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Quantum Mechanics – Applications and Atomic Structures
Jeong Won KangDepartment of Chemical Engineering
Korea University
Applied Statistical Mechanics Lecture Note - 3
Subjects
Three Basic Types of Motions (single particle) Translational Motions Vibrational Motions Rotational Motions
Atomic Structures Electronic Structures of Atoms One-electron atom / Many-electron atom
Free Translational Motion
ψψ Edxd
m=− 2
22
2
ψψ EH =
All values of energies are possible (all values of k) Momentum Position See next page
2
22
2 dxd
mH −=
ikxikxk BeAe −+=ψ
OR
Solutions
mkEk 2
22=
kpe xikx +=→ kpe x
ikx −=→−
Probability Density
B = 0
A = 0
A = B
ikxAe=ψ
ikxBe−=ψ
kxAcos2=ψ
22 A=ψ
22 B=ψ
kxA 22 cos4=ψnodes
A particle in a box
m
Consider a particle of mass m is confined between two walls
LxxVLxV
==∞=<<=
and 0for 0for 0
ψψψ ExVdxd
m=+− )(
2 2
22
kxDkxCk cossin +=ψ mkEk 2
22=
All the same solution as the free particlebut with different boundary condition
A particle in a box
Applying boundary condition
Normalization
0)(0)0(
====
Lxx
ψψ
kxDkxCk cossin +=ψ
0=D0sin =kLC πnkL =
1,2,3,...n )/sin()( == LxnCxn πψ
12
)/(sin2
0
22
0
2 === LCdxLxnCdx
LL
n πψ
2/12
=
LC 1,2,3,...n )/sin(2)(
2/1
=
= Lxn
Lxn πψ mL
hnEn 8
22
=
n cannot be zero
A particle in a box- Properties of the solutions
n : Quantum number (allowable state)
Momentum
1,2,3,...n )/sin(2)(2/1
=
= Lxn
Lxn πψ
Lnk)-e(e
LiLxn
Lx -ikxikx
nππψ =
=
= 2
21)/sin(2)(
2/12/1
kpe xikx +=→ kpe x
ikx −=→−
Probability : 1/2 Probability : 1/2
A particle in a box- Properties of the solutions
n cannot be zero (n = 1,2,3,…) Lowest energy of a particle (zero-point energy) If a particle is confined in a finite region (particle’s
location is not indefinite), momentum cannot be zero
If the wave function is to be zero at walls, but smooth, continuous and not zero everywhere, the it must be curved possession of kinetic energy
mLhE
8
2
1 =
Wave function and probability density
)/(sin2)( 22 LxnL
x πψ
=
With n→∞, more uniform distribution: corresponds to classical prediction
“correspondence principle”
Orthogonality and bracket notation
0'* = τψψ dnn
1 0' == nnnn
)'( 0''* nnnndnn ≠== τψψ
'' nnnn δ=
Orthogonality
Dirac Bracket Notation
<n| “BRA”|n’> “KET”
Kronecker delta
Wave functions corresponding to differentenergies are orthogonal
Orthonormal=Orthogonal + Normalized
Motion in two dimension
ψψψ Eyxm
=
∂∂+
∂∂− 2
2
2
22
2 )()(),( yYxXyx =ψ
XEdx
Xdm X=− 2
22
2
YEdx
Ydm Y=− 2
22
2
YX EEE +=
1
1
2/1
1
sin2)(1 L
xnL
xX nπ
=
2
2
2/1
2
sin2)(2 L
ynL
yYnπ
=
2
2
1
1
2/1
2
2/1
1, sinsin22),(
21 Lyn
Lxn
LLyxnn
ππψ
= m
hLn
LnE nn 2
2
22
22
21
21
, 21
+=
Separation of Variables
Solution
The wave functions for a particle confined to a rectangular surface
2,2 21 == nn1,1 21 == nn 1,2 21 == nn2,1 21 == nn
Degeneracy
Degenerate : Two or more wave functions correspond to the same energy If L1=L2 then
2
2
2,1 85mLhE =
Ly
Lx
Lyx ππψ 2sinsin2),(2,1
=
Ly
Lx
Lyx ππψ sin2sin2),(1,2
= 2
2
1,2 85mLhE =
the same energywith differentwave functions
Degeneracy
Ly
Lx
Lyx ππψ 2sinsin2),(2,1
=
Ly
Lx
Lyx ππψ sin2sin2),(1,2
=
degeneratenot are functions wave the, if 21 LL ≠The degeneracy can be traced to the symmetry of the system
Motion in three dimension
ψψψψ Ezyxm
=
∂∂+
∂∂+
∂∂− 2
2
2
2
2
22
2
3
3
2
2
1
1
2/1
3
2/1
2
2/1
1,, sinsinsin222),,(
321 Lyn
Lyn
Lxn
LLLzyxnnn
πππψ
=
mh
Ln
Ln
LnE nnn 2
2
23
23
22
22
21
21
,, 321
++=
Solution
Tunneling
If the potential energy of a particle does not rise to infinity when it is in the wall of the container and E < V , the wave function does not decay to zero
The particle might be found outside the container (leakage by penetration through forbidden zone)
Use of tunneling
STM (Scanning Tunneling Microscopy)
5 nm
fu
Si(111)-7x7
Vibrational Motion
Harmonic Motion
Potential Energy
Schrödinger equation
kxF −=
2
21 kxV =
ψψ Ekxdxd
m=+− 2
2
22
21
2
m1 m2
k : force constant
21
21
mmmm
+=μ
reduced mass
Solutions
αxy =
Wave Functions
Energy Levels
4/12
=
mkα
,...2,1,0 )21(
2/1
=
=+= v
mkvEv ωω
2/2
)( yvvv eyHN −=ψ
Energy Levels,...2,1,0 )
21(
2/1
=
=+= v
mkvEv ωω
1 ω=−+ vv EE
21
0 ω=E
Reason for zero-point energy the particle is confined : position is not
uncertain, momentum cannot be zero
2
21 kxV =
Potential energy
Quantization of energy : comes from B.C
±∞== xv at 0ψ
Wave functions
Quantum tunneling
The properties of oscillators
Expectation values
Mean displacement and mean square displacement
Mean potential energy and mean kinetic energy
Virial Theorem If the potential energy of a paticle has the form V=axb, then its mean
potential and kinetic energies are related by
dxvv vv∞
∞−
Ω=Ω=Ω ψψ ˆˆ *
2/12
)()
21(
mkvx +=0=x
2/12
)()
21(
21
21
mkvkxV +==
VbEK =2
vEV21= vK EE
21=
Rotation in Two Dimensions : Particles on a ring
A Rotational motion can be described by its angular momentum J J : vector Rate at which a particle circulates Direction : the axis of rotation
A particle of mass m constrained to move in a circular path of radius r in xy-plane V = 0 everywhere Total Energy = Kinetic Energy
locityangular ve : inertial ofmoment : 2
ω
ωmrII
IJ=
=
IJE
prJmpE
z
z
2
2/
2
2
=
±==
Rotation in Two Dimension
Wave Functions
Energy Levels
,...2,1,0 ±±=lm
Im
IJE lz
ml 22
222 ==
2/1)2()(
πφψ
φl
l
im
me=
Schrödinger eqn.
)()2( φψπφψll mm =+ψ
φψ
22
2 2
IEdd =
B.C
The wave function have to be
single-valued
Wave functions
States are doubly degenerate except for ml = 0
The real parts of wave functions
Rotation in Three dimension
A particle of mass m free to move anywhere on the 3D surface or a sphere
Colatitude (여위도) : θ Azimuth (방위각) : φ
Rotation in Three dimension
llllml −−−= ,...,2,1,
harmonics spherical : ),(, φθlmlY
Wave Functions
Energy Levels
,...,2,1,0=l
IllE
2)1( +=
Schrödinger eqn.
ψψ Em
=∇− 2
2
Wave functions
(2l + 1) fold degeneracy
Orbital momentum quantum number
Magnetic Quantum number
llllml −−−= ,...,2,1,
,...,2,1,0=l
Location of angular nodesWave functions
Angular Momentum
The energy of a rotating particle Classically, Quantum mechanical
Angular MomentumMagnitude of angular momentum Z-component of angular momentum
IJE2
2
=
,...2,1,0 2
)1(2
=+= lI
llE
{ } 2/1)1( hll +=
lm=
Space Quantization
The orientation of rotating body is quantized : rotating body may not take up an arbitrary orientation with respect to some specified axis
Space Quantization
The vector model
lx, ly, lz do not commute with each other. (lx, ly, lz are complementary observables)
uncertainty principle forbids the simultaneous, exact specification of more than one component (unless l=0).
-If lz is known, impossible to ascribe values to lx, ly.
lx = h/2πi {-sinφ(∂/∂θ) -cotθ cosφ(∂/∂ϕ)}
ly = h/2πi {cosφ(∂/∂θ) -cotθ sinφ(∂/∂ϕ)}
ly = h/2πi (∂/∂ϕ)
Angular momentum L (lx, ly, lz)
The Vector Model
The vector representing the state of angular momentum lies with its tip on any point on the mouth of the cone.
Experiment of Stern-Gerlach
Otto Stern and Walther Gerlach (1921) Shot a beam of silver atoms through an inhomogeneous magnetic field Evidence of space quantization
The classically expected result
The Observed outcomingusing silver atoms
Spin
Stern and Gerlach observed two bands using silver atoms (2) The result conflicts with prediction : (2l+1) orientation (l
must be an integer) Angular momentum due to the motion of the electron
about its own axis : spin Spin magnetic number : ms
ms =s, s-1,…,-s Spin angular momentum
Electron spin s = ½ Only two states
866.043 2/1
=
=
Fermions and Bosons
Electrons : s=1/2 Photons : s=1
Particles with half-integral spin (s=1/2)Elementary particles that constitute matters Electrons, nucleus
Fermions
Particles with integral spin (s=0,1,…)Responsible for the forces that binds fermions Photons
Bosons
Atomic Structureand Atomic Spectra
Electronic Structure of Atoms One Electron Atom : hydrogen atomMany-Electron Atom (polyelectronic atom)
Spectroscopy Experimental Technique to determine electronic structure of atoms. Spectrum
• Intensity vs. frequency (ν), wavelength (λ), wave number (ν/c)
Topics
Structure and Spectra of Hydrogen Atoms
Electric discharge is passed through gaseous hydrogen , H2 molecules and H atoms emit lights of discrete frequencies
Spectra of hydrogenic atoms
Balmer, Lyman and Paschen Series ( J. Rydberg) n1 = 1 (Lyman) n1 = 2 (Balmer) n1 = 3 (Paschen) n2 = n1 + 1, n1 +2 , … RH = 109667 cm -1 (Rydberg constant)
Ritz combination principle The wave number of any spectral line is the difference between two terms
−= 2
221
11~nn
RHν
21~ TT −=ν2n
RT Hn =
The wave lengths can be correlated by two integers Two different states
Spectra of hydrogenic atoms
Ritz combination principle Transition of one energy level to another level with emission of energy as photon
Bohr frequency condition
21 hcThcThvE −==Δ
원자에서 방출되거나 흡수된 electromagnetic radiation 은
주어진 특정 양자수로 제한된다.
따라서 원자들은 몇 가지 주어진 상태만을 가질 수 있음을
알 수 있다.
남은 문제는 양자역학을 이용하여 허용된 에너지 레벨을
구하는 것이다.
The Structure of Hydrogenic Atoms
Coulombic potential between an electron and hydrogen atom ( Z : atomic number , nucleus charge = Ze)
Hamiltonian of an electron + a nucleus
reZeV
0
2
4π−=
rZe
mmVEEH N
Ne
enucleusKelectronK
0
22
22
2
,, 422ˆˆ
πε−∇−∇−=++=
Electron coordinate
Nucleus coordinate
Separation of Internal motion
Full Schrödinger equation must be separated into two equations Atom as a whole through the space Motion of electron around the nucleus
Separation of relative motion of electron from the motion of atom as a whole (Justification 13.1)
rZeH
0
22
2
42 πεμ−∇−=
eNe mmm1111 ≈+=
μ
The Schrödinger Equation
ψψ EH = ψψπε
ψμ
Er
Ze =−∇−0
22
2
42
YllY )1(2 +−=Λ
2
2
0
2
2)1(
4 rll
rZeVeff μπε
++−=
ERRVdrdR
RdrRd
eff =−
+− 2
2 2
22
μ
),()(),,( φθφθψ YrRr =
Separation of Variable : Energy is centrosymmetricEnergy is not affeced by
angular component (Y)(Justification 13.2)
R : Radial Wave EquationNew solution required
Y : All the same equation as in Chap.12 (Section 12.7, the particle on a sphere)Spherical Harmonics
The Radial Solutions
0
2aZr=ρ
ERRVdrdR
RdrRd
eff =−
+− 2
2 2
22
μ
Solution
nln
l
lnln eLn
NrR 2/,,, )()( ρρρ −
=
r)in lexponentia (decaying r)in polynomial()( ×=rR
...3,2,1 with 32 222
02
42
=−= nne
eZEnπ
μAllowed Energies
Radial Solution
2
20
04
ema
e
πε=Reduced distance Bohr Radius
Associated Laguerre polynomials
Hydrogenic Radial Wavefunctions
The Radial Solutions
Atomic Orbital and Their Energies
Quantum numbers
n : Principal quantum number (n=1,2,3,…)• Determines the energies of the electron
an electron with quantum number l has angluar momentum
An electron with quantum number ml has z-component of angular momentum
lmln ,,
...3,2,1 with 32 222
02
42
=−= nne
eZEnπ
μ
{ } 1,..,1,0 with 1)l(l 1/2 −=+ nl
l±±±= ,...,2,1,0m with m ll
Shell
Sub Shell
Bound / Unbound State Bound State : negative energy Unbound State : positive energy (not quantized)
Ryberg const. for hydrogen atom
Ionization energy Mininum energy required to remove an electron
form the ground state
32 22
02
4
eehcR H
H πμ−=
32 22
02
4
1e
ehcRE HH π
μ=−=
Energy of hydrogen at ground state n=1
HhcRI =
Ionization Energy
The energy levels
Shells and Subshells Shell
n = 1 (K), 2 (L), 3 (M), 4(N) Subshell ( l=0,…. , n-1)
l = 0 (s), 1 (p), 2 (d), 3(f), 4(g), 5 (h), 6 (i) Examples
n = 1• l = 0 only 1s (1)
n = 2• l = 0, 1 2s (1) , 2p (3)
n = 3• l = 0, 1, 2 3s (1), 3p (3), 3d (5)
number of orbitals in nth shell : n2
n2 –f old degenerate
ml s are limited to the value -l,..,0, …+l
Ways to depicting probability density
Electron densities (Density Shading)
Boundary surface (within 90 % of electron probability)
Radial Nodes
Wave function becomes zero ( R(r) = 0 ) For 2s orbital :
For 3s orbital :
Zar /2 0=
Zar /90.1 0= Zar /10.7 0=
0
0
Radial Distribution Function
Wave Function probability of finding an electron in any region
Probability Density (1s)
Probability at any radius r = P (r ) dr
Radial Wave Function : R(r )
Radial Distribution Function : P(r) dr 을 곱하면 확률이 된다. 1s orbital 에 대하여 ,
0/22 aZre−∝ψ
22 )()( rRrrP =
2ψ
224)( ψπrrP =
0/2230
34)( aZreraZrP −=
Radial Distribution Function
Bohr radius
Wave function Radial Distribution Function
p-Orbital
Nonzero angular momentum l > 0
p-orbital l = 1
• ml = 0
• ml = + 1
• ml = -1
lr∝ψlr∝ψ
zpψ
ypψxpψ
D-orbital
n = 3l = 2ml = +2, +1, 0,1,2
Spectroscopic transitions and Selection rules
Transition (change of state)
All possible transitions are not permissible Photon has intrinsic spin angular momentum : s = 1 d orbital (l=2) s orbital (l=0) (X) forbidden
• Photon cannot carry away enough angular momentum
...3,2,1 with 32 222
02
42
=−= nne
eZEnπ
μn1, l1,ml1 n2, l2,ml2
PhotonhvE =Δ
Selection rule for hydrogenic atoms
Selection rule Allowed Forbidden
Grotrian diagram
1,0 ±=Δ lm1±=Δl
Structures of Many-Electron Atoms
The Schrödinger equation for many electron-atoms Highly complicated All electrons interact with one another Even for helium, approximations are required
Approaches Simple approach based on H atom
• Orbital Approximation Numerical computation technique
• Hartree Fock self consistent field (SCF) orbital
Pauli exclusion principle
Quantum numbersPrincipal quantum number : n Orbital quantum number : l Magnetic quantum number : ml
Spin quantum number : ms
Two electrons in atomic structure can never have all four quantum numbers in common
All four quantum number “Occupied”
Penetration and Shielding
Unlike hydrogenic atoms 2s and 2p orbitals are not degenerate in many-electron atoms electrons in s orbitals generally lie lower energy
than p orbital electrons in many-electron atoms experiences
repulsion form all other electrons shielding
Effective nuclear charge, shielding constant σ−= ZZeff
constant shielding : chargenuclear effective :
σeffZ
Penetration and Shielding
An s electrons has a greater “penetration” through inner shell than p electrons
Energies of electrons in the same shell s < p < d < f
Valence electrons electrons in the outer most shell of an atom in its
ground state largely responsible for chemical bonds
The building-up principle
The building-up principle (Aufbau principle ) The order of occupation of electrons 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s There are complicating effects arising from electron-electron
repulsion (when the orbitals have very similar energies) Electrons occupy different orbitals of a given subshell
before doubly occupying any one of them Example : Carbon
• 1s2 2s2 2p2 1s2 2s2 2px12py
1 (O) 1s2 2s2 2px2 (X)
Hund’s maximum multiplicity principle An atom in its ground state adopts configuration with the greatest
number of unpaired electrons
favorable unfavorable
3d and 4s orbital
Sc atom (Z=21) [Ar]3d3
[Ar]3d24s1
[Ar]3d14s2
3d has lower energy than 4s orbital
3d repulsion is much higher than 4s (average distance from nucleus is smaller in 3d)
Ionization energy
Ionization energy I1 : The minimum energy required to remove an electron from many-
electron atom in the gas phase I2 : The minimum energy required to remove a second electron from
many-electron atom in the gas phase
Li :2s1
Be :2s2
B : 2s2p1
Self-consistent field orbital
Potential energy of electrons
Central difficulty presence of electron-electron interaction Analytical solution is hopeless Numerical techniques are available
• Hartree-Fock Self-consistent field (SCF) procedure (HF method)
+−=i ji ijoio re
ere
ZeV'
,
22
421
4 επεπ
Schrödinger equation for Neon Atom1s22s22p6 , 2p electrons
)(
)()()(
4
)()()(2
)(4
)(2
122
1212
22*
22
12212
2*
22
121
2
1221
2
rE
rdr
rre
rdr
rre
rr
Ze
rm
pp
ii
pi
o
pi
ii
o
po
pe
ψ
ψτψψ
πε
ψτψψπε
ψπε
ψ
=
−
+
−
∇−
Sum over orbitals (1s, 2s)
Kinetic energy of electron
Potential energy (electron – nucleus)
Columbic operator (electron – electron charge density)
Exchange operator (Spin correlation effect)
HF Procedure
There is no hope solving previous eqn. analytically Alternative procedure (numerical solution)
Assume 1s, 2s orbital
Solve 2p
Solve 2s
Solve 1s
Compare E andwave function
Converged Solution
Example
Orbitals of Na using SCF calculation
Quiz
What is a degeneracy ? Explain four quantum number.Why transition between two states are not always
allowed ? Explain difficulties of solving Schrödinger
equations for many-electron atoms.