Start with Helium: He+ same as h but with Z=2

• Start with Helium: He+ - same as H but with Z=2

• He - 2 electrons. No exact solution of S.E. but can use H wave functions and energy levels as starting point. We’ll use some aspects of perturbation theory but skip Ritz variational technique (which sets bounds on what the energy can be)

• nucleus screened and so Z(effective) is < 2

• “screening” is ~same as e-e repulsion (for He, we’ll look at e-e repulsion. For higher Z, we’ll call it screening)

• electrons are identical particles. Will therefor obey Pauli exclusion rule (can’t have the same quantum numbers). This turns out to be due to the symmetry of the total wave function

have kinetic energy term for both electrons (1+2)

• have kinetic energy term for both electrons (1+2)

• V12 is the e-e interaction. Let it be 0 for the first approximation, that is for the base wavefunctions and then treat it as a (large) perturbation

• for the unperturbed potential, the solutions are in the form of separate wavefunctions

The total wave function must be antisymmetric

• The total wave function must be antisymmetric

• but have both space and spin components and so 2 choices:

• have 2 spin 1/2 particles. The total S is 0 or 1

• S=1 is spin-symmetric S=0 is spin-antisymmetric

There are symmetric and antisymmetric spatial wavefunctions which go with the anti and sym spin functions. Note a,b are the spatial quantum numbers n,l,m but not spin

• There are symmetric and antisymmetric spatial wavefunctions which go with the anti and sym spin functions. Note a,b are the spatial quantum numbers n,l,m but not spin

• when two electrons are close, the antisymmetric state is suppressed (goes to 0 if exactly the same point). symmetric state is enhanced

• “Exchange Force” S=1 has the electrons (on average) further apart (as antisymmetric space). So smaller repulsive potential and so lower energy

• if a=b, same space state, must have symmetric space and antisymmetric spin S=0 (“prove” Pauli exclusion)

V terms in Schrod. Eq.:

• V terms in Schrod. Eq.:

• Oth approximation. Ignore e-e term.

• 0.5th approximation: Guess e-e term. Treat electrons as point objects with average radius (for both n=1) a0/Z (Z=2). electrons on average are ~0.7ao apart with repulsive energy:

V terms in Schrod. Eq.:

• V terms in Schrod. Eq.:

• First approximation: look at expectation value of e-e term which will depend on the quantum states (i,j) of the 2 electrons and if S=0 or 1

• Assume first order wavefunction for ground state has both electrons in 1S (spin S=0)

• expectation value

do integral. Gets 34 eV. The is measured to be 30 eV  E(ground)=-109+30=-79 eV

• do integral. Gets 34 eV. The is measured to be 30 eV  E(ground)=-109+30=-79 eV

• For n1=1, n2=2. Can have L2=0,1. Either S=0 or S=1. The space symmetrical states (S=0) have the electrons closer  larger and larger E

• formally

• first two terms are identical ”Jnl” as are the second two “Knl”. Both positive definite. Gives

L=0 and L=1 have different radial wavefunctions. The n=2, L=1 has more “overlap” with the n=1,L=0 state  electrons are closer  larger and larger E

• L=0 and L=1 have different radial wavefunctions. The n=2, L=1 has more “overlap” with the n=1,L=0 state  electrons are closer  larger and larger E

• usually larger effect than two spin states. Leads to Hund’s rules (holds also in other atoms)

can’t solve S.E. exactly  use approximations

• can’t solve S.E. exactly  use approximations

• Hartree theory (central field model) for n-electron atoms. Need an antisymmetric wave function (1,2,3 are positions; i,j,k…are quantum states)

• while it is properly antisymmetric for any 1  j practically only need to worry about valence effects

have kinetic energy term for all electrons

• have kinetic energy term for all electrons

• and (nominally) a complete potential energy:

• Simplify if look at 1 electron and sum over all the others

•  ~spherically symmetric potential as filled subshells have no (theta,phi) dependence. Central-Field Model. Assume:

“ignore” e-e potential..that is include this in a guess at the effective Z

• “ignore” e-e potential..that is include this in a guess at the effective Z

• energy levels can depend on both n and L

• wavefunctions are essentially hydrogen-like

• guess at effective Z: Ze(r)

• solve (numerically) wave equation

• fill energy levels using Pauli principle

• use wavefunctions to calculate electrons’ average radii (radial distribution)

• redetermine Ze(r)

• go back to step 2

electron probability P(r). Electrons will fill up “shells” (Bohr-like)

• electron probability P(r). Electrons will fill up “shells” (Bohr-like)

• assume all electrons for given n are at the same r with a Vn for each n. Know number of electrons for each n. One can then smooth out this distribution to give a continuous Z(r)

First shell n=1 1S Z1 ~ 18-2 = 16 Second shell n=2 2S,2P Z2 ~ 18-2-8 = 8 Third shell n=3 3S,3P Z3 ~ 18-10-8/4 = 4

• First shell n=1 1S Z1 ~ 18-2 = 16 Second shell n=2 2S,2P Z2 ~ 18-2-8 = 8 Third shell n=3 3S,3P Z3 ~ 18-10-8/4 = 4

• for the outermost shell…good first guess is to assume half the electrons are “screening”. Will depend on where you are in the Periodic Table

• guess energy levels.

After some iterations, get generalizations

• After some iterations, get generalizations

• for the innermost shell:

• Outermost shell

• atoms grow very slowly in size. The energy levels of the outer/valence electrons are all in the eV range

If one removes an electron from an inner orbit of a high Z material, many transitions occur as other electrons “fall”

• If one removes an electron from an inner orbit of a high Z material, many transitions occur as other electrons “fall”

• as photon emitted, obey

• “large” energies so emit x-ray photons. Or need x-ray photons (or large temperature or scotch tape) to knock one of the electrons out

• study of photon energy gives effective Z for inner shells

Only valence electrons need to be considered (inner “shield” nucleus)

• Only valence electrons need to be considered (inner “shield” nucleus)

• Filled subshells have L=0 and S=0 states (like noble gases).

• Partially filled subshells: need to combine electrons, make antisymmetric wavefunctions, and determine L and S. Energy then depends on J/L/S

• if >1/2 filled subshell then ~same as treating “missing” electrons like “holes”. Example: 2P no. states=6 1 electron  5 electrons 2 electrons  4 electrons 3 electrons 6 electrons (filled) (can prove by making totally antisymmetric wavefunctions)

Follow “rules”

• Follow “rules”

• for a given n, outer subshell with the lowest L has lowest E (S
   larger effective Z

• For a given L, lowest n has lowest energy (smaller n and smaller )

• no rule if both n,L are different. Use Hartree or exp. Observation. Will vary for different atoms

• the highest energy electron (the next state being filled) is not necessarily the one at the largest radius (especially L=4 f-shells)

Filled shell plus 1 electron. Effective Z 1-2 and energy levels similar to Hydrogen

• Filled shell plus 1 electron. Effective Z 1-2 and energy levels similar to Hydrogen

• have spin-orbit coupling like H

Helium: 1s2 ground state. 1s2s 1s2p first two excited states. Then 1s3s 1s3p and 1s3d as one electron moves to a higher energy state.

• Helium: 1s2 ground state. 1s2s 1s2p first two excited states. Then 1s3s 1s3p and 1s3d as one electron moves to a higher energy state.

• combine the L and S for the two electrons to get the total L,S,J for a particular quantum state. Label with a spectroscopic notation

• Atoms “LS” coupling: (1) combine Li give total L (2) combine Si to give total S; (3) combine L and S to give J

• (nuclei use JJ: first get Ji and then combine Js to get total J)

• as wavefunctions are different, average radius and ee separation will be different  2s vs 2p will have different energy (P further away, more shielding, higher energy)

• S=0 vs S=1 S=1 have larger ee separation and so lower energy

• LS coupling similar to H  Lower J, lower E

D=degeneracy=number of states with different quantum numbers (like Sz or Lz) in this multiplet

• D=degeneracy=number of states with different quantum numbers (like Sz or Lz) in this multiplet

Transitions - First Order Selection Rules (time dep. Pert. Theory)

• Transitions - First Order Selection Rules (time dep. Pert. Theory)

• Metastable 2S with S=1 state - long lifetime

Look at both allowed states (which obey Pauli Exclusion) and shifts in energy

• Look at both allowed states (which obey Pauli Exclusion) and shifts in energy

• Carbon: 1s22s22p2 ground state. Look at the two 2p highest energy electrons. Both electrons have S=1/2

• combine the L and S for the two electrons to get the total L,S,J for a particular quantum state

2p13s1 different quantum numbers  no Pauli Ex. 12 states = 6 ( L=1, S=1/2) times 2 (L=0, S=1/2)

• 2p13s1 different quantum numbers  no Pauli Ex. 12 states = 6 ( L=1, S=1/2) times 2 (L=0, S=1/2)

• where d=degeneracy = the number of separate quantum states in that multiplet

2p13p1 different quantum numbers  no Pauli Ex. 36 states = 6 ( L=1, S=1/2) times 6 (L=01 S=1/2)

• 2p13p1 different quantum numbers  no Pauli Ex. 36 states = 6 ( L=1, S=1/2) times 6 (L=01 S=1/2)

• where d=degeneracy = 2j+1

2p2 state: same L+S+J combinations as the 2p13p1 but as both n=2 need to use Pauli Exclusion reject states which have the same

• 2p2 state: same L+S+J combinations as the 2p13p1 but as both n=2 need to use Pauli Exclusion reject states which have the same

• not allowed. Once one member of a family is disallowed, all rejected

• But many states are mixtures. Let:

Need a stronger statement of Pauli Exclusion principle (which tells us its source)

• Need a stronger statement of Pauli Exclusion principle (which tells us its source)

• Total wavefunction of n electrons (or n Fermions) must be antisymmetric under the exchange of any two electron indices

• 2 electrons spin part of wavefunction:S=0 or S=1

• So need the spatial part of the wavefunction to have

• other states not allowed (but are in in 2p3p)

Use 2p2 how to build up more complicated wave functions.

• Use 2p2 how to build up more complicated wave functions.

• All members of the same multiplet have the same symmetry

• redo adding two spin 1/2  S=0,1 . Start with state of maximal Sz. Always symmetric (all in same state)

• other states must be orthogonal (different Szi are orthogonal to each other. But some states are combinations, like a rotation, in this case of 45 degrees)

• S=0 is antisymmetric by inspection

Adding 2 L=1 is less trivial  2,1,0

• Adding 2 L=1 is less trivial  2,1,0

• L=2. Values in front of each term are Clebsch-Gordon coeff. Given by “eigenvalues” of stepdown operator

• other states are orthogonal. Given by CG coefficients/inspection

• L=1 is antisymmetric by inspection. Do stepdown from L=1 to L=0

• L=0 symmetric by inspection. Also note orthogonality: A*C=1-2+1 and B*C=1-1

If all (both) electrons are at the same n. Use Hund’s Rules (in order of importance)

• If all (both) electrons are at the same n. Use Hund’s Rules (in order of importance)

• minimize ee repulsion by maximizing ee separation  “exchange force” where antisymmetric spatial wavefunction has largest spatial separation

•  maximum S has minimum E

• different L  different wavefunctions  different effective Z (radial distributions)  maximum L has minimum E

•  minimum J has minimum E unless subshell >1/2 filled and then  minimum J has maximum E