I S=1 and I=1: How Can L be Anything Other than 0?

[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!

 

[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 $L = 0,1,2$.

 

[Plaetean]

Thank you!

 

[misko]

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 $L = 0,1,2$.

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?

 

[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.

 

[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?

 

[vanhees71]

The (one and only) bound state of the deuteron is not a pure s wave but has some admixture of a d wave as can be seen from the fact that this nucleus has a electric quadrupole moment.

https://en.wikipedia.org/wiki/Deuterium#Nuclear_properties_.28the_deuteron.29