Recent content by joycey

I have been searching for a description of how these diodes operate. I am reasonably familiar with solid state physics and the physics of semiconductors and think I have worked out how these should work (being two back-to-back Schottky diodes) based on the Fermi levels and band-bending, but I'm...

By that definition (very different than the one I have learned) these operators do commute. Something I am quite certain they should not. Your definition leads to, \begin{array}{rl} [\hat{U_t}(a),\hat{U_p}]f(x)&=\hat{U_t}(a)\hat{U_p}f(x)-\hat{U_p}\hat{U_t}(a)f(x)\\...

lalbatros, To manipulate it in such a manner, you assume the particle is at x, not x+a. The parity operator should effectively reflect around the axis, meaning something that is at x+a must be x+a to the left, or, at -x-a.

lalbatros, I think you made a mistake in going from g(-x') to f(-x-a). If we define x' \equiv x-a, as you have done to go from f(x-a) \Rightarrow g(x'), then shouldn't we find that -x'=-(x-a)=-x+a?

I just want to ensure that I understand you correctly. Especially since I wrote out the statements in two different orders. To recap how I interpret things, f(x)=<x|f> \begin{array}{rl} \hat{A}f(x)&=<x|\hat{A}|f>\\ &=\int dx' <x|\hat{A}|x'><x|f> \end{array} Now, if I let...

In fact, \hat{U_t}(a)|x> = |x+a> \hat{U_t}(a) represents the translation operator, translating the current x-coordinate. <x|\hat{U_t}(a) = <x|\hat{U_t}^\dagger(-a) = <x-a| hence, my statement that \hat{U_t}(a)f(x) = <x|\hat{U_t}(a)|f> = <x-a|f> = f(x-a) So, if the current x...

Hey Marlon, Thanks for the answer, but that's not what I was trying to get at. I realize that either way I evaluate these the commutation relation will be non-zero, I just get a different relation depending on the method. Which is what I'm wondering here. \begin{array}{rl}...

I'm having difficulty trying to figure out which of the following is the correct method to properly evaluate the effect of the operators on f(x). Given that, \hat{A}f(x)=<x|\hat{A}|f> If the polarity operator, \hat{U_p}, and the translation operator, \hat{U_t}(a), act as...