The one-electron atom H(Z =1), He + (Z =2), Li 2+ (Z =3), ...,U 91+ (Z = 92), ... Electronic Hamilton operator (for point-like clamped nucleus) * : c H el = b T e + b V en = - ~ 2 2m e ∇ 2 - Ze 2 κ 0 r (139) Determination of stationary bound states, i.e. solutions h r | ψ i = ψ(r) with E< 0, to the time-independent Schr¨ odinger equation (with c H el from above): c H el - E ψ(r)= - ~ 2 2m e ∇ 2 - Ze 2 κ 0 r - E ! ψ(r)=0 (140) Firstly, we remove the fundamental constants by switching to ‘atomic units’. This reduces the mathematical work to pure numbers, and eliminates quantities which have experimental uncertainties. * The finite mass of the nucleus can be taken into account by switching from the electron mass m e to a reduced mass μ, where μ -1 = m e -1 + m N -1 and m N is the nuclear mass. FAQC — D. Andrae, Theoretical Chemistry, U Bielefeld — 2004-06-30 2/ 48-5 1. Historical introduction 2. The Schr¨ odinger equation for one-particle problems 3. Mathematical tools for quantum chemistry 4. The postulates of quantum mechanics 5. Atoms and the ‘periodic’ table of chemical elements 6. Diatomic molecules 7. Ten-electron systems from the second row 8. More complicated molecules FAQC — D. Andrae, Theoretical Chemistry, U Bielefeld — 2004-06-30 1/ 48-5
24
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Atoms and the ‘periodic’ table of chemical elements · 2p 2 1 0 1 2 q Z 6 x 2e x=2 3s 3 0 2 1 3 q Z 3 x(3 3x+x2=2)e x=2 3p 3 1 1 1 3 q Z 24x 2(4 x)e x=2 3d 3 2 0 1 3 q Z 120x
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The one-electron atom
H (Z = 1), He+ (Z = 2), Li2+ (Z = 3), . . ., U91+ (Z = 92), . . .
Electronic Hamilton operator (for point-like clamped nucleus)∗:
Hel = Te + Ven = − ~2
2me∇2 − Z e
2
κ0 r(139)
Determination of stationary bound states, i.e. solutions 〈 r |ψ 〉 = ψ(r)
with E < 0, to the time-independent Schrodinger equation (with Helfrom above):
(Hel −E
)ψ(r) =
(− ~2
2me∇2 − Z e
2
κ0 r−E
)ψ(r) = 0 (140)
Firstly, we remove the fundamental constants by switching to ‘atomic
units’. This reduces the mathematical work to pure numbers, and
eliminates quantities which have experimental uncertainties.
∗The finite mass of the nucleus can be taken into account by switching from the electron mass me
to a reduced mass µ, where µ−1 = me−1 +mN
−1 and mN is the nuclear mass.
FAQC — D. Andrae, Theoretical Chemistry, U Bielefeld — 2004-06-30 2/ 48-5
1. Historical introduction
2. The Schrodinger equation for one-particle problems
3. Mathematical tools for quantum chemistry
4. The postulates of quantum mechanics
5. Atoms and the ‘periodic’ table of chemical elements
6. Diatomic molecules
7. Ten-electron systems from the second row
8. More complicated molecules
FAQC — D. Andrae, Theoretical Chemistry, U Bielefeld — 2004-06-30 1/ 48-5
Now the Schrodinger equation reads
(Hel − E
)ψ(r) =
(− 1
2∇2 − Z
r− E
)ψ(r) = 0 (141)
which is transformed from cartesian coordinates (x, y, z) to spherical
coordinates (r, θ, φ), with 0 ≤ r <∞, 0 ≤ θ ≤ π, and 0 ≤ φ ≤ 2π.
Laplace operator ∆ and squared angular momentum operator l2 (in
atomic units) in spherical coordinates:
∆ = ∇2 =1
r2∂
∂rr2∂
∂r− l2
r2=
1
r
∂2
∂r2r − l2
r2(142)
l2 = −
1
sin θ
∂
∂θsin θ
∂
∂θ+
1
sin2 θ
∂2
∂φ2
(143)
Separation ansatz for the state function:
ψ(r) = ψ(r, θ, φ) = R(r)Y (θ, φ) (144)
(this is always possible for ‘central fields’, i.e. V = V (r)).
FAQC — D. Andrae, Theoretical Chemistry, U Bielefeld — 2004-06-30 4/ 48-5
Atomic unitsa
Physical quantity Symbol (name) Value in SI unitsb
mass me 9.1093826(16) × 10−31 kgcharge e 1.60217653(14) × 10−19 Cangular momentum, action ~ 1.05457168(18) × 10−34 J sel. permittivity 4πε0 κ0 1.112650056 . . . × 10−10 F m−1
length κ0~2/(mee2) = ~/(αmec) a0 (bohr) 5.291772 × 10−11 m
time ~/Eh = ~/(α2mec2) 2.418884 × 10−17 s
velocity a0Eh/~ = αc 2.187691 × 106 m s−1
linear momentum ~/a0 = αmec 1.992852 × 10−24 kgm s−1
force Eh/a0 8.238723 × 10−8 Nenergy e2/(κ0a0) = α2mec
2 Eh (hartree) 4.359744 × 10−18 J
power Eh2/~ 1.802378 × 10−1 W
charge density e/a03 1.081202 × 1012 C m−3
el. current eEh/~ 6.623618 × 10−3 Ael. potential Eh/e 2.721138 × 101 Vel. capacitance κ0a0 5.887891 × 10−21 Fel. resistance ~/e2 4.108236 × 103 Ωel. field strength (E) Eh/(ea0) 5.142206 × 1011 V m−1
el. displacement (D) e/a02 5.721476 × 101 C m−2
el. dipole moment ea0 8.478353 × 10−30 C mel. quadrupole moment ea0
2 4.486551 × 10−40 C m2
el. polarizability (ea0)2/Eh 1.648777 × 10−41 C2 m2 J−1
FAQC — D. Andrae, Theoretical Chemistry, U Bielefeld — 2004-06-30 15/ 48-5
The ‘periodic’ table of the chemical elements
(2004)
1
2
3
4
5
6
7
1
1
H
3
Li
11
Na
19
K
37
Rb
55
Cs
87
Fr
2
4
Be12
Mg
20
Ca
38
Sr
56
Ba
88
Ra
3
21
Sc
39
Y
∗
∗∗
∗ 57
La58
Ce59
Pr60
Nd61
Pm62
Sm63
Eu64
Gd65
Tb66
Dy67
Ho68
Er69
Tm70
Yb71
Lu∗∗ 89
Ac90
Th91
Pa92
U93
Np94
Pu95
Am96
Cm97
Bk98
Cf99
Es100
Fm101
Md102
No103
Lr
4
22
Ti
40
Zr
72
Hf
104
Rf
5
23
V
41
Nb
73
Ta
105
Db
6
24
Cr
42
Mo
74
W106
Sg
7
25
Mn
43
Tc
75
Re
107
Bh
8
26
Fe
44
Ru
76
Os
108
Hs
9
27
Co
45
Rh
77
Ir
109
Mt
10
28
Ni
46
Pd
78
Pt
110
Ds
11
29
Cu47
Ag
79
Au
111
X
12
30
Zn
48
Cd80
Hg112
(X)
13
5
B
13
Al
31
Ga
49
In
81
Tl
113
14
6
C
14
Si
32
Ge
50
Sn
82
Pb114
(X)
15
7
N
15
P
33
As
51
Sb
83
Bi
115
16
8
O
16
S
34
Se
52
Te
84
Po116
(X)
17
9
F
17
Cl
35
Br
53
I
85
At
117
18
2
He
10
Ne
18
Ar
36
Kr
54
Xe
86
Rn118
(?)
FAQC — D. Andrae, Theoretical Chemistry, U Bielefeld — 2004-06-30 18/ 48-5
Degree of degeneracy¶:
m degeneracy: gl =l∑
m=−l1 = 2l+ 1 (160)
l degeneracy (without spin): gn =n−1∑
l=0
gl = n2 (161)
l degeneracy (spin included): gsn = 2 gn = 2n2 (162)
The value gsn essentially determines the length of the rows (‘periods’)
in the table of chemical elements (2, 8, 18, 32), whereas the value
2 gl = 2(2l+1) determines the block structure of the ‘periodic’ table
(s-, p-, d-, and f-block for l = 0,1,2,3, respectively).
¶ The degeneracy with respect to l, eq. (161), is a special property of the one-electron atom (withpoint-like nucleus), and is not present in many-electron atoms. For example, the 2s and 2p statesof a one-electron atom are degenerate, i.e. they have the same energy, but the ‘orbital energies’for the 2s and 2p orbitals in any state of a many-electron atom are always different.
FAQC — D. Andrae, Theoretical Chemistry, U Bielefeld — 2004-06-30 17/ 48-5
FAQC — D. Andrae, Theoretical Chemistry, U Bielefeld — 2004-06-30 20/ 48-5
The variation principle at work (I)
The ground state 1s1 2S of the one-electron atom
Definition of the system under consideration:
H = T + V , T = − 1
2∆ , V = Vnuc(r) = − Z
r
A set of trial functions‖:
1. Slater function (ζ > 0):
ψS = NS exp (−ζr)
2. Gauß function (α > 0):
ψG = NG exp (−αr2)
3. Lorentz function (a > 0):
ψL = NL
[1 + (ar)2
]−1
4. Preuß function (c > 0):
ψP = NP
[1 + cr
]−2
‖Refs.: C. Zener, Phys. Rev. 36 (1930) 51, J. C. Slater, Phys. Rev. 36 (1930)57 (Slater function) — S. F. Boys, Proc. R. Soc. London A 200 (1950) 542,H. Preuß, Z. Naturforsch. A 11 (1956) 823 (Gauß function) — H. Preuß, Z.Naturforsch. A 13 (1958) 439 (Preuß function).
FAQC — D. Andrae, Theoretical Chemistry, U Bielefeld — 2004-06-30 19/ 48-5
1. The Slater function ψS = NS exp (−ζr), ζopt =?:
〈rk〉 = N2S 4π
∫ ∞0
dr rk+2e−2ζr = N2S 4π
Γ(k+ 3)
(2ζ)k+3(k > −3)
〈1〉 = 〈r0〉 = N2Sπ
ζ3≡ 1 ⇒ NS =
√ζ3/π
〈rk〉 = Γ(k+ 3)
2k+1
1
ζk
〈T 〉 = +1
2N2
S4π
ζ
∫ ∞0
dx
(d
dxx e−x
)2
=1
2N2
S4π
ζ
∫ ∞0
dx (1− x)2e−2x =1
2ζ2
〈V 〉 = − Z ζ
〈E 〉 = 1
2ζ2 − Z ζ ⇒ d〈E 〉
dζ= ζ − Z = 0 ⇒ ζopt = Z
〈E 〉Z2
= x
(1
2x− 1
)(x = ζ/Z) ⇒ Emin = − 1
2Z2
FAQC — D. Andrae, Theoretical Chemistry, U Bielefeld — 2004-06-30 22/ 48-5
Estimate for the ground state energy:
〈E 〉 = 〈H 〉 = 〈ψ | H |ψ 〉 =∫ψ(r) H ψ(r) dr = 〈T 〉+ 〈V 〉
〈T 〉 = − 1
2〈ψ |∆ |ψ 〉 = +
1
2〈∇ψ |∇ψ 〉 , 〈V 〉 = − Z 〈r−1〉
Mathematical preliminaries:
• The beta function:
B(a, b) =Γ(a) Γ(b)
Γ(a+ b)= B(b, a)
B(a, b) =∫ 1
0ta−1(1− t)b−1 dt =
∫ ∞0
ta−1
(1 + t)a+bdt (a > 0, b > 0)
• Useful integral formulas:∫ ∞0
xs−1 exp (−p xn) dx =1
n
Γ(s/n)
ps/n(p > 0, s/n > 0)
∫ ∞0
xs−1
(1 + xn)pdx =
1
nB(p− s/n, s/n) (s/n > 0, np− s > 0)
FAQC — D. Andrae, Theoretical Chemistry, U Bielefeld — 2004-06-30 21/ 48-5
3. The Lorentz function ψL = NL
[1 + (ar)2
]−1, aopt =?:
〈rk〉 = N2L 4π
∫ ∞0
rk+2 dr
(1 + (ar)2)2= N2
L4π
ak+3
1
2B(
1− k2,k+ 3
2)
〈1〉 = 〈r0〉 = N2Lπ2
a3≡ 1 ⇒ NL =
√a3/π
〈rk〉 = 2
π
1
akB(
1− k2,k+ 3
2) (−3 < k < 1)
〈T 〉 = +1
2N2
L4π
a
∫ ∞0
dx
(d
dx
x
1 + x2
)2
=1
2N2
L4π
a
∫ ∞0
dx(1− x2)2(1 + x2)4
=1
4a2
〈V 〉 = − 2Z a/π
〈E 〉 = 1
4a2 − 2Z
a
π⇒ d〈E 〉
da=
1
2a− Z 2
π= 0 ⇒ aopt =
4Z
π〈E 〉Z2
= x
(1
4x− 2
π
)(x = a/Z) ⇒ Emin = − 4
π2Z2
FAQC — D. Andrae, Theoretical Chemistry, U Bielefeld — 2004-06-30 24/ 48-5
2. The Gauß function ψG = NG exp (−αr2), αopt =?:
〈rk〉 = N2G 4π
∫ ∞0
dr rk+2e−2αr2 = N2G 4π
1
2
Γ(k+32 )
(2α)k+32
(k > −3)
〈1〉 = 〈r0〉 = N2G
(π
2α
)3/2≡ 1 ⇒ NG =
(2α
π
)3/4
〈rk〉 = 2√π
Γ(k+32 )
(2α)k/2
〈T 〉 = +1
2N2
G4π√α
∫ ∞0
dx
(d
dxx e−x
2)2
=1
2N2
G4π√α
∫ ∞0
dx (1− 2x2)2e−2x2 =3
2α
〈V 〉 = − 2Z√
2α/π
〈E 〉 = 3
2α− 2Z
√2α
π⇒ d〈E 〉
dα=
3
2− Z
√2
πα= 0 ⇒ αopt =
8Z2
9π
〈E 〉Z2
= x
(3
2x− 2
√2
π
)(x =
√α/Z) ⇒ Emin = − 4
3πZ2
FAQC — D. Andrae, Theoretical Chemistry, U Bielefeld — 2004-06-30 23/ 48-5
Variation of the ground state energy 〈E 〉 of the one-electron atom
FAQC — D. Andrae, Theoretical Chemistry, U Bielefeld — 2004-07-14 29/ 48-5
For the spatial part f(r1, r2) a suitable choice must be made. For S
states we can simplify further to f(r1, r2, r12), or equivalently f(s, t, u)
with s = r1 + r2, t = r1 − r2, and u = r12.
Variational results for the helium ground state 1s2 1S a.
f(s, t, u) −Eopt/a.u.
e−ζr1e−ζr2 = e−ζs 2.8477
ϕ(r1)ϕ(r2) 2.8617 b
e−ζr1e−ηr2 + e−ηr1e−ζr2 2.8757 c
e−ζs+cu 2.8896
e−ζs (1 + cu) 2.8911 d
e−ζs+cu cosh (at) 2.8994e−ζs
(c0 + c1u+ c2t2 + c3s+ c4s2 + c5u2
)2.9032
exact 2.9037 e
a E. A. Hylleraas, Z. Phys. 54 (1929) 347
b C. Froese Fischer: The Hartree-Fock method for atoms. Wiley, New York, 1977
c C. Eckart, Phys. Rev. 36 (1930) 878
d W.-K. Li, J. Chem. Educ. 64 (1987) 128
e K. Frankowski, C. L. Pekeris, Phys. Rev. 146 (1966) 46
FAQC — D. Andrae, Theoretical Chemistry, U Bielefeld — 2004-07-14 32/ 48-5
The two-electron atom
H− (Z = 1), He (Z = 2), Li+ (Z = 3), . . ., U90+ (Z = 92), . . .
Hamilton operator (in atomic units):
Hel = − 1
2∇2
1 −1
2∇2
2 −Z
r1− Z
r2+
1
r12= h1 + h2 +
1
r12(169)
General structure of state functions for two-electron systems∗∗:
Φ(x1,x2) = f(r1, r2)Θ(σ1, σ2)
Spin part Θ = |SMS 〉: Singlet (S = 0, para-He) or triplet (S = 1,
ortho-He)
|00 〉 = (αβ − βα)/√
2
|11 〉 = αα |10 〉 = (αβ + βα)/√
2 |1− 1 〉 = ββ
∗∗ . . . as long as the Hamilton operator does not act on the spin of the particles.
FAQC — D. Andrae, Theoretical Chemistry, U Bielefeld — 2004-07-14 31/ 48-5
Electron configuration of neutral atoms in the ground state (designated as 2S+1LJ)
1 H 1s12S1/2
2 He 1s21S0
3 Li [He] 2s1 2S1/2
4 Be [He] 2s2 1S0
5 B [He] 2s2 2p1 2P1/2
6 C [He] 2s2 2p2 3P0
7 N [He] 2s2 2p3 4S3/2
8 O [He] 2s2 2p4 3P2
9 F [He] 2s2 2p5 2P3/2
10 Ne [He] 2s2 2p6 1S0
11 Na [Ne] 3s1 2S1/2
12 Mg [Ne] 3s2 1S0
13 Al [Ne] 3s2 3p1 2P1/2
14 Si [Ne] 3s2 3p2 3P0
15 P [Ne] 3s2 3p3 4S3/2
16 S [Ne] 3s2 3p4 3P2
17 Cl [Ne] 3s2 3p5 2P3/2
18 Ar [Ne] 3s2 3p6 1S0
19 K [Ar] 4s1 2S1/2
20 Ca [Ar] 4s2 1S0
21 Sc [Ar] 3d1 4s2 2D3/2
22 Ti [Ar] 3d2 4s2 3F2
23 V [Ar] 3d3 4s2 4F3/2
24 Cr [Ar] 3d5 4s1 7S3
25 Mn [Ar] 3d5 4s2 6S5/2
26 Fe [Ar] 3d6 4s2 5D4
27 Co [Ar] 3d7 4s2 4F9/2
28 Ni [Ar] 3d8 4s2 3F4
29 Cu [Ar] 3d10 4s1 2S1/2
30 Zn [Ar] 3d10 4s2 1S0
31 Ga [Ar] 3d10 4s2 4p1 2P1/2
32 Ge [Ar] 3d10 4s2 4p2 3P0
33 As [Ar] 3d10 4s2 4p3 4S3/2
34 Se [Ar] 3d10 4s2 4p4 3P2
35 Br [Ar] 3d10 4s2 4p5 2P3/2
36 Kr [Ar] 3d10 4s2 4p6 1S0
37 Rb [Kr] 5s1 2S1/2
38 Sr [Kr] 5s2 1S0
39 Y [Kr] 4d1 5s2 2D3/2
40 Zr [Kr] 4d2 5s2 3F2
41 Nb [Kr] 4d4 5s1 6D1/2
42 Mo [Kr] 4d5 5s1 7S3
43 Tc [Kr] 4d5 5s2 6S5/2
44 Ru [Kr] 4d7 5s1 5F5
45 Rh [Kr] 4d8 5s1 4F9/2
46 Pd [Kr] 4d10 1S0
47 Ag [Kr] 4d10 5s1 2S1/2
48 Cd [Kr] 4d10 5s2 1S0
49 In [Kr] 4d10 5s2 5p1 2P1/2
50 Sn [Kr] 4d10 5s2 5p2 3P0
FAQC — D. Andrae, Theoretical Chemistry, U Bielefeld — 2004-07-14 34/ 48-5
The many-electron atom
A part of the knowledge of the state functions of the one-electron
atom can be transferred to the many-electron atom, if the following
assumptions are made††:1. Central field approximation: The electrons in the many-electron
atom are assumed to move in an effective central field Veff,l(r),
so that the orbitals can be written as ψ(r) = R(r) Y (θ, φ), with
Y (θ, φ) = Ylm(θ, φ).
2. Equivalence restriction: The radial parts are assumed to be
independent of the magnetic quantum number m: R(r) = Rnl(r).
The resulting set of radial functions Pnl(r) = r Rnl(r) has to be de-
termined for every state of the many-electron atom, e.g.
- He ground state (singlet): 1s2 1S → P10(r)
- He excited states (singlet or triplet): 1s1 2s1 1,3S → P10(r), P20(r)
- Li ground state (doublet): 1s2 2s1 2S → P10(r), P20(r)
†† In addition to the approximation of the many-electron state function as a Slater determinant (anantisymmetrized product of spin orbitals), or a linear combination thereof.
FAQC — D. Andrae, Theoretical Chemistry, U Bielefeld — 2004-07-14 33/ 48-5
LS terms for electron configurations p2 and p3
p2: 3P (9), 1D (5), 1S (1)(62
)= 15 = 9 + 5 + 1
p3: 4S (4), 2D (10), 2P (6)(63
)= 20 = 4 + 10 + 6
Energy levels of neutral tetravalent atoms from the p-, d-, and f-block
(C, Ti, Ce), within an energy range above lowest ground state level:
∆E = 1eV ≈ 8065.55 cm−1 (Eh = 2h cR∞)
FAQC — D. Andrae, Theoretical Chemistry, U Bielefeld — 2004-07-14 36/ 48-5