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How does the Pauli Exclusion Principle in this problem?

  1. Oct 28, 2017 #1
    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. jcsd
  3. Oct 28, 2017 #2

    kuruman

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    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.
     
  4. Oct 28, 2017 #3
    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##?
     
  5. Oct 28, 2017 #4

    kuruman

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    I' ll have to draw you a diagram. Stay tuned.
     
  6. Oct 28, 2017 #5

    kuruman

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    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.

    p levels.png
     
  7. Oct 28, 2017 #6
    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.
     
  8. Oct 28, 2017 #7

    kuruman

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    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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