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| classes:2009:fall:phys4101.001:q_a_1026 [2009/10/28 20:18] – cwu | classes:2009:fall:phys4101.001:q_a_1026 [2009/12/15 13:54] (current) – x500_choxx169 |
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| ===Hardy 10/28 20:20=== | ===Hardy 10/28 20:20=== |
| My understanding of why the <math><dp/dt></math> is not equal to <math><\partial p/\partial t></math> is that these two things has totally different meanings mathematics. <math><dp/dt></math> is the time deriavate on the operator <math>\hat{P}<\math>, in the other words a function of <math>\hat{P}<\math>. But <math><\partial p/\partial t></math> is explained as the product of two operators <math>\hat{H}\hat{p}</math>. Since <math>\hat{H}</math> and <math>\hat{p}</math> are all Hermitian operators, they commute with each other. | My understanding of why the <math><dp/dt></math> is not equal to <math><\partial p/\partial t></math> is that these two things has totally different meanings mathematics. <math><dp/dt></math> is the time deriavate on the operator <math>\hat{P}</math> , in the other words a function of <math>\hat{P}</math>. But <math><\partial p/\partial t></math> is explained as the product of two operators <math>\hat{H}\ and\ \hat{p}</math>. Since <math>\hat{H}</math> and <math>\hat{p}</math> are all Hermitian operators, they commute with each other. |
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| ==== prest121 10/25 3:30 pm ==== | ==== prest121 10/25 3:30 pm ==== |
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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). |
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| | === 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==== |
| === 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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