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classes:2009:fall:phys4101.001:q_a_1109 [2009/11/12 20:18] – myersclasses:2009:fall:phys4101.001:q_a_1109 [2009/11/30 09:13] (current) – x500_bast0052
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-===== Nov 09 (Mon)  =====+===== Nov 09 (Mon) Legendre closure, Radial WF and spherical Bessel =====
 **Return to Q&A main page: [[Q_A]]**\\ **Return to Q&A main page: [[Q_A]]**\\
 **Q&A for the previous lecture: [[Q_A_1106]]**\\ **Q&A for the previous lecture: [[Q_A_1106]]**\\
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 ===Pluto 4ever 11/9 8:33PM=== ===Pluto 4ever 11/9 8:33PM===
 I basically did the same thing by using integration by parts l times that way you can get the <math>(\frac{d}{dx})^l^'(x^2-1)^l^'</math> to reduce to (2l')!. From there I expanded out <math>(x^2-1)^l</math> and noticed that when I took the integral of the first term from -1 to 1 I got <math>\frac{2}{2l+1}</math>. From there I used some reasoning to get a final answer. I basically did the same thing by using integration by parts l times that way you can get the <math>(\frac{d}{dx})^l^'(x^2-1)^l^'</math> to reduce to (2l')!. From there I expanded out <math>(x^2-1)^l</math> and noticed that when I took the integral of the first term from -1 to 1 I got <math>\frac{2}{2l+1}</math>. From there I used some reasoning to get a final answer.
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 ====Esquire 11/12/09 AD (Information Age) 8:15pm==== ====Esquire 11/12/09 AD (Information Age) 8:15pm====
 Can a multidimensional hermitian operator be expressed as a non-square matrix? Such as a 3X27 matrix?  Can a multidimensional hermitian operator be expressed as a non-square matrix? Such as a 3X27 matrix? 
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 +===Spherical Harmonic  11/12/09===   
 +er.. chicken....
 +I believe that in fact the operator has to be square -- the rules of matrix multiplication say that a 1x4 * 4x4 give 1x4.  We need to preserve the space coordinates -- if it were a 4x3... we'd get a 1x3 and completely wipe out a space coordinate.... so unless you have an operator that shrinks space down a dimension... I do believe we need square operators of 1x4, 4x4 etc.  or... maybe there's a space creating operator... 4x27 ... STring theory'd be happy. 
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 +===Devilin===
 +I'm also pretty sure it needs to be square in order to take it's determinant. 
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 +===Esquire (Age of Aquarius) 11/12/09===
 +I believe that such operators need to be square so that eigenvalues and eigenvectors can be found. This is done via the determinent method, which only is doable with square matrices. 
  
  
classes/2009/fall/phys4101.001/q_a_1109.1258078700.txt.gz · Last modified: 2009/11/12 20:18 by myers

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