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© 2001 by Chapman & Hall/CRC
CHAPTER 6

Radial Problems and
Spin-orbit Coupling


6.1 SUSY and the Radial Problems
The techniques of SUSY can also be applied to three and higher
dimensional quantum mechanical systems. Consider the three-
dimensional problem ¬rst. The time-independent Schroedinger equa-
tion having a spherically symmetric potential V (r) can be expressed
in spherical polar coordinates (r, θ, φ) as (¯ = m = 1)
h

1 1‚ ‚
r2

2 r2 ‚r ‚r
1 ‚ ‚
+2 sin θ
r sin θ ‚θ ‚θ
‚2
1
+2 2 u(r, θ, φ)
r sin θ ‚φ2
+V (r)u(r, θ, φ) = Eu(r, θ, φ) (6.1)

Equation (6.1) is well known [1] to separate in terms of the respective
functions of the variables r, θ, and φ by writing the wave function as
u(r, θ, φ) ≡ R(r)˜(θ)¦(φ) where R(r) is the radial part and the an-
gular part ˜(θ)¦(φ) is described by the spherical harmonics Y (θ, φ).


© 2001 by Chapman & Hall/CRC
The radial equation has the form
11 d dR l(l + 1)
r2
’ + V (r) ’ E + R=0 (6.2)
2 r2 dr 2r2
dr
in which the ¬rst-order derivative term can be removed by making a
further transformation R ’ χ(r)/r. As a result (6.2) can be reduced
to
1 d2 χ l(l + 1)
’ + V (r) ’ E + χ=0 (6.3)
2 dr2 2r2
which is in the form of a Schroedinger equation similar to that of the
one-dimensional problems and can be subjected to a supersymmetric
treatment. However, for the radial equation r ∈ (0, ∞) it constitutes
only a half-line problem.
To look into how SUSY works in higher-dimensional models let
us start with the speci¬c case of the Coulomb potential. We distin-
guish between two possibilities according to n, the principal quantum
number, which is ¬xed, and l, the angular momentum quantum num-
ber, which is allowed to vary [2-4] or as n varies but l is kept ¬xed
[5-7].

Case 1 Fixed n but a variable l
2
Here V (r) = ’ Ze . So the radial equation is
r

1 d2 1 l(l + 1)
’ ’ En ’ + χnl (r) = 0 (6.4)
2 dr2 2r2
r
1
where χnl (0) = 0, En = ’ 2n2 and r has been scaled1 appropriately.
From (2.29) we ¬nd that the underlying superpotential for (6.4)
satis¬es
1
W2 ’ W
V+ =
2
1 l(l + 1) 1
=’+ + (6.5)
2r2 2(l + 1)2
r
The solution of (6.5) can be worked out as
1 l+1
W (r) = ’ (6.6)
l+1 r
h2 1
1 ¯
The transformation used is of the form r ’ r.
µ Ze2




© 2001 by Chapman & Hall/CRC
implying that the supersymmetric partner to V+ is
1
W2 + W
V’ ≡
2
1 (l + 1)(l + 2) 1
=’+ + (6.7)
2r2 2(l + 1)2
r
Whereas the Bohr series for (6.5) starts from (l+1) with energies
2
1 ’2 , we see from (6.7) that the lowest level for V
2 (l + 1) ’ n ’
begins at n = l + 2 with n ≥ l + 2. SUSY therefore o¬ers a plausible
interpretation of the spectrum of H+ and H’ which corresponds to
the well known hydrogenic ns ’ np degeneracy. In particular if we
set l = 0 we ¬nd H+ to describe the ns levels with n ≥ 1 while H’ is
is consistent with the np levels with n ≥ 2. The main point to note
[2-4] is that SUSY brings out a connection between states of same n
but di¬erent l.
What happens if we similarly apply SUSY to the isotropic oscil-
lator potential V (r) = 1 r2 ? We ¬nd then the Schroedinger equation
2
to read
1 d2 1 l(l + 1) 3
+ r2 +
’ ’ n+ χnl (r) = 0 (6.8)
2 dr2 2 2r2 2
where n is related to l by

n = l, l + 2, l + 4, . . . (6.9)

From (6.8) the superpotential W (r) and the associated super-
symmetric partner potentials can be ascertained to be
l+1
W (r) = r ’
r
1 2 l(l + 1) 3
V+ (r) = r+ ’ l+
2r2
2 2
1 (l + 1)(l + 2) 1
V’ (r) = r2 + ’ l+ (6.10a, b, c)
2r2
2 2
It is here that we run into a di¬culty. The undersirable feature of
(6.10b) and (6.10c) is that these only furnish a connection between
levels l and (l + 1) which is not borne out by (6.9). This counter-
example serves as a pointer that SUSY cannot be applied naively to
higher-dimensional systems.


© 2001 by Chapman & Hall/CRC
It has been argued that supersymmetric transformations are ap-
plicable to the radial problems only after the latter have been trans-
formed to the full-line (’∞, ∞). In the following we see that such
an exercise brings out an entirely di¬erent role of SUSY; it is found
[5,6] to relate states of the same l but di¬erent n and nuclear charge
Z. Interestingly it does also explain the energy jump of two units in
the isotropic oscillator system.

Case 2 Fixed l but a variable n

A relevant transformation which switches r ∈ (0, ∞) to x ∈
(’∞, ∞) is given by r = ex . As a result an equation of the type
(6.3) gets transformmed to

1 d2 Ψ
+ {V (ex ) ’ E} e2x
’ 2
2 dx
12
1
+ l+ Ψ=0 (6.11)
2 2

where χ(r) ’ ex/2 Ψ(x). Note that E no longer plays the role of an
eigenvalue in (6.11).
Corresponding to the Coulomb potential, (6.11) can be written
as

1 d2 Ψ
+ ’ex ’ En e2x
’ 2
2 dx
12
1
+ l+ Ψ=0 (6.12)
2 2

which describes a full-line problem for the Morse potential. From
(6.12) the superpotential W (x) and V± (x) follow straightforwardly

ex 1
W (x) = + ’n
n 2
2
e2x 1 1
’ ex +
V+ (x) = ’n
2n2 2 2

2
e2x 1x1 1
V’ (x) = 2 ’ 1 ’ e+ ’n (6.13a, b, c)
2n n 2 2


© 2001 by Chapman & Hall/CRC
Having obtained the x-dependent expression for V’ we can now
transform back to our old variable r to get the SUSY partner equa-
tion. We ¬nd in this way [7]


1 d2 1 1
’ ’ En ’ 1 ’
2 dr2 n r
l(l + 1)
+ χnl (r) = 0 (6.14)
2r2


The nontrivial nature of the mappings r ’ ex ’ r is evident
from the result that the coe¬cient of 1 term in (6.14) has undergone
r
a modi¬cation by an n-dependent factor compared to (6.4). To inter-
1
pret (6.14) we therefore need to rede¬ne 1 ’ n r as the new variable
2
1
by dividing (6.14) throughout by the factor of 1 ’ n . This also
necesitates rede¬ning the nuclear charge Z by bringing it out explic-
1
itly in (6.14): Z ’ Z 1 ’ n . We thus ¬nd a degeneracy to hold

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