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--- classes:2009:fall:phys4101.001:lec_notes_1007 [2009/10/07 18:43] – kuehler
+++ classes:2009:fall:phys4101.001:lec_notes_1007 [2009/10/07 18:55] (current) – kuehler
@@ Line 43: / Line 43: @@
 with <math>k^2=-\frac{2mE}{\hbar^2}</math>
+as opposed to the situation of <math>k^2=\frac{2mE}{\hbar^2}</math>
+which is defined by the complex equation <math>\psi(x)=e^{(-ikx)}</math>.
+Normally, this would be the case for E>0 thus as k increases so does the energy. Yet, here we are dealing with E<0 so what ultimately happens is that as k gets larger positively the energy gets larger negatively.
 We end up getting:
 <math>{\small^+_-} e^{-2ak}=2bk-1</math>, where <math>b=\frac{\hbar^2}{2m\alpha}</math>
+Here a specifies how quickly the slope for the first solution decays, and b deals with the depth of the potential well.
 There may be 2 values for k, depending on the values of a and b.