Recent content by souda64

Well, the phase factor ##e^{i\phi(x)}## is not a Schwartz function either. I think that the rest of this conversation (thanks to all contributors) resolves the "problem" with the relation ##f(x)\delta(x)=f(a)\delta(x)## which is valid for all distributions but not necessarily for all tempered...

Here is something I wanted to point out. Your suggestion that ##f(x)\delta(x-a)=f(a)\delta(x-a)## should imply that $$\frac{d}{dx}\left[f(x)\delta(x-a)\right]=f(a)\delta^\prime(x-a)$$ which is not true for any (even smooth) function ##f(x)##. A simple example is ##f(x)=\sqrt{x-a}+1##...

Thanks! That makes sense... However, if ##\delta(x)## is understood as a distribution, then the sifting property should work only when calculating an integral with the distribution, and this is a part of definition of ##\delta(x)## distribution. Otherwise, what if ##f(x)## in your example has a...

Really? I always thought it's true only if you integrate over ##x## (sifting property)... If that is always true, that resolves the problem with coordinate representation of the propagators. But I still feel a bit uncomfortable about the whole thing. Thanks though!

I guess, I'm taking back the last argument about ##(x+1)\delta(x)## simply because ##(x+1)\delta(x) = x\delta(x) + \delta(x) = \delta(x)## because ##x\delta(x) = 0##. So, the Dirac delta function actually absorbs any phase factor ##e^{i\phi(x)}## with ##\phi(0)=0##. Right? As to the coordinate...

Okay, I see now what you mean. You mean that as long as $$\int\limits_{-\infty}^{\infty} f(x) \psi(x-a)dx = f(a)$$ for any f(x), the distribution ##\psi(x-a)## is a delta function (distribution). So, you mean that a delta function multiplied by an arbitrary phase factor ##e^{i\phi(x)}## is still...

Thanks a lot for your input! I'm still confused though... What do you mean by "reexpressing ##f(x)\delta(x-a)##"? The product ##e^{i\phi(x,x^\prime)}\delta(x-x^\prime)## is still NOT a delta function, even if ##\phi(x,x)=0##. What am I missing?

I get the following phase factors (##e^{i\phi(x,x^\prime)}##) Free particle with mass ##m##: $$\phi(x,x^\prime) = \dfrac{m\left(x^2-x^{\prime 2}\right)}{2t}$$ Harmonic oscillator with frequency ##\omega##: $$\phi(x,x^\prime) = \omega\left(x^2-x^{\prime 2}\right)\cot\omega t$$

Here is a mystery I'm trying to understand. Let ##\hat{U}(t) = \exp[-i\hat{H}t]## is an evolution operator (propagator) in atomic units (\hbar=1). I think I'm not crazy assuming that ##\hat{U}(-t)\hat{U}(t)=\hat{I}## (unit operator). Then I would think that the following should hold \left\langle...