Recent content by frerk

So I have to think about an standing wave which occurs between mirrors 1, 2, 3 and 4? (Because the entire cavity is made of all the mirrors). And the birefringent plate "choses" the waves, which can build up such a standing wave. (Results out of destructive and constructive interference)...

Hello :) I have 2 questions about the beam propagation of an Ti:sapphire-laser: Picture 1: The cw argon ion laser brings the beam to mirror (1). Then...? Then the beam (100% of it), go to the birefringent plate (4), gets then reflectet to the first mirror again (1) , which reflects it to the...

That is almost exactly how this task is written. Yes, right, that ist just the order. I try to explain it with a similar easy example: the energy E for an atom is E = \hbar \omega . We know that \hbar = 10^-34 and E is about 1eV = 10^-19 J. So a typical value for omega is 10^-15. And at the...

hello :) 1. Homework Statement I have to show typical values for length (amplitude) x and energy E of a quantum linear harmonic oscillator Homework Equations maybe this one: E = (\frac{1}{2} + n) \hbar\omega with n = 0, 1, 2, ... But here are 2 unknown variables: E and omega. \hbar =...

hey. thank you for your answer. Yes, right. that brings to to the result I want. Is there a rule, why I have to multiply the result of the dot product with the idendity matrix? Because the other terms include a Pauli Matrix and the result of the dot produkt must adapt to that structure?

Homework Statement Hey :-) I just need some help for a short calculation. I have to show, that (\sigma \cdot a)(\sigma \cdot b) = (a \cdot b) + i \sigma \cdot (a \times b) The Attempt at a Solution I am quiet sure, that my mistake is on the right side, so I will show you my...

Yes, I have also googled a lot, and they almost use the ladder operators... Sure :smile: First half makes very much sense for me, and I have to read more about the second half :smile: Yes, for me too. You helped me a lot :smile: greetings from germany :smile:

Thank you for your recognition :redface: I see that. Just the "minus" is wrong I think. \rightarrow \bar{\zeta} = \int_{-\infty}^{+\infty} (\frac{m\omega}{\pi\hbar})^{1/4} \sqrt{2} \zeta e^{-0,5\zeta^2} e^{+iE_1t/\hbar} \cdot \zeta \cdot (\frac{m\omega}{\pi\hbar})^{1/4} e^{-0,5\zeta^2}...

That is the same, I wanted to say with that, just graphical: I make an example. I calculate the expectation value for the position for ## \psi_1 ## which is: \psi_1(\zeta) = (\frac{m\omega}{\pi\hbar})^{1/4} \frac{1}{\sqrt{2^11!}} H_1(\zeta) e^{-0,5\zeta^2} =...

In my first part, I just checked, whether \bar{x} = 0 for steady states. And yes, \psi(x,t) = \phi(x)(Et/\hbar) is not a solution of time-dependent SE. And also yes, I wanted to write \psi(x,t) and not \psi(x,r) , but the second time I wrote it I just copied it, so the mistake happened...

Sorry for not answering, I had to make a break with that... ok. here is my steady state: \psi(x,r) = \phi(x)e^{-iEt/\hbar} \bar{x} = \int_{-\infty}^{+\infty} dx \phi^*(x)e^{iEt/\hbar} x \phi(x)e^{-iEt/\hbar} = = \int_{-\infty}^{+\infty} |\phi(x)|^2 x And this has to be zero. Because an...

I already checked it: \frac{d}{dt} \bar{p} = -\langle \frac{dU}{dx} \rangle\quad\quad\quad U = \frac{1}{2}m\omega^2x^2 I recognize nothing. you mean, when I compare it with: \frac{d}{dt} \bar{x} = \frac{1}{m} \bar{p} = \frac{\hbar}{im} \frac{\bar{d}}{dx} ? I don`t know...

oh. no. \frac{d}{dt} \bar{x} = \frac{1}{m} \bar{p} = \frac{\hbar}{im} \frac{\bar{d}}{dx} So I write it explicit: \frac{d}{dt} \int_a^b \psi^*(x,t) x \psi(x,t) dx = \frac{\hbar}{im} \int_a^b dx \psi^*(x,t) \frac{d}{dx} \psi(x,t) And now, go on...

hello :-) here is my problem...: 1. Homework Statement For a linear harmonic oscillator, \hat{H} = \frac{\hat{p}^2}{2m} + \frac{1}{2} m \omega^2x^2 a) show that the expectation values for position, \bar{x}, and momentum \bar{p} oscillate around zero with angular frequency \omega. Hint...

thank you both :-)