Can total angular momentum j be negative?

[skate_nerd]

Homework Statement

I'm just stuck on one part of a larger problem. I need to find the range of total angular momentum values for an electron in a j-j coupling scheme.

Homework Equations

j= l + and - 1/2

The Attempt at a Solution

The electrons here are in a 5d 6s configuration. So for the second electron, l=0. This means j for the second electron is 0 plus and minus 1/2, so -1/2 and +1/2. This formula for j is what my book says to use with j-j coupling, but it seems to imply that j can be negative, and if that were the case, couldn't then J be a complex number? (Recall J=root(l(l+1))*hbar)
Just a little stumped here, and I want to get this right so I don't screw up the rest of the problem. Thanks for any hints

 

[TSny]

$j$ cannot be negative. The $l = 0$ case is a little special. You only get $j = l + 1/2$ in this case. $j = l - 1/2$ is ignored when $l = 0$

 

[skate_nerd]

I appreciate the response! Cheers

 

[DrClaude]

The $l = 0$ case is a little special. You only get $j = l + 1/2$ in this case. $j = l - 1/2$ is ignored when $l = 0$

I would disagree that $l=0$ is a special case. When summing angular momenta $\hat{j}_1$ and $\hat{j}_2$ into $\hat{J} = \hat{j}_1 + \hat{j}_2$, the quantum number $J$ can take the values
$$
J = \left| j_1 - j_2 \right|, \left| j_1 - j_2 \right| + 1, \ldots, j_1 + j_2
$$
The absolute value prevents $J$ from being negative, whatever the relative values of $j_1$ and $j_2$.

 

[TSny]

I would disagree that $l=0$ is a special case.

Yes, you are right.

From the OP it appears that the textbook might have written $j = l \pm \frac{1}{2}$ when combining the orbital and spin angular momentum of a single electron. Hopefully it was made clear that this doesn't hold for $l = 0$.