Recent content by redtree

The larger goal is to consider the symmetries of Lagrangian densities in both X and K. Certainly, that does include elements of QFT, but it also includes the symmetries of basis transformations in both X and K, which is why I raised the question in this forum. My goal is to see if formulations...

In general, statistical theories do not formulate observations or measurements as delta functions. Why such a formulation is necessary or even proper in quantum theory is unclear to me. First of all, the assumption that any particular observation (or eigenstate) should be represented by a delta...

From Auletta, Gennaro, Mauro Fortunato, and Giorgio Parisi. Quantum mechanics. Cambridge University Press, 2009. See section 2.2.5 Momentum representation, equation 2.148: $$\hat{p} \tilde{\psi}(p) = p \tilde{\psi}(p)$$ (The same formulation can be found in Fitzpatrick, Richard. Quantum...

Thank you for your response. Why do you assume that ##\textbf{p} \in K## is a point relative to the origin, let alone one represented by a delta function? I have made neither of those assumptions. A vector may or may not include the origin as one of its endpoints. If I wanted to represent a...

For ##\hbar =1##, the choice of K space or P space is a matter of convention. I also choose to use the Fourier transform convention ##e^{2 \pi i k x}## instead of ##e^{i p x}##. I tried to be explicit about all this. I apologize, but I don't know what you mean in saying my calculation is...

How is the mapping ##m## part of the Fourier conjugate relationship? It does not respect the Fourier conjugate relationship between the spaces X and K. This is trivial to demonstrate. Assuming ##\hbar =1 \rightarrow \textbf{p} = \textbf{k} = \{ k_0, \vec{k} \}##, where ##\vec{k} = \vec{p}##...

I am considering K space as the Fourier conjugate space of X space, not a completely independent space, and I am requiring that dynamics in K space respect that Fourier conjugate relationship between X and K spaces. My opinion is that this exercise is both trivial and useful, and I want to see...

What I really want is the Poincare-type group in K-space, which of course is trivial to construct from the Lorentz group in K-space. The article you quote is interesting but suffers from the following problem: Given a 4-momentum ##\textbf{p} \in K##: $$\textbf{p} = \{E_0,p_1,p_2,p_3 \} =...

My reference is: Schwichtenberg, Jakob. Physics from symmetry. Springer, 2018. In this text, matrices are utilized for the generators of the Poincare group and these matrices are 4x4 or 2x2. In any case, the generators, particularly the boosts, are related to complex rotations over an angle...

Can someone share a paper or chapter from a textbook if they know a good one? I'm curious to see the explicit form of these matrices. In position space, the generators of boosts act on the rapidity, which can be related to velocity in X. Assuming the generators of boosts in K act on rapidity in...

It’s not the convolution theorem in that only $$\hat{f}_2$$ is Fourier transformed. I was told by that $$\hat{f}_1$$ cannot be moved into the integral $$\int_{-\infty}^{+\infty} dx$$ and so the equation is not accurate. I disagreed and so posted the question. It seems an identity to me too.

I was wondering if the following is true and if not, why? $$ \begin{split} \hat{f}_1(\vec{k}) \hat{f}_2(\vec{k}) &= \hat{f}_1(\vec{k}) \int_{\mathbb{R}^n} f_2(\vec{x}) e^{-2 \pi i \vec{k} \cdot \vec{x}} d\vec{x} \\ &= \int_{\mathbb{R}^n} \hat{f}_1(\vec{k}) f_2(\vec{x})...

Got it; thanks!

Pin Groups are the double cover of the Orthogonal Group and Spin Groups are the double cover of the Special Orthogonal Group. Both sets of the double cover are considered to be groups, but it seems that only one of the sets of the double cover actually contains the identity element, which means...

First off, I am trying to write the following in Einstein notation, where ##R=\mathrm{diag}[1,-1,1]##, then ##R^T R = \mathbb{1}_{\dim{R}}##.