Introduction to Quantum Mechanics AEP3610 Professor Scott Heinekamp

1
Making more complicated atoms starting with He

• Pauli Exclusion Principle allows for at most 2
  electrons in a spatial energy eigenstate y(r)
  (spin-up, spin-down)
• the nucleus provides Coulomb potential energy of
  charge Ze
• the overall electron wavefunction is y(r)csm
• all spin operators commute with all space
  operators
• lets start to make atoms by filling in
  electrons
• He n 1, s state, with 2 electrons in it,
  y(r) R10(r)Y00(q,f), we put electron 1 into
  the space state and electron 2 into the same
  space state and then build the symmetric
  combination

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The ground state of He

• how else might we do it? Build the antisymmetric
  space y(r)

• one refers to this as the paired-spin (updown)
  2-electron state

The first excited state of He

• Knock 1 electron up to n 2, l 0 and leave
  electron 2 where it was, in n 1, l 0 (so one
  is 1s and the second is 2s)
• spin states are both TBD at the moment
• now, should we build the symmetric space combo
  or the anti?
• the unexcited electron is nearer the nucleus and
  is said to shield the nuclear charge from the
  excited electron, thereby reducing the Coulomb
  attraction with the excited electron, thereby
  raising (making less negative) the energy

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Parahelium the three first excited states

• in the antisymmetric space combo, the electrons
  are further apart, so shielding is less
  successful? this is the lower-energy solution
• therefore spin is symmetric ? parahelium comes
  as triplets
• the space wavefunction for parahelium

• and of course the full wavefunction is to be
  multiplied by one of the three symmetric
  two-electron spinors s 1, m 1, 0, 1
• we will shortly adjust the notation to
  accommodate both L and S

Orthohelium the sole second excited state

• in the symmetric space combo the electrons are
  closer together, so the shielding is more
  complete, and the excited electron is not bound
  so tightly the effect is small but sufficient
  that the energy is larger
• therefore spin is antisymmetric ? orthohelium
  comes as a singlet
• the space function for orthohelium (multiply by
  spinor s 0 m 0)

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Finding ground states for more-electron atoms

• the n 1 shell is now full with its two
  electrons up down ?
• more precisely, the n 1 shell is full, as is
  its l 0 orbital, with the helium ground state
  spatial and spin-singlet wavefunction
• one can say that both 1s orbitals are occupied
• now let L be the total orbital ang mom quantum
  number and S be the total spin quantum number, so
  here we have L 0 and S 0
• the next electron has to go into the 2s orbital
  but it can choose l 0 2s or l 1 2p.
• Which to choose? SHIELDING gives the way!!
• in general, the lowest l electrons are closest
  to the nucleus and so they are not shielded so
  well, so their energy is lower ? s, then p
• Li and Be thus fill up the 2s orbital
• Li is hydrogen-like plus it has the third
  electron in the 2s orbital
• to generate its fully antisymmetric wavefunction
  we resort to the Slater determinant
• for He, the Slater determinant is that of a 2 x
  2 matrix

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The Slater determinant for He and Li

• for helium ground state the determinant captures
  both terms

• for lithium ground state we can express full
  wavefunction fw as a determinant to capture all
  six terms (put third electron up in 2s)

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The overall antisymmetry is tangled up in
spin/space

• the 2s electron was chosen to be up arbitrarily
• not at all expressible separably!!
• there is a certain cyclicity in the spin
  arrangements
• beyond element three it gets rapidly more
  baroque
• notice, though, that if you exchange any
  coordinate with any other, you get a sign flip
  perfect!!
• the remaining elements for n 2 fill up as
  Hunds rules govern, getting you to element 10
  (Ne) ? Now the n 2 shell is full and all three
  orbitals (1s 2s 2p) are filled and, in
  addition, are said to have no unpaired spins
• thus, the TOTAL spin is zero S 0, and the
  TOTAL L 0 too!
• how did we get there, in more detail?

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Term symbols and spectroscopic notation

• for historical reasons we use term symbols to
  denote the ground state electron configurations
  2S1LJ
• L is the total orbital angular momentum quantum
  number, so

• it is denoted by letters S, P, D,
  F(confusing!!!)
• S is the total spin angular momentum quantum
  number, so

• the z-component quantum numbers obey the usual
  rules range from (e.g. for L) ML L, (L
  1), ., (L 1), L
• similar for z-components of total spin quantum
  number MS
• multiplicity is 2S1 which is the number of MS
  possibilities
• finally, J is the sum of spin plus orbital
  quantum number and it and MJ too obey the same
  kinds of rules as L and S and ML and MS
• MJ can range from L S up to L S but not
  our concern here
• the trick is sorting out what J is going to be
  in terms of L and S

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Hunds rules for filling electron levels

• All states must be antisymmetric in exchange of
  any spin pair!
• singlet spins antiparallel ? paired (so
  antisymmetric in spin)
• triplet spins parallel ? unpaired (so
  antisymmetric in space)
• First rule other things being equal, the state
  with highest S has the lowest energy this means
  that spins tend to fill in by lining up
• also called the bus seat rule
• physical cause shielding is worst in this
  case.. a tricky quantum mechanics problem
• Second rule For a given spin configuration,
  the state with the largest L will have the lowest
  energy
• physical cause electrons orbiting in opposite
  directions tend to bump into each other more
  often, thus more repulsion energy
• Third rule if a subshell (n, l) is up to and
  including half filled, then the lowest energy
  state has J L S if it is more than
  half-filled, the lowest energy state has J L
  S
• physical cause spin-orbit coupling (a magnetic
  effect)

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Applying Hunds rules to simple atoms Z 1 to 6

• in general, empty or filled shells tend to have
  the simplest term symbols noble gases always
  have 1S0? S L J 0
• closed shells (or subshells) will be denoted by
  the atom and so only the subsequent electrons
  need to be analyzed
• both electron configuration (ShellSubshell) and
  term symbol
• H (1s) ? one s electron so S ½ , L 0 ?
  2S1/2
• He (1s)2 ? two (singlet/paired) s electrons S
  0, L 0 ? 1S0
• Li (He)(2s) ? one s electron so S ½ , L 0 ?
  2S1/2
• Be (He)(2s) 2 ? two (singlet/paired) s
  electrons S 0, L 0 ? 1S0
• now the fun begins!!
• B (Be)(2p) ? one p electron S ½ , L 1 ?
  2P1/2 H3
• C (Be)(2p)2 ? two (triplet/unpaired) p
  electrons S 1, L 1 ? 3P0
• Huh?? Why not L 2 to obey H2 (would imply m
  1 for both)? But we cant have both of the p
  electrons have m 1, since they are in the same
  spin state ? one has m 1, one has m 0 to
  optimize H2 consistent with PEP so via H3 J
  L S 1 1 0

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Continuing along the shell n 2 up to Z 10

• N (Be)(2p)3 ? three (unpaired, so spin
  symmetric) p electrons S 3/2 , L 0 m 1,
  0, 1 ? 4S3/2
• catching on yet? This one was the most
  extreme..
• O (Be)(2p)4 ? four (two paired, two not) p
  electrons S 1, L 1 m 1, 0 ? 3P2 used
  version II of H3 here to get J
• F (Be)(2p)5 ? five (four paired, one not) p
  electrons S 1/2 , L 1 m 1 ? 2P3/2 used
  version II of H3 here to get J
• Ne (Be)(2p)6 ? six (all paired) p electrons S
  0, L 0 ? 1S0
• we have now filled the second shell all the way
  up to Z 10
• from here on out to Z 18 (Ar) we fill 3s shell
  and then 3p shell just as before and all term
  symbols are the same.. Ar is closed shell
• then, instead of filling 3d subshell, 4s fills
  next, up to Z 20 (Ca) which is a closed
  subshell ? 1S0
• now the 3d shells start to fill the
  transition metals have begun, starting with Z
  21 (Sc) and ending with closed-subshell Z 30
  (Zn)
• lets do these 10 and call it a day..

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Continuing along the shell n 3 up to Z 27

• Sc (Ca)(3d) ? one d electron S ½ , L 2 ?
  2D3/2 H3
• Ti (Ca)(3d)2 ? two (triplet/unpaired) d
  electrons S 1, L 3 ? 3F2 m 2, 1 due to
  PEP, as for carbon
• V (Ca)(3d)3 ? three (all unpaired so spin
  symmetric) d electrons S 3/2, L 3 m 2, 1,
  0 ? 4F3/2
• the next one (Z 24, Cr) is very anamalous!
  one of the 4s electrons moves into the 3d
  orbital, so we have (here, and for Cu)
• Cr (Ar)(4s)(3d)5 ? one unpaired s electron and
  five (all unpaired so spin symmetric) d
  electrons S 3, L 0 m 2, 1,0 1, 2 ? 7S3
  used version II of H3 here to get J
• Mn (Be)(3d)5 ? five (all unpaired so spin
  symmetric) d electrons S 5/2 and L 0 why?
  ? 6S5/2 halfway there!
• Fe (Be)(3d)6 ? six (two paired so spin
  antisymmetric, four not)
• easiest way to deal with it is to work from Ms
  and add one d electron with opposite spin S 2,
  L 2 ? 5D4
• Co (Be)(3d)7 ? Like Mn but with two opposite d
  electrons S 3/2, L 3 m 2, 1? 4F9/2

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Most of the rest of the story

• interlude these large numbers of of unpaired
  electrons give rise to ferromagnetism (exhibited
  by Fe, Co and Ni) the spins magnetic moments
  communicate to nearby spins ? they line up too ?
  whole sample is macroscopically spin-aligned a
  huge magnetic dipole!
• Ni (Be)(2p)8 ? Like Mn but with three opposite
  d electrons S 1, L 3 m 2, 1, 0? 3F4
• Cu (Ar)(4s)(2p)10 ? Like Cr so d subshell is
  closed S ½ thanks to the s electron L 0 m
  2, 1, 0, 1, 2 ? 2S1/2
• Zn (Be)(2p)10 ? d subshell is closed S 0, L
  0 ? 1S0
• now the 4p subshell start to fill with exactly
  the same pattern as previously seen for the
  filling of a p subshell up to Z 36 (Kr)
• next comes subshell 5s 4d transitions (with a
  Cr-like anamoly) 5p up to Z 54 (Xe)
• next comes subshell 6s 4f (one 5d leaks in at
  first the lanthanides 5d transitions 6p up
  to Z 86 (Rn)
• similar thing for 7s, 5f (the actinides) stop
  at Z 92 (U) why?
• most horrid term symbol is Neptunium
  Ra(5f)4(6d)1 ? 6L11/2

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Most of the rest of the story