Sarahisme

hmm ok maybe i have something here... can you just use the formula: [math] KE \geq \frac{\hbar^2}{8(\Delta x)^2 m} [/math] as for part © i guess that heating up say a sodium salt which glows yellow when put in a bunsen flame is an example of 'easy' excitment of an electron to the next excited state.?

Hey all, i can do part (a) (or at least i think i can, i get: 9.4eV but for part (b), i can't find a formula to use. any ideas guys/gals? and for part © i caluclated that you would need to give ~28eV worth of energy to the an elecron to excite it to the first excited state. i am unsure whether this is said to be 'easy'.... do you think 'easy' involves cmplicated equipment or me poking it with my finger? sorry bout the 1000 questions Sarah

phew, ok. so are their any pairs in these sets (other than cos and sin if evaluated between 0 and 2pi) which are orthogonal, because i don't think there is, i just want to make sure i am not doing somethign wrong here

ok, nevermind, think i got it! yay! although i am a little unsure what the 'uncertainty' in the energy refers to? you work it out by taking sqrt(<E^2>-<E>^2) is it the uncertainty in your measurement? or does it affect your measurement or what? i am little confused

ok i have been told that they none are othorgoanl sets. but i know that (-x+1) & (x^2 - 4x +2)/2 is orthogonal. but i am can't seem to show it.... i just get inifinity terms...

ok i am pretty sure that i've got parts (a) and © all i am having trouble now is with part (b) (when will it return to orginal state) and (d).

ok here is what i got part ©: <x> = [math] \frac{a}{2} - \frac{32a}{9 \pi^2 \sqrt(10)}cos(\frac{3 \pi^2 \hbar t}{2ma^2}) [/math] and so angular frequency is: [math] \omega = \frac{3 \pi^2 \hbar}{2ma^2} [/math] and the amplitude of oscillation is : [math] \frac{a}{2} - \frac{32a}{9 \pi^2 \sqrt(10)} [/math] is that anywhere near the right answer?

oh ok, i thought when it said sketch the wavefunction, it meant sketch the spatial part , because i thought you couldnt really sketch the time-dependent part because of the imaginary stuff. also for part (b), do you think it means "returns to its original form" means Psi(x,0) ? if so, doesnt that never happen because the time-dependent part needs to be one, and this only happens once (at t = 0 )?

k yeah that makes sense. so how do you think this would look for Psi(x,t): [math] \Psi(x,t) = (\sqrt(\frac{2}{5a})sin(\frac{\pi x}{a}))e^{-i( \pi^2 \hbar/(2ma^2))t} + (\sqrt(\frac{8}{5a})sin(\frac{2 \pi x}{a}))e^{-i(4 \pi^2 \hbar/(2ma^2))t} [/math] but i still have the problem of when i go to sketch at the two specfied times..... i get the feeling this is still wrong??

so is all i have to do is say that: [math] \Psi(x,t) = (\sqrt(\frac{2}{5a})sin(\frac{\pi x}{a}) + \sqrt(\frac{8}{5a})sin(\frac{2 \pi x}{a})) e^{-i(n^2 \pi^2 \hbar/(2ma^2))t} [/math] but then, wouldnt the sketches be the same? :S i am having real trouble with this problem, lol

hmm now i have now tried the other bits of the question, seems i can't quite get any of it

ok, thanks for that Tom!

Hey, i am little puzzled by this question.... i get that none of the sets (a,b or c) are orthogonal when taken over -infinity to infinity. am i doing something wrong, i would have thought at least one of them would be orthogonal...??? this is the problem in question: the inner product i think i am meant to be using is: <f|g> = ?f*g where the thing on the left (f*) is complex conjugate of f. the combinations i tried were: (a) cos(x) & 3sin(x) (b) 1 & (-x+1) © x * 4x^{3} i would have thought that cos(x) & 3sin(x) would be orthogonal...?? or is it that i am not meant to be integrating from -infinty to infinity? (that is should i just be going from 0 to 2pi?) -a puzzled Sarah :S

Hey all, i am a little confused with this problem.. well i have only tried part (a) so far, so that is what my current question is concerning. my problem is that when i work out c_n i get that it is some multiple of sin(n*pi) which is obviously zero for all n. in which case i cant work out Psi(x,t)?! i have put the integrals into maple and everything but it gives me the same thing... 0 any ideas anyone? -sarah

i guessed the solution [math] g(t) = Ae^{t} [/math] and after transforming back i got the GENERAL solution to be somethign which has TWO arbitrary constants: [math] y(t) = \frac{1}{\frac{-1}{2A}e^{-t} + Ce^{-t(1 + 2A)}} + Ae^{t} [/math] where C is constant of integration. and A is any real number. what do you think?

lol, sorry i got that 'show' part out now. i put a plus instead of a minus at one point and everything went crazy, but its all good now!

Hey everyone, i am having a bit of trouble trying to do this question. how do i show that they are linearly dependent ? i tried using the wronskian but this doesnt work does it? (because just because the wroskian is 0 on the interval it doesnt neccessarily mean that the two functions are dependent, right?) this is what i have done so far: (a) for 0<t<1 |t| = t so f(t) = t^2.t = t^3 = g(t) therefore f(t) and g(t) are linearly dependent on 0<t<1 for -1<t<0 |t| = -t so f(t) = t^2.-t = -t^3 = -g(t) therefore f(t) and g(t) are linearly dependent on -1<t<0 for -1<t<1 because f(t) is a different multiple of g(t) on 0<t<1 than on for -1<t<0, therefore f(t) & g(t) are linearly independent on for -1<t<1 also for part (b) i get W(f,g) = {t^6}/|t| which is obviously not zero for all -1<t<1 ?? :S ========================= does that makes sense to you guys?? because i am not to sure it makes sense to me! any help would be lovely -Sarah

hey all, just having a bit of trouble with this problem... i have done a little bit on it, just thought i'd let you guys look at it first, because i havent been able to type up my solution yet. Cheers guys & gals Sarah

hmm ok i think i begining to understand all this a bit more. So for [math] \Psi_I [/math] , it is a solution if k=w=0 or if A = 0. both are degenerate not physical cases. just to confirm... there is only those 2 cases for [math] \Psi_I [/math] ? so i did a similar thing for [math] \Psi_{II} [/math]: i got, that for [math] \Psi_{II} [/math] to satify the TDSE you need: [math] \frac{\hbar^2 k^2}{2m} Asin(kx - \omega t) = -i \hbar \omega Acos(kx - \omega t) [/math] for x = 0, t = 0, you get that w = 0 and for (kx - wt) = pi/2 you get, k = 0 so the only 2 possible cases for [math] \Psi_{II} [/math] to be a solution to teh TDSE is when either: A = 0 or when k=w=0 what do you guys think? i didnt really understand the eigenstates stuff, so yeah. do you think what i have is an acceptable answer? because it seems a bit of a degenerat kind answer...???

ok thanks for that, i'm off to bed, but i'll have a look in the morning! thanks you very much once again Severian

plz delete this post, accidently posted same thing twice!

for kx-wt = pi/2 ----> itwA = 0 for kx-wt = pi/4 ----> [math] \frac{\hbar^2 k^2}{2m} A = it \omega A [/math] k, sorry, not to sound too stupid, but i am still stuck. :S

ok, so i do: [math] \frac{\hbar^2 k^2}{2m}cos(kx - \omega t) = it \omega sin(kx - \omega t) [/math] then pick x=0 and t=0, and so get: [math] \frac{\hbar^2 k^2}{2m}cos(0) = it \omega sin(0) [/math] which leads me too: [math] \frac{\hbar^2 k^2}{2m} = 0 [/math] but how does that show anything? i think i am little lost