0
$\begingroup$

Wikipedia has a nice article outlining Clebsch-Gordan coefficients.

For example, to my understaning, this table tells us how to combine two particles, each having a maximum total angular momentum $1$ into one wavefunction with maximum angular momentum $2$:

enter image description here

Take the first column from the last table. It tells us, I believe:

$|2,0\rangle = \sqrt{\frac{1}{6}} |1,1\rangle |1,-1\rangle +\sqrt{\frac{2}{3}}|1,0\rangle|1,0\rangle+\sqrt{\frac{1}{6}} |1,-1\rangle|1,1\rangle$

How I interpret this:

The total angular momentum of a particle which arises from such a combination of wavefunctions of two other particles will have total angular quantum number 2 (so total angular momentum $\sqrt{j(j+1)\hbar^2}=\sqrt{2(2+1)\hbar^2}$), but $0$ around the $z$ axis (as $m_j$, what I understand to be the angular momentum around the $z$ axis, is $0$).

So the constituent particle's angular momentum is not aligned with each other, in fact they are antialigned enough so that the total z-directional angular momentum will be 0.

Is this interpretation of what's going on correct? My concern is that there are no tables for $m=-1,-2$. If my interpretation of the situation is correct, I see no reason why I couldn't produce a combined particle with these $m$ values, if I can do it for $m=0,1,2$.

$\endgroup$
3
$\begingroup$

The Wikipedia article says the following:

For brevity, solutions with $M < 0$ and $j_1 < j_2$ are omitted. They may be calculated using the simple relations $$ \langle j_{1},j_{2};m_{1},m_{2}\mid j_{1},j_{2};J,M\rangle =(-1)^{J-j_{1}-j_{2}}\langle j_{1},j_{2};-m_{1},-m_{2}\mid j_{1},j_{2};J,-M\rangle .$$ and $$ \langle j_{1},j_{2};m_{1},m_{2}\mid j_{1},j_{2};J,M\rangle =(-1)^{J-j_{1}-j_{2}}\langle j_{2},j_{1};m_{2},m_{1}\mid j_{2},j_{1};J,M\rangle.$$

In other words, the Clebsch-Gordon coefficients for a negative value of $m$ are the same (up to a sign) as those for the corresponding positive value of $m$, so long as you switch the signs of $m_1$ and $m_2$ as well.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.