How does the Pauli Exclusion Principle in this problem?

[whatapples]

Homework Statement

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

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?

Homework Equations

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?

 

[kuruman]

I think the assumption here is that they are all $p$-electrons. Furthermore, they should all have the same m_s=+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 M_S = 3/2; two up and one down for M_S = 1/2; two down and one up for M_S = -1/2; all three down for M_S = -3/2. Note that in all of these states there is one electron in each of the three m_l 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 m_l = 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 ⁶S_5/2.

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

 

[whatapples]

I think the assumption here is that they are all $p$-electrons. Furthermore, they should all have the same m_s=+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 M_S = 3/2; two up and one down for M_S = 1/2; two down and one up for M_S = -1/2; all three down for M_S = -3/2. Note that in all of these states there is one electron in each of the three m_l 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 m_l = 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 ⁶S_5/2.

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

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 $L=2$?

 

[kuruman]

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

 

[kuruman]

Here is the schematic diagram as promised. It shows two of the four states in the quartet ⁴S. The other two, S_z = -1/2 and S_z = -3/2 are obtained simply by flipping all the arrows by 180^o so I skipped them. Note that L_z = m_l1 + m_l2 + m_l3 = 0 for all 4 states.

M_L = 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.

 

[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.

 

[kuruman]

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.

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