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## Hot answers tagged mathematical-physics

4

Yes, they can. It's not great form, but they can. For a simple example, consider the hamiltonian for a 1D particle in a potential $V(x)$, $$\hat H=\frac1{2m}\hat{ p}^2+V(\hat{x}).$$ This holds equally well if the potential is a delta function at $x_0$, $V(x)=\kappa \delta(x-x_0)$, in which case $$\hat H=\frac1{2m}\hat{ p}^2+\kappa\delta(\hat{x}-x_0).$$ ...

4

For the simplest case of interaction, the quadratic one, the interacting representation is roughly speaking the Fock one, but with a different mass (representations with different masses are proved to be inequivalent in the second volume of the Reed-Simon book). For $(\varphi^4)_3$, with spatial cut off $g(x)$, the construction is a bit more involved (and ...

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The equivariant moment map has several applications. Its meaning is that it provides an encoding of how the Lie group $G$ acts on the phase space, and it gives you a way to find the observables corresponding to the conserved quantities/generators of the symmetry $G$: It also defines the process of symplectic reduction to a reduced phase space. Given that $$... 3 Essentially, I would like to prove$$ \sum_k f(k) \to \int f(k) \rho dE \tag{1}$$where$$ \rho = \frac{dk}{dE} \tag{2}$$is the density of states and k \to \infty. As mentioned in the comments, you need to introduce a measure on the LRS to get the dimensions to work out. To put it another way, your f(k) on the LHS can't be the same as your ... 3 The answer is no, I am afraid. As you may well know, the self-adjoint Laplace operator -\Delta on L^2(\mathbb{R}) has purely absolutely continuous spectrum \mathbb{R}^+. Now let V\in L^{\infty}(\mathbb{R},\mathbb{R}^+) be an arbitrary bounded positive function. Then -\Delta_x +V(x), where V acts as a multiplicative operator is self-adjoint and ... 2 Indeed using a rigged Hilbert Space is almost certainly what you should use and there are tutorials on it on arxiv (I'll add them when/if/I can find them again). However it isn't necessarily going to help you understand what an average physicist is doing on the one hand, and on the other hand there were already many things you were doing that not rigorous ... 2 Your final expression is correct except that V should be in front since it does not commute with Θ_0 in general. In potential scattering {Θ_0}V is often a compact operator and for large positive imaginary part of z=E+iℏε its norm becomes arbitrarily small so the series converges. In certain cases one can do things a little differently by ... 2 For an analytic function f, f(A) can be defined as the Maclaurin series, as described in the OP. However, when A is an element of a C*-algebra rather than just a Banach algebra, one can also consider continuous functions over the spectrum of A, which can still be defined as limits of certain polynomials of A (as a consequence of the ... 2 First, we will compute \text{div}\,F. The partial derivatives are given by$${{\partial F}\over{\partial x_i}} = q {\partial\over{\partial x_i}}\left({{x_i}\over{r^3}}\right) = q\left({1\over{r^3}} - {{3r^2 {{x_i}\over{r}} x_i}\over{r^6}}\right) = 0.Thus, $\text{div}\,F = 0$ away from the origin. Consider now a ball $B$ of radius $r$ centered at the ...

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are measure zero sets physically relevant (meaning that they are found as objects of interest among physicists or play an important part in some physical models), They come up naturally. For instance if you study the actual dynamics of putting an eigenstate of $\sigma_x$ into a z oriented Stern-Gerlach device you'll see the Schrödinger equation predicts ...

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A helpful yet elementary answer may do the trick, If you are familiar with the Euler-Lagrange equation then it will be straight forward and you can skip ahead a little. If not then you have to accept that there is an equation in physics that generalises classical mechanics called the Euler-Lagrange equation. For a particle moving in one dimension under a ...

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Technically speaking, manifolds are by definition topological spaces, which resemble locally an inner-product space. Since there are vector spaces (with a dot product) of infinite dimension, then there shall be infinately-dimensional manifolds as well. The infinity of the dimension is not a problem for the tensors as well - each multi-linear function over a ...

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I am not sure if you mean $x$ one or two dimensional variable. I don't have the article at hand but I guess we probably mean $D=1$?? The argument goes as follows: you have to use Bochner's theorem, which says that $W(x-y):=W_2(x,y)$ is positive distribution if and only if the Fourier transformation yields is a positive measure $\tilde W(p) d^Dp$. In this ...

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Ref.1 is already in eq. (3.1) considering a functional integral over the scalar field $\phi:M\to \mathbb{R}$. Here $M$ is spacetime. For a rigorous treatment of functional integrals, Ref. 1 points in the beginning of Section 3 to its Ref. [3.2]. In this answer we will just take an intuitive heuristic approach, and try to construct the functional integral as ...

1

In quantum field theory, one often considers meager subsets of the naively defined space of possible states. Indeed this is one of the Wightman Axioms: the Hilbert space of physical states is separable (has countable basis). This is a manifestation of the intuitive assertion "we must only put a finite number of particles into a finite quantisation volume". ...

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