The sum of S and L quantum numbers

[annaphys]

I am reading the textbook Magnetism and Magnetic Materials by Coey and I am confused about how they grouped the terms and how they ended up getting the sums of L and S. My confusion lies in the two red boxes. Also, how is D even considered here when we have up to $2p_1$? And why would the spin totals change on the S and D in the first box...were dealing with fermions so S should always equal to 1/2. The last box is really confusing-- I don't see how L=1,S=1 , L=2,S=0 and L=S=0.

Thanks in advance!

 

[annaphys]

Forgot the photo

 

[PeroK]

Have you studied addition of angular momenta in QM?

 

[annaphys]

Yea

 

[DrClaude]

Yea

Then you should know that if you have two p electrons with $l=1$ each, the possible values of $L$ are
$$
L = |l_1 - l_2 |, \ldots, l_1 + l_2 = 0, 1, 2
$$
so you can have a D term. Likewise,

were dealing with fermions so S should always equal to 1/2.

if you have two $s=1/2$ electrons, total spin can be $S=0$ or $1$ (remember the singlet and triplet states).

And why would the spin totals change on the S and D in the first box...

While two electrons with $l=1$ and $s=1/2$ can give rise to total $L$ corresponding to S, P, and D and total spin $S=0$ and $1$, since you have two equivalent electrons (they are both 2p), you need to take into account of the Pauli exclusion principle. This is what is done at the bottom, where all allowed configurations are listed, leading to only $^1S$, $^3P$ and $^1D$ being allowed.

If instead the configuration had been 2p3p, then there would be no exclusion principle to consider (since the electrons already differ by the quantum number $n$) and there would have been 6 terms, $^1S$, $^1P$, $^1D$ $^3S$, $^3P$, and $^3D$.

 

[annaphys]

Hi Thanks for the help. I'm really not understanding your reasoning as to why there is $^1S$, $^3P$ and $^1D$ when the spin is 1/2. If possible, I'd appreciate if we can work through a couple examples from the photo. I mean it get due to the total spin being either zero or one but I am not seeing why the total spin and L are the values they are.

So the first red box, could someone explain why the first row is L=S=0 M_L/M_S = (0,0) and which one of the fifteen options in table 4.3 is this one?

 

[TeethWhitener]

when the spin is 1/2

The spin isn’t 1/2: there are two electrons in the carbon 2p orbitals. If their spins are aligned, $S=\frac{1}{2}+\frac{1}{2}=1$; if their spins are anti-aligned, $S=\frac{1}{2}-\frac{1}{2}=0$.

 

[annaphys]

The spin isn’t 1/2: there are two electrons in the carbon 2p orbitals. If their spins are aligned, $S=\frac{1}{2}+\frac{1}{2}=1$; if their spins are anti-aligned, $S=\frac{1}{2}-\frac{1}{2}=0$.

Yea but what is then the difference between m_s and s then? The spin of a fermion is 1/2 and then the total spin of a two-fermion system is always 1 since S = s_1 + s_2

 

[PeroK]

Yea but what is then the difference between m_s and s then? The spin of a fermion is 1/2 and then the total spin of a two-fermion system is always 1 since S = s_1 + s_2

The sum of two spin 1/2 systems is a product of spin 0 and spin 1 systems. There are four basis states: one is the singlet and corresponds to spin 0 (and is anti-symmetric); three states have spin 1 (the triplet) and each is symmetric.

You need to revise the addition of AM.

All these states ultimately are formed by the addition of orbital and spin angular momenta.

 

[vela]

Yea but what is then the difference between m_s and s then? The spin of a fermion is 1/2 and then the total spin of a two-fermion system is always 1 since S = s_1 + s_2

What @TeethWhitener said was misleading.

If you have two electrons, you can enumerate the possible states as $|\uparrow \uparrow \rangle$, $|\uparrow \downarrow \rangle$, $|\downarrow \uparrow \rangle$, and $|\downarrow \downarrow \rangle$, and these four states have $M_S = 1, 0, 0,\text{ and }{-1}$ respectively. For the first and last states, you must have $S=1$, but for the middle two, it's not clear whether $S=0$ or $S=1$. In fact, you can't map one to $S=0$ and the other to $S=1$. That's because the two $M_S=0$ states are linear combinations of $|\uparrow \downarrow \rangle$ and $|\downarrow \uparrow \rangle$. The symmetric combination has $S=1$ while the antisymmetric combination has $S=0$.

I agree with @PeroK that it would be a good idea for you to review the addition of angular momentum before trying to understand how to go from Table 4.3 to the breakdowns given in the red boxes. There's a change of basis going on that I think is the source of your confusion.

 

[DrClaude]

Yea but what is then the difference between m_s and s then? The spin of a fermion is 1/2 and then the total spin of a two-fermion system is always 1 since S = s_1 + s_2

To add to the good replies above, let me remind you also that spin is a vector quantity. Two vectors of length 1/2 do not always add up to one vector of length 1.