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classes:2009:fall:phys4101.001:q_a_0928 [2009/09/27 21:51] – czhangclasses:2009:fall:phys4101.001:q_a_0928 [2009/09/30 10:55] (current) – myers
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-===== Sept 28 (Mon) Free particle and wave packet(?)=====+===== Sept 28 (Mon) SHO wrapping up, Free particle and wave packet=====
 **Return to Q&A main page: [[Q_A]]**\\ **Return to Q&A main page: [[Q_A]]**\\
 **Q&A for the previous lecture: [[Q_A_0925]]**\\ **Q&A for the previous lecture: [[Q_A_0925]]**\\
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 ===chavez 9:35 9/25=== ===chavez 9:35 9/25===
-Angular frequency for (deep) water waves is <math>\omega = \sqrt{\frac{g }{2}}</math> and the phase velocity is given by <math>v_{phase}=\frac{\omega}{k}=\sqrt{\frac{g}{2k}}</math>. The group velocity is given by <math>v_{group}=\frac{\delta\omega}{\delta k}=\frac{1}{2}\sqrt{\frac{g}{2k}}=\frac{v_{phase}}{2}</math>.+Angular frequency for (deep) water waves is <math>\omega = \sqrt{\frac{g k}{2}}</math> and the phase velocity is given by <math>v_{phase}=\frac{\omega}{k}=\sqrt{\frac{g}{2k}}</math>. The group velocity is given by <math>v_{group}=\frac{\delta\omega}{\delta k}=\frac{1}{2}\sqrt{\frac{g}{2k}}=\frac{v_{phase}}{2}</math>. 
 + 
 +=== Can 9:45 9/27 === 
 +I didn't see the water wave problem on page 64, it is talking about the analytical methods. 
 +Anyway, a question for chavez, if <math>\omega = \sqrt{\frac{g }{2}}</math>, then shouldn't <math>v_{phase}=\frac{\omega}{k}=\sqrt{\frac{g}{2k^2}}</math>, is there a typo in the expression for the frequency somewhere? 
 + 
 +=== chavex 10:14 9/27 === 
 +Ah thanks for catching my mistake. It should have been <math>\omega = \sqrt{\frac{g k}{2}}</math>. I'll edit that into my original post.
  
  
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 === Schrodinger's Dog - 11:40 - 09/27/09 === === Schrodinger's Dog - 11:40 - 09/27/09 ===
 You can't evaluate it, you can numerically, unless of course you have the trivial case where the error function has the limits 0 to <math> \infty </math>, where you get the error function being 1. For very large upper limits and very low upper limits, I suppose you can assume the value of 1 for large upper limits and use the Trapezoid rule to approixmate the integral in very low limits. You could even write a rough program evaluating the integral using the Trapezoid rule, which I think would give you a sufficient solution.  You can't evaluate it, you can numerically, unless of course you have the trivial case where the error function has the limits 0 to <math> \infty </math>, where you get the error function being 1. For very large upper limits and very low upper limits, I suppose you can assume the value of 1 for large upper limits and use the Trapezoid rule to approixmate the integral in very low limits. You could even write a rough program evaluating the integral using the Trapezoid rule, which I think would give you a sufficient solution. 
 +
 +=== Blackbox - 23:49 - 09/27/09 ===
 +Yes, I had the same problem when I tried to calculate the expectation value in homework #1 and #3. For example, some of numerical integration can be solved by substituting "x^2" into "t" but some others are not easily done by integration rules. Should we memorize some specific integration results? 
 +
 +=== Esquire -10:54 09/30/09 ===
 +Here is a very nice site which discusses just about everything you need to know about the properties of the error function.
 +
 +[[http://functions.wolfram.com/GammaBetaErf/Erf/introductions/ProbabilityIntegrals/05/]]
  
 ==== Daniel Faraday 9/27 11am ==== ==== Daniel Faraday 9/27 11am ====
 Homework Question: on the question that's on the discussion sheet about energy scales in eV, what do we use for the size of the well for the neutron in the nucleus? Homework Question: on the question that's on the discussion sheet about energy scales in eV, what do we use for the size of the well for the neutron in the nucleus?
  
-==Schrodinger's Dog 9/27 2:07am==+===Schrodinger's Dog 9/27 2:07am===
 Well, I would guess the the rest mass of a neutron, but I was wondering what problem you are talking about? I didn't see this in our homework.  Well, I would guess the the rest mass of a neutron, but I was wondering what problem you are talking about? I didn't see this in our homework. 
 +
 +===Pluto 4ever 9/27 10:31pm===
 +If you are referring to the second half of the discussion problem then you just have to use the one half nanometer scale as it says in the problem.
 +
 +====Hydra   9/27 11:00pm ====
 +Can somebody show me why the gaussian wave packet has the minimum uncertainty? It makes sense intuitively, I just want to see quantitatively.
 +
 +===Schrodinger's Dog 9/27 3:03am===
 +Find <math>\sigma_x</math> and <math>\sigma_p</math>, multiply them together and you should get minimum uncertainty (i.e. <math>\frac{\hbar}{2}</math>).
 +
 +//**Yuichi**// You can check out section 3.5.2.
 +
 +
 +
 +==== time to move on ====
 +
 +
 +It's time to move on to the next Q_A: [[Q_A_0930]]
 +
 +
  
 **Return to Q&A main page: [[Q_A]]**\\ **Return to Q&A main page: [[Q_A]]**\\
classes/2009/fall/phys4101.001/q_a_0928.1254106299.txt.gz · Last modified: 2009/09/27 21:51 by czhang

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