# Tag Info

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The no-go results from Algebraic and Constructive QFT you mention deal with related but slightly different matters. (Edit: the previous version of the following paragraph was slightly misleading - Haag's theorem is actually stronger than I stated before; see below for details) Haag's theorem (which actually slightly predates the inception of Algebraic ...

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But can the eigenstates of the position observable be individually thought of as delta functions? Yes they can in a sense but it is rather inaccurate. First, kets and functions(distributions) are somewhat different things although they share most of their properties. If the ket $|x'\rangle$ satisfies $$\hat{x} |x'\rangle = x' |x'\rangle,$$ then we ...

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Neuneck's answer is the pithiest description of how you get normalisable states as superpositions of non-normalisable states, but the following is more of a "why" these states happen. Hopefully, you should see that this discussion is independent of the number of dimensions. Practically speaking, the reason why there are always such states it is because ...

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Hilbert's Sixth problem is not the same as finding the theory of everything and then making the maths rigorous. This is a very common misconception, and has led to people thinking that making renormalisation in QFT rigorous was the main thing to do. But in fact Hilbert stated explicitly that it would be just as important to axiomatise false physical ...

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You've forgotten one crucial thing when you've written your superposition: the separate $\psi_{k,\ell,m}(x,y,z)$ are eigenfunctions of the Hamiltonian with different eigenvalues. The superposition will no longer be an eigenstate because of this. In fact, by taking an appropriate superposition, you would be able to get any function you like (in your case, ...

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All our sophisticated mathematical tools - Derivatives and Integrals, Fourier Transforms, Groups and Representations, Riemann Tensors, Kähler manifolds, etc. are merely descriptive techniques. What exists is what exists, independent of how we try to describe it. New mathematical ideas often help us see known phenomena more clearly, or deal with the ...

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It is possible to say something more precise than Martin's answer (that is correct however). The key-point is that self-adjoint operators are closed operators. An operator $A: D(A) \to H$, with $D(A) \subset H$ a linear subspace of the Hilbert space $H$ is said to be closed if, for every sequence of vectors $f_n\in D(A)$ such that (1) $f_n \to f \in ... 3 Since I'm not an expert on spectral theory, this will only be a partial answer, however, I believe that this question, is mathematically much more involved than you think. First of all, let's review the finite dimensional case: We have two Hermitian matrices$A,B\in\mathcal{M}_d$and they commute if and only if their spectral projections commute, i.e. they ... 3 First, if you take the fundamental representation (representation$N$) of$SU(N)$made of$N$objects$\varphi^i$, the transformation law is :$\varphi^i \to U^i{}_j \varphi^j$. By taking the complex conjugate, you get :$\varphi^{*i} \to (U^*)^i{}_j \varphi^{*j}= (U^\dagger)^j{}_i \varphi^{*j}$. Now, looking at the last expression with$U^\dagger$, one ... 3 Ok, I think there is a mistake here: A general tensor$\varphi^i$transforms as: $$\varphi^i\rightarrow U^i_{\phantom{1}j}\varphi^j$$ whereas$\varphi_i$transforms as: $$\varphi_i\rightarrow (U^\boldsymbol{\ast})_i^{\phantom{1}j}\varphi_j$$ Where did you find these equations? The unitary matrix element in the second line should not be a complex ... 3 It's a matter of definition. According to Hirzebruch/Scharlau, Einführung in die Funktionalanalysis (1971), Definition 21.10: An orthonormal system$\{x_i\}_{i\in I}$that meets the [proven equivalent] conditions of this theorem is called a Hilbert basis or just basis of$X$. Wikipedia calls it an orthonormal basis, which is a Schauder basis if your ... 2 The scattered states are indeed non-normalizable. This is because a plane wave is an unphscial state (which you can for example see by trying to calculate the Heisenberg uncertainty, which will read$\Delta x \cdot \Delta p = \infty \cdot 0 = ??$). In order to create a physical state, you need to specify boundary conditions, i.e. a physical wavefunction at ... 2 Both$\mathbb{R}^3$and$S^3$are rank 1 symmetric spaces explicitly, as a homogeneous spaces they are given by: $$\mathbb{R}^3 = ISO(3)/SO(3)$$ and $$S^3 = SO(4)/SO(3)$$ The significance of their being rank-1 symmetric spaces is that there is only one "two-point" invariant on them, i.e., any function of two points$r_1$and$r_2$invariant under the ... 1 Are there any physical meaning or application of this property? Why nature gives this property to the widely used Fermi-Dirac Function? The property you found (Taylor expansion$\neq$original function on any interval around the point) has nothing to do with physics or nature, and is not particularly connected to the Fermi-Dirac distribution. It may be ... 1 There is absolutely physical meaning in it -- the sum is what gives you the final probability of the particle being at a particular location. But, often knowing just the single number is not illustrative enough to get much information from, which is why we look at the eigenfunctions individually. Quantum mechanics isn't my area, but I'll tie it back to ... 1 Okay, so a general observable acting on$|x\rangle$won't give you$x' |x\rangle$. Only the position operator, acting on the state$|x'\rangle$will give us$x'|x'\rangle$, where the x' is a label for the state, think of it as a number, not a variable. Just because the state$|x'\rangle$is an eigenstate of the position operator, it does not mean that it ... 1 I do not know if my answer can be classified as physics but Julia in her question wrote that the students are generally very interested in technology, engineering, physics and computer science and so I offer and answer that maybe can be considered technology or computer science. The pinhole or projective camera is a purely geometric model that ... 1 I would use the Van der Waals equation. Here you are, in the German wiki: http://de.wikipedia.org/wiki/Van-der-Waals-Gleichung Also, there's lot of laws expressed as$a = b/c$. They may be too simple for your purposes, but they can be useful to show the application. For example, Ohm's law$I=V/R$, where$I$is current intensity,$V$voltage and$R$is the ... 1 As you and Christoph already pointed out, the difference comes from the fact that these contiuous "bases" do not belong to the respective Hilbert space themselves. This is why they are not actually bases at all in the general sense. Rather, they are useful mathematical tools to expand any states actually belonging to the Hilbert space. As they obey certain ... 1 There seems to be some evidence Gödel pondered this. He wrote: "But, despite their remoteness from sense experience, we do have something like a perception of the objects of set theory, as is seen from the fact that the axioms force themselves upon us as being true. I don’t see any reason why we should have less confidence in this kind of perception, i.e. ... 1 It took me an embarrassingly long time to understand Qmechanic's construction, so I thought I would add this elaboration, as an aid to anyone as perplexed as I was. I think this integral (3') is best understood not as a surface integral but as a line integral, just not of your garden-variety gradient of a scalar function ($-\boldsymbol{\nabla} U\$ , like ...

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This is an experimentalist's answer: I do believe that an axiomatic model , note "model", of nature can be found, but as an experimentalist I am wary of claims that "we have now wrapped up physics and only details have to be mopped up" which was the claim before quantum mechanics rocked the science in the beginning of the twentieth century. One should be ...

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Yes it is possible. As J. Bell eloquently wrote, Quantum Mechanics, together with a finite cut-off QED, explains all of chemistry and nearly everything in Physics. It was axiomatised by Weyl and Dirac by 1930. There are only six axioms, which is certainly a finite number. Five would be better still..since most physicists no longer believe in the ...

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