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--- classes:2009:fall:phys4101.001:q_a_1021 [2009/10/22 23:06] – x500_szutz003
+++ classes:2009:fall:phys4101.001:q_a_1021 [2009/10/23 10:34] (current) – jbarthel
@@ Line 16: / Line 16: @@
 === Zeno 10:45 10/21 ===
 I am! Just don't forget how complicated scattering states can be... and that the transmission coefficient isn't always |F/A|^2. Otherwise, I think this should be a very manageable test. If it's anything like the sample exam I'll be just fine ;)
+=== Dark Helmet 10/22 ===
+Depending on your degree of sarcasm, i am either just as excited as you or much less excited than you. :)
+===Captain America 10-23===
+I'm pumped!
 ====Hydra====
@@ Line 168: / Line 174: @@
 ===Spherical===
-Maybe, as Griffiths says, I shoulda been a mathematician...
+Maybe, as Griffiths says, I shoulda been a mathematician...  I realize he's not explicitly mentioning a delta function, it just sounds like what he refers to "could be" a delta function as well...  but ya.  It's just not. Is a delta function "square integrable"? -- Yes because it has an 'area' yes?
+===liux0756===
+The 'isolated points' the footnote related is talking about points with finite values. Of course delta function is not square integrable, because <math>\int_{-\infty}^{+\infty} \delta(x)^2 dx = \int_{-\infty}^{+\infty} \delta(x) \delta(x) dx =\delta(0) = + \infty </math> (equation [2.113])
 ====Daniel Faraday 10/22 10:30pm====
 So, for the finite square well problem in the sample quiz, we are asked to calculate the first two values of //l//. Is this doable without a computer? If so, how?
+===Green Suit 10/23 8:00===
+I think this is as simple as looking at the graph provided and looking for where the lines cross. For symmetric z= 2, 4 (approximately) with z=la then l= 2/a, 4/a. For asymmetric z= 3, 5 (approximately) then l= 3/a, 5/a. I think. It's then up to you to figure out if those numbers are sensible.
+====Sherical chicken====
+In the homework we just turned in, question 3.4a doesn't really make sense.  A priori it says "<math>\alpha</math> is imaginary, and ^Q is hermitian.  Under what conditions is alphaQ hermitian?" --
+The answer is "Alpha is real".  isn't this contradicting what we  were told to assume?
+===Daniel Faraday 10/23 730am===
+It doesn't say that <math>\alpha</math> is imaginary, it says that it is //complex//. So it can be represented by <math>\alpha = a + bi</math>. So you solve the problem to find that <math>b=0.</math>
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