Recent content by Mohandas

Nice answer Fredrik. Some additional points: While F=ma determines x(t) for all future time, the Schrödinger equation determines Psi(x,t) for all future time. So while Newtons F=ma returns x(t), the S.E returns a wave function which is spread out in space. At this point it is hard to see...

In which case the only state is the ground state, with 100% probability to find E=hw/2 at t=0 ?

But how can there be a general case for a? The time independent SE is only valid for a=0, so what are the other Psi_n(x) that you can tack exp(-i*E_n*t/h_bar) to? I'm not sure this makes any sense, but I am too tired to care now..

in b) Psi(x,t) for a=0 is just (eq. 4) * exp(-iEt/h_bar), where eq. 4 is the first eq on this page. ?

Actually in 1a) i assume that E = hw/2 in one of the steps to show that 'a' must be zero. But then they say 'find E for a=0'.. .. ? The way i see it, it's not a solution of the time ind. SE because it can't be solved for any other case than a=0, but again, I am assuming the E=hw/2 Still...

Yo dude. Did you find N? I just copied some crap from the book and got (pi*h_bar/mk)^1/4, assuming i could normalize the first eq. on this page for a=0. Kinda hard to think straight right now :))

\psi(x,0) = N exp[-\alpha(x-a)^2] is a solution to the time-independent SE at time t = 0 for the potential \ V(x) = (1/2)*m\omega^2x^2 where N is a constant and \alpha = m\omega/(2\hbar). I'm asked to show that the solution is valid only if a = 0. I'm a little at loss as to...