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classes:2009:fall:phys4101.001:q_a_1026 [2009/10/26 16:17] – ludemanclasses:2009:fall:phys4101.001:q_a_1026 [2009/12/15 13:54] (current) – x500_choxx169
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 ===nikif002 10/16 2:50=== ===nikif002 10/16 2:50===
 Yeah, I wrote it all in LaTeX. Making pretty things is fun. Theorem 3.71 asks for an explicit time dependence, so I think <math><\delta p/\delta t>=0</math> (I fixed all the derivatives to be partial - sorry for the typo. All derivatives inside the expected value should be partial). That's why I say it's actually pointless. Yeah, I wrote it all in LaTeX. Making pretty things is fun. Theorem 3.71 asks for an explicit time dependence, so I think <math><\delta p/\delta t>=0</math> (I fixed all the derivatives to be partial - sorry for the typo. All derivatives inside the expected value should be partial). That's why I say it's actually pointless.
 +
 +===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}\ and\ \hat{p}</math>. Since <math>\hat{H}</math> and <math>\hat{p}</math> are all Hermitian operators, they commute with each other.
  
 ==== 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).
 +
 +=== 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.
 +
  
 ====Green Suit 10/26 4:11pm==== ====Green Suit 10/26 4:11pm====
 In the note at the bottom of page 104 Griffiths writes, "any complex number is an eigenvalue of the operator <math>\hat{p}</math>, but only real numbers are eigenvalues of the hermitian operator <math>\hat{p}</math>.  Can anyone paint a picture of what he means here. In the note at the bottom of page 104 Griffiths writes, "any complex number is an eigenvalue of the operator <math>\hat{p}</math>, but only real numbers are eigenvalues of the hermitian operator <math>\hat{p}</math>.  Can anyone paint a picture of what he means here.
 +
 +===Blackbox 10/26 7:00pm===
 +My understanding is the hermitian operators have a special condition, just like the definition of 3.16 on p97. In order to satisfy this condition, the eigenvalues of the momentum operator should be real numbers. If eigenvalues or the momentum operator are not real number, this definition is not satisfied. In other words, the imaginary part of an eigenvalue will be changed.
 +
 +=== 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.
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classes/2009/fall/phys4101.001/q_a_1026.1256591862.txt.gz · Last modified: 2009/10/26 16:17 by ludeman

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