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9

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 ...


5

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 ...


3

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 ...


2

An antilinear map (scalars are factorized "out of the map" as their complex conjugates) $C:\mathscr{H}\to \mathscr{H}$ is called a conjugation if it preserves the norm of $\mathscr{H}$ and $C^2=\mathrm{id}$ (the identity operator). Theorem [von Neumann]. Let $A$ be a symmetric operator. If there exists a conjugation $C$ such that $C: D(A)\to D(A)$ and ...


2

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 ...


2

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. ...


2

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.


2

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 ...


2

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 $T^{\ast}Q$ ...


2

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.


1

$\newcommand{\bra}[1]{\langle #1 \vert}$ $\newcommand{\ket}[1]{\vert #1 \rangle}$ 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))\\ ...


1

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 ...


1

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 ...


1

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 ...


1

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 ...


1

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} ...


1

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 ...


1

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 ...


1

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 ...


1

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 ...


1

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 ...


1

The inverse of the Wigner map is the Weyl quantization map. Let $a(x,\xi)$ be a function of the phase space (i.e. the symbol, in mathematical terms). If $a$ is real-valued, then $(a)^{Weyl}$ is symmetric; Let $g(x,\xi)\in C^\infty(\mathbb{R}^{2d}; \mathbb{R}_+^*)$ such that $\partial_{(x,\xi)}^\alpha g=O(g)$ for any $\alpha\in \mathbb{N}^{2d}$ and ...



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