# Angular momentum quantum numbers

1. Apr 21, 2007

### BeauGeste

For angular momentum quantum numbers j, l, and s must it be true that $$m_s, m_l < m_j$$?
It would seem that it is true because I assume that $$m_s +m_l = m_j$$, but I have not actually seen that written down anywhere and am curious.
Thanks.

2. Apr 22, 2007

### Parlyne

No. It is not generally true. Remember that m tells you the component of angular momentum along some certain direction. This means that it can be either positive or negative. Consider, then, the case where $$m_s = -\frac{1}{2}$$. It is hopefully clear that $$m_l = m_j + \frac{1}{2}$$.

3. Apr 22, 2007

### BeauGeste

ok, I think I meant the magnitude of the m's. i.e.
$$|m_s|, |m_l| \leq |m_j|$$.
For example, take the situation where $$m_j = -1/2$$, j=3/2, s=1/2. The orbital and spin angular momenta magnetic numbers can add to this for two cases:
1. $$m_s = -1/2, m_l = 0$$.
2. $$m_s = 1/2, m_l = -1$$.
I would argue from above that case 2 is not a viable option because $$|m_l| \nleq |m_j|$$.
What do you think?

Last edited: Apr 22, 2007
4. Apr 22, 2007

### Parlyne

If you're using magnitudes, you can't have negative numbers. However, even if you want to compare the magnitudes of $$m_s$$, $$m_l$$, and $$m_j$$, you'll find that there are states where $$|m_j|$$ is smaller than either $$|m_s|$$ or $$|m_l|$$. All this requires is that the spin and orbital angular momenta have their z components in opposite directions. This is something we should expect from normal vector analysis (i.e. it has nothing to do with quantum mechanics specifically). If I add two vectors which have projections in opposite directions along the z-axis, I should expect that the magnitude of the z component of the resultant vector must be smaller than at least that of one of the two vectors I added; and, it may be smaller than both.

5. Apr 22, 2007

### BeauGeste

darn, I wrote that wrong. Hopefully this makes my question more clear:

if j=3/2 then possible $$m_j$$ values are -3/2 to 3/2 by ones. if j=3/2 and we're dealing with an electron then l = 1 and s = 1/2. $$m_l$$ values are 1,0, and -1.
If $$m_j = - 3/2 (+3/2)$$ then of course $$m_l = -1 (+1)$$ and $$m_s = -1/2 (+1/2)$$ respectively. Those are the only possibilities to form $$m_j$$ for the maximum values of $$m_j$$.
Now when $$m_j = \pm 1/2$$ it seems ambiguous as to what $$m_{l,s}$$ are. For instance:
1. $$m_s = -1/2, m_l = 0$$
2. $$m_s = 1/2, m_l = -1$$
both give $$m_j = -1/2$$.

Is that ok or is one of them not correct?

6. Apr 22, 2007

### Staff: Mentor

The state with $m_j = -1/2$ is a linear combination of the two states with $(m_s = -1/2, m_l = 0)$ and $(m_s = +1/2, m_l = -1)$. The coefficients of the linear combination are called Clebsch-Gordan coefficients:

http://farside.ph.utexas.edu/teaching/qm/lectures/node47.html

To relate the notation on that page to your example, substitute $l$ for its $j_1$, $s$ for its $j_2$, $m_l$ for its $m_1$, $m_s$ for its $m_2$, and finally $m_j$ for its $m$.

7. Apr 23, 2007

### Parlyne

Adding to what jtbell said, you should also be aware that there are $$j=3/2$$ states arising from $$l=2,\ s=1/2$$. Here, there are 10 possible $$|m_l, m_s\!\!>$$ states; but, only 8 of them give allowed values of $$m_j$$.

Last edited: Apr 23, 2007