- #1
epsilonjon
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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:
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|}
As a first attempt, he ignores the electron interaction energy altogether, the Schrödinger 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):
\psi(r_1 , r_2) = \psi_{nlm}(r_1) \psi_{n'l'm'}(r_2)We can construct symmetric and anti-symmetric combinations in the usual way:
\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)]
and
\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)].
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 spatial 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, \psi_0 = \psi_{100}(r_1) \psi_{100}(r_2), 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 Schrödinger equation should separate to give
\psi(r_1 , r_2, r_3) = \psi_{nlm}(r_1) \psi_{n'l'm'}(r_2) \psi_{n''l''m''}(r_3)
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, \psi_0 = \psi_{100}(r_1) \psi_{100}(r_2) \psi_{100}(r_3), 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! :)
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:
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|}
As a first attempt, he ignores the electron interaction energy altogether, the Schrödinger 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):
\psi(r_1 , r_2) = \psi_{nlm}(r_1) \psi_{n'l'm'}(r_2)We can construct symmetric and anti-symmetric combinations in the usual way:
\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)]
and
\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)].
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 spatial 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, \psi_0 = \psi_{100}(r_1) \psi_{100}(r_2), 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 Schrödinger equation should separate to give
\psi(r_1 , r_2, r_3) = \psi_{nlm}(r_1) \psi_{n'l'm'}(r_2) \psi_{n''l''m''}(r_3)
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, \psi_0 = \psi_{100}(r_1) \psi_{100}(r_2) \psi_{100}(r_3), 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! :)
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 \psi(r_1,r_2,r_3) \chi(s_1,s_2,s_3)?