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classes:2009:fall:phys4101.001:q_a_1007

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Oct 7 (Wed) delta-function potential well - reflection/transmission

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Green Suit

What makes the wide/deep well and the shallow/narrow well, as Griffiths puts it the “two limiting cases of special interest”?(pg 80) What about the wide/shallow well and the narrow/deep wall??

David Hilbert's hat - 10/6 3pm

The wide/deep well and the shallow/narrow well correspond with the endpoints of the graph on page 80 - the two cases you're asking about are sort of like “mixed” cases: it seems like if a potential was wide it would also be deep (for a generally “large” potential), or if it was shallow it would also be narrow (small). This is described by what Griffiths calls z0, which has terms for both a and square root of V0. It seems like as z0 and tan(z) intersect, only the product of V0 and a determine specific points, not one or the other. So in the limit of high or low z0, the well must be both wide and deep, or shallow and narrow, or a z0 with either a or V0 dominating the other.

Put another way, if you want to look at a well that is both wide and shallow, you have to decide - is it primarily wide, or primarily shallow? Which term dominates z0, the width or the depth? The question of a wide/shallow well is kind of ambiguous in that sense: in general, depending on which term is more important, the question can have more than one answer.

Ekrpat - 12:54 - 1172

In discussion, Justin mentioned that the first excited state of the double delta potential we worked on only sometimes exists. I believe he said it was dependent on the spacing <math>a</math> (and <math>\alpha</math>?). How can you determine whether this second solution exists for a particular <math>a</math> and <math>\alpha</math>?

liux0756 17:31 10/06/2009

If you solve the wave function and apply the boundary conditions, the following relationship can be achieved:<math>\frac{\hbar^2k}{m\alpha}-1=\pm exp(-2ak)</math>. Then sketch the left-hand-side and right-hand-side respectively to solve k. Note k must be positive. When <math>\frac{\hbar^2}{m\alpha}<2a</math> (line a), there are two possible values for k, and in this condition the second solution exists. When <math>\frac{\hbar^2}{m\alpha}\ge2a</math> (line b or c), there is only one possible value for k, in this condition no excited state exists.


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classes/2009/fall/phys4101.001/q_a_1007.1254868721.txt.gz · Last modified: 2009/10/06 17:38 by x500_liux0756