Which value of c would you choose?

Sunday, November 30, 2008

For Tuesday: more about the periodic table, states in solids, U and the Schro eqn.

So far we have been mostly considering an idealized periodic table, but actually, as some people pointed out, the d-levels don't even start in the correct row, etc. We can discuss this a little more on Tuesday. Perhaps the key things that survive from the idealized picture are the s-p degeneracies (which are v. important!) and the fact that there are always 6 p states, 10 d states, etc. in a group. Anybody have an idea as to why the d-states are shifted out of their expected row? (comments are welcome)

Here is a problem that might be good preparation for understanding electron states in crystals:
Consider a potential that consists of 4 equal-strength and equally-spaced attractive delta functions all in a line (in 1 dimension). Suppose they are relatively far apart*. How many different (orthogonal) low-energy (one-electron) states are there and what do they look like?
(You could warm up with the 2 delta function problem, and then maybe try 3 or 6 delta functions. 3 is a little tricky. *can you define "relatively far apart"?

Is there anything else we should be working on or thinking about for this last week? [After this week it will be just you and the Schrodinger equation* traveling through the world together for the rest of your life. I hope you will be ready for that. (*and a normalization condition)]

4 comments:

Zack Schlesinger said...

What would be a way to construct a wave-function that might be appropriate for an infinite chain of identical attractive potentials (all equally spaced along a line in 1 dimension). [they could be delta functions, for example, or finite sq wells... suppose you start with the gs for one of them...)

Derek said...

for infinite attractive delta function potentials the wave function should just become an infinite sum of gs for a single delta function. i.e. sum over n exp(|x+nd|/a) where d is the spacing. seems like this is the same way we looked at the two delta functions.

Derek said...

should be exp[-|x-nd|/a]

Zack Schlesinger said...

Can you come up with another one?