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##.