som
DESCRIPTION
Structure of MatterTRANSCRIPT
-
wehaveto
take
into
accountonlytheterm
ei
tbecause
wearedealingwithab
sorption
(theother
term
eitleadsto
fast
oscillations,wou
ldberelevantwhen
wewantto
treat
stim
ulatedem
issionandcompute
M21).
=
M12
= 2 H
1=
eA0
2m
e2|p
|1(3.195)
2|p
|1=
2 i
1d3r=
ime
2|r
|1(3.196)
=
M12
=i 2
eA02|r
|1=
E0 22|er
|1(3.197)
Because
eristhedipoleop
erator,M
12iscalled
dipole-m
atrix-elementan
dtheab
ove
approxim
ationdipole
approxim
ation.
=
|M12|2
=e2 4|2
|E0r
|1|2
(3.198)
ifE
0=
E0ex,
=
M2 12
=e2E
2 0
4|2
|x|1
|2(3.199)
Butthesystem
issymetric=
(x;y;z)areidentical
=
|2|x
|1|2
=1 3|2
|r|1
|2(3.200)
=
M2 12
=e2E
2 0
12|2
|r|1
|2(3.201)
1 2 0E
2 0:averaged
energydensity
forplanewaveatthefrequency
=
u(
)=
1 2 0E
2 0(
)forspectralenergydensity,an
dwith
(
)=
(E
)dE d
=(E
)(3.202)
=
(E
)=
2u(
)
0E
2 0
(3.203)
Therefore,withW
1
2=
B12u(
)wehave
B12
=e2
3 02|2
|r|1
|2(3.204)
A12
=e23
3 0c3|2
|r|1
|2(3.205)
Indipoleapproxim
ation:
|2|r
|1|2=
| 2r1d3r|2
=0
=
transitiondipole
allowed
(3.206)
if|2
|r|1
|2=
0=
transitiondipole
forbidden(3.207)
Oscillatorstrengh(see
chapter2)isproportionnalto
|2|r
|1|2 .
80
3.4
Molecu
les
Source:
Fox,Chapter8.
3.4.1
H2molecu
le
Letsstart
withtheeasiestmolecule
H2.
Weexpectedthatthetw
oprotonssharethetw
oelectronsin
thegroundstate
(low
estenergy).
Typicalforcovalent
bond,i.e.
co-valence,sharingvalence,sharing
outere
.
Moleculesareusuallytreatedin
Born-O
ppenheimer
approxim
ation,whichmeans:
weseparate
electronic
andnuclearmotion,justied
bythehugedierence
inmasses.
For
ethenuclei
arenotmoving,wecantreatrelative
positionsof
nuclei
asparametersandsolveelectronic
problem
En(R
),withR
-positionof
nuclei.
Nuclei
seepotentialcreatedbye
,wesolveproblem
fornuclei
insecondstep.
Problem
forthetw
oe
(electron1an
d2)
(x
1,y
1,z
1,x
2,y
2,z
2)
(3.208)
H=
2
2me(
2 1+2 2
)
e2
4 0
( 1 r 1l+
1 r 1r+
1 r 2l+
1 r 2r
1
|r 1r 2|)
(3.209)
e.g.r 1
lisdistance
betweenelectron1andleftnucleus.
Complicated6-dim
ensionalproblem! 81
-
Approxim
ationap
proach
tondgroundstate:ifthetw
onuclei
are
farapart,we
havetw
ohydrogen
atoms:
l(r)
=100(|r
r lp|)
(3.210)
r(r)
=100(|r
r r
p|)
(3.211)
withr lpandr r
pare
positionof
leftan
drightprotonsrespectively.
(r
1,r
2)
=l(r 1)
r(r
2)
(3.212)
Electron1is
aroundleft
protonandelectron2is
aroundrightproton,electronsare
independent.
Whatabouttheprobabilitydensity?Problem,wehave6dim
ensions!
Probability
density
tondeither
e:
(r)
=
|(r,r
2)|2
d3r 2
+
|(r
1,r)|2
d3r 1
(3.213)
with,e.g., |
(r,r
2)|2
d3r 2
isprobabilityto
ndelectron2,
whateverelectron1is
doing!
For
(r
1,r
2)=
l(r 1)
r(r
2),theprobabilitydensity
lookslike:
Ofcourse,
exchangingtheelectronsisphysicallyequivalent:
(r
1,r
2)
=r(r
1)
l(r 2)
(3.214)
Note
thatr(r
1)
r(r
2)or
l(r 1)
l(r 2)arenotrelevantto
ndagroundstate
(2e
aroundoneproton).
82
Stateswheretheelectronsare
nolonger
assigned
toonespecicproton(shared
electronstates)
canbefoundaslinearcombination:
(r
1,r
2)
=al(r 1)
r(r
2)+b
r(r
1)
l(r 2)
(3.215)
Norm
alization(
randlare
realandnorm
alized,aandbare
real):
(r 1,r
2)
(r1,r
2)d
3r 1d3r 2
(3.216)
=a2+b2
+2ab
l|r2
! =1
(3.217)
a bcanbechosenfreely.
Compute
expectationvalueforenergy H
forallratio
a bandproton-protondis-
tances(variationalapproach,possible
because
weknow
100(r)=
e
r/a0
a3 0
).Integrals
haveto
beevaluatednumerically.
Approxim
ationof
ground-state:thecongurationwendwiththelowestenergy.
Itturnsou
t:groundstate
a=
b
E=
H=
2ER
13.6eV
3.2eV
exactvalue4.52eV
(3.218)
Distance
betweenprotons:
0.87A(0.74A
)
83
-
Symmetricstate
s:sw
appingr 1
andr 2
changes
nothing.
Importantother
possibility-antisymmetricstate
a:sw
appingr 1
andr 2
changes
sign.
Generic
phenomena:lettw
osystem
swithidenticaleigenfunctionsinteract:
withoutinteraction:degeneracy
ofsym/antisymmetricstate,
withinteraction:degeneracy
getslifted
Esym0.
Compatible
withBloch:acanbechosenarbitrarily!
V0=
0
n(x)=
Ansin[k(x
na)]+B
ncos[k(x
na)]
for(n
1)a