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Number of electrons in each shell

  1. Jun 10, 2012 #1
    Hi.

    I'm currently working through Griffith's Introduction to QM, and have gotten to the section on the periodic table. I'll explain my understanding a little bit...

    Before this he's been looking at the Hamiltonian for helium:

    [tex]H = \left[ \frac{- \hbar ^2}{2m}\nabla _1 ^2 - \frac{1}{4 \pi \epsilon _0} \frac{2e^2}{r_1} \right] + \left[ \frac{- \hbar ^2}{2m}\nabla _2 ^2 - \frac{1}{4 \pi \epsilon _0} \frac{2e^2}{r_2} \right] + \frac{1}{4 \pi \epsilon _0} \frac{e^2}{|r_1 - r_2|}[/tex]
    As a first attempt, he ignores the electron interaction energy altogether, the Schrodinger equation separates, and solutions can be written as products of hydrogen wavefunctions (only replacing e^2 with 2e^2 in the Bohr radius and energies):

    [tex]\psi(r_1 , r_2) = \psi_{nlm}(r_1) \psi_{n'l'm'}(r_2)[/tex]We can construct symmetric and anti-symmetric combinations in the usual way:

    [itex]\psi(r_1,r_2) = A[\psi_{nlm}(r_1) \psi_{n'l'm'}(r_2) + \psi_{nlm}(r_2) \psi_{n'l'm'}(r_1)][/itex]

    and

    [itex]\psi(r_1,r_2) = A[\psi_{nlm}(r_1) \psi_{n'l'm'}(r_2) - \psi_{nlm}(r_2) \psi_{n'l'm'}(r_1)][/itex].

    We know that electrons are fermions, so the wavefunction has to be anti-symmetric under exchange of particles. However, it is the total wavefunction (position and spin) which must be anti-symmetric, so we can have symmetric spacial wavefunctions as long as they're with anti-symmetric spinors, etc. For example, this means that we can have both electrons in the n=1 state, [itex]\psi_0 = \psi_{100}(r_1) \psi_{100}(r_2)[/itex], so long as this is combined with the anti-symmetric spin configuration (singlet state).

    If we now go on to lithium (Z=3), once again ignoring the electron interaction energy, the Schrodinger equation should separate to give

    [tex]\psi(r_1 , r_2, r_3) = \psi_{nlm}(r_1) \psi_{n'l'm'}(r_2) \psi_{n''l''m''}(r_3)[/tex]
    Now this is the bit I don't understand: Griffiths says that "by the Pauli exclusion principle only two electrons can occupy any given orbital (one with spin up, one with spin down - or more precisely, the singlet configuration). There are n^2 hydrogenic wavefunctions (all with the same energy) for a given n, so the n=1 shell has room for 2 electrons, the n=2 shell holds 8, and in general the nth shell can accommodate 2n^2 electrons."

    I understand that each n state has degeneracy n^2 due to the orbital angular momenta, but the rest of that argument does not make much sense to me. For instance, how do we know that we cannot have the 3 electrons each in the n=1 state, [itex]\psi_0 = \psi_{100}(r_1) \psi_{100}(r_2) \psi_{100}(r_3)[/itex], so long as it is combined with an anti-symmetric spinor (as before). Do we not need to work out the possible spin configurations for three spin 1/2 particles before we can even start to work this out? Or am I missing the point entirely?

    Thanks for any help! :)
     
  2. jcsd
  3. Jun 10, 2012 #2

    Matterwave

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    Electrons are spin 1/2 and so there are only 2 basis spin states. You can't anti-symmetrize 3 spin states when there are only 2 basis spin states.

    Adding 3 spins together, you cannot get a "singlet" state.
     
  4. Jun 11, 2012 #3
    Ah okay, I think I see that. So for 3 or more particles the spin wavefunction is always symmetric, meaning that (for fermions) the spacial wavefunction has to be anti-symmetric?

    Assuming no spin-position-coupling, the total wavefunction is [itex]\psi(r_1,r_2,r_3) \chi(s_1,s_2,s_3)[/itex]. But if I use the slater determinant method to construct an anti-symmetric spacial wavefunction for 3 particles in states a, b and c, I get

    [itex]\psi(r_1,r_2,r_3) = \psi_a(r_1)\psi_b(r_2)\psi_c(r_3) - \psi_a(r_1)\psi_c(r_2)\psi_b(r_3) - \psi_b(r_1)\psi_a(r_2)\psi_c(r_3) + \psi_b(r_1)\psi_c(r_2)\psi_a(r_3) + \psi_c(r_1)\psi_a(r_2)\psi_b(r_3) - \psi_c(r_1)\psi_b(r_2)\psi_a(r_3)[/itex]

    which means that if any two of the states are the same (e.g. 2 particles in the n=1 hydrogenic state), the wavefunction is zero :confused: I think the answer is going to be that they don't cancel because the spins are different, but I don't see how this works when the total wavefunction is [itex]\psi(r_1,r_2,r_3) \chi(s_1,s_2,s_3)[/itex]?

    Sorry if i'm making a bit of a mountain out of a molehill in understanding this, I just want to get it clear in my mind.

    Thanks :-)
     
    Last edited: Jun 11, 2012
  5. Jun 12, 2012 #4

    mfb

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    ... in the exchange of at least 2 particles.

    You forgot the spin here, and I think in the general case you cannot factor spatial and spin component like that. If any two of the states are the same, the electrons in these states have to have different spin.
     
  6. Jun 13, 2012 #5
    But then how do you know that the spin state being symmetric under exchange of particles implies the spacial wavefunction is anti-symmetric? I thought this came from the fact that the total wavefunction is [itex]\psi(r)\chi(s)[/itex]?
     
  7. Jun 13, 2012 #6

    mfb

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    The interesting case here are two electrons in one state (now called G) and one electron in another state (E).

    Neglecting prefactors, the two electrons in G can be described as
    [tex]\psi_G(1)\psi_G(2) \left(\chi_u(1) \chi_d(2) - \chi_u(2) \chi_d(1)\right) := \theta(1,2)[/tex]
    A third electron has its own state and polarisation, here called "u".
    [tex]\psi_E(3)\chi_u(3)=\Psi(3)[/tex]
    These two can be combined to an antisymmetric function via
    [tex]\theta(1,2)\Psi(3) - \theta(1,3)\Psi(2) - \theta(3,2)\Psi(1)[/tex]
    While I got this by hand, algorithms should work as well.

    To see that this expression is antisymmetric:
    An exchange of 1<->3 swaps the first and the last summand (which have opposite sign) and swaps the sign at the middle summand (as theta is antisymmetric).
    In a similar way, 1<->2 is fine.
    The exchange 1<->2 swaps the sign of the first part. If you swap the order of electrons in the thetas of the other summands, you get a sign swap there as well. This one is a bit tricky to see.

    I do not think that you can separate spin and spatial components in this function. Only the combination of both parts make the function antisymmetric.
     
  8. Jun 16, 2012 #7
    That definitely makes it clearer. Thanks for your help :)
     
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