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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