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classes:2009:fall:phys4101.001:q_a_1026

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classes:2009:fall:phys4101.001:q_a_1026 [2009/10/28 20:22] – cwuclasses: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). 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.
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 ====Green Suit 10/26 4:11pm==== ====Green Suit 10/26 4:11pm====
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 === Mercury 10/26 11:00pm === === 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. 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.
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classes/2009/fall/phys4101.001/q_a_1026.1256779349.txt.gz · Last modified: 2009/10/28 20:22 by cwu

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