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| classes:2009:fall:phys4101.001:q_a_0923 [2009/09/22 19:03] – x500_wuxxx528 | 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 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. |
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| ====Hardy 9/22 19:02==== | ====Hardy 9/22 19:02==== |
| I do not quite understand why the <math>a_-\psi_n(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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| | === prest121 9/22/2009 21:22 === |
| | It has to be zero because of quantization of energy. If you apply the lowering operator <math>a_-</math> to a wavefunction repeatedly, you will travel down the "energy ladder" in discrete increments. The <math>a_-</math> operator is our method of "stepping down" a rung of energy. At some point, you have to reach a state of zero energy, since we can't have negative energy. The ground state <math>\psi_0</math> of a wavefunction is defined as the state with the smallest amount of energy a particle can possess in a binding potential (<math>\frac{\hbar\omega}{2}</math> for the simple harmonic potential). Since <math>\psi_0</math> is the lowest energy state with nonzero energy, applying the lowering operator <math>a_-</math> will step down a rung on the energy ladder to zero energy. |
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| | === 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. |
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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==== |
| | Would the Algebraic Method (with or without ladder operators) work with potentials other than the SHO? The spring potential is an interesting classical problem, but there are still so many more. Will some form of the Algebraic Method work in general? Or must other potentials be solved analytically? |
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| | ==== 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. |
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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]] |