I Deuteron Spin

1. Jan 18, 2016

Plaetean

My notes say:

http://i.imgur.com/Z0v7Psi.png

The particular part I don't understand is that we know I = 1, and the text says:

for S = 1, I can be equal to 1 for L = 0, 1, 2.

If we know I = 1 and S = 1, how can L be anything other than 0?

Thanks as always!

2. Jan 18, 2016

blue_leaf77

Addition of quantum angular momentum $\mathbf{I} = \mathbf{L} + \mathbf{S}$ yields, for the possible value of $I$,
$$I = |L-S|,|L-S|+1,\ldots,L+S-1,L+S$$.
If you plug in $S=1$ to the above equation, you should see that $I$ can be unity only if $L = 0,1,2$.

3. Jan 18, 2016

Thank you!

4. Jul 6, 2016

misko

Can you elaborate how to look at this kind of problem when solving?
How exactly do you plug in $S=1$ in this equation and get only 0,1,2 as solutions for L?

5. Jul 7, 2016

blue_leaf77

Given $L = S = 1$, the possible total angular momenta are
$$I = |1-1|,|1-1|+1,1+1 = 0,1,2$$
To put it in a simple way, the min total angular momenta is $|L-S|$ while the maximum is $L+S$. Between these values the angular momenta are spaced uniformly at unit increment.

6. Jul 8, 2016

ChrisVer

In general I don't like what these notes say... For S=0 indeed you have antiparallel spins... but that is too vague since you can have antiparallel spin configuration for S=1 too....
In particular the spin states are (I give it as $|S, S_z>$ in the LHS and $|pn>$ spins in the RHS ):
S=1
$|1,1> = |\uparrow_p \uparrow_n>$
$|1,0> =\frac{1}{\sqrt{2}}\Big( |\uparrow_p \downarrow_n> +|\downarrow_p \uparrow_n> \Big)$
$|1,-1> = |\downarrow_p \downarrow_n>$
S=0
$|0,0> =\frac{1}{\sqrt{2}}\Big( |\uparrow_p \downarrow_n> -|\downarrow_p \uparrow_n> \Big)$

As you can see, with adding two spins 1/2 you can either get either total spin 1 (triplet) or 0 (singlet)...in an extension of your question, how did the addition of two 1/2s resulted to something that is 0?
What will happen if you add three spins 1/2?

7. Jul 10, 2016