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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.
For the 2×1 system (|3 0>) example, first we find the 2×1 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>
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.
In class when we found the<math> L_{-}=\sqrt{2}\to</math> , what equation yields the <math> \sqrt{2}</math>