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 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).

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

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

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.

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??

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.

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 )?

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?

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, 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

nm, got it.

 

thanks once again for all your help Atheist!

 

-sarah

 

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)?

OK, I see now that the h-bar takes out one of the "2pi's" so I am seeing the construction.

Once someone said: "If you don't get confused with quantum mechanics... then you didn't understand it"

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.