Timeline for self-dual integral transform
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 8, 2021 at 4:11 | comment | added | Myridium | A self-dual integral transform is analogous (equivalent, even) to a unitary operator on a finite-dimensional space. These are 'coordinate transformations' in the sense that one basis of orthonormal functions is transformed to another. Therefore any transformation between function bases (e.g. wavelets are a basis, Fourier functions are a basis, dirac-delta distributions are a basis, probably many others) will be unitary. Measures between functions (which I think are analogous to symmetric postiive definite matrices) will have their own sets of self-dual operators I think. | |
| Feb 2, 2019 at 9:01 | comment | added | Kphysics | Clever answer. The Fourier transform itself is used to obtain the Schroedinger propagator, by starting with $ U(t) = e^{-itk^2/2}$, then the Fourier gives the propagator $K = \frac{1}{{2\pi i t} e^{ix^2/t}$. And yes, $U(-t) * U(t) = 1$, but the Fourier transform itself not an example of this phenomenon as far as i can tell. | |
| Jan 28, 2019 at 21:25 | comment | added | Willie Wong | You get for free a lot of examples from (time reversible) PDEs, especially since you allow a "sign flip". For example, consider the free propagator for the linear Schrodinger equation $U(t) = e^{it \Delta}$ which can be expressed as an integral operator. You have that the backwards propagator is $U(-t) = \overline{U(t)}$. Similar things happen for the wave and Klein-Gordon equations. | |
| Jan 28, 2019 at 17:14 | history | edited | Kphysics | CC BY-SA 4.0 |
deleted 2 characters in body
|
| Jan 28, 2019 at 15:50 | review | First posts | |||
| Jan 28, 2019 at 15:52 | |||||
| Jan 28, 2019 at 15:45 | history | asked | Kphysics | CC BY-SA 4.0 |