1. Aug 27, 2011

### Jim Kata

Say I wanted to tensor $$|1,-1> \otimes |1,0>$$ Then looking at the Clebsch Gordons I get $$|1,-1> \otimes |1,0> = \frac{1}{\sqrt {2}}|2,-1> - \frac{1}{\sqrt{2}}|1,-1>$$
When I try to do this another way I run into a problem that I don't understand.

$$|1,0> = \frac{1}{\sqrt{2}} (|\frac{1}{2}, \frac{1}{2}> \otimes |\frac{1}{2},-\frac{1}{2}> + |\frac{1}{2},-\frac{1}{2}> \otimes |\frac{1}{2},\frac{1}{2}>)$$

So

$$|1,-1> \otimes |1,0> = |1,-1> \otimes ( \frac{1}{\sqrt{2}} (|\frac{1}{2}, \frac{1}{2}> \otimes |\frac{1}{2},-\frac{1}{2}> + |\frac{1}{2},-\frac{1}{2}> \otimes |\frac{1}{2},\frac{1}{2}>))$$

Going through this I get

$$|1,-1>\otimes|\frac{1}{2},\frac{1}{2}>\otimes|\frac{1}{2},-\frac{1}{2}> =(\frac{1}{\sqrt{3}}|\frac{3}{2},-\frac{1}{2}> -\sqrt{\frac{2}{3}}|\frac{1}{2},-\frac{1}{2}>)\otimes|\frac{1}{2},-\frac{1}{2}>=\frac{1}{2}|2,-1> +(\frac{\sqrt{3}}{6}-\sqrt{\frac{2}{3}})|1,-1>$$

similarly I get

$$|1,-1>\otimes|\frac{1}{2},-\frac{1}{2}>\otimes|\frac{1}{2},\frac{1}{2}>=\frac{\sqrt{3}}{2}|2,-1>-\frac{1}{2}|1,-1>$$

but when I add these two I don't get

$$|1,0> = \frac{1}{\sqrt{2}} (|\frac{1}{2}, \frac{1}{2}> \otimes |\frac{1}{2},-\frac{1}{2}> + |\frac{1}{2},-\frac{1}{2}> \otimes |\frac{1}{2},\frac{1}{2}>)$$

What am I doing wrong? Where does my reasoning break down?