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A closed extension $A_c$ of an operator $A$ is an operator whose action is the same as $A$, the domains satisfy $D(A_c)\supset D(A)$ and $A_c$ is closed. Given that, the smallest closed extension of a symmetric (densely defined) operator is its double adjoint $A^{**}$. We call it the closure of $A$, and denote it by $\overline{A}$. An operator $A$ is ...


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No, because Haag's theorem states that there is no map between the free and interacting Hilbert spaces such that the fields and their commutation relations on one space are unitarily mapped onto the fields and their commutation relations on the other space. That is, the space of states of the interacting theory is as a representation of the commutation ...


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The algebraic approach gives the better idea of what the states and observables of a quantum theory are, and this holds in infinite dimensional systems as well. In the modern mathematical terminology, observables of quantum mechanics are the elements of a topological $*$-algebra, and states are objects of its topological dual that are positive and have norm ...


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I) Disclaimer: In this answer we will use the (traditional) physicist's definition of tensors using indices and their transformation properties under coordinate transformations. Moreover, let us suppress time dependence $t$ for simplicity. II) Let the manifold $Q$ be the configuration space. The Lagrangian $L:TQ\to \mathbb{R}$ transforms as a scalar ...


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You're right, there is no eigenfunction. The eigenfunctions of a self-adjoint operator form a complete basis for the Hilbert space, but this is simply not true for symmetric operators. Therefore if an operator is not self-adjoint, it may not have any eigenfunctions.


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This doesn't directly answer your question of orthogonality, but may still address your concern. I need to point out that you seem to be working on a scattering problem and resonant states. Resonant wave functions DO NOT belong to the Hilbert space. They are not even eigenfunctions of the Hamiltonian in an usual sense. You already know that they do not ...


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Some tensors have more intuitive pictures than others. For instance, simple antisymmetric tensors tensors can actually represent oriented subspaces with a magnitude. So that's the clear inheritor of the oriented-1-dimensional-subspace with magnitude. And then you can imagine little oriented planes, little oriented 3-volumes in 4d (for relativistic ...


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I'll just throw out two simple examples of tensors that you already know: 1) the dot product, itself, is a tensor. It takes two vectors as input, and spits out a number 2) Now, consider a general fluid. It might have viscosity, and shear and everything. It's completely general. Now, consider yourself to be something living in this fluid. You move a ...


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There is another definition. Let's say $v$ and $w$ are part of the vector spaces $V$ and $W$. Now, let's consider bilinear maps $f$ from $V$ and $W$ to vector spaces such as $Z$. Bilinear means that $ f(v_1+v_2, w) = f(v_1, w) + f(v_2, w)$ $ f(v, w_1 + w_2) = f(v, w_1) + f(v, w_2)$ $f(cv, w) = cf(v, w) = f(v, cw)$ This maps are important for some reason. ...


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That is only one definition my friend, but there is another. But to understand $v \otimes w$, you must first understand $V \otimes W$ (where $V$ and $W$ are the vector spaces for $v$ and $w$ respectively.) First, consider, for all the vectors $v$ and $w$, the abstract symbols $v \otimes w$, and use this as a basis for a free vector space. Now, using ...


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I really recommend 4 books in 3 steps 1)Mathematics for physics, Michael Stone Paul Goldbart 2)Modern Mathematical Physics, Peter Szekeres 3)Geometry for Physics, T. Frankel with An introduction to Manifolds, Loring W. Tu After all, The Road to Reality, Roger Penrose


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Sufficient conditions for a discrete energy spectrum: In one-dimension Sturm-Louville theory implies that the spectrum is purely discrete provided that the system is restricted to a finite interval [a,b] (with appropriate boundary conditions). I assume (but don't know) this can be extended to the case of a system restricted to finite volume in higher ...


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Comments to the question (v1): Let there be given an $n$-dimensional manifold $M$ with a smooth vector field $X\in \Gamma(TM)$. If $(x^1, \ldots, x^n)$ is some local coordinates on $M$, then the vector field takes the form $$\tag{A} X~=~X^i(x)\frac{\partial}{\partial x^i},$$ and one may study the autonomous first-order ODE $$\tag{B} ...


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Consider the drift diffusion equation $$\dfrac{\partial}{\partial t}\psi=\mu\dfrac{\partial}{\partial x}\psi+\kappa^2\dfrac{\partial^2}{\partial x^2}\psi.$$ Dimensional analysis tells us that $\mu$ is a characteristic length per time (drift velocity) while $\kappa$ is a characteristic length per square root of time. This small factoid has curious ...


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Your question is "Is there an infinite series of higher derivatives of position for this to work?" Answer: No. Acceleration can jump from zero to something. When it does, its derivative is not defined, so the series of position derivatives stops after the second one. From the question: "A change is velocity is acceleration, so the value of the ...


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I may add a few recent works that use coadjoint orbits to better understand the space of solutions of 2+1 gravity. With a cosmological constant you get Virasoro group coadjoint orbits, and without a cosmological constant you get BMS$_3$ coadjoint orbits. These are the symmetries of the spaces of solutions of the corresponding gravitational theories. The ...


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It happens because of quantum tunneling. In the first scenario, we cannot possibly find the particle beyond the barrier. in the second, however, the particle can quantum tunnel over.


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For an infinite well (actually infinitely high barriers) the probability to find an electron in the barrier vanishes. Therefore the wavefunction in the barrier has to be 0. For barriers with a finite height, the commonly used, but actually wrong boundary conditions require the wavefunction and the first derivative to be continuous. This relies on the wrong ...


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When people study continuum mechanics they usually do so at first in $\mathbb{R}^3$ where we have usually implied the usual metric tensor $(g_{ij}) = \operatorname{diag}(1,1,1)$ and the Levi-Civita connection associated with it. In that case vectors and covectors are equivalent: the metric tensor induces the musical isomorphism and allows one to convert ...


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My favorite reference for these sorts of things that straddle physics and geometry is Frankel's "The geometry of physics". In the chapter on harmonic forms, you will find what he refers to simply as "Hodge's Theorem". It's a little more general than you need, because it applies to general $p$-forms, and you only need functions (0-forms). So I'll ...


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The definition suggested by joshphysics and clarified by Qmechanic already exists in the literature under then name of representation operator. This is discussed in, e.g., Sternberg's Group Theory and Physics, as well as the somewhat more elementary text An Introduction to Tensors and Group Theory for Physicists by Jeevanjee.


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Comments to the question (v2): On one hand, let there be given a configuration space $(Q,g)$ endowed with a metric $g$. (As ACuriousMind points out in a comment, there is a 1-1 correspondence between a metric $g$ and the kinetic term in a Lagrangian.) On the other hand, note that the canonical symplectic 2-form $\omega$ on the cotangent bundle $T^{\ast}Q$ ...


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No, but you are most likely to get one from the kinetic term of the Lagrangian itself. In most cases one requires it to be a convex function in the $\dot q$ variables. You then get a metric if such kinetic term is quadratic in $\dot q$ (and of course sensible kinetic energy is positive-definite). The metric and symplectic structures on a manifold are ...


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$\newcommand{\bra}[1]{\langle #1 \vert}$ $\newcommand{\ket}[1]{\vert #1 \rangle}$ In the position basis $\hat{p}^2$ acts as $$ -\nabla^2 \psi(x) = \bra{x}\hat{p}^2\ket{\psi}$$ From \begin{align} \bra{\psi}\hat{p}^2\ket{\psi}&=\int\mathrm{dr}\,\bra{\psi}\ket{r}\bra{r}\hat{p}^2\ket{\psi}\\ &=\int\mathrm{dr}\,\psi(r)^*(-\nabla_r^2\psi(r))\\ ...


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The physical observables both in classical and quantum mechanics are thought to have, in modern mathematical terms, an involutive algebra structure. This point of view has been developed first and foremost for quantum mechanics. The observables of quantum mechanics are a suitable $C^*$-algebra. By Gel'fand-Naimark isomorphism any $C^*$ algebra is ...


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Yes I'd agree that reference frames are sections in the frame bundle. For physical intuition, consider the notion of frames in classical GR. An observer, at any given point in spacetime, would measure things in an orthogonal basis. That is, to the observer, the time he observes should be orthogonal to the spacial distances he observes, and spacial ...


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you can either evalute the integral numerically or search for a good coordinate transformation to evaluate it "by hand". This transformation is the one to spherical coordinates, though. Also you could try this: \begin{align} Q & = 2 \pi A \int_{-\infty}^{\infty}\int_0^{\infty} r \cdot H(R^2-r^2-z^2)~dr dz \\ & = 2 \pi A ...


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The extra derivative in Polchinski comes from the following version of the Fundamental Lemma of Calculus of Variation (FLCV): $$\tag{1} \left[ \forall g : ~~\int_a^b\! dx~ g(x) ~=~0 \quad\Rightarrow \quad \int_a^b\! dx ~f(x) g(x) ~=~0\right]\quad\Rightarrow \quad f^{\prime}~=~0.\quad $$ FLCV (1) states in words: If it is true that for all functions ...


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The question of AdS (in)stability is indeed a hot topic in current research of the AdS/CFT correspondence. It is a field that ties together many interesting subjects: Gravity in AdS (i.e a confining box), thermalization in QFTs, the theory of non-linear differential equations and their perturbative treatment, turbulence etc. This explains the explosion of ...


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If I had to describe the purpose of these series as succinctly as possible it would be these three words: to handle complexity. What do I mean by this? Lets suppose we need to describe some function $\mathbb{R}^M\to\mathbb{R}^N$. The cardinality of the set of all such functions is $2^{\beth_1} = 2^{\left(2^{\aleph_0}\right)}$: the cardinality of the ...


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I do not think Maclaurin series has any special physical significance that is different from the Taylor series. But then for some functions it is desirable to have a series expansion at the origin even though it is singular at the origin. Maclaurin series requires that the function is analytic everywhere within the circle of convergence so it can not be ...


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Here's a supplement to ACuriousMind's excellent answer, which: Goes into more depth about the meaning of "walking in the direction specified by [an] operator." Connects this to the more familiar operation of exponentiating a complex number. Gives another take on the physical meaning of exponentiation. Sorry it's so long! If you don't feel like reading ...



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