relativistic quantum chemistry - start [ctcc...
Post on 30-Mar-2018
223 Views
Preview:
TRANSCRIPT
Relativistic quantum
chemistry
Trond Saue
Laboratoire de Chimie et Physique Quantiques
CNRS/Université de Toulouse (Paul Sabatier)
118 route de Narbonne, 31062 Toulouse (FRANCE)
e-mail: trond.saue@irsamc.ups-tlse.fr
Relativistic Quantum Mechanics I Trond SAUE
The DIRAC code
P rogram• forA tomic• andM olecular
D irectI terativeR elativisticA ll-electronC alculations
• Web site: http://wiki.chem.vu.nl/dirac
• Wave functions: [HF, MP2, RASCI, MCSCF, CCSD(T), FSCCSD]
+ DFT [LDA, GGAs, hybrids]
• HF/DFT: Electric and magnetic properties: expectation values, linear and quadratic response functions,
single excitation energies
Relativistic Quantum Mechanics I Trond SAUE
Periodic table (1871)
eka-aluminium:gallium (1875)
eka-boron: scandium (1879) eka-silicon:germanium (1886)
Relativistic Quantum Mechanics I Trond SAUE
Golodschmidt and Einstein in Norway 1920
Relativistic e�ects
• scalar e�ects
• spin-orbitinteraction
Lorentz factor:
γ =1√
1− v2
c2
Lanthanide contractionV.M. Goldschmidt, T. Barth, G. Lunde:Norske Vidensk. Selsk. Skrifter I Mat. Naturv. Kl. 7, 1 (1925)D. R. Lloyd, J. Chem. Ed. 63 (1986) 503
• La3+ - Lu3+ (117.2 - 100.1 pm)
• Ca2+ - Zn2+ (114 - 88 pm)
• Cu (138 pm) < Au (144 pm) < Ag (153 pm)
P.S.Bagus et al., Chem. Phys. Lett. 33 (1975) 408
Relativistic Quantum Mechanics I Trond SAUE
Ionisation energy of gold
O. Fossgaard, O. Gropen, E. Eliav and T. Saue, J. Chem. Phys. 119 (2003) 9355
Relativistic Quantum Mechanics I Trond SAUE
Electron a�nity of gold
O. Fossgaard, O. Gropen, E. Eliav and T. Saue, J. Chem. Phys. 119 (2003) 9355
4 Gold and caesium are extremes on the electron a�nity scale � 2.309 eV vs. 0.472 eV.
4 CsAu is a semi-conductor with a CsCl crystal structure in the solid state;
it forms an ionic melt. The oxidation state of gold is -I.
Relativistic Quantum Mechanics I Trond SAUE
Spectroscopic constants of CsAu and homologues
O. Fossgaard, O. Gropen, E. Eliav and T. Saue, J. Chem. Phys. 119 (2003) 9355
Method re (pm) ωe (cm−1) ωexe (cm
−1) Dcove (eV) µ (D)
CsAu CCSD(T) rel 326.3 89.4 0.21 2.52 11.73
nrel 357.1 67.9 0.08 1.34 11.05
nrel-ps 376.3 59.9 0.13 1.17 9.47
Exp.[1]a (320) (125) 2.58±0.03
Exp.[1]b - - - 2.53±0.03 -
CsAg CCSD(T) rel 331.6 88.0 0.17 1.51 10.69
nrel 345.9 78.5 0.02 1.26 10.89
CsCu CCSD(T) rel 319.8 101.6 0.09 1.36 10.34
nrel 327.7 97.1 0.18 1.31 10.88
1) B. Busse and K. G. Weil, Ber. Bunsenges. Phys. Chem. 85(1981) 309
Relativistic Quantum Mechanics I Trond SAUE
Scalar relativistic e�ects: hydrogen-like atoms
• In atomic units the average speed of the 1s electron is equal to the nuclear charge
v1s(au) = Z
• The relativistic mass increase of the 1s electron is thus determined by the nuclear charge
m = γme =me√
1− Z2/c2
• The Bohr radius is inversely proportional to electron mass
a0 =4πε0~2
m
• Relativity will contract orbitals of one-electron atoms, e.g.
� Au78+: Z/c = 58%⇒18% relativistic contraction of the 1s orbital
Relativistic Quantum Mechanics I Trond SAUE
Scalar relativistic e�ects: polyelectronic atoms
• The e�ect of the other electrons is e�ectively to screen the nuclear charge:
• The relativistic contraction of orbitals will increase screening of nuclear charge and thusindirectly favor orbital expansion.
• In practice we �nd:
� s, p orbitals : contraction� d, f orbitals : expansion
Relativistic Quantum Mechanics I Trond SAUE
The colour of gold
The colours of silver and gold arerelated to the transition between the(n−1)d and ns bands. For silver thistransition is in the ultraviolet, givingthe metallic luster. For gold it is inthe visible, but only when relativistice�ects are included.
Relativistic Quantum Mechanics I Trond SAUE
Two contrasting neighbours: Gold and mercury
L. J. Norrby, J. Chem. Ed. 68(1991) 110
1064◦C Mp. -39◦C19.32 kJ/mol ∆Hf 13.53 kJ/mol9.29 J/Kmol ∆Sf 9.81 J/Kmol19.32 g/cm3 ρ 13.53 g/cm3
426 kS/m Conductivity 10.4 kS/mdimer Gas phase monomer
[Xe]4f145d106s1 [Xe]4f145d106s2
pseudo halogen pseudo noble gas
Without relativistic e�ects mercury would probably not be a liquid at room temperature !
Relativistic Quantum Mechanics I Trond SAUE
Metal-water interaction
C. Gourlaouen, J.-P. Piquemal, T. Saue and O. Parisel, J. Comp. Chem. 27 (2006) 142
[Ag(H2O)]+:
electrostatic interaction
bonding dominated by charge-dipoleinteraction
[Au(H2O)]+:
orbital interaction
relativistic stabilisation of the Au 6s orbitalinduces charge transfer and covalent bonding
Relativistic Quantum Mechanics I Trond SAUE
Spin-orbit interaction
hso =1
2m2c2s · [(∇V )× p]
V = −Zr→
Z
2m2c2r3s · l
The spin-orbit interaction is not the interaction between spin andangular momentum of an electron. � An electron moving alone in spaceis subject to no spin-orbit interaction !
The basic mechanism of the spin-orbit interaction is magnetic induction:
An electron which moves in a molecular �eld will feel a magnetic �eld in its rest frame,in addition to an electric �eld. The spin-orbit term describes the interaction of the spin ofthe electron with this magnetic �eld due to the relative motion of the charges.
This operator couples the degrees of freedom associated with spin and space and thereforemakes it impossible to treat spin and spatial symmetry separately.
Relativistic Quantum Mechanics I Trond SAUE
Atomic oxygen
The ground state con�guration is 1s22s22p4 which in a non-relativistic framework(LS-coupling) gives rise to three states:
Term L S Possible J values3P 1 1 2,1,01D 2 0 21S 0 0 0
Energy levels (http://physics.nist.gov/PhysRefData/Handbook/Tables/oxygentable1.htm):
Term J Level (cm−1)3P 2 0.000
1 158.2650 226.977
1D 2 15867.8621S 0 33792.583
Relativistic Quantum Mechanics I Trond SAUE
Atomic oxygen emissions in atmospheric aurora
Transition Wavelength(Å) Type Lifetime(s)
Green line 1S0 → 1D2 5577 E2 0.75Red line 1D2 → 3P2 6300 M1 110
Relativistic Quantum Mechanics I Trond SAUE
Molecular oxygen
http://webbook.nist.gov/chemistry/
Term Te(cm−1)
X3Σ−g 0.0a1∆g 7918.1b1Σ+
g 13195.1
The ground state of the oxygen molecule is a triplet. A magnetic interaction such asspin-orbit interaction is required for interaction with singlet states. This is crucial for life !
Relativistic Quantum Mechanics I Trond SAUE
Theoretical model chemistries
The Hamiltonian, relativistic or not, has the same generic form
H =∑i
h(i) +1
2
∑i6=j
g(i, j) + VNN ; VNN =1
2
∑K 6=L
ZKZLRKL
Computational cost: xNy
Relativistic Quantum Mechanics I Trond SAUE
The theory of special relativity
Inertial frames :a coordinate system in which a particle, in theabsence of any forces, move in a straight line(Newton's �rst law). Two inertial frames arerelated by uniform relative velocity.
1. The principle of relativity: The in-variance of physical laws in the universe.
2. The invariance of the speed c oflight.
Relativistic Quantum Mechanics I Trond SAUE
The principle of relativity
The laws of physics are the same in all inertial frames. This implies:
4 Homogeneity of space: no preferred point in space
4 Isotropy of space: no preferred direction in space
4 Homogeneity of time: no preferred point in time
Relativistic Quantum Mechanics I Trond SAUE
Choice of reference frame
Speed of boat with respect to the river bank: 3 km/h
Speed of water with respect to the river bank: 7 km/h
Relativistic Quantum Mechanics I Trond SAUE
Galilean transformation
x = x− vty = y
z = z
t = t
Velocity transformation:
ux = ux − vuy = uy
uz = uz
The laws of Newtonian mechanics, but not the laws of electrodynamics (Maxwell'sequations), are invariant under the Galilean transformation.
Relativistic Quantum Mechanics I Trond SAUE
Invariance of the speed of light c
In an inertial frame, the speed of light c is constant, whether emitted by an object inmovement or at rest.√
(x2 − x1)2 + (y2 − y1)2 + (z2 − z1)2
(t2 − t1)= c =
√(x2 − x1)2 + (y2 − y1)
2 + (z2 − z1)2
(t2 − t1)
Invariance of the interval:
s12 =√c2(t2 − t1)2 − (x2 − x1)2 − (y2 − y1)2 − (z2 − z1)2 = s12
Relativistic Quantum Mechanics I Trond SAUE
Simultaneity: a relative concept
Observer in the train:
tb = ta
Observer on the ground:
tb < ta
Two events that are simultaneous in one inertial frame are generally notso in another inertial frame.
Relativistic Quantum Mechanics I Trond SAUE
Time dilation
Observer in the train:
∆t =h
c
Observer on the ground:
∆t =
√h2 + (v∆t)
2
c
∆t = γ∆t > ∆t; γ =1√
1− v2/c2
Clocks in movement go slower.
Relativistic Quantum Mechanics I Trond SAUE
Reciprocity
The observer on the ground comparesone clock in the train with two clocks on
the ground.
The observer in the train compares oneclock on the ground with two clocks in
the train.
Relativistic Quantum Mechanics I Trond SAUE
Length contraction
Observer in the train:
∆t =2∆x
c
Observer on the ground:
∆t1 =∆x+ v∆t1
c; ∆t2 =
∆x− v∆t2c
∆t = ∆t1 + ∆t2 = 2∆x
cγ2 = γ∆t
∆x = γ∆x
An object in movement is contractedin the direction of movement(but not in other directions !).
Relativistic Quantum Mechanics I Trond SAUE
Lorentz transformation I
x = d+ vt
d =
x; (Galilei)
γ−1x; (Lorentz)
x = γ (x− vt)
x = γ(x+ vt
)t = γ
(t− v
c2x)
Relativistic Quantum Mechanics I Trond SAUE
Lorentz transformation II
x = γ (x− vt)y = y
z = z
t = γ(t− v
c2x)
The Lorentz transformation can be generalized to velocity in any direction. We note thatonly space coordinates parallel to velocity are modi�ed and so we may write
r‖ = γ(r‖ − vt
); r⊥ = r⊥
Noting that we may write r‖ =(r · v)v
v2, we can then combine these relations to
r = r + (γ − 1)(r · v)v
v2− γvt
t = γ
(t− (r · v)
c2
)Relativistic Quantum Mechanics I Trond SAUE
Velocity transformation
Special case: v = (v, 0, 0):
ux =dx
dt=
γ (dx− vdt)γ(dt− v
c2dx) =
ux − v1− uxv/c2
uy =dy
dt=
dy
γ(dt− v
c2dx) =
uyγ (1− uxv/c2)
uz =dz
dt=
dz
γ(dt− v
c2dx) =
uzγ (1− uxv/c2)
General case:
u =u + (γ − 1)u·v
v2 − γvγ(1− u·v
c2
)
Relativistic Quantum Mechanics I Trond SAUE
Particles and �elds
H = Hp +Hint +Hf
Hp Hamiltonian of particle
Hf Hamiltonian of external �eld
Hint Hamiltonian of interaction
In the following we will consider the construction of our Hamiltonian.
Relativistic Quantum Mechanics I Trond SAUE
The principle of least action
For a system of N particles there exists a function, the Lagrangian L(x, x, t), such thatthe action integral
S[x] =
∫ tb
ta
L(x, x, t)dt; x =dx
dt
is minimized on the actual trajectory x (t) of the physical system.
The Lagrangian is a function of the 3N coordinates (collectively denoted by x), and theircorresponding total time derivatives (velocities) x and time t .
Constraints on movement (e.g. particle on a sphere) may be introduced
by the use of generalized coordinates.
The action S is a functional of the 3N coordinates.
Relativistic Quantum Mechanics I Trond SAUE
Functional derivative
4 Di�erential of a function
f(x) ⇒ df =df
dxdx
4 First variation of a functional
E [ρ(r)] ⇒ δE =
∫∫∫ (δE
δρ (r)
)δρ (r) dτ
Relativistic Quantum Mechanics I Trond SAUE
Variational calculus I
We seek the minimum:
δS
δx= 0
δS =
∫ tb
ta
(δS
δx(t)
)δx(t)dt
=
∫ tb
ta
[L(x+ δx, x+ δx, t)− L(x, x, t)] dt; δx(ta) = δx(tb) = 0
=
∫ tb
ta
[(∂L
∂x
)δx+
(∂L
∂x
)δx
]dt = 0
;
Relativistic Quantum Mechanics I Trond SAUE
Variational calculus II
By partial integration
∫ tb
ta
(∂L
∂x
)δxdt =
(∂L
∂x
)δx
∣∣∣∣tbta︸ ︷︷ ︸
=0!
−∫ tb
ta
d
dt
(∂L
∂x
)δxdt
one obtains
δS =
∫ tb
ta
[(∂L
∂x
)− d
dt
(∂L
∂x
)]δxdt = 0
Relativistic Quantum Mechanics I Trond SAUE
The Euler-Lagrange equations
(∂L
∂x
)− d
dt
(∂L
∂x
)= 0
4 Generalized momentum: p =∂L
∂x→
(∂L
∂x
)− dpdt
= 0
4 In general: L = T − U
T (x) − kinetic energy
U(x, x) − generalized potential
4 Generalized force: F = −∂U∂x
+d
dt
(∂U
∂x
)→ F =
d
dt
(∂T
∂x
)4 The Lagrangian is chosen so as to give trajectories in accordance with experiment.
Relativistic Quantum Mechanics I Trond SAUE
Choosing the Lagrangian: free particle
4 Homogeneity of space: no dependence on position r
4 Isotropy of space: no dependence on direction of movement
4 Homogeneity of time: no dependence on time
L(r,v, t) → L (v) ∼ v2
4 (Non-relativistic) free particle: L = T = 12mr2 ⇒ d
dt(mr) = ma = 0
4 Momentum
p =∂L
∂v= mv
Relativistic Quantum Mechanics I Trond SAUE
Relativity and 4-vectors
A simple way to see if a physical law is relativistic or not is to see
if it can be expressed in terms of 4-vectors:
• 4-position: rµ = (x, y, z, ict) = (r, ict); rµrµ = r2 − c2t2 = −s2 (the interval)
• Proper time τ : ds =√c2dt2 − dx2 − dy2 − dz2 = cdτ
Note thatcdτ = cdt
√1− v2/c2
so that cdτ = cdt when v = 0,meaning that the proper time is the time in the rest frame of the moving particle
• 4-velocity: vµ =drµdτ
= γdrµdt
= γ(v, ic); vµvµ = −c2
• 4-momentum: pµ = mvµ = γ (mv, imc) ; pµpµ = −m2c2
Relativistic Quantum Mechanics I Trond SAUE
Choosing the Lagrangian: relativistic free particle
4 Action integral S[xµ] =∫ τbτa
Λ(xµ, vµ)dτ
4 Non-relativistic free particle: LNR = 12mv
2
4 Relativistic free-particle: Λ(xµ, vµ) = mvµvµ = −mc2
4 We select a speci�c frame
cdτ =√c2dt2 − dx2 = cdt
√1− v
2
c2= cdtγ−1
which gives
S[xµ] =
∫ tb
ta
L(xµ, vµ)dt; L(xµ, vµ) = Λ(xµ, vµ)γ−1 = −mc2γ−1
4 Connection to non-relativistic theory:
LR = −mc2√
1− v2/c2 = −mc2 +1
2mv2 +
1
8mv2v
2
c2+ . . .
Relativistic Quantum Mechanics I Trond SAUE
The Hamiltonian
Legendre transformation: H(x, p, t) = xp− L(x, x, t)
In�nitesimal:
dH =
(∂H
∂x
)dx+
(∂H
∂p
)dp+
∂H
∂tdt
or, using the Legendre transformation :
dH = xdp+ pdx−
(∂L
∂x
)dx+
(∂L
∂x
)︸ ︷︷ ︸
=p
dx+∂L
∂tdt
= −
(∂L
∂x
)︸ ︷︷ ︸
p
dx+ xdp− ∂L∂tdt
= −pdx+ xdp− ∂L∂tdt
Relativistic Quantum Mechanics I Trond SAUE
Hamilton's equations
x =∂H
∂p; p = −∂H
∂x;
∂L
∂t= −∂H
∂t
4 3N second-order di�erential equations (Euler-Lagrange) are replaced by 6N+1 �rst-orderdi�erential equations (Hamilton).
4 for an isolated system, the Hamiltonian represents the total energy of the system
4 non-relativistic free particle:
HNR = p · v − L =1
2mv2 =
p2
2m
Relativistic Quantum Mechanics I Trond SAUE
Hamiltonian for a relativistic free particle
4 Lagrangian and momentum
LR = −mc2γ−1; p =∂L
∂v= γmv
4 Hamiltonian
H = p · v − L = γmv2 + γ−1mc2 = mc2γ
(v2
c2+ γ−2
)︸ ︷︷ ︸
=1
= mc2γ
4 Elimination of velocity
H2 = m2c4γ2 = m2c4 +m2c4(γ2 − 1
)= m2c4 + c2p2
H = mc2√
1 +p2
m2c2= mc2︸︷︷︸
rest mass
+p2
2m− p4
8m3c2+ . . .︸ ︷︷ ︸
kinetic energy
Relativistic Quantum Mechanics I Trond SAUE
Pauli spin matrices
representation matrices of the operator 2s in the basis {|α〉 , |β〉} :
σx =
[0 11 0
]; σy =
[0 −ii 0
]; σz =
[1 00 −1
]
Relativistic Quantum Mechanics I Trond SAUE
Mathematical toolkit
4 Einstein summation convention: any repeated index implies a summation, for instance
A ·B = AiBi
δii = 3
4 Levi-Cevita symbol:
εxyz = εyzx = εzxy = 1
εxzy = εyxz = εzyx = −1
εijk = 0 otherwise
for instance
A×B = eiεijkAjBk
σiσj = δij + iεijkσk
Relativistic Quantum Mechanics I Trond SAUE
A very important formula !
(σ ·A) (σ ·B) = A ·B + iσ · (A×B)
(σiAi) (σjBj) = AiBj (δij + iεijkσk) = AiBi + iσkεkijAiBj
Relativistic Quantum Mechanics I Trond SAUE
Quantization
1. Determine the Lagrangian L who, according to the principle of least action, gives thecorrect equations of motion.
LNRp =1
2mv2; LRp = −mc2γ−1
2. Determine momentum: p =∂L
∂r; ⇒ pNR = mv, pR = γmv
3. Construct the Hamiltonian by a Legendre transformation
HNRp =
p2
2m; HR
p =√m2c4 + c2p2
4. Quantize by replacing classical variables by quantized operators
pµ =
(p,i
cE
)→ pµ =
(−i∇,
i
c
{i∂
∂t
})= −i
(∇,−i
c
∂
∂t
)︸ ︷︷ ︸
4-gradient
= −i∂µ
Relativistic Quantum Mechanics I Trond SAUE
The Klein-Gordon equation
4 Classical expression
E2 = m2c4 + c2p2; E ∈⟨∞,−mc2
]∪[+mc2,∞
⟩4 Quantization: the Klein-Gordon equation[
− 1
c2∂2
∂t2− p2
]ψ = (mc)2ψ
Problem: the integral∫ψ∗ψdr is time-dependent.
Relativistic Quantum Mechanics I Trond SAUE
The Dirac equation I
Factorization
[i
c
∂
∂t+ (σ · p)
] [i
c
∂
∂t− (σ · p)
]φ1︸ ︷︷ ︸
mcφ2
= (mc)2φ1
gives a system of two coupled equations[i
c
∂
∂t− (σ · p)
]φ1 = mcφ2 (a)[
i
c
∂
∂t+ (σ · p)
]φ2 = mcφ1 (b)
Relativistic Quantum Mechanics I Trond SAUE
The Dirac equation II
We take linear combinations
i
c
∂
∂t[φ1 + φ2]− (σ · p) [φ1 − φ2] = mc [φ1 + φ2] (a+ b)
−ic
∂
∂t[φ1 − φ2] + (σ · p) [φ1 + φ2] = mc [φ1 − φ2] (b− a)
and introduce the large ψL and small ψS components
ψL = [φ1 + φ2] ; ψS = [φ1 − φ2] .
which then gives i
c
∂
∂t− (σ · p)
(σ · p) −ic
∂
∂t
[ ψLψS
]= mc
[ψL
ψS
].
Relativistic Quantum Mechanics I Trond SAUE
The Dirac equation III
This 4-component equation can be expressed in Lorentz covariant form
(iγµ∂µ −mc)ψ = 0; γµ = (βα, iβ) , ψ =
[ψL
ψS
]through introduction of the Dirac matrices
α =
[0 σσ 0
], β =
[I2 00 −I2
]
We obtain the conventional form of the Dirac equation by multiplication by βc from the left
[βmc2 + c (α · p)
]ψ = i
∂
∂tψ
Relativistic Quantum Mechanics I Trond SAUE
Maxwell's (microscopic) equations
(SI-based atomic units)
4 The homogeneous pair:
∇ ·B = 0
∇×E +∂B
∂t= 0
4 The inhomogeneous pair includes sources:the charge density ρ and current density j (c is the speed of light)
∇ ·E = 4πρ
∇×B− 1
c2∂E
∂t=
4π
c2j
4 Boundary conditions must be introduced:E and B go to zero at in�nity
Relativistic Quantum Mechanics I Trond SAUE
Maxwell's equations: the stationary case
(time-independent electric and magnetic �elds)
4 The homogeneous pair:
∇ ·B = 0
∇×E = 0
4 The inhomogeneous pair:
∇ ·E = 4πρ
∇×B =4π
c2j
4 Static case: no currents (j = 0)
4 A useful formula∇× (∇× F ) = ∇ (∇ · F )−∇2F
Relativistic Quantum Mechanics I Trond SAUE
Stationary case: electric �eld
∇2E = −∇ (∇ ·E)︸ ︷︷ ︸4πρ
+∇× (∇×E)︸ ︷︷ ︸=0
= −4π∇ρ
Charles-Augustin de Coulomb
(1736 - 1806)
Each component of the electric �eld ful�lls the Poisson equa-
tion:∇2
Ψ (r, t) = −4πf (r, t)
with solutions
Ψ (r1, t) =
∫f (r2, t)
r12
dτ2
Generalized Coulomb's law (electrostatics)
E(r1) =
∫ ∇2ρ (r2)
r12dτ2 =
∫r12ρ (r2)
r312
dτ2
Point charge:
ρ(r2) = QAδ (r2 − rA(t)) ⇒ E(r1) =QAr1A
r31A
Relativistic Quantum Mechanics I Trond SAUE
Stationary case: magnetic �eld
∇2B = −∇ (∇ ·B)︸ ︷︷ ︸=0
+∇× (∇×B)︸ ︷︷ ︸4πc2
j
= ∇× 4π
c2j
Jean-Baptiste Biot
(1774-1862)
Biot-Savart law (magnetostatics):
B(r) = − 1
c2
∫r12 × j(r2)
r312
dτ2
It would be tempting to insert the expression for a moving point charge
j (r2) = QAr′Aδ (r2 − rA(t))
but this is wrong, since a moving charge is not static. More generally,
the Biot-Savart law applies to steady currents
∇ · j = −∂ρ
∂t= 0
Relativistic Quantum Mechanics I Trond SAUE
The Helmholtz theorem
The general solution to
∇2F = −∇ (∇ · F ) + ∇× (∇× F )
isF(r) = −∇s(r) + ∇× v(r)
where
s (r1) =1
4π
∫ ∇2 · F (r2)
r12dτ2; v(r1) =
1
4π
∫ ∇2 × F (r2)
r12dτ2
The divergence and curl of F must go to zero faster than 1r2 ; otherwise the above integrals
blow up in the limit.
This results show that we can reconstruct a vector function from knowledge of itsdivergence and curl combined with proper boundary conditions.
Relativistic Quantum Mechanics I Trond SAUE
The meaning of divergence: �ux
The integral of the divergence ∇· of a vector function F over some volume V is related toits �ux through the surface S enclosing the volume through the divergence theorem∫
V
(∇ · F ) dV =
∫S
F · ndS
where the unit normal vector n points outwards from the surface and dS is an in�nitesimalsurface element.
We can accordingly de�ne the divergence of a vector function F in a point by a limitingprocess in which the surrounding volume above tends towards zero, that is
∇ · F = lim∆V→0
1
∆V
∫S
F · ndS
Relativistic Quantum Mechanics I Trond SAUE
Continuity equation
∇×B− 1
c2∂E
∂t=
4π
c2j
⇓
∇ · (∇×B)︸ ︷︷ ︸=0!
− 1
c2∂
∂t(∇ ·E)︸ ︷︷ ︸
4πρ
=4π
c2(∇ · j)
⇓
∇ · j +∂ρ
∂t= ∂µjµ =
(∇,−i
c
∂
∂t
)︸ ︷︷ ︸
4-gradient
(j, icρ)︸ ︷︷ ︸4-current
= 0
Integral form :∫V
(∇ · j) dV =∫Sj · ndS = −
∫V
∂ρ
∂tdV
Current density is the movement of charge density. The continuity equation says that if charge is moving
out of a di�erential volume (i.e. divergence of current density is positive) then the amount of charge within
that volume is going to decrease, so the rate of change of charge density is negative. Therefore the continuity
equation amounts to a conservation of charge.
Relativistic Quantum Mechanics I Trond SAUE
The meaning of curl: circulation
The integral of the curl ∇× of a vector function F over some surface S is related to itscirculation around a curve C enclosing the surface by the curl theorem∫
S
(∇× F ) · ndS =
∮C
F · tdl
where t is a unit tangential vector and dl an in�nitesimal line element.
We can accordingly de�ne the curl of a vector function F in a point by a limiting processin which the surrounding volume above tends towards zero, that is
(∇× F ) · n = lim∆S→0
1
∆S
∮C
F · tdl
Relativistic Quantum Mechanics I Trond SAUE
Helmholtz decomposition
Any vector function F (di�erentiable) who goes to zero faster than1
rwhen r →∞ can be
expressed as the sum of the gradient of a scalar and the curl of a vector
F(r) = −∇s(r) + ∇× v(r)
Longitudinal component (�parallel�):
FL = −∇s(r); ∇× FL = 0
Transversal component (�perpendicular�):
FT = ∇× v(r); ∇ · FT = 0
Relativistic Quantum Mechanics I Trond SAUE
Maxwell's equations: homogeneous pair
1. ∇ ·B = 0 impliesB = B⊥ = ∇×A(r) and B‖ = 0
2. ∇×E +∂B
∂t= 0 then becomes ∇×
(E +
∂A
∂t
)= 0 and one may write
E +∂A
∂t= −∇φ(r)
• Longitudinal component: E‖ = −∇φ−∂A‖
∂t
• Transversal component: E⊥ = −∂A⊥∂t
With the introduction of the scalar potential φ and the vector potential A,the homogeneous pair of Maxwell's equations is automatically satis�ed.
Relativistic Quantum Mechanics I Trond SAUE
Maxwell's equations: inhomogeneous pair
1. ∇ ·E = 4πρ becomes
∇2φ+∂
∂t(∇ ·A) = −4πρ
or
[∇2 − 1
c2∂2
∂t2
]φ+
∂
∂t
[(∇ ·A) +
1
c2∂φ
∂t
]= −4πρ
2. ∇×B− 1
c2∂E
∂t=
4π
c2j becomes
[∇2 − 1
c2∂2
∂t2
]A−∇
[(∇ ·A) +
1
c2∂φ
∂t
]= −4π
c2j
Relativistic Quantum Mechanics I Trond SAUE
Reminder: 4-vectors
A simple way to see if a physical law is relativistic or not is to see
if it can be expressed in terms of 4-vectors:
• 4-position: rµ = (x, y, z, ict) = (r, ict); rµrµ = r2 − c2t2 (the interval)
• 4-velocity: vµ = γ(v, ic); vµvµ = −c2
• 4-gradient: ∂µ =
(∇,−i
c
∂
∂t
); ∂µ∂µ = ∇2 − 1
c2∂2
∂t2= �2 (d'Alembertian)
• 4-potential: Aµ =
(A,
i
cφ
)• 4-current: jµ = (j, icρ)
Relativistic Quantum Mechanics I Trond SAUE
Maxwell's equation: 4-vector notation
4 We start from: [∇2 − 1
c2∂2
∂t2
]φ+
∂
∂t
[(∇ ·A) +
1
c2∂φ
∂t
]= −4πρ[
∇2 − 1
c2∂2
∂t2
]A−∇
[(∇ ·A) +
1
c2∂φ
∂t
]= −4π
c2j
4 This can be written more compactly as
�2φ+∂
∂t(∂µAµ) = −4πρ
�2A−∇ (∂µAµ) = −4π
c2j
and �nally compacted into
�2Aβ − ∂β(∂αAα) = −4π
c2jβ
Relativistic Quantum Mechanics I Trond SAUE
Gauge transformations
4 B = ∇×A implies that the longitudinal component A‖ of the vector potential can bemodi�ed without changing B, that is
A → A′ = A + ∇χ
4 However E‖ = −∇φ−∂A‖
∂timplies that a modi�cation of A‖ requires a corresponding
modi�cation of the scalar potential
φ → φ′ = φ− ∂χ∂t
4 Lorentz covariant form : Aµ → A′µ = Aµ + ∂µχ; Aµ =
(A,i
cφ
)︸ ︷︷ ︸
4-potential
4 The electric and magnetic �elds are gauge invariant.
Relativistic Quantum Mechanics I Trond SAUE
Lorentz gauge ∂µAµ = 0
4 Maxwell's equations simpli�es to
�2Aβ = −4π
c2jβ
4 Each component of the 4-potential satis�es the inhomogeneous wave equation
�2Ψ (r1, t) = −4πf (r1, t)
with general solution
Ψ (r1, t) =
∫f (r2, tr)
r12dτ2; Retarded time: tr = t− r12
c
Relativistic Quantum Mechanics I Trond SAUE
Coulomb gauge: ∇ ·A = 0
4 Maxwell's equation:
∇2φ = −4πρ(∇2A− 1
c2∂2A
∂t2
)−∇ 1
c2∂φ
∂t= −4π
c2j
4 The scalar potential is the solution of the Poisson equation
φ(r1, t) =
∫ρ(r2, t)
r12dτ2
and describes the instantaneous Coulomb interaction.
4 Problem (?): The theory of relativity does not allow instantaneous interactions.
4 Retardation is hidden in the solution for the purely transversal vector potential
A(r1, t) =4π
c2
∫j(r2, tr)
r12dτ2
Relativistic Quantum Mechanics I Trond SAUE
Gauge freedom
4 Uniform electric �eld E :
a) φ(r) = −r ·E A(r) = 0
b) φ(r) = 0 A(r) = Et
4 In general, the operator
(H − i ∂
∂t
)and not the Hamiltonian H alone is gauge invariant.
Relativistic Quantum Mechanics I Trond SAUE
Potentials of a moving charge I
4 In the rest frame:A′x = 0, A′y = 0, A′z = 0, φ′ =
q
r′
4 Lorentz transformation: (A↔ r, φ/c2 ↔ t):
Ax = γv
c2φ′, Ay = 0, Az = 0, φ = γφ′
4 Problems:
1. The potentials of the rest frame are expressed by the variables of the frame of themoving charge.
2. The interaction is not instantaneous.
Relativistic Quantum Mechanics I Trond SAUE
Potential of a moving charge II
4 Assume that one measures at time t = t′ = 0 the potential at a distance r′ in the frameof the moving charge.The interaction signal was accordingly emitted at time
t′ = −r′
c
4 In the rest frame the signal has travelled a distance r and was therefore emitted at time
t = −rc
4 The two times are connected thorough
t′ = γ(t− v · r
c2
)= γ
(−rc− v · r
c2
)= −r
′
c
Relativistic Quantum Mechanics I Trond SAUE
Potentials of a moving charge III
4 We accordingly obtain the relation
r′ = −ct′ = γ(r +
v · rc
)
4 Liénard-Wiechert potentials:
A =v
c2φ; φ =
q(r +
v · rc
)ret
4 In Coulomb gauge and at moderate velocities v
A =q
2cr
[v
c+
(v · r) r
r2c+ . . .
]; φ =
q
r
Relativistic Quantum Mechanics I Trond SAUE
Maxwell's equations: Lorentz covariance
Maxwell's equations can be derived from the Lagrangian density
L = jαAα −c2
16πFαβFαβ
in which appears components of the electromagnetic tensor
Fαβ = ∂αAβ − ∂βAα; ∂µ =
(∇,−i
c
∂
∂t
),
The Euler-Lagrange equations
∂L∂Aα
− ∂α(
∂L∂ (∂βAα)
)= 0
gives Maxwell's equations in Lorentz covariant form
∂αFαβ = �2Aβ − ∂β(∂αAα) = −4π
c2jβ
Relativistic Quantum Mechanics I Trond SAUE
Particles and �elds
4 Complete Hamiltonian
H = Hparticles +Hinteraction +H�elds
4 Fields speci�ed:
Non-relativistic limit
(iγµ∂µ −mc)ψ = 0 →(p2
2m− i ∂
∂t
)ψ = 0
Dirac equation Schrödinger equation
4 Particles (sources) speci�ed:
Non-relativistic limit
�2Aµ − ∂µ(∂νAν) = −4π
c2jµ → ???
Maxwell's equations
Relativistic Quantum Mechanics I Trond SAUE
The non-relativistic limit of electrodynamics
∇ ·B = 0 ∇ ·B = 0
∇×E +∂B
∂t= 0 c→∞ ∇×E = 0
∇ ·E = 4πρ ⇒ ∇ ·E = 4πρ
∇×B − 1
c2∂E
∂t=
4π
c2j ∇×B = 0
In the strict non-relativistic limit there are no magnetic �elds and no e�ects of retardation !
The Coulomb gauge bears its name because it singles out the instantaneous Coulombinteraction, which constitutes the proper non-relativistic limit of electrodynamics and whichis the most important interaction in chemistry.
All retardation e�ects as well as magnetic interactions are to be considered corrections of aperturbation series of the total interaction (in 1/c2).
Relativistic Quantum Mechanics I Trond SAUE
top related