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classes:2009:fall:phys4101.001:q_a_0923 [2009/09/23 07:30] – x500_razi0001classes:2009:fall:phys4101.001:q_a_0923 [2009/09/26 23:37] (current) – yk
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 ====Esquire 9/22 12:54==== ====Esquire 9/22 12:54====
 Near the beginning of Section 2.3.2 of the book ("Analytic Method") I am not following the jump between equation 2.76 and 2.77. How does one "'peel off'" the <math>h(\xi)</math> factor? Near the beginning of Section 2.3.2 of the book ("Analytic Method") I am not following the jump between equation 2.76 and 2.77. How does one "'peel off'" the <math>h(\xi)</math> factor?
 +
 +===Captain America 9/23 10:39===
 +I believe what he is trying to do here is say that the <math>\psi(\xi)</math> is equal to something "( )" times the exponent.  Then he replaces the "( )" with <math>h(\xi)</math> and wants to solve for the <math>h(\xi)</math>, which he states is easier than solving for <math>\psi(\xi)</math> itself.
  
 ====liux0756 9/22 13:18==== ====liux0756 9/22 13:18====
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 <math>G(x_0)>0</math>, or else <math>G(x)\equiv0</math>, can't be normalized. Then <math>G(x_1)+G(x_2)\ge2G(x_0)>0</math>. <math>G(x_0)>0</math>, or else <math>G(x)\equiv0</math>, can't be normalized. Then <math>G(x_1)+G(x_2)\ge2G(x_0)>0</math>.
 Now let <math>x_1\rightarrow-\infty</math>, <math>x_2\rightarrow+\infty</math>, in order to satisfy the above eqation <math>G(x_1\rightarrow-\infty)</math>,<math>G(x_2\rightarrow+\infty)</math> cannot be 0 at the same time, so the integral <math>\int |\psi(x)|^2\,dx=\int G(x)\,dx</math> will go to infinity, cannot be normalized. Now let <math>x_1\rightarrow-\infty</math>, <math>x_2\rightarrow+\infty</math>, in order to satisfy the above eqation <math>G(x_1\rightarrow-\infty)</math>,<math>G(x_2\rightarrow+\infty)</math> cannot be 0 at the same time, so the integral <math>\int |\psi(x)|^2\,dx=\int G(x)\,dx</math> will go to infinity, cannot be normalized.
-====Andromeda====+====Andromeda 16:50 9/22====
 is there any relation between Hermite polynomial and Legendre polynomial???  is there any relation between Hermite polynomial and Legendre polynomial??? 
  
-==Schrodinger's Dog==+===Schrodinger's Dog 21:11 9/22===
 No, although they both are recursive relations of sorts, they aren't related in any way. But, Hermite Polynomials  are special cases of Laguerre polynomials, if your interested in looking into that.  No, although they both are recursive relations of sorts, they aren't related in any way. But, Hermite Polynomials  are special cases of Laguerre polynomials, if your interested in looking into that. 
 +
 ====Hardy 9/22 19:02==== ====Hardy 9/22 19:02====
 I do not quite understand why the <math>a_-\psi_0(x)</math> should be zero. Can it be some value between zero and <math>\frac{1}{2}\hbar\omega</math>? I do not quite understand why the <math>a_-\psi_0(x)</math> should be zero. Can it be some value between zero and <math>\frac{1}{2}\hbar\omega</math>?
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 === Daniel Faraday 7:20am 9/23 === === Daniel Faraday 7:20am 9/23 ===
-I don’t think about it in terms of energy. Thinking about it that way led me to the same question you asked. Here's how I think about now, after asking the Prof about it:  We know that for any wavefunction there is some ground state, <math>\psi_0</math>, which is a stationary state; it’s the bottom rung on the ladder. Anywhere below this ground state the wavefunction cannot exist at all. So, if we step down one step from <math>\psi_0</math>, we’ll get zero (no wavefunction). That is, <math>a_-\psi_0(x) = 0</math>. Then we solve and magically get a value for <math>\psi_0</math> for any wavefunction in a harmonic oscillator, which is cool.+I don’t think about it in terms of energy. Thinking about it that way led me to the same question you asked. Here's how I think about now, after asking the Prof about it:  We know that for any wavefunction there is some ground state, <math>\psi_0</math>, which is a stationary state; it’s the bottom rung on the ladder. Anywhere below this ground state the wavefunction cannot exist at all. So, if we step down one step from <math>\psi_0</math>, we’ll get zero (no wavefunction). That is, <math>a_-\psi_0(x) = 0</math>. Then we solve and magically get a value for <math>\psi_0</math> for the wavefunction of a harmonic oscillator, which is cool.
  
 ==== Zeno 9/22 9:15==== ==== Zeno 9/22 9:15====
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 ==== Daniel Faraday 7:20am 9/23 ==== ==== Daniel Faraday 7:20am 9/23 ====
 I noticed at the beginning of the power series method for the harmonic oscillator, Griffiths assumes that x is very large (bottom of p.51). But aren’t we usually looking at small x in a harmonic oscillator? How is Griffiths defining ‘very large x’ so that the solution is still useful and valid? I noticed at the beginning of the power series method for the harmonic oscillator, Griffiths assumes that x is very large (bottom of p.51). But aren’t we usually looking at small x in a harmonic oscillator? How is Griffiths defining ‘very large x’ so that the solution is still useful and valid?
 +
 +=== joh04684 11am 9/23 ===
 +I'm also confused by this...Isn't //x// supposed to represent the position of the oscillator?  Why are we only looking at large displacements?
 +
 +==== time to move on ====
 +
 +
 +It's time to move on to the next Q_A: [[Q_A_0925]]
classes/2009/fall/phys4101.001/q_a_0923.1253709022.txt.gz · Last modified: 2009/09/23 07:30 by x500_razi0001

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