Recent content by ognik

Thanks drvrm. Could you check that link please, I get not found?

Ah, assumptions :-(, the expression is only 0 at x=0, looks like a capital M ... For ## \delta = 1 ##, the ##max^2## is ## 2 \times 0.664 ##, but I would need to know this in terms of ## \delta ## to use FWHM. slightly off-topic to help me get a feel for these things, what would be an approx...

I want to find FWHM and thus the values of ##k_x## at 1/2 max of ## |psi|^2 ## - its the method this section of the text seems to be focused on. I don't expect it at 0, but using the Sinc function it workes out to 0 - which I'm sure is wrong, but can't see my mistake. I don't think I know that...

Frustratingly I can't find the source I found this in, but this was in a section on wave packets and the uncertainty principle - and my apologies for using \nu instead of v ... ##\Delta x_0 \Delta p_0 \ge \frac{\hbar}{2} ## ## \therefore \Delta v_0 \ge \frac{\hbar}{2 m \Delta x_0 } ## ##...

I was trying the expansion based on advice from a similar problem, that using sync wouldn't allow an analytic solution - see https://www.physicsforums.com/threads/fourier-analysis-of-wave-packet.855706/ ## \psi(x, 0) = -\int^{k_x - \frac{\delta}{2}}_{k_x - \delta} e^{i k_x x} dk_x +\int^{k_x...

Hi - I am effectively self-taught with this forum my only source of interaction and assistance, so would really, really appreciate any and all help making progress please ...

Homework Statement I don't understand how the following works:Homework Equations I follow ## \Delta \nu_0 = \frac{\hbar }{2 m \Delta x_0} ## easily, but then ## \Delta x = \frac{\hbar }{2 m \Delta x_0} t ## ;leaves me puzzled - I understand multiplying through by t, but how does ## \nu_0 t =...

If I plot ## x^4 ## I get an upside down parabola, centre 0, instead of being periodic as I would expect - so I suppose that would need a few additional terms in the expansion? May I assume my solution of ## \psi(x,0) = \frac{1}{x} e^{i x \bar{k_x}} [ 4 sin (\frac{x \delta}{2}) - 2 sin (x...

Sorry, I was cutting & pasting and didn't notice I had left the ##k_x, i ## in, had it right on paper ... So I think my solution is correct up to this: ## \psi(x,0) = \frac{1}{x} e^{i x \bar{k_x}} [ 4 sin (\frac{x \delta}{2}) - 2 sin (x \delta)] ##; - The exponent will vanish when I find ##...

Have been going round in circles with this. ## \bar{k_x} ## is the value of ## k_x ## where ## | \phi(k_x) |^2 ## is a maximum, and the average momentum ## p_x = \hbar \bar{k_x} ## But ## \Delta p_x = \hbar \Delta k_x ## is from ## 2 * (\bar{k_x} - k_x) ## where we find ## k_x ## at ##...

Sorry, I meant the ones I have marked as 'solved'

Hi - another feature I would find useful is to filter all my threads - leaving out the ones I have closed... is that possible, or worthwhile as a suggestion for the webmaster? Thanks Ognik

Homework Statement ## \phi(k_x) = \begin{cases}\phantom{-} \sqrt{2 \pi},\; \bar{k_x} - \frac{\delta}{2} \le k_x \le \bar{k_x} + \frac{\delta}{2} \\ - \sqrt{2 \pi},\; \bar{k_x} - \delta \le k_x \le \bar{k_x} - \frac{\delta}{2} \:AND \: \bar{k_x} + \frac{\delta}{2} \le k_x \le \bar{k_x} +...

Thanks, I think that 'feel' for the max will be useful, and I've learned a lot else through your help. I get 5.3 after fixing that denominator (should have been 6, I mustn't take short cuts in my algebra)

Thanks (divided one side by 2 then got distracted...). I quite often see Taylor truncated to 2 terms, is that a reasonable rule of thumb for OM? Or just a convenience because of the complexity the 3rd term usually adds? Also, If you wouldn't mind checking my algebra here...my final answer...