How does the Pauli Exclusion Principle in this problem?

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1. Oct 28, 2017

whatapples

1. The problem statement, all variables and given/known data

This is not a homework problem. It's an example in a text book.

3 electrons.

For $S=3/2$, we have that
$$m_{s_1} = m_{s_2} = m_{s_3} = 1/2$$

Therefore by the Pauli Exclusion principle,

$$m_{l_1} \neq m_{l_2} \neq m_{l_3}$$

and they take the values $-1,0,1$ respectively since each electron has $l=\{0,1\}$. I understand so far. Then it says that $^4P$ and $^4D$ and $^4F$ are excluded because of this. This is where I got confused. Why is this?And why is it that $^4S$ is allowed?

2. Relevant equations

3. The attempt at a solution

We know that the $s$ quantum number is the same for all and that $l$ is the same for two of the electrons but those two can always have a different $m_l$. So $L=0$ is the only $L$ allowed? Why?

2. Oct 28, 2017

kuruman

I think the assumption here is that they are all $p$-electrons. Furthermore, they should all have the same ms=+½. This means you can only put one of them in each of the three $p$-orbitals. Thus, the four states in the quartet of S = 3/2 are all three up for MS = 3/2; two up and one down for MS = 1/2; two down and one up for MS = -1/2; all three down for MS = -3/2. Note that in all of these states there is one electron in each of the three ml states giving a total orbital angular momentum of zero*. Hence the quartet S state. To get a D state (L = 2) you need more orbital angular momentum. That means flipping the spin of one of the electrons and doubling up the ml = 1 orbital, but then it would not be a quartet state because of the flipped spin. Similar considerations apply to the other orbitals. For example the ground state of Fe3+ with 5 $d$-electrons is a sextet 6S5/2.

__________________________
*One can write the wavefunctions of these states as Slater determinants of one-electron states.

3. Oct 28, 2017

whatapples

Thanks for the reply. Yes it is assumed that they are all p-electrons. Forgot to mention that.

I still don't understand something. Why does $M_L=0$ imply that $L=0$?

If $L=1$ that could be had from $l_1=1, l_2=1, l_3=1$ with $||1-1|+1|$ and $m_{l_1} = 0, m_{l_2} = 1, m_{l_3} = -1$ so where is the violation?

Also in the $L=2$ case cannot you just see that $l_1=1, l_2=1, l_3=1$ cannot ever result in $L=2$?

4. Oct 28, 2017

kuruman

I' ll have to draw you a diagram. Stay tuned.

5. Oct 28, 2017

kuruman

Here is the schematic diagram as promised. It shows two of the four states in the quartet 4S. The other two, Sz = -1/2 and Sz = -3/2 are obtained simply by flipping all the arrows by 180o so I skipped them. Note that Lz = ml1 + ml2 + ml3 = 0 for all 4 states.

ML = 0 does not necessarily imply that L = 0. The point to be made here is that if you have a half-filled shell (3 $p$-electrons, 5 $d$-electrons, 7 $f$-electrons, etc.) and demand maximum spin multiplicity (3/2, 5/2, 7/2 etc.), then the total angular momentum must be zero. A half-filled shell has the same spherically symmetric spatial distribution as a completely filled shell.

6. Oct 28, 2017

whatapples

Thanks. The last two sentences make the whole situation very easy to understand.

I think what the book was trying to say is that if you calculate all the possible $M_L$ levels and then take the maximum $M_L$, then since that is just the projection of $L$ you can take that as the value of $L$. So if $\max M_L=0$ then $L=0$.

What I learned from your drawing is that $M_S=1/2$ is a superposition of all possible arrangements and not just one such arrangement at random.

7. Oct 28, 2017

kuruman

That is correct. The three electrons are indistinguishable, you can't say that Fred has ml=1, Maria ml = 0 and Ahmed ml=-1.