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--- classes:2009:fall:phys4101.001:q_a_1026 [2009/12/15 13:52] – x500_choxx169
+++ classes:2009:fall:phys4101.001:q_a_1026 [2009/12/15 13:54] (current) – x500_choxx169
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 Correct me if I'm wrong, but I think the Dirac-delta function comes from the fact that the integral is zero unless p=p'. The way I think about it is to rewrite the exponential in terms of cosine and i*sine--the integral of these two from -∞ to +∞ is zero, so unless p=p' (making the integrand 1), the integral must be zero. This is what Griffiths calls "Dirac orthonormality"--it's not true orthonormality because p can also have complex values, which is not taken into consideration in this calculation (and true orthonormality has a Kronecker delta, not a Dirac delta).
+=== Aspirin ===
+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.