Recent content by Davio

Hiya, thank you for your help. I managed to meet up with some friends and we managed to do it. Your help was invaluable. Thanks :-D

Yay, thank you, I got it, me and my friend are very grateful!

Taking the solution with no time dependance: Cexp ikx + D exp -ikx and then I substite into the equation: I will do left hand side then right hand side. -(hbared)/2m d^2 phi over dx^2 = ( -k^2 Cexp ikx + k^2 Dexp - ikx) . -(hbared)/2m = (right hand side) E. phi E= left hand side divided by...

\frac{1}{1 + Exp (-(E-U))/KT} \frac{1}{KT} . exp (-(e-u)/KT ) I have edited the previous post to reflect what I think is the correct derivation. The factor which I tried is \1 + Exp (-(E-U))/KT however what I don't understand is, surely multiplying by 1, will just result in the same thing? I...

Hi part A just says E smaller than 0.. oh ohh I've done that part already, 3 things: must be continuous and single valued function have continuous first derivative unless potential goes to infinity. and have a finite normalization integral. Do i just take away the t's for the time...

Posting on behalf of Zophixan: <N> = 1 / exp (E-U)/KT +1 from Zg = 1 + Exp (-(E-U))/KT where <N> = kt d (ln Zg) / dU He left out some brackets, hence why it makes no sense. Unfortunately i can't do the question either. however here is my working for you to ponder. Bear with me, this is my...

Just to confirm, is E negative because dirac is negative? I solved equation to get phi = Cexp i (Kx- (ET/ hbar))+ Dexp i (Kx- (ET/ hbar)) how do i know how to get it for x>0 and x<0? I would have thought them to be the same but I'm worried the question asks for solutionS, and not solution.

"The Dirac delta can be loosely thought of as a function on the real line which is zero everywhere except at the origin, where it is infinite" WOuld it be 0? and if that is the case, I Would use just a normal schoinger equation without the V or rather the V set as 0 and solve as normal?

hmmm, the dirac potential i mentioned. Is it an infinite well? or are you just explaining why I'm wrong :-s

hmm, the only thing i can think of which has a "discontinuety" is the infinite well, which allows discontinueties due to the infinite potential. I thought wavefunctions had to by definition be continuous?

Sigh, a mod has deleted my mates account, which was a valid account made from my computer hence the IP address. Regardless, I'll post on his behalf now anyway. He says: Sorry, I'm really tired today , I will check my working, I really appreciate your help. I say: I'm going to have a crack at...

Ah I see. Um.. this opens a new kettle of fish now... how would I find a general solution to the Schrödinger equation for that?! I would use the right hand side picture from before? How does this new information alter my general method of solving Schrödinger equations? Thanks.

It's now considered a past paper. Quite a few people are doing a retest, we are betting that whatever paper he will give us, is going to be VERY similar to this paper. My tactic is to do this entire paper and then the rest of the past papers and see what happens. heh, take home examinations...

Hiya, its just a beginners course to quantum physics. We were expecting an OK paper, but put it this way, my mates who are graduates think the paper is ridicolous! The tutors petitioned against this paper apparently. Interestingly the dirac function was bought up, apparently it was litually on...

ps I#m posting on behalf of Zophixan as I discovered I know him! haha