# Clebsch-Gordan coefficients symmetry relation

1. Feb 23, 2016

### Happiness

Why are $<j_1j_2m_1m_2|jm>$ and $<j_2j_1m_2m_1|jm>$ negative of each other when $j_1+j_2-j$ is odd as given below?

I would expect $<j_1j_2m_1m_2|jm>$ and $<j_2j_1m_2m_1|jm>$ to always have the same sign since nature doesn't care which particle we label as particle 1 and which as particle 2.

From (6.274), isn't it true that
$|j_1j_2m_1m_2>\,=|j_1m_1>|j_2m_2>\,=|j_2m_2>|j_1m_1>\,=|j_2j_1m_2m_1>$?

And hence
$<j_1j_2m_1m_2|=\,<j_2j_1m_2m_1|$?

And so $<j_1j_2m_1m_2|jm>$ and $<j_2j_1m_2m_1|jm>$ should always have the same sign.

$j$ is the general angular momentum quantum number and $m$ is the associated magnetic quantum number.

2. Feb 23, 2016

Staff Emeritus
Think of a cross product: a x b is - b x a, even though nature doesn't care which particle we call a and which we call b.

3. Feb 23, 2016

### Happiness

Why does the order in the direct product matter?

From (6.276),
$J_z \psi_{j_2 j_1 m_2 m_1}=(m_1+m_2)\hbar\psi_{j_2 j_1 m_2 m_1}$

It seems like the order doesn't matter. Could you give an example where the negative sign is necessary when the order is flipped around?

4. Feb 23, 2016

Staff Emeritus
The important thing the C-G coefficients is their relative sign. That's where the SU(2) algebra is captured. You will need the sign only when you add terms.