5.73 Lecture #28
28 - 1
Hydrogen Radial Wavefunctions
The Hydrogen atom is special because it has electronic states and properties that scale
with n and l in a simple and global way. This is “structure” that is more than a
collection of unrelated facts. H serves as our model for “electronic structure” of manyelectron atoms, molecules, and possibly solids.
By showing how E, ⟨rσ⟩ (size and shapes), ⟨nl|r|n′l′⟩ (general matrix element) scale
with n and l, it tells us the kind of behavior to look for in more complex systems.
*
*
as a perturbation on H (quantum defects)
as a hint of relationships useful for extrapolation, assignment,
for recognizing when something behaves differently from naive
expectations.
TODAY
1.
2.
3.
4.
Simplified Radial Equation
Boundary conditions at r → 0 and r → ∞
qualitative features of Rnl(r)
n-scaling of ⟨rσ⟩
5. mathematical form of Rnl(r)
6. regular and irregular Coulomb functions
updated September 19,
5.73 Lecture #28
28 - 2
For any central force problem
 pˆ 2
lˆ 2 
H= r +
2  + V (r )
 2µ 2µr 
√ , l√2 , l√ commute, so spherical harmonics, Y m θ , φ ,
We know that H
)
z
l (
√ with eigenvalues h 2 l( l + 1).
are eigenfunctions of H
ψ ( r, θ , φ ) = R( r )Ylm (θ , φ )
trial form for separation of ψ
2
2
√ ψ =  p r + l + V ( r ) Y m (θ , φ ) R ( r ) = Eψ
H

 l
2
 2µ 2µr

Operate on the Ylm (θ , φ ) angular wavefunction and move it through to left.


 p 2 h 2 l ( l + 1)

√
m
r

+ V ( r ) R ( r ) = Eψ
Hψ = Yl (θ , φ )
+
2
 2µ 144
2µr4
2444
3 

Vl ( r )


so we can take Y (θ, φ) out of the Schrödinger Equation and we are left with
a 1-D radial equation where the only trace of the angular part is the
l -dependence of V (r), the effective potential energy function.
m
l
l
Since the differential equation depends on l, R(r) must also depend on l,
thus Rn (r) is the radial part of ψ, and it will generally be an explicit
function of two quantum numbers, n and l.
l
Usually n specifies the number of radial nodes and l the number of angular
nodes, but a special numbering convention for Hydrogen (and hydrogenic
ions) causes a slight distortion of this rule.
updated September 19,
5.73 Lecture #28
28 - 3
The radial equation, when the explicit differential operator form of Pr2
is derived and inserted, has the form
  h2 1 d 2
−
  2µ r dr 2

Tr


r +

 h2 l(l + 1)

+
(
)
V
r

  R ( r) = E R ( r)
2
µ
2
r
nl nl

  nl

Vl

It is customary to simplify this equation by replacing Rnl ( r ) by
R nl ( r ) =
1
u ( r)
r nl
*
*
*
insert
1
r
1
u ( r)
r nl
equation looks simpler
volume element looks simpler
behavior as r → 0 seems more familiar
u n (r ) in place of Rn (r ) and then multiply through on left by r
l
 h 2 d2
+
−
2
 2µ dr
l
2
(
h l l
+ 1)
2µr 2

+ V (r ) − E n  u n (r ) = 0

l
l
looks like ordinary 1-D Schrödinger Equation.
Boundary condition:
u n (r ) → 0
l
as
WHY? Because for all l > 0, Vl(0) → ∞.
r→0
exactly as if V (r ) = ∞ r ≤ 0, but of course r < 0 is impossible, so we had better
be careful about behavior of u n (r ) and Rn (r ) as r → 0
l
l
updated September 19,
5.73 Lecture #28
28 - 4
Note also that d 3r = r 2 sin θdrdθdφ
Rn*′ ′ Rn r 2 dr = un*′ ′ (r )un (r )dr
l
l
l
l
r2 cancelled.
So volume
element looks
just as in 1-D
problem
Return to special situation as r → 0.
Why do we care? It turns out that s-orbitals have Rns(0) ≠ 0 and that in ESR
one measures “Fermi-contact” hyperfine structure which is the spin-density
at specific nuclei. It is a direct measure of the ns atomic orbital character
in each molecular orbital!
CTDL, p. 781
What is the worst possible divergence of Rnl(r) as r → 0?
For r → 0, Rnl(r) will be dominated by rs where |s| is as small as
possible. This is the most strongly divergent part of Rnl(r), which
is all we need to be concerned with as r → 0.
Let Rnl ~ Crs, where this is a good approximation at r → 0.
Plug this definition into Schrödinger Equation
d2
dr
2
rRnl ( r ) =
Tr = −
2
d2
dr
2
Cr s +1 = (s + 1)(s)Cr s −1
2
1 d
2µ r dr 2
h
HRnl ( r ) = −
h
2
2µ
C(s + 1)(s) r
if V( r ) ∝
s −2
+
2
( + 1)
h l l
2µ
1
r
Cr s −2 + V( r )Cr s − E nl Cr s = 0
As r → 0 V( r ) rarely diverges
 more rapidly than 1 / r , thus


V(r)C r s gives r s-1 .






*
Then, in the limit r → 0, the coefficients of the rs–2 term (i.e. the
most rapidly divergent term) must be = 0
–(s + 1)s + l(l + 1) = 0
*
This excludes the stronger divergence of the centrifugal barrier term in Vl(r).
updated September 19,
5.73 Lecture #28
satisfied if s =
28 - 5
s = − (l + 1)
or
l
verify second possibility:
s(s + 1) = ( − l − 1)( − l − 1 + 1) = −( l + 1)( − l ) =
In other words R nl ( r ) → r l
(
l l
OR (if s = ±(l + 1))
well behaved
at r → 0
+ 1)
1
as r → 0
r l +1
disaster even if
l = 0
Actually both of these possibilities satisfy the differential equation for
1
V( r ) = (known as the Coulomb – or H atom Hamiltonian), but
r
the one that diverges as r → 0 cannot satisfy the r → 0 boundary condition
for the H atom.
** Regular and Irregular Coulomb wavefunctions – we will return to
these later in the context of Quantum Defect Theory.
So for now we insist that
Rn (r ) → r
l
l
as
r→0
Rns (0) ≠ 0 special situation for R
ns(r)
Rnl>0(0) = 0
unl (0) = 0
for all l
(no special case for uns(r))
updated September 19,
5.73 Lecture #28
28 - 6
For Hydrogen
V (r ) = +
l
2
( + 1) e 2
−
2µr 2
r
q2
e ≡
4 πε °
h l l
2
µ=
V (r )
mem p
me + m p
≈ me
l
0
0
r
•
V0 (r )
V (r )
l
What do we know from our study
of 1-D problems?
WKB
ψ envelope ∝
p −1/ 2
~
h
λ
=
2 2p(r )
# of nodes, placement of nodes,
degeneracy, behavior at inner
and outer turning points,
location of inner and outer
turning points
•
ASK QUESTIONS
shape of un (r )
l
1st lobe, last lobe
h
of action acquired
2
(associated with tunneling into nonclassical region)?
inner vs. outer part of u nl ( r ) – where is the extra

 recall

∫
r > (E)
r < (E)
p( r )dr =
h
2

( n + 1 /2) 

updated September 19,
5.73 Lecture #28
28 - 7
E nl = −
Find that
ℜ
2
n
At turning point Vl ( r ) = E nl
−
ℜ
=
2
( + 1)
h l l
−
ℜ=
2
−
e
r±
2µr±2
n2
solve for r± as function of n and l
[
r± = a0 n 2 ± n(n 2 − l(l + 1))
1/ 2
]
−
a0 =
e4 me
2 h2
ℜ
=
n2
2
( + 1) − r± e 2 2µ
2µr±2
h l l
2µℜr±2
= h2l(l + 1) − r± e 2 2µ
2
n
Use Quadratic
formula to find r±(n)
h
2
Bohr radius
e 2m e
when l << n, where are r+ and r– ?
  l(l + 1)  1/ 2 
2
= a0 n 1 ± 1 −

n2  
 
Use this equation for the turning points to construct qualitatively correct
cartoons of Rnl(r) in crucial regions.
surprising systematic degeneracy
etc.
3s ————
3p ————
2s ————
2p ————
3d ————
En
l
1s ————
Because of pattern, we use n to label degenerate groups
ℜ
n2
hence n is not # of radial nodes.
En = −
l
updated September 19,
5.73 Lecture #28
28 - 8
orbital
# of radial nodes
1s
2s
0
1
2p
3s
0
2
3p
3d
1
0
(because it is lowest solution to
l
= 1 equation)
# radial nodes = (n − 1) − l
# angular nodal surfaces
total # nodes n − 1
l
n
degeneracy
1
1
2
1 + (2 l + 1) = 4
3
1+3+5=9
…
n2
n
n - scaling of r σ
two limits:
σ<0
determined near
inner turning
point
vs.
σ>0
outer turning point
~ n −3
Bohr model rn = a0 n 2
(see argument on
next page)
r σ ∝ a0σ n 2 σ
l
Expectation values of rσ vs. transition moments and off--diagonal matrix
elements of rσ. Stationary phase.
updated September 19,
5.73 Lecture #28
28 - 9
inner region
3% of IP
n=6
En = −
ℜ 109737cm −1
=
≈ 3000cm −1
36
36
n=1
p2
T=
> IP in the “inner region.”
2µ
variation of T n = 6 to n = ∞ < 3%
variation of p
< 1.5%
h

 independent of n ! Because p is large and
p

fractional change of p vs. n is negligible.
location of innermost node

deBroglie λ ~
e– comes into core region fast and leaves fast — ∆t same for all n
time inside 2( v λ)
=
h
one period
E n − E n +1
−1
fraction of time inside core?
E n = – ℜ n2
2δℜ
 h 
E n +δ 2 − E n −δ 2 = 3
λ/v
 p 
n
2

 p m  4 mℜ
= 
≈ 2 3
h
pn
2ℜ n3
n-independent
probability of finding
e– inside core ∝ n–3 !
updated September 19,
5.73 Lecture #28
28 - 10
fraction of time inside ∝
n −3
~
amplitude of ψ n ∝ n −3/ 2 inside core region
l
Basis of all Rydberg scaling
1st node does not shift with n

inner lobe 
Astonishingly important!
−3/2 
amplitude
in
first
lobe
scales
as
n



all n , n ′ matrix elements of r σ where σ < 0 scale as ( nn ′ )
−3/2
!
Some matrix elements scale this way even when σ > 0.
McQuarrie, page 223
 (n − l − 1)!   2 
Rn (r ) = − 
 

3
 2n (n + l )!   na0 
1/ 2
[
l
l
+ 3/ 2
 2r 
r e −r na0 L2n ++11 

 na0 
l
]
l
normalization
exponential
→ 0 as r → ∞
Regular and Irregular Coulomb functions (E ≤ 0)
un (r) ≡ rR(r)
r→0
l
regular
f(E,l,r)
∝
r +1
l
associated
Laguerre functions
(polynominals)
r→∞
u(ν,l,r)sin πν – v(ν,l,r)eiπν, which is an
increasing exponential except when ν is
a positive integer. Need some other way
to satisfy r → ∞ boundary condition
when ν is not an integer
irregular
g(E,l,r)
∝
r–( )
l
–u(ν,l,r)cos πν + v(ν,l,r)eiπ(ν+1/2), which
blows up.
* u(ν,l,r) is an increasing exponential as
r→∞
* v(ν,l,r) is a decreasing exponential as
r→∞
(see Gallagher, page 16)
updated September 19,
5.73 Lecture #28
28 - 11
T.F. Gallagher, Rydberg Atoms, page 25
r
1 2
[3n − l(l + 1)]
2
r2
n2
[5n2 + 1 − 3l(l + 1)]
2
1r
1 n2
1 r2
1
n (l + 1 / 2 )
1 r3
n (l + 1)(l + 1 / 2)l
3
1
3
1 r4
3n 2 − l(l + 1)
2 n 5 (l + 3 / 2)(l + 1)(l + 1 / 2)l(l − 1 / 2)
1 r6
35n 4 − 5n 2 [6l(l + 1) − 5] + 3(l + 2)(l + 1)l(l − 1)
8n 7 (l + 5 / 2)(l + 2)(l + 3 / 2)(l + 1)(l + 1 / 2)l(l − 1 / 2)(l − 1)(l − 3 / 2)
Note!
all r σ
σ < −1
scale as n −3!
σ>0
scale as n 2 σ !
updated September 19,