Differences

This shows you the differences between two versions of the page.

--- classes:2009:fall:phys4101.001:q_a_1202 [2009/12/02 10:41] – jbarthel
+++ classes:2009:fall:phys4101.001:q_a_1202 [2009/12/06 21:49] (current) – ely
@@ Line 8: / Line 8: @@
 **Main class wiki page: ** [[home]]
 ---------------------------------------
+====Captain America 12-02 10:39====
-===Captain America 12-02 10:39===
 Can anyone explain how to find the value you want on the Clebsch-Gordon table?  I read the bottom of page 187, but I'm still confused when given a problem how to find out which part of the table to look at (the example is 2 X 1 with the total spin of 3 and the z component 0).  How do you get those four values out of a problem?  Should be easy, but I'm having difficulties with it.
-==== Zeno 12/2 ====
+===Devlin ===
+I'm also having some serious issues with this table.
-What makes the 3-body problem --or the n-body problem for n>2-- so difficult? I've heard of the issue in modern physics, astro, and quantum, but I've never seen the problem set up or how there would arise a mathematical barrier. I guess I'll look it up or try to set it up myself. I'm sure the problem will present itself quickly.
+===prest121 12/2 5:50pm===
+For the 2x1 system  (|3 0>) example, first we find the 2x1 table.  Then, we find the column with 3 0 for the header (notice it is shaded for this example).  Remember that each value in the table actually has a square root.  Each coefficient in the shaded column corresponds to the coefficient for finding the spin 2 particle in with z-component given by the furthest left number, and the spin 1 particle with z-component given by the next number.  For this example, we get
+<math>|3\quad0> = \sqrt{\frac{1}{5}}|2\quad1>|1\quad-1>+\sqrt{\frac{3}{5}}|2\quad0>|1\quad0>+\sqrt{\frac{1}{5}}|2\quad-1>|1\quad1></math>
+Here's another example.  For a system consisting of a spin 3/2 particle and a spin 1/2 particle, with total angular momentum 2 and z-component 1, we get
+<math>|2\quad1> = \sqrt{\frac{1}{4}}|\frac{3}{2}\quad\frac{3}{2}>|\frac{1}{2}\quad\frac{-1}{2}>+\sqrt{\frac{3}{4}}|\frac{3}{2}\quad\frac{1}{2}>|\frac{1}{2}\quad\frac{1}{2}></math>
+===Andromeda 12/02 7:34pm===
+the book is talking about 2 ways of reading this table. if you read it vertically (like griffiths doing at the bottom of pg. 187) you will be expressing the total combined state in terms of combinations of states for each particle in the system. the second way  he talks about in the next page (reading the table horizontally) will tell you the possibilities for the combined state if you know the state of each particle.
+==== Zeno 12/2 ====
+What makes the 3-body problem --or the n-body problem for n>2-- so difficult? I've heard of the issue in modern physics, astro, and quantum, but I've never seen the problem set up or how there would arise a mathematical barrier. I guess I'll look it up or try to set it up myself. I'm sure the problem will present itself quickly.
+=== The Doctor 12/3 1:31 AM ===
+The variety of divergent trajectories with various Lyapunov exponents makes the problem too difficult for undergraduate courses.  So says wikipedia.
+===Dark Helmet 12/6 21:46===
+My understanding is that it just gets really really complicated because the interaction of body 1 with body 2 changes how they interact with body 3 which then changes again how body 1 and 2 react, etc etc.  However, i think this problem only applies when doing it by hand considering the large number of N body simulations that are being done by computer(dark matter collapse, what-have-you).
+====Hydra 12/2====
+In class when we found the<math> L_{-}=\sqrt{2}\to</math> , what equation yields the <math> \sqrt{2}</math>
+===Andromeda 12/2 2:38 PM===
+I think equation 4.121: <math>A_lm=\hbar\sqrt{(l+-m)(l-+m+1)}</math>
+====Blackbox 12/3====
+On page 164, the equation between 4.112 and 4.113, why L_L+ becomes zero? Is this the reason that L+*ft is zero?
+===Captain America 12/4 10:57===
+Yes, I believe this is the case.  The <math>L_+</math> operator should act on <math>f_t</math> first, and as equation 4.110 on page 163 points out, it is 0.

User Tools

Differences

Page Tools