Transcript Introduction to Quantum Mechanics AEP3610 Professor Scott

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
• let’s 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
y (r )  12 [ R10 (r1 )Y00 (q1 ,f1 ) R10 (r2 )Y00 (q 2 ,f2 )  R10 (r2 )Y00 (q 2 ,f2 ) R10 (r1 )Y00 (q1 ,f1 )] 
 R10 (r1 )Y00 (q1 ,f1 ) R10 (r2 )Y00 (q 2 ,f2 ) no sign flip if Pˆexch acts on this thing!!
• therefore the spin state must be antisymmetric: the electrons must
1
be in the singlet state: ground state
    
c  00 
00
2
The ground state of He
• how else might we do it? Build the antisymmetric space y(r)
y (r )  12 [ R10 (r1 )Y00 (q1 ,f1 ) R10 (r2 )Y00 (q 2 ,f2 )  R10 (r2 )Y00 (q 2 ,f2 ) R10 (r1 )Y00 (q1 ,f1 )]  0
• that’s not good  ground state of He is a spin singlet, symmetric
in space (separable!) y (r ) c  1 R10 (r1 )Y00 (q1 ,f1 ) R10 (r2 )Y00 (q 2 ,f2 )   
2
• one refers to this as the paired-spin (up&down) 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

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:
y (r ) 
1
[ R10 (r1 )Y00 (q1 ,f1 ) R20 (r2 )Y00 (q 2 ,f2 )  R10 (r2 )Y00 (q 2 ,f2 ) R20 (r1 )Y00 (q1 ,f1 )]
2
• 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)
y (r ) 
1
[ R10 (r1 )Y00 (q1 ,f1 ) R20 (r2 )Y00 (q 2 ,f2 ) 
2
R10 (r2 )Y00 (q 2 ,f2 ) R20 (r1 )Y00 (q1 ,f1 )]
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
The Slater determinant for He and Li
• for helium ground state the determinant captures both terms:
full wavefunction 
:
1
2

R10 (r1 )Y00 (q1 ,f1 ) R10 (r2 )Y00 (q 2 ,f2 ) 
1
y (1)y 100(2)
2! 100
 
12
 
12

12
 
12

 y (1) 
y 100 (1)  
1
100

1
1
det
y 100 ( 2) 
2!
y 100 ( 2)  
2
2

• for lithium ground state we can express full wavefunction fw as a
determinant to capture all six terms (put third electron up in 2s):
fw 


y
(
1
)

y
(
1
)

y
(
1
)

100
1
s
100
1
s
200
2
s

1
1
1
1
det y 100 ( 2 ) 1 s
y 100 ( 2 ) 1 s
y 200 ( 2 )  2 s 

2
2
2
3!
y (3) 
y 100 (3) 1 s
y 200 (3)  2 s 
1s
 100
3
3
3

1 


y 200 ( 2 )y 100 (3)  2 s 1 s
y 100 (1) 1 s y 100 ( 2 )y 200 (3) 1 s  2 s

1 
23
23 
3! 

1 


y 100 ( 2 )y 200 (3) 1 s  2 s
y 100 (1) 1 s y 200 ( 2 )y 100 ( 3)  2 s 1 s

1
23
23




3!
1 



y 100 ( 2 )y 100 (3) 1 s 1 s
y 200 (1)  2 s y 100 ( 2 )y 100 (3) 1 s 1 s

1 
23
23 
3! 
The overall antisymmetry is tangled up in spin/space
fw 
1
y (1)y 100(2)y 200(3)
3 100
1s 1s  2 s
123

1
y (1)y 200(2)y 100(3)
3 100
1s  2 s 1s

1
y (1)y 100(2)y 100(3)
3 200
 2 s 1s 1s

123
123
1
y (1)y 200(2)y 100(3)
3 100
1s  2 s 1s
123

1
y (1)y 100(2)y 200(3)
3 100
1s 1s  2 s
123

1
y (1)y 100(2)y 100(3)
3 200
 2 s 1s 1s
123
• 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 Hund’s 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?
Term symbols and spectroscopic notation
• for historical reasons we use term symbols to denote the ground
state electron configurations: 2S+1LJ
• L is the total orbital angular momentum quantum number, so
orbital angular momentummagnitude L  L(L  1)2
• it is denoted by letters S, P, D, F…(confusing!!!)
• S is the total spin angular momentum quantum number, so
spin angular momentummagnitude S  S (S  1)2
• 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 2S+1 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
Hund’s 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 halffilled, the lowest energy state has J = L + S
• physical cause: spin-orbit coupling (a magnetic effect)
Applying Hund’s 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 can’t 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
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)
• let’s do these 10 and call it a day..
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
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 spin’s 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
Most of the rest of the story
•