Bradley j. nartowt; qualitative methods in physics; Friday, March 29, 2013; assignment assignment #3 Problem 1 - Charge density distribution d istribution in the Thomas-Fermi model: consider the Thomas-Fermi scaling form for the 1 electrostatic potential inside a heavy atom,
( r )
Z r
f br 3 Z
keZ
f b 3 Z r ; b 1.13
r
(1.1)
Find the scaling scaling form for the local local electron electron density density n n (r ) . Use atomic units, defined by,
me
e
1 4 0
k 1; re
e2
1
(
1
2
4 0 me c
2
2
137
) ; Eh
(
1 4 0
)
2
me e 4
2
me c2 2 ;
(1.2)
2 4 (r) 4 k e n(r) . While you do so, be 2 (r 1 ) 4 3 (r ) . The other term will use 2 f ( r ) r 2 f 1r r 2 (rf ) .
A summary of our computation: just use the Poisson equation, careful to recall
r 2 f 1r r 2 ( rf ) , in order to save labour, 1 2 1 2 1 2b 3 Z f (b 3 Z )2 f 2 3 3 r f (rf ) rf (br Z ) r r 1 f b Z (rf ) r r 2 r r 2 r
Get a messy step out of the way: Carry out the Laplacian
(1.3)
The scaling form we need to get: Thus, our density has the following scaling form,
2 1 Zk e Z 2 1 f 1 2 f 2 1 f 2 n f br 3 Z r 4 4 k e 4 ke r r r (1.4) 2 3 3 3 Z 1 2b Z f (b Z ) f 2(b Z ) f 1 3 3 (b 3 Z ) 2 f n( r ) 4 (r ) f Z (r ) f 2 2 4 r r r 4 r Interestingly, when we compute number density of electrons, the SI units naturally become atomic units, (1.2). Boundary condition/general solution to (1.4): to make good physical sense, the function f must be smooth for all r 0 . Integrate both sides of (1.4), lim 0
0
dr , and you get the boundary condition on the scaling-function 1/3
Estimate the local density density at distances distances of the the order of the Thomas-Ferm Thomas-Fermii radius: aTF ~ Z
f (0) 0 .
. It can be shown (c.f.,
Appendix I, up to (2.9)) that the semiclassical approximation of the radial-eigenfunctions to the Schrödinger equation,
un ( r ) r Rn ( r ) , with the potential V
Z / r i 1 r ri Z
1
, with the ri arranged spherically-symmetrically
(permitting an expansion in spherical-harmonics) produces the following “Schrödinger/Laplace” equation,
2
8 2 3
3/ 2
C
3/ 2
1 d 2 2
r dr
( r ); C
8 2 3
;
(1.5)
This Schrödinger-Laplace equation (1.5) can then be re-written entirely in terms of the scaling function, (1.1), as
f
1 x
f 3/ 2 , (c.f., Appendix II, up to (2.12)) with the indicated approximate solution (see (2.16)),
f ( x 1) 144 x 3 Bx 3.77
144 x (1 Cx 3
0.77
)
n
144 x (1 (12 3
2 / 3 0.77
)
x
0.77
)
3. 3.90
;
Z r
f ( x); x b 3 Z r ;
putting (1.6) (1.6) into the scaling-form for the density, density, (1.4), and and effecting effecting the 2nd derivative at r ~ aTF
Z 1/3 , and throwing
out the delta-function because r 0 , 1
Note the units: use the Gaussian prescription
1 4 0
1 , followed by energy Ry me4 / 2 , and length
(1.6)
a B
2 / me2 .
n( r ~
1 3 Z
Z (b 3 Z )2 1.132 Z 7/3 10 )~ f ( x ~ b) (13.582.0920 x ) 1 2 4 ( 3 Z )
4
10 (13.581.203 x )
4
0.77 5.9
x
6.54
4
1
10 (13.1.728 58 x )
3
0.77 4.9 5.77
x
0.77 3.9 5
x ~ b 1.13
x
0.022Z 7 /3
(1.7)
1/3
Determine the asymptotic behavior of n(r) at distances Z r Z . Transforming this into a condition on x, we have the conditions 1~1.13 Z 2/ 3 x and x 1.13 ~ 1 (e.g., we are really considering x 1 and x 1 ). We thus series-expand the closed-form expression shown in (1.6) (with C, n undetermined). It is actually not illuminating to show this, but here you go, 0.77 1 x 1/3 2/3 1/3 2/3 (1 22 0.773 )3.9 (221/332/3 x0.77 ) 2 (1 22 0.773 )3.9 (221/332/3 x0.77 )2 x x O5 ( 1 ) 7/ 3 n( x 1) ~ Z x x1.54 (1 221/332/3 )3.9 (221/332/3 x0.77 )2 0.77 x 1 221/332/3 8.8 0.77 4555 2 ((1 ) x ( ( x )) O 1/3 2/3 22 3 14.7 2.31 x0.77 x5 (1 ) x 0.77 n( x 1) ~ Z 7/ 3 x 1/3 2/3 1/3 2/3 22 3 10.8 2.31 176 22 3 9.8 1.54 1790 2 2 (1 0.77 ) x ( 5 O ( x)) (1 0.77 ) x ( 5 O ( x))) x x x x
4/3
7/ 3
Perhaps more illuminating: a plot of n / Z , and a plot of / Z
(1.8)
(1.9)
1.13 f ( x ) / x , we have,
4 3 Z
73 n Z
1.2
0.005 1.0
x 1
2
3
4
0.8
5 0.6
0.005 0.4
0.2
0.010
x
;
1
2
3
4
5
(1.10)
Appendix I – The Schrödinger/Laplace nonlinear equation (from Migdal’s text)
We need the potential to get the density. Potential related to density by Poisson equation. note that in atomic units, r n(r ) ,
2V 4 ;
V
V (r , r1 , r2 ,..., r Z ) V (r , ri ) Z / r i 1 r ri Z
1
;
(r ) n n (r );
(2.1)
This potential is averaged over motion of electrons. Write Schrödinger’s equation in atomic units,
2 2( E V ) 0; n m Rn (r )Y m ( , ) u n (r )Y m / r ;
(2.2)
Let the potential V be sphericaly symmetric. Then,
2deg
n m 2 n m 2
nm
un (r )Ym r
2
2 n
un (r) 2 r
2
m
Y m
2
2 n
un (r) 2 2 1
2
r
4
(2.3)
Semiclassical approximation and Bohr quantization: let the radial wavefunctions be approximated using the WKB approximation. In passing, we also write down the Bohr quantization rule,
un (r )
r
an
cos( kn dr
k n
4
r1
); kn
V (r) (
2[ En
r
1 2
2
) ] 2r
2 En V ;
k
n
dr
( n 12 );
(2.4)
r 1
Putting (2.4) into (2.3), we get the density to be expressed as,
2 1 an 2
2 1 an 2 2 1 an 2 2 (r ) ~ 2 cos ( k n dr ) ~ 2 cos 2 2 2 2 4 r kn 4 4 r k 4 r kn n n n n r 1 2 1 1 En 2 1 1 E n 2 1 2 an 2 ~ 2 2 2 n kn r 2 2 2 n kn r 2 4 n kn r 4 n n r
2
1
1 2
(2.5)
Now, change the sum into integration,
En
n
n
E n dn dEn ; n
(2.6)
Putting (2.6) into (2.5), and introducing the explicit k n of (2.4), we get 2,
(r ) ~
1 2 2 r 2
2
2 1
0
1
2
2
Emin
k n r
(2 1) 2
dEn
0 V
2
2 r
1/2 2
r
2
2 En
E min
1
E min V
The V is the effective potential, which is V ( r )
V
(2 1)dE n
0
1
2
( 1)
2 mr
2
2r
2
V
1
(2 1) 2 r 2
2
0 V
2
( 21/r 2)
2
2 En V |0E min
(2 1) 2
2 2 r 2
( 1)
(2.7)
0
for a hydrogen atom. Its derivative is
. We can introduce this to (2.7), and get,
(r )
1 1 2 2
1
2
2 3
(2 1)
2V
2
2r
V 2 2V 1
V
(2V )3/ 2 |V max min
1 3 2
(2 0)
3/ 2
1
V max
2
2V dV
V min
( 2 V )3/2
(2.8)
1 3 2
( 2 V ) 3/ 2
(2 V )
3/ 2
3 2
;
Now, using V , we can put Poisson’s equation (2.1) as,
( ) 4 2
( 2 ( )) 3 2
3/2
2
8 2 3
3/2
C
3/ 2
2
1 d
2
r dr
( r )
(2.9)
Boundary conditions: looking at the original potential (2.1), we have,
Z Z 1 V ( r 0) ( r 0) r i 1 r ri
Z Z 0 ; r r
V (r
) 0 (r );
(2.10)
Change of variables: to solve the nonlinear (1.5), we introduce Zf / r (see (1.1)); this rids the Laplacian of (2.9) 1 stnd th derivative terms, and leaves only 2 and 0 derivatives,
2
Notice we use
Emin V
0 ; recall, (2.4) says that the in tegrand is a cosine function. For this region of integration, the cosine is
of an imaginary argument, which produces exponential decay.
Zf C ( )3/ 2 r
1 d2 r dr
(r
2
Zf r
)
Z
Z
f f C
r
r
f 3/ 2
(2.11)
Second change of variables: now introduce 3 x b 3 Z r (again, see (1.1)) ; we see it is convenient to set 0
C 2/ 3 , so
(2.11) is,
f
d
2
dr
f
2
d
2
x
d (b3 Z )
2
f
Z
C
x b3 Z
f 3/ 2
b 2 Z 2/3 f ( x) C
x
2/3
0 C
1
bZ 4/6
f
C 4/3 Z 2/3 f CC1/3 Z 4/6
3/ 2
x
f 3/ 2 ( x) (2.12)
3/ 2
f f
x
This function (2.12) is independent of atomic number, so it’s universal. It’s called the Thomas-Fermi equation. Thomas Fermi equation and boundary values: consider the (nonlinear) differential equation,
1
f
x
f 3/ 2 ; f (0) 1; f ( ) 0;
Compute f ( x 1) by assuming inverse-power law 4 f Ax
0
(2.13)
; just put this into (2.13) and see that,
x 1 0 ( 0 1) Ax
0 2
1
( Ax 0 ) 3/ 2
x
(2.14)
0 3 0
0 3 0 ( 0 1) Ax (3 1 2 4)/ 2 A 144 0 ( 0 1) A 0
0
Getting a higher order: we could get another order by saying f ( x 1) 144 x expansion f ( x
1)
f 0 f 1 into the original (2.13), using ( f
)3/2
3
f 0 f 1 . Putting this perturbative
f 3/ 2 32
f
O2 ( ) , and using ansatz
Bx 0 , we get, ( f
) f
3
3
144 x
Choose the 0
3.77
2
x
f ( x 1) 144 x
Bx
3/ 2
x
1 x
( f )
0
3/ 2
3 12
0 ( 0 1) x 2 0
2 x2
x
1
f x
0
3/ 2
3 2
f
O
2
( ) ~
0 ( 0 1) 18 0
f
3/ 2
32
f
x
1
2
73
(2.15)
3.8, 4.8 ;
root, for finitude at x . This finally gives, 3
Bx
3.77
144 x3 (1 Cx 0.77 )n
;
f (0) f ~ x3 (0 144 0.77 n Cx ) f (0) 1 144 / C
3 0.77n 0 n 3.90
3.90
1 C (12 )
(2.16)
2/3 0.77
Two-dimensional/cylindrical hydrogen atom5: Suppose that a “nucleus” is an infinitely-thin line of positive charge per unit length. A two-dimensional hydrogen atom is formed by a “nucleus” and an electron which is free to move along the line but is confined in the transverse direction. (This situation is common for, e.g., charged polymer molecules or dislocations in solids.)
3
The book uses in place of my
0 ; this is potentially confusable with the fine str ucture constant mentioned in our atomic units
(1.2), although it is n ever used throughout this work… 4 The inverse power law will easily show the leading-behaviour as x ≫ 1. 5 More details on this interesting problem from: K. Eveker et al. Am. J. Phys. 58, 1183 (1990); T. Garon et al., ibid. 81, 92 (2013).
(a) Determine the appropriate set of “atomic units” ( Ry, a B ,... ); describing a bound state in such an atom. Preliminary: consider atomic units for 3D atom,
me
1
e
4 0
e2
1
k 1; re
4 0 me c
(
1
2
2
2
137
) ; Eh
(
1 4 0
)
2
me e 4
me c2 2 ;
2
(1.1)
We are to eventually get the semiclassical eigenenergies of the Schrödinger equation in a cylinder, at the center of which is a two-dimensional proton 6. First, we develop appropriate units. In SI-units, the Schrödinger equation appears as,
1 2 2 ln E ; 2 ; 2 2 0 0 2 2 z
2 2
( e )
1
2
(1.2)
Some numerical quantities we will need,
Z
1;
e 2r p
; r p
proton-radius
1.65 2
1015 m;
(1.3)
Bohr radius: ratio of quantum (zero point) force to Coulomb-force. In two dimensions,
T0 ~
( / a B ) 2
; U0 ~
e / (2 0 ) a B
U0
; T0
( / aB )2
e / (2 0 ) aB
a B
2
2 0
vs. aB
e
2
4 0
e
2
(1.4)
The log-length-scale: now we need to find 0 . Reproducing some work from K. Eveker et al. (1990),
ln
0
ln z ln ;
2 q 2
2
; q
2
e 2 0
; 0
1
1 2 e 2 2 0
2
0
e
;
(1.5)
Rydberg: using the virial theorem,
E
E
T
V
1 2
V
V
a 1 e V ~ ln B 2 2 2 0 0
1
e
8 0
ln(
42 0
) Ry e
(1.6)
0.529 A and that Ry 13.6 eV ; let’s compute (1.4), (1.5), and (1.6) vs. these, 0 1.47 1013 m a B 4.36 1026 m Ry 4.03 1012 J 17 4 6 (1.7) 8.25 10 ; 2.79 10 ; 1.85 10 ; a B 5.29 1010 m aB 5.29 1010 m Ry 2.18 1018 J
We know that a B
(b) Using the Bohr-Sommerfeld quantization condition for angular momentum, find the spectrum of the bound states. Separating variables: put in the usual ansatz
( ) ( ) z ( z ) , and then (1.2) becomes separated,
2 1 1 e [ ] ( E ln ) z z z z 2 2 2 0 0 2 1 1 e ] ( E ln ) ; [ ] [ z 2 2 2 0 0 z
6
Potential a distance away from an infinite-linear-charge-density
is ( )
E d
(1.8)
Q2 / z d 0 2 encl
0
0
ln
0
.
As usual, a subset of the set of all solutions to (1.8) is formed by taking
z / z 2 ~
1 4 r p 2
/ m 2
and similarly we take 7
; the generality we seem to lose is regained by the completeness of the spectrum of each integer-
indexed set of eigenfunctions (which a Fourier-series takes advantage of, though we don’t need this). Writing a bunch of characteristic-wavevectors for the semiclassical orbit-integral: We are left with the cylinder equation, from which we can read off an appropriate set of units,
2 2 e 2 2 2 2 2 E m2 2 ln 2 0 0 2 2 2 E e 0 1 m 0 2 2 ln 2 0 2 0 2 1 m 2 2 2 ln 2 0 0
2
(1.9)
The original Schrödinger equation (1.2) with the units indicated in (1.9) then appears as,
2 2 ln 2
(1.10)
Let’s find an order of magnitude estimate for the wavevectors in (1.10). using the quantities (1.7), we can measure in Bohr-radii,
2
a B
e
2
0
and E
2
1 a B
E ( Ry) ; we get,
; 2 2
2 E 1 Ry 28.9 2 aB 2
1
E; 2 ~
4 rp
2
1.603 104
1 aB
; 0
2
3.38 1012 a B ; (1.11)
2 k 2 2k 2 p 2 ; using this to find the eigen-energies by integrating this over an orbit using k d (n 12 ) , for turning points [1, 2 ] , and using the 2 2 fact that bound states have E E 0 , we get , and we get, The orbit integral: The semiclassical approximation says
2
( n ) 1 2
2
2 d ln 2 a B d 1 K 2 ln 2
2
1
1
) 2 a
2 a B 1 K ln
( n 12 Here, we noted that 0
B
2
2
0
K
2
2
e K 0 erf K ln 1 2
0 0
2
(1.12)
1
2
( K )2 ln 12 e K K 0 erf K 2 ln 0
0
2
1
3.38 1012 a B aB , whereas ~ a B , so K 2 ln 1 . 0
Since this is a bound state, you have K
/
28.9 / 2 /
E 1 (note that E 1 , by construction).
Turning points: The turning points are found by considering the classical Hamiltonian (or Lagrangian) with a logarithmic potential, and the familiar effective-potential term for an orbiting body,
L
1 2
2
The solutions of Veff 7
1 2
( ) 2
e 2 0
ln
0
1
1
m
2
2
2 ( 2
)2
e 2 0
ln
1
2 Veff V ;
0 2
(1.13)
V are the turning points,
Note that the continuous/unbound z-direction should have a wavevector whose inverse is on the order of the proton-diameter, (1.3).
m2 2 2 1,2 2
e 1,2 2 0
ln
1,2 ln 2
0
0 1,2
m 2 2 0 e
2
1,2 ln 0 1,2 2m a B '
2
2m 2 a B 2 1,2 2 ln
3.38 1012 a B
The graphical solution to (1.14) for m = 1, 2, 3, … and 0 2
m2
0
2m2 0;
(1.14)
1,2
appears as,
2
1.0
0.5
x 0.5
1.0
1.5
2.0
2.5
0.5
(1.15)
1.0
The function that approximately gives these turning points vs. m is,
1,2 Integrating from 1
1,2 (m) (0.2787 m 0.063) a B 2 ( m) 0.2787 m 0.063 M 0 m M1
0
(1.16)
to this turning point 8, and noting that though (1.14) is transcendental, it does provide a 2
substitution for the quantity ln
0
, which appears twice in (1.12); thus, the (1.12) becomes,
( M m M ) 1 K ln M m M 1 e K K erf K 2 ln M m M 0 1 0 2 ( n 12 ) 2 a B 1 K 2 ln1 1 e K K erf K 2 ln1 0 2 0 0
2
1
0
0
1
0
2
2 a B (M 0 m M1 ) 1 K ln M m M 0
1
0
1 2
2
Ke K
1 erf 0 erf K
(1.17) M m M 1
K 2 ln 0 0
1 0
2
Re-introducing the quantities (1.4) through (1.7) to this (1.17) we see that the term that is ~ 0 is, by far , the most dominant term, since 0
( n 12 )
3.38 1012 a
2 a B a B
0
B
, and we have
(M m M ) K ln 0
1
2
2 K
0 e
0 M 0 m M 1
1 erf 1 . Retaining the log-term, (1.17) is, 2
12 Ke K
1 0 2 erf K
2 1 a B 0 2 O3 ( K ) erf K K
(n 12 ) 12 3/ 2 E 2 2 0
2
28.9
(1.18)
2
I am almost certain that I screwed something up, because 0 is just such a huge number. However, in any event, (1.18) says the energy is quadratic in n , which is the well-known condition for free electrons. That’s kind of boring, but corrections higher-order in 0 are just so tiny…at least, I think they are…
8
Migdal sets his r min = 0 when effecting the quasiclassical approximation; see QM 11 – 193 – 3rd (unnumbered) equation.
Two-dimensional Thomas-Fermi atom: Suppose that Z-1 more electrons are added to the atom of Problem 2 in such a
way that the density of electrons, n(r ) ; satisfy the electroneutrality condition, e n ( r ) d r . Here, n(r ) is the 2
number of electrons per unit volume which depends only on the coordinate transverse to the line, r . (a) Construct the Thomas-Fermi model for such an atom. Note: I used r
. It is easier notation…
Recall: the wavefunction for the cylinder equation,
z n 2 z z
( ) ( ) z ( z );
1
e im e im 2 m ; 2 aTF
sin n z;
r p
(1.1)
Let the potential be cylindrically-symmetric, and let there be bound states in the z-direction. Use the radial wavefunction (1.12) developed in the appendix, for which S ( )
m,n 2 z 2 2
2
n 2deg
m, n
n~
R( ) ; we get,
2
m, n
1
r p
2
1 r p 2
S
2 0; m , n
2
m ,n
m , n n
1
S m, n ( )
m ,n
sin 2
2 E m, n
2
z
2 n z 4 cos2 m ~ 2rp r p
1
2 E
r p 2
m, n
2
r p
2
Sm ,n ( ) sin 2
m, n
S
2
0; m ,n
m ,n
n z 4 cos2 m 2r p
12 21 4 21
(1.2)
E m,n n
2
1
2
m ,n
Now, change the sum into integration, and also recall the effective-potential term,
Em, n E m,n n n n dn dEm,n ;
Veff ( )
m2 2 2
2
Veffm ( );
(1.3)
This (2.6) makes (1.2) appear as,
n~
0
1 r p 2
2
m E m min mmax
2 r p 2
2 2
E m ,n
m
2
2
0
dEm ,n
dm
2 r p
mmax 2rp 2
2 2
m
Veff
m
2 V
mmax eff
rp 2
(1.4)
~
2 rp 2
Now, using V ~ V eff max , we can have Poisson’s equation appearing as, m
2
4 2
2
1/ 2 ;
r p
aTF
; ~ 1 2 ~ 2 ; The Thomas-Fermi model;
(1.5)
(b) Find the scaling forms for the electrostatic potential, (r ) ; and local density, n(r ) . Use the change of variables
2
, and, recalling (1.9), the radial Laplacian becomes, 4 5/ 2
4 2 aTF 2 1/4 r p
3
With this scaling form, the density (1.4) similarly becomes,
4 2 aTF 2 15/ 4 r p
4 2
(1.6)
2
n~
2
2
r p
3/4
1/2
2
(1.7)
r p
. Using the electroneutrality condition, we have Z ~ n e aTF , and using the (c) Determine the Thomas-Fermi radius, aTF 2
Poisson/Schrödinger equation (1.5) we have
1/2 ~ 2 ~ / aTF 2 ~
2 4 n ~ 4
Z ne Z ~ e 4 eaTF 22 e 2 2 Z ~ ~ aTF 4 4
Z
Z 2
eaTF
~ 2 ~ / aTF 2 ; putting these together, we have,
1/ 2
ne
aTF
~ 1/4
~
eaTF 4 4
aTF ~
4
4 Z e
(1.8)
Appendix – semiclassical treatment of radial eigenfunction
If you use the substitution S ( )
R( ) (note: S then has no units), the Laplacian transforms as, 1
R (r ) ( (
2
1
S ( )
)) ( (
( S S
1
1 2
S
S 12 S
1
))
1
2
1 1 2 2
S
1
3/2
(1.9)
S S ) 5/ 2 4
Also, recall abbreviations from previous problem; modify them as,
2
Z e
2
0
2
1 a B
; 2 2
2 E
28.9
2
Thus, Schrödinger’s equation for potential U ( )
Z 2 0
1 aB
E ; 2
n
n L
n 2rp
; 0
12 aB ; 3.38 10
(1.10)
ln( / aTF ) appears as,
2 S S 3S 1 S S S 2 m 2 2 0 3/ 2 4 5/ 2 2 3/ 2 ln aTF 2 2 2 m 14 2 2 S S ln 2 ; a TF
(1.11)
Now use the ansatz S S / i , where ( ) ~ / aTF in general. This renders (1.11) separable as a 1 order ODE, st
but the complex-number i is there to introduce both oscillation and exponential behaviour,
S
S0 e S0 e
i
2
2 2 ln
m 2 1/4 i 2
2
d
S0 e
2 m iaTF 1/4 iaTF ( 2 2 ) 2 ln 2 d aTF
m 2 iaTF 1/4 iaTF 12 2 ( 2 k 2 12 k 2 k 2 ln ) ln 2 a TF
S0 e
m2 1/4 iaTF 12 2 ( 2 k 2 12 k 2 k 2 ln ) ln 2 aTF
(1.12)
1
(4) Bound states in 2D: Find an approximate ground-state energy in a weak, attractive potential U (r) in 2D.
, and that max U and typical length-scale a of U (r) are such that, Assume that 0 rU ( r ) dr
max U Draw a schematic of this potential,
max U
U max
2
a 2
~ 0 ;
(1.1)
(1.2) Radial Schrödinger equation in 2D appears as,
2
1 2 2 2 U (r ) E 2 1
(1.3)
R m2 / ; 2 R U (r ) R E R 2 R ;
m
2
2
1
On changing variables as R S , and dimensionless x
2 S m2 S 2 4 5/ 2 2
U ( )
S
S
2 E
x / k 0 , we get 9,
d 2 1 U E 2 4 m2 x2 1 S 0 E dx S
In the ground state, we’d have m 1 . The next step is to compare a 1 4
m2 x
2
U ( x ~ ) E
x
m
2
14
E
U
2mE
(1.4)
, and effect (1.1),
E
(1 14 )
U
3 2
E / U
(1.5)
Enforcing the Bohr-quantization condition, we compute the bound state energies by using the series-expansion of the potential U ( r ) k 0 xU (0) O ( x) , and writing, 1
2
n dx 1 2
x
2 dx x0
1 4
x m2 U (r ) 1 2 dx 2
x 1 4
E
x0
m xU (0) 1 2 2 x k 0 E 2
1 4
m2 k01 xU (0) O 2 ( x) 1 x2 E
4 Fr k0 E
x
3
3
3
4 F r k0 E
x
3
3 tanh
1
1
4U (0) x3 3 k 0 E
x
(1.6)
3 3 k 0 E r x3 4 F
x0
The turning points x0 , x are given by the classical 2D Hamiltonian for an orbiting body,
E
9
p2 2
2
m2
2 r
2
p 0
U ( r ) r m
This is the Ricci differential equation, w ( x n( n 1) 1) w 0 w 2
1 2m E U ( r )
Axjn ( x ) Bxy n ( x ) .
(1.7)
