5.73 Lecture #8
Rydberg Klein Rees
8-1
Last time: WKB quantization condition for bound eigenstates of
almost general V(x) — Connection into bound region from
left and right
h
x (E)
∫x −+(E) pE (x ′)dx′ = 2 (n + 1/ 2)
pE (x) = [ 2m(E − V(x))]
1/2
En without ψ n !
timing of w.p. as
it moves on V(x)
But where do we get V(x)?
Certainly not from femtochemistry!
From FREQUENCY DOMAIN SPECTROSCOPY
E v,J → V(x)
RKR method
Next time: Numerical Integration of 1-D Schr. Eq. — see handouts
Then begin working toward matrix picture
Need background in Ch.2 of CTDL
pages 94-121 soon, pages 121-144 by next week
Postulates and theorems not to be covered except as needed for
solving problems.
Today:
E v,J → spectroscopic notation
A(E, J) =
E
V(x)
xe
x
dV
Equilibrium:
= 0 → xe
dx
∂A ∂A
1
1
,
→ x + (E) − x − (E) and
−
x + (E) x − (E)
∂E ∂J
WKB QC applied to
∂A ∂A
,
← G( v),B(v) used to determine x ± (E).
∂E ∂J
Long Range Theory: Ultra Cold Collisions: Atom in Molecule
updated 9/18/02 8:57 AM
5.73 Lecture #8
8-2
Rydberg Klein Rees
Someday you will discover that the energy levels of a diatomic
molecule are given by
E evJ / hc = T e +
electronic
[
G(v)
vibration
+
F v (J)
rotation
cm-1
]
= νe + Y 00 + ω e (v + 1 / 2) − ω e x e (v + 1 / 2)2 +…
G(v)
+ [Be − α e (v + 1 / 2)+…]J(J + 1) − DJ 2 (J + 1)2
B(v)
RKR requires only G(v) and B(v) to get VJ(x)
h 2 J(J + 1)
where V J (x) = U(x) +
J-dependent
bare
2µx 2
effective
potential
potential
x ≡ R − Re
µ=
centrifugal barrier
(actually rotational
kinetic energy)
m1m2
m1 + m2
We are going to derive V 0 (x) directly from G(v), B(v) data.
This is the only direct spectrum to potential inversion
method! WKB quantization is the basis for this.
x + (E v )
∫x (E
−
v
)
pE v ( x′)dx′ = (h / 2)(v + 1 / 2)
v = 0,1,…# of nodes
In this equation, what we know (Ev) and what we want (V(x) and x at turning points)
are hopelessly mixed up. There is a trick!
A(E, J) ≡ ∫
x + (E,J)
x − (E,J)
[E ± V J (x′)]dx′
x–(E)
x+(E)
area at E
updated 9/18/02 8:57 AM
5.73 Lecture #8
8-3
Rydberg Klein Rees
but, still, we know neither V J (x) nor x ± (E, J)! !
∂A
∂A
and
are numerically evaluable
∂J
∂E
integrals (via WKB QC) involving only E v,J info
data input
here
∂A
∂A
2 . independently,
and
determine
∂J
∂E
 1
1 
2 eqs. in 2
x + (E, J) − x − (E, J)] and 
−
[

unknowns give
 x + (E, J) x − (E, J) 
Roadmap: 1 . Show that
turning points
Do #2 first because it is so easy
∂A
∂  x + (E,J)
=
∫
∂ E ∂ E  x − (E,J)
x (E,J)
= ∫x +(E,J) 1dx′
−

h 2 J(J + 1)  
E – U( x ′ ) −
 dx′ 
2µx′ 2  

+
0
+
contributions from
0
∂x ± (E, J)
∂E
are zero because integrand is 0
at turning points
∂A
= x + (E, J) − x − (E, J)
∂E
!
∂A
∂  x + (E,J)
=
∫
∂ J  x − (E,J)
∂J

h 2 J(J + 1)  
E – U( x ′ ) −
2  dx′ 
2µx′  

h 2 x + (E,J) 2J + 1
dx′ +
012
+ 30
=−
∫
2µ x − (E,J) x′ 2
integrand = 0 at x ±

h2(2J + 1)  1
∂A
1
=+
−
 x (E, J) x (E, J) 
∂J
2µ
−
 +

So, if we can evaluate these derivatives from EvJ data, we have VJ(x)!
updated 9/18/02 8:57 AM
5.73 Lecture #8
8-4
Rydberg Klein Rees
some clever manipulations to put A(E,J) into convenient form
(see nonlecture notes on pages 8-5,6,7)
A(E, J) = ∫
[E ± V J (x′)]dx′
x + (E,J)
x − (E,J)
1/2
 2h 2 
A(E, J) = 2

 µ 
1/2


E ± E′vJ 
∫v(E
,J) 
{

min 3
1424

data 
data
v(E,J)
dv
skipped steps are shown
on pages 8-5, 6, 7.
this integral could be evaluated at any E, but we really only
∂A
∂A
and
. Evaluate these derivatives at J = 0.
want
∂E
∂J
 2h2 
∂A
= 2

∂E
 µ 
1/ 2
−1 / 2
 1  v( E, J )
E – E ′vJ ] dv + 0 + 0
[
∫
 2  v ( E min , J )
integrand = 0
at upper limit
1 Y
v(E min , J) = − − 00
2 ωe
lower limit
independent of E
defined so that G(vmin) = 0

 G(v) = Y 00 + ω e ( v + 1 / 2)


0 = G(v min ) = Y 00 + ω e ( v min + 1 / 2)


 Y

− 00 = v min + 1 / 2

 ωe


Y
1

 v min = − 00 −
ωe 2


for J = 0
[vmin is slightly different from –1/2]
E′v,J = G(v)
∂ A  2h 2 
=
∂ E  µ 
1/ 2
data
v(E)
−1/ 2
∫−1/ 2−Y 00 / ω e [ E – G(v)] dv ≡ 2f(E)
data
evaluate this integral numerically at any E.
[Singularity at upper limit fixed by change of variable: Zeleznik JCP 42, 2836 (1965).]
updated 9/18/02 8:57 AM
5.73 Lecture #8
Rydberg Klein Rees
8-5
 2h 2  ⌠ v(E)
∂A
∂E
=
[E − G(v)]−1/ 2 dv + 0 + 0


∂ J J=0  µ  ⌡−1/ 2−Y 00 / ω e
∂J
E = BJ J(J + 1)
∂E
= Bv (2J + 1)
∂J
 2h 2 
∂A
∴
=
∂ J J=0  µ 
∂E
= Bv
∂ J J=0
1/ 2
v(E)
data
⌠
 h2 
−1/ 2
[E − G(v)] Bvdv ≡ −  2g(E)

 2µ 
⌡−1/ 2−Y 00 / ω e
data
(again, a nonfatal singularity at upper limit)
f(E) and g(E) are “Klein action integrals” which are jointly determined by
empirical G(v) and B(v) functions.
Nonlecture derivation of this useful form of
 2h 2 
A(E,J) = 2

 µ 
1/ 2
v(E ,J)
∫v(E
min
1/ 2
E − E′vJ ] dv
[
,J )
A(E, J) = ∫x +( E,J ) [ E − V J (x ′ )]dx′
x (E,J)
Begin here:
−
b
integral identity
let
2 ⌠  x − a  1/2
b−a= 
dx
π ⌡a  b − x 
b= E
a = VJ ( x )
x = E′vJ
 x - a  E vJ − VJ ( x )
=
so that 
 b- x
E − E vJ
∴
(E,J)
A(E, J) = ∫xx +(E,J)
[ b − a]dx′
−
Now insert the integral identity
updated 9/18/02 8:57 AM
5.73 Lecture #8
8-6
Rydberg Klein Rees
x (E,J)
⌠ +
 2 ⌠ b x − a 1/ 2 
A(E, J) = 
  
 dx dx′

π
b
−
x


⌡


a
⌡x − (E,J)
put in values of a, b, and x
x (E,J)
1/ 2
⌠ +
 ⌠E

 E′ − V (x ′ ) 
2

J
 
=
 dE′vJ  dx′
 vJ

 π 

E
−
E′

vJ



⌡x − (E,J)  ⌡V J (x)
reverse order of integration and recognize WKB QC in disguise
E
1/2
⌠
 ⌠ x + (E,J)

 E′vJ − V J (x ′ ) 
2

=

 dx′ dE′vJ



π
E
−
E′

vJ



⌡V J (x)  ⌡x − (E,J)
numerator of dx′ integral is QC — insert QC and then integrate by parts.
denominator is independent of x′, so insert QC
x (E,J)
x
+
⌠ +
E′ − V(x ′ )]1/2 dx′ = (2µ)−1/2 ⌠
[

 p(x ′ )dx′
⌡x − (E,J)
⌡x −
= (2µ)−1/2
h
(v + 1/ 2)
2
E
⌠


v(E
,
J)
+
1/
2
′
−1/2 h 
 2
dE′

∴ A(E, J) =
(2µ )
1/2
 vJ

 π
2
E
−
E′
vJ

⌡E min 
(
)
**
** integrate by parts
(
)−1/2
1/2
f = −2(E − E′vJ )
g = [ v(E′vJ , J) + 1/ 2]
f′ = E − E′vJ
g′ =
dv
,
dE′
A(E, J) =
(not a typo because variable is E′vJ not E)
which is known from E vJ
 2h 2 
+

 µ 
14243
E′=E
fg E′=E
min
=0 at both limits
1/ 2
E
dv
⌠
2( E − E′ )1/ 2
dE′

⌡E min
dE′
(caution: f and g here
are not Klein’s action
integrals)
updated 9/18/02 8:57 AM
5.73 Lecture #8
8-7
Rydberg Klein Rees
** change variables from dE′ to dv′
dv
dE′
dE′
dv =
limits of integration become ⌠

finished:
 2h 2 
A(E, J) = 2

 µ 
1/ 2
v(E,J)
⌡v( E min ,J )
v(E,J)
⌠
[E − E′vJ ]1/ 2 dv

⌡v( E min ,J )
we have two independent evaluations of f(E) and g(E)
one leads to
x + (E, 0) − x − (E, 0) = 2f(E)
1
1
−
= ±2g(E)
x + (E, 0) x − (E, 0)
[
]
pair of turning
2 1/2
(E,
0)
=
f(E)
/
g(E)
+
f(E)
± f(E)
x
±
points
from quadratic
formula
so we get a pair of turning points at each E. Not restricted to E’s with integer v’s!
V(x)
connect the dots!
Robert LeRoy: modern, n-th generation RKR program at
http://theochem.uwaterloo.ca/~leroy/
Download program, instructions, and sample data.
( )
RKR does not work for polyatomic molecules because
E−V Q
r
~
does not determine the multicomponent vector P
updated 9/18/02 8:57 AM
5.73 Lecture #8
Long-Range Molecules
8 - 8
Rydberg Klein Rees
W.C. Stwalley CPL 6, 241 (1970)
(weak perturbation of atomic properties)
R.J. Bernstein & R.LeRoy JCP 52, 3869 (1970)
soft turning point
hard
turning
point
What does ψ (x) look like at very high v?
* lots of nodes (v nodes)
* small lobe at inner turning point. Why?
* large lobe at outer turning point. Why?
H int: Force = –
dV(x)
dx
at sufficiently large v, it is certain that ψ (x) is dominated by
o u t e r-most lobe and any expectation value of a function of x, such as
V(x), will be dominated by the outer turning point region. Since the
vibrational Schrödinger equation contains V(x), it is evident that Ev
at high v should be determined primarily by the long range part of
V(x) (and insensitive to details near x e and at the inner turning
point).
What do we know about covalent bonding?
ATOMIC ORBITAL OVERLAP IS REQUIRED!
NO OVERLAP at large x, V(x) determined by properties of isolated
a t o m s: dipole moment, polarizability — return to this later when we
do perturbation theory.
It is always possible to predict the longest range term in V(x) = C n x −n
where the longest range term is the one with SMALLEST n.
updated 9/18/02 8:57 AM
5.73 Lecture #8
8-9
Rydberg Klein Rees
Quick review of the Long-Range Theory
x+(vD) = ∞
x–(vD)
ε v ≡ E vD − E v
x–(v)
vD is noninteger v
of “level” at “top” of
potential
x+(v)
D
εv is binding
energy of v-th level
xe
V0 (x ) = U(x ) = −Cnx −n
at J = 0
at long range (large x)
U(∞ ) = 0 ≡ E v D
U( x e ) = −De
x + (v ) = ( −Cn E v )
1/n
x + ( vD ) = ∞
[E
v
= V( x + (v)) = −Cn x −n
+
]
binding energy: εv = E v D − E v = Cnx −n
+
How many levels are there in potential?
x (v
) =∞
+
D
h
( vD + 1 / 2) = ⌠

2
⌡ x− (v D )
p D (x ′)dx ′.
Now we do not know v D , Cn , or D, but we do know n and know that
E v will be primarily determined by long-range part of V(x) near v D .
So, for any Ev we expect that it will be possible to derive a
relationship between
( v D − v)
and
(E v
D
− Ev
# of levels below highest bound level
)
binding energy
by some clever tricks you may discover on Problem Sets #4 and 5,
we find
n−2
v D − v = a n ε v2n
Tells us how to plot Ev vs. v to extrapolate to vD and then to obtain accurate value
of De from a linear plot near dissociation.
updated 9/18/02 8:57 AM
5.73 Lecture #8
8 - 10
Rydberg Klein Rees
Power
n=1
2
3
3
of longest range term in V(x):
charge - charge
charge - dipole
charge - induced dipole
dipole - dipole
(also transition dipoles)
5 dipole - induced dipole
5 quadrupole - quadrupole
6 induced dipole - induced dipole
+,– point charges (e.g. H atom)
—
+
H + H
Na 2 S + Na 2P
( )
( )
—
I( P3/ 2 ) + I( 2 P3/ 2 )
2
( )
( )
Na 2 S + Na 2 S
not only is the limiting n known, but also Cn is known because it is
calculable from a measurable property of the free atom. Many molecular
states are described at long range by the same Cn 's ! Ultra-cold collisions
now used to determine V(x) to very large x. Now best route to the
properties of separated atoms!
Mostly, long-range theory has been used as a guide to extrapolation to
accurate dissociation energy (relevant to ∆H°f ). Now Bose condensates.
Molecule trapping.
x −1 and x −2 potentials have ∞ number of bound levels.
x −3 , x −5 , x −6 potentials
have finite number, and the number of levels breaks off more abruptly as
n increases.
n →∞
high n
low n
i.e., # of bound levels
action integral affected more by wider classical ∆ x region than by
deeper ∆ E binding region because p ∝ (E–V(x))1 / 2
updated 9/18/02 8:57 AM
5.73 Lecture #8
8 - 11
Rydberg Klein Rees
This means (equation at bottom of 8-9) that if we plot (given that we can predict n
with certainty)
vD
v
correct D: STRAIGHT LINE
wrong D: guess for Ev D too high
wrong D: guess for Ev D too low
0
(Ev
D
n−2
− E v 2n
guessed
n
n−2
2n
3
5
6
7
1/6
3/10
1/3
5/14
)
known
it is possible to determine D and v D very
accurately
much better than Birge − Sponer plot,which is valid only for a Morse potential
Birge-Sponer: ∆G vs. v
∆G(v+1/2)=G(v+1) – G(v) •
• ••
long linear
extrapolation to vD
•
v
vD
G( v) = ω e (v + 1 / 2) − ω e x e (v + 1 / 2) 2
∆G(v + 1 / 2) = G(v + 1) − G(v) = ω e − ω e x e (2v + 2) decreasing to 0 as v increases
Morse Potential
when ∆G(v + 1 / 2) = 0, ω e = ω e x e (2v + 2)
2
ωe
V0 ( x ) = D[1− e − Ax ]
vD =
−1
v D is noninteger # of bound vibrational levels
2ω e x e
0
for Morse
D = G( v D ) =

1  ω 2e
− ω e xe 

4  ω e xe

= ( v D + 1)
ω e ω e xe
ω
−
≈ ( v D + 1) e
2
4
2
But Morse inevitably has incorrect long-range form
updated 9/18/02 8:57 AM
5.73 Lecture #8
8 - 12
Rydberg Klein Rees
Which is longer range? Morse or Cnx–n? Take ratio of binding energy at large x.
lim
lim
−C n x −n
−C n x −n
=
x → ∞ D 1 − e −Ax 2 − D x → ∞ De −2Ax − 2De −Ax
[
]
lim −C n x −n e 2Ax
x → ∞ D − 2De Ax
lim C n −n Ax
=
x e →∞
x → ∞ 2D
=
dominant term
This means that Morse binding energy gets small faster than Cnx–n for any n.
Morse
–Cnx–n
G(v+1) – G(v) will get small faster for Morse. Plot ∆G(v + 1/2) vs. v.
G(v+1) – G(v)
0
v
Morse
vD
True
vD
linear Birge-Sponer
Dissociation energy usually underestimated by linear Birge-Sponer
extrapolation. Long-range plot of correct power of E v D − E v gives more
accurate dissociation energy.
updated 9/18/02 8:57 AM