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Ok, so I asked this question a while back, but didn't get a answer to it, but found it with help from Yuichi. So my questions was:
“How does Griffths goes from <math>a_{j+2} \approx \frac{2}{j}a_{j} </math> to <math>a_j \approx \frac{C}{(j/2)!}</math>, where C is some constant, on the bottom page 53 and top of page 54?”
Well, what Griffth simply did was a large about substitution. Griffths basically just used the formula <math>a_{j+2} \approx \frac{2}{j}a_{j} </math> , and looked at vaules for j-2, j-4,…etc, so we get <math>a_{j} \approx \frac{2}{j-2}a_{j-2} </math> , <math>a_{j-4} \approx \frac{2}{j-4}a_{j-4} </math>, etc. We then subsitute <math>a_{j} \approx \frac{2}{j-2}a_{j-2} \approx \frac{2}{j-2}…\frac{2}{1}a_{1} </math>. <math> a_{1}=C</math>, since it is the first term. Using factorials allows us to obtain a closed expression for <math>\frac{2}{j-2} …\frac{2}{1}=\frac{1}{(j/2)!}</math>. Combining the fact that <math>a_{1}=C </math>, using the factorial expression, and disregarding <math>a_{j+2} </math> term, we find that <math>a_j \approx \frac{C}{(j/2)!}</math>.
Sorry if this is messy, but this it I guess. Back to MXP lab :P.
In page 64, water waves are mentioned, however, I cannot understand the description in the textbook why the group velocity is one half of phase velocity, for water.
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>.
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?
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.
How do we evaluate an integral of which has a solution containing the erf error function?
I assume you're referring to the integral of <math>e^{-cx^2}</math> or something similar? I would try using a definite integral that doesn't contain the erf in the solution. Wikipedia seems to be a good source for that; search for “list of integrals of exponential functions.”
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.
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?
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/
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?
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.
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.
Can somebody show me why the gaussian wave packet has the minimum uncertainty? It makes sense intuitively, I just want to see quantitatively.
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.