1\}$. This map is actually defined on$\mathbb{R}^n\cup\{\infty\}$, it takes the point$\infty$to the north pole of the unit sphere. Moreover, ... 0 I found this paper in Numdam's (a mathematical journal compilation) archive, which encompasses all that you talked about and I found it clear, with some references to Witten as well. This paper goes much further in detail, but I did not read all of it. And this might help if you aren't bothered by learning by forum posts? 0 It has to be " Statistical Mechanics and Thermodynamics " by Claude Garrod". You can use the text by Macquarie as a supplement. For renormalization group and advanced concepts, use " Statistical Physics of Fields" by Kardar. 2 First, let's answer the questions precisely as you worded them: The point spectrum is always discrete in the sense that it consists of at most countably many points. This is true by proving the following results: a) the space spanned by all eigenvectors is a closed subspace of the Hilbert space, hence we have an orthonormal system of eigenvectors, b) two ... 0 I found a paper that helped me a bit in understanding how things work. So here is what I understood. Given a random variable we define a formal power series and a formal derivation such that: $$\frac{\partial}{\partial U}\left(\sum_{n=0}^\infty a_nU^n\right)=\sum_{n=0}^\infty (n+1)a_{n+1}U^n$$ The "usual" Wick product$:\ :$is ... 4 (1) Yes, the point spectrum is countable in your hypotheses: otherwise the operator would have an uncountable set of pairwise orthogonal vectors since eigenvectors of a self-adjoint operator with different eigenvalues are orthogonal. This is impossible because, in every Hilbert space, every set of (normalized) orthogonal vectors can be completed into a ... 5 Whilst it is certainly true that Quantum Probability Theory (QPT) is an entirely different framework from Classical (Kolmogorovian) Probability Theory (CPT) (specifically because the event structure is non-Boolean and the random-variable structure is non-commutative), we can still identify enough formal similarity to borrow the classical terminology. In ... 0 Maybe you can be interested in another interpretation of Hermitian matrices. In a recent paper we have proposed to see them as gambles on a quantum experiment. We have then enforced rational behaviour in the way a subject accepts/rejects these gambles by introducing few simple rules. These rules yield, in the classical case, the Bayesian theory of ... 1 A quantum system can be described by a set of evolving quantum mechanical observables. This is not the same as describing a system in terms of a stochastic quantity described by a single number chosen at random. A quantum system really does have multiple values of any unsharp observable, see https://arxiv.org/abs/quant-ph/0104033. Those different versions ... 7 I'm going to try to explain why and how density operators in quantum mechanics correspond to random variables in classical probability theory, something none of the other answers have even tried to do. Let's work in a two-dimensional quantum space. We'll use standard physics bra-ket notation. A quantum state is a column vector in this space, and we'll ... 0 Quantum mechanics starts with wave functions living in Hilbert space. But later for Born's interpretation, the wave function need to be of unit energy No, Hilbert space is a Hilbert space knowing nothing about external constructions to itself like "probabilities as observable values". Normalization of state vectors is an example of such external ... 6 I believe it is misguided to think that classical probability makes sense any more than quantum mechanics, with its "peculiar" probability calculations, makes sense. I'm going to be slightly mischievous here and make a friendly attack your first paragraph: does really make sense? Of course it makes perfect sense as a measure-theoretic definition, but how ... 11 Quantum mechanics is indeed a probability theory, but it is a non-commutative probability theory. So it is not just a matter of having signed/complex measures, but really of having a non-commutative probabilistic framework. Quantum mechanics was developed, historically, before non-commutative probability theories and I think that people in probability ... 9 You could certainly model any one quantum observable as a random variable. The problem comes in when you have multiple observables, which you might attempt to model as classical random variables with some joint distribution. From this joint distribution, you can compute various probabilities (like$\textrm{Prob}(Y\neq X)$, for example), according to the ... 1 This functional analysis textbook is quite good, it even has a chapter dedicated on some of the mathematical foundations of quantum mechanics without using this name. This book is quite compact though, the classic text by Walter Rudin is more detailed. Prerequisites are topology and measure theory. 4 Preliminaries: If you "limit" your description of quantum mechanics to$L_2$Hilbert spaces, all your bases will be discrete, both bounded or unbounded. You can have Hilbert spaces of any cardinality, but the one in "standard" quantum mechanics is$L_2$, the space of square integrable functions, which has a countable cardinality,$\aleph_0$. In this case ... 0 Disclaimer: I'm an engineer, and my knowledge of higher-level math is probably lacking here. But I'll try to guess what you mean by the question. It's not very clear as-is what boundary conditions you're using, since an open boundary on the plane doesn't really qualify as a boundary for the heat equation itself (I'm assuming here that by the words "open ... 2 I) Apart from a full proof of the Gutzwiller's formula in the context of the Feynman path integral (FPI), then OP is essentially asking if the FPI knows about the metaplectic correction/Maslov index and caustics? The physics lore is that when the FPI is set up and interpreted properly, it does contain these semiclassical phase factors. II) In practice, let ... 3 The infinities of set theory aren't the same kind of infinities addressed with regulatisation or renormalisation. For example, if$\kappa$is cardinal (be it finite or transfinite, which for some reason is the name used rather than infinite) then$2^\kappa>\kappa>0$, but the "infinities" we regularise or renomalise are$\pm\infty$, satisfying$2^{\...

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Let the state $|\psi'\rangle = |\psi\rangle + V|\psi'\rangle$, where $V$ is the perturbing or potential that induces scattering. We are looking at a process that propagates a scattering influence, and propose the Green's function $(E - H)G = 1$ for $G = G(x,x')$. We can then see $$(H' - E)|\psi'\rangle = (H' - E - V)|\psi'\rangle + (E - H)GV|\psi'\rangle$$...

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One solution would be to keep track of the number of cells with a certain number of receptors. In other words, instead of $P(m,t)$ you would consider $P(\{m_k\}_{k=0}^\infty,t)$, where $k$ is an index for the number of receptors on each cell (and $\sum_{k=0}^{k_{max}} m_k=m$). Depending on the details of cell division, you would have some $k$ dependent ...

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