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| classes:2009:fall:phys4101.001:q_a_0923 [2009/09/22 22:00] – prestegard | classes: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? |
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| | ===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. |
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| ====liux0756 9/22 13:18==== | ====liux0756 9/22 13:18==== |
| <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??? |
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| ==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. |
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| === Anaximenes - 21:34 - 09/22/09 === | === Anaximenes - 21:34 - 09/22/09 === |
| <math>\(a_{-}\)\psi_{0}\(x\)</math> must be non-normalizable (read: identically 0) or it wouldn't be <math>\psi_{0}</math> but <math>\psi_{1}</math> instead. | <math>\(a_{-}\)\psi_{0}\(x\)</math> must be non-normalizable (read: identically 0) or it wouldn't be <math>\psi_{0}</math> but <math>\psi_{1}</math> instead. |
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| | === 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 the wavefunction of a harmonic oscillator, which is cool. |
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| ==== Zeno 9/22 9:15==== | ==== Zeno 9/22 9:15==== |
| ==== prest121 9/22/2009 21:50 ==== | ==== prest121 9/22/2009 21:50 ==== |
| With regards to the end of the analytical solution to the harmonic oscillator potential (p.54): I understand that the power series must terminate to prevent the asymptotic behavior as x goes to infinity. After this is where I get confused. We choose some arbitrary n so that the coefficients for all higher terms are 0. I see how K = 2n + 1 fits into this requirement from Equation 2.81. My question is, why can we decide that the power series terminates at arbitrary n for a corresponding energy <math>E_n</math>? I see how the quantization falls out of the math, but the logic of it all isn't clicking. | With regards to the end of the analytical solution to the harmonic oscillator potential (p.54): I understand that the power series must terminate to prevent the asymptotic behavior as x goes to infinity. After this is where I get confused. We choose some arbitrary n so that the coefficients for all higher terms are 0. I see how K = 2n + 1 fits into this requirement from Equation 2.81. My question is, why can we decide that the power series terminates at arbitrary n for a corresponding energy <math>E_n</math>? I see how the quantization falls out of the math, but the logic of it all isn't clicking. |
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| | ==== 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? |
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| | === 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? |
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| | ==== time to move on ==== |
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| | It's time to move on to the next Q_A: [[Q_A_0925]] |