# Homework Help: 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.
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