Sarahisme Posted March 21, 2006 Share Posted March 21, 2006 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 Link to comment Share on other sites More sharing options...

timo Posted March 21, 2006 Share Posted March 21, 2006 I fail to find a c_n in the text. But since I have a guess what it´s supposed to be: I think the poblem assumes that you know the (energy-) eigenstates of the particle (after all, they are given - you only have to identify them). Link to comment Share on other sites More sharing options...

Sarahisme Posted March 21, 2006 Author Share Posted March 21, 2006 hmm now i have now tried the other bits of the question, seems i can't quite get any of it Link to comment Share on other sites More sharing options...

Sarahisme Posted March 21, 2006 Author Share Posted March 21, 2006 I fail to find a c_n in the text. But since I have a guess what it´s supposed to be: I think the poblem assumes that you know the (energy-) eigenstates of the particle (after all, they are given - you only have to identify them). 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 Link to comment Share on other sites More sharing options...

timo Posted March 21, 2006 Share Posted March 21, 2006 Psi is a superposition of two eigenstates with diferent energies (different n in your case). So you cannot multiply the whole wavefunction by a common phase to get the time-developement. Link to comment Share on other sites More sharing options...

Sarahisme Posted March 21, 2006 Author Share Posted March 21, 2006 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?? Link to comment Share on other sites More sharing options...

timo Posted March 21, 2006 Share Posted March 21, 2006 I can´t see any problem sketching that function except for that your piece of paper lacks a dimension so you can´t plot a R->C (time -> value of wf) function on it. Either plot real and imaginary parts seperately or plot psi². I can´t see how the wavefunction should look the same for both times. At t=0, the exponential factor gives 1 for both, at the other time, the first eigenfunction will be multiplied by -1 and the second by 1 -> I doubt that this results in the same function. EDIT: ^^ And putting that together, the wavefunction will be real-valued for both times, anyways so there´s even less of a problem. Link to comment Share on other sites More sharing options...

Sarahisme Posted March 21, 2006 Author Share Posted March 21, 2006 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 )? Link to comment Share on other sites More sharing options...

Sarahisme Posted March 21, 2006 Author Share Posted March 21, 2006 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? Link to comment Share on other sites More sharing options...

Sarahisme Posted March 22, 2006 Author Share Posted March 22, 2006 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). Link to comment Share on other sites More sharing options...

Sarahisme Posted March 22, 2006 Author Share Posted March 22, 2006 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 Link to comment Share on other sites More sharing options...

Sarahisme Posted March 23, 2006 Author Share Posted March 23, 2006 nm, got it. thanks once again for all your help Atheist! -sarah Link to comment Share on other sites More sharing options...

Norman Albers Posted April 5, 2006 Share Posted April 5, 2006 so how do you think this would look for Psi(x' date='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] How do you get the squares up in the argument of the exponents (i-omega-t, per mode)? Link to comment Share on other sites More sharing options...

Norman Albers Posted April 6, 2006 Share Posted April 6, 2006 OK, I see now that the h-bar takes out one of the "2pi's" so I am seeing the construction. Link to comment Share on other sites More sharing options...

nicobudini Posted April 7, 2006 Share Posted April 7, 2006 Once someone said: "If you don't get confused with quantum mechanics... then you didn't understand it" Link to comment Share on other sites More sharing options...

Norman Albers Posted April 7, 2006 Share Posted April 7, 2006 At the age of 57, I can say that every few years or so I have again approached quantum mechanics for the n'th time. Each time I feel I am ready to see deeper and this is so. I have learned to be able to start reading things I cannot even understand, because I am well prepared to go further with the necessary work. I skip around, go back, repeat, etc. Now the value of 'n' is roughly 8, and I have done significant work in E&M field theory. I am really gonna get stuff together this time. Link to comment Share on other sites More sharing options...

## Recommended Posts

## Create an account or sign in to comment

You need to be a member in order to leave a comment

## Create an account

Sign up for a new account in our community. It's easy!

Register a new account## Sign in

Already have an account? Sign in here.

Sign In Now