Differences

This shows you the differences between two versions of the page.

--- classes:2009:fall:phys4101.001:q_a_1026 [2009/12/15 13:40] – x500_choxx169
+++ classes:2009:fall:phys4101.001:q_a_1026 [2009/12/15 13:54] (current) – x500_choxx169
@@ Line 54: / Line 54: @@
 === Aspirin ===
-I think it is helpful when you see page 70. <math> f(x)\delta(x-a) = f(a)\detla(x-a), (because the product is zero anyway except at the point a.) Furthermore, through Eq. 2.113, you can get Eq. 3.35.
+It is helpful when you see page 70.<math>f(x)\delta(x-a)=f(a)\delta(x-a)</math>,(because the product is zero anyway except at the point a.)  Furthermore, using Eq.2.113 <math>\int f(x)\delta(x-a)dx=f(a)\int\delta(x-a)dx=f(a)</math>,you can get Eq. 3.35.
 ====Green Suit 10/26 4:11pm====
@@ Line 64: / Line 65: @@
 === Mercury 10/26 11:00pm ===
 For an operator to be Hermitian, all the eigenvalues must be real (and conversely, if all the eigenvalues are real, the operator is Hermitian, or at least I think that's what Yuichi said in class). Therefore, if we consider <math>\hat{p}<math> a Hermitian operator, we can only consider its real eigenvalues. <math>\hat{p}<math> also has complex eigenvalues, but with these values the eigenfunctions are non-normalizable and blow up at ±∞ (i.e., they do not exist in Hilbert space), meaning that the operator cannot be Hermitian.
 --------------------------------------