Recent content by Dyad

Yes, it must have to do with the allowed functions. If you choose an energy that is not an eigenvalue, you're wavefunction won't satisfy the boundary conditions--it'll blow up. Indeed, when you start doing problems, you'll see that the explicit calculation of the eigenvalues is done by...

I don't quite understand the rest of the question. Perhaps you can show us Pauling's text/picture. Let's still have a crack at it. let's say c > a. Then we know from the time-ind. Schrodinger equation (TISE) that the wavefunction is concave up and will stay that way for any c > a...

The first part (assumption the the wavefunction can be decomposed into X(x)*T(x)) is a standard trick for solving partial differential equations. The resulting time-ind. Schrodinger equation is an eigenvalue problem. By solving it, we get all the different eigenfunctions which we assume form a...

Definitely. It's been demonstrated by an MIT group last year: http://web.mit.edu/newsoffice/2007/wireless-0607.html

Ok, but just be careful; this doesn't always work. Check the Fourier transform of the unit step function; this method doesn't give you the total answer. Problems arise at f = 0. (To do this problem, might try writing the step function in terms of the sgn function.)

What if you had x(t) = \int_{-\infty}^{\infty} je^{j2\pi f t}e^{-b|f|}df with b real and greater than zero? This is a doable calculation. Then, after you're done, what happens when you take a certain limit involving a?

You have a constant times the sign (not sine) function: sgn(f) = 1 for f >0, and -1 for f<0. Your problem, as you point out is convergence. Why not try multiplying your function by decaying exponentials, say exp(af) or exp(-af)?