# Tag Info

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Comments to the question (v2): Noether's (first) Theorem is really not about Lie groups but only about Lie algebras, i.e., one just needs $n$ infinitesimal symmetries to deduce $n$ conservation laws. If one is only interested in getting the $n$ conservation laws one by one (and not so much interested in the fact that the $n$ conservation laws often ...

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The "signed arc-length" is not used in relativity and I give reasons why. You are free to call and denote it in any way you like. $s$ and $\ell$ are interchangeably used to denote arc-length of space-like paths $g(\gamma',\gamma')>0$ in relativity and $\tau[\gamma]$ is used for "proper time" of $g(\gamma',\gamma')<0$ time-like paths but the notation ...

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To specify a Gel'fand triple $(\Phi^*,\mathscr{H},\Phi)$ it is sufficient to specify the Hilbert space $\mathscr{H}$ and the topological vector space $\Phi\subset \mathscr{H}$. The necessary requirement is that the imbedding of $\Phi$ into $\mathscr{H}$ is continuous with respect to the topology of $\Phi$, so this gives the imbedding of ...

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When I said we don't actually do path integrals, what I meant to say is that we can do some very specific path integrals and the way we do them is rather ad-hoc. In other owrds, it's highly nontrivial and not straightforward. To show this, I'll do a very general path integral for you. (I had most of this typed up already for another reason.) The vacuum ...

2

Eigenvectors exist only for the point spectrum of an operator. For any other point of the spectrum one can only find a sequence of vectors for which $(A-\lambda I)u_n\to0$, where $A$ is said operator, and $\lambda$ is a point in the spectrum which is not an isolated point. So in this case there is a sequence of approximate eigenvectors. With a bit of extra ...

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A self-adjoint operator $T$ on $L^2(\mathbb{R})$ has in its spectrum three different kinds of subspectra: A discrete point spectrum, a continuous spectrum, and a singular spectrum. The latter is physically discarded. The point spectrum consists of the eigenvalues of $T$, that is, the spectral values for which true eigenvectors in $L^2(\mathbb{R})$, and ...

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Item 1. has answered affirmative by Phoenix87 in the comments. Item 2. has been answered affirmative by yuggib in the other answer. The answers to item 3. and 4. have been given without proof by Phoenix87 in the comments, namely that the $\mathbb R$- and $\mathbb R^3$-Schwartz-space are isomorphic as rigged Hilbert spaces. A nice proof of this fact is given ...

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Same as with the symplectic form: $\omega(v) = (u_\omega,v)$ defines the isomorphism between 1-forms and vector fields. When the metric is Euclidean the dual basis to an orthonormal basis corresponds to the basis itself.

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I have never seen that second definition before. The first definition is standard. For a Riemannian manifold the metric tensor is positive definite, that is $$g(u,u)>0\quad\forall u\ne0$$ We have the standard relation (tensor product omitted) $$\mathrm{d}s^2=g_{ij}\mathrm{d}x^i \mathrm{d}x^j$$ Let $t\in\mathbb{R}$ be a curve parameter and ...

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No, the Hilbert space is not spanned by continuous "eigenfunctions" because they are not eigenfunctions at all! A self-adjoint operator $T$ on $L^2(\mathbb{R})$ has a point spectrum, a continuous spectrum and a singular spectrum. The latter is physically irrelevant. The point spectrum consists of the values $\lambda_i$ for which a true eigenvector ...

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The trace class operators form a Banach space. There is a concept of (countable) basis for Banach spaces that is called Schauder basis. Not every Banach space has a Schauder basis, but it is true e.g. for the space of compact operators (the case of $\mathcal{K}(l^2)$ is given explicitly in the wikipedia article). Since the trace class operators are ...

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You do not obtain the rules for the infinite-dimensional case by "proving" them from the finite-dimensional rules. Rather, you know that you need to have a Hilbert space, which is a complex vector space with an inner product, essentially. If you now search for infinite-dimensional Hilbert spaces that could possibly be used in quantum mechanics, you find ...

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The $e^{\vec{x}\cdot(\sum \vec{k})}$ which leads to momentum conservation at each vertex when we go to momentum space feynman diagrams is a perturbative result at heart. Your model as it stands is not cast in a way conducive of perturbation theory.. Let's try this instead, if we want to have something amenable to perturbation. (if $p<1$), then let's do a ...

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Let $D$ be the subspace of Schwartz functions $\psi$ of rapid decrease $\mathscr{S}$ such that their primitive $\Psi$ is in $\mathscr{S}$. Then $A:D\to D$ (easy to see calculating the primitve with integration by parts); $A$ defined on $\mathscr{S}$ and $B:D\to \mathscr{S}$. Hence on $D$ both $AB$ and $BA$ are well-defined, and [A,B]\psi=B\psi\;,\; ...

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Choosing a gauge usually means implicitly performing a gauge transformation such that a given condition holds, such as choosing a gauge $\partial_\mu A^\mu = 0$ in electrodynamics. A local trivialization of a $G$-principal bundle $G \to P \overset{\pi}{\to} M$ is given by an open covering $\{U_i\}$ of $M$ and diffeomorphisms \$\{\phi_i : U_i \times G \to ...

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