# Homework Help: Stupid question about Clebsch Gordan

1. Feb 15, 2008

### malawi_glenn

[SOLVED] Stupid question about Clebsch Gordan

I know that

$$\sum _{m_1,m_2}|<j_1m_1,j_2m_2|jm>|^2 = 1$$

But is

$$\sum _{m_1,m_2}|<j_1m_1,j_2-m_2|jm'>|^2 = 1$$ ?

I have tried to justify this by using a CG calculator and so on, but I just cant figure out why :S

I think that doesent matter, since you sum over all m_2 and m_1 so that m_1 + m_2 = m'.

Does anyone have a hint or clue?

Last edited: Feb 15, 2008
2. Feb 15, 2008

### George Jones

Staff Emeritus
Maybe an example will help. Suppose m_1 = -1, 0 , 1. Write out the two sums over m_1 explicitly.

3. Feb 15, 2008

### pam

If you use -m_2, then you can only have m_1-m_2=m'.
CG are not defined (or are zero) unless m_1+m_2=m.

4. Feb 15, 2008

### malawi_glenn

those things I know.

Geroge Jones:

$$\sum _{a=-1}^1a = \sum _{a=1}^{-1}(-a) 5. Feb 15, 2008 ### malawi_glenn those things I know. Geroge Jones: [tex] \sum _{a=-1}^1a = \sum _{a=1}^{-1}(-a)$$

6. Feb 15, 2008

### kdv

As you say, since you are summing over all possible m_2 anyway, it does not matter at all if the label is $- m_2$ or $m_2$. Unless I am missing something...

7. Feb 15, 2008

### malawi_glenn

Thats what I am wondering too :)

I was looking in Greniers book "nuclear models" page 93 and 94, and saw that he must have uses this property. And this is a sub problem for my HW, so therefor I aksed here.

8. Feb 15, 2008

### kdv

The two expressions are equal. I was wondering if something had made you doubt it and if I was maybe not understanding the notation.

9. Feb 15, 2008

### malawi_glenn

No you have understand my question correct. I will mark this as solved now.

10. Feb 15, 2008

### George Jones

Staff Emeritus
This is what I was trying to get at, but I accidentally wrote m_1 instead of m_2.

11. Feb 15, 2008

### pam

I'm sorry, but I'm still puzzled. If that isn't a misprint, you can't have
<m_1,-m_2|m'> if m_1+m_2=m'.

12. Feb 15, 2008

### malawi_glenn

no of coruse not, But the other guys understood my question;)