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between states of same l but di¬erent n and Z. More speci¬cally,
while (6.4) is concerned with states possessing quantum numbers n, l,
2 4
and energies ’ Z2 me , (6.14) acconts for states with quantum num-
2
n¯ h
1
bers (n ’ 1), l and nuclear charge Z(1 ’ n ) having same energies. In
this way the model may be used to establish supersymmetric inter-
atomic connections between states of iso-electronic ions under the
simultaneous change of the principal quantum number and nuclear
charge.
We now consider the isotropic oscillator problem as an applica-
tion of this scheme.
Here the Schroedinger equation is given by (6.8) along with the
energy levels (6.9). By following the prescription of transforming the
half-line (0, ∞) to (’∞, ∞) we employ x = 2lnr to get


1 d2 Ψ 1 1 3x
+ e2x ’
’ n+ e
2 dx2 8 4 2
12
1
+ l+ Ψ=0 (6.15)
8 2


© 2001 by Chapman & Hall/CRC
Equation (6.15) gives2
ex 1 1
W (x) = ’ n+
2 2 2
2
1 2x 1 3x1 1
V+ (x) = e’ n+ e+ n+
8 4 2 8 2
12
1 2x 1 1x1
V’ (x) = e ’ n’ e+ n+ (6.16a, b, c)
8 4 2 8 2
As required we need to transform the Schroedinger equation for
V’ back to the half-line to ¬nd the suitable supersymmetric partner
to (6.8). We ¬nd

1 d2 1 l(l + 1)
+ r2 +

2 dr2 2 2r2
3
’ n+ + 2 χnl (r) = 0 (6.17)
2
It is easy to see that because of the presence of an additional
factor of 2 in (6.17) the di¬erence in the energy levels between (6.8)
and (6.17) is consistent with (6.9).


6.2 Radial Problems Using Ladder Opera-
tor Techniques in SUSYQM
Radial problems can also be handled using ladder operator tech-
niques [8-10]. The advantage is that explicit forms of the superpo-
tential are not necessary. Let us introduce for convenience the ket
notation to express the radial equation (6.3) in the form

1 d2 l(l + 1)
Hl |N, l > ≡ ’ + V (r) + |N, l >
2 dr2 2r2
= ElN |N, l > (6.18)

Here N denotes the radial quantum number for the Coulomb
problem n = N + l + 1 while for the isotropic oscillator case n =
2N + l, N = 0, 1, 2, . . . etc.
2
(n + 1 )2 ’ (l + 1 )2
1
The corresponding eigenvalue of V+ is so that for
8 2 2
n ≥ l (and l ¬xed) the lowest level is zero.



© 2001 by Chapman & Hall/CRC
Consider an operator A given by
N
A= ±l |N , l >< N, l| (6.19)
N

which obviously maps the ket |N, l > to |N , l >. Then

A+ = βlN |N, l >< N , l | (6.20)
N

where ±l = (βlN )— .
N

The representations (6.18) and (6.19) allow one to interpret A+
and A as raising and lowering operators, respectively
N
A|N, l > = ±l |N ’ i, l + j > (6.21)
A+ |N ’ i, l + j > = βlN |N, l > (6.22)

where we have set N = N ’ i and l = l + j. Furthermore it follows
that
A+ A|N, l >= |±l |2 |N, l >
N


AA+ |N ’ i, l + j >= |±l |2 |N ’ i, l + j >
N
(6.23a, b)
where i, j = 0, 1, 2 etc.
The next step is to carry out a factorization of the Hamiltonian
according to
A+ A = H l + F
AA+ = Hl+j + G (6.24a, b)
where F and G are scalars independent of the quantum number N .
Combining (6.18) with (6.24) we arrive at the result

|±l |2 = ElN + F
N
(6.25)

Moreover

(Hl+j + G) A|N, l > = AA+ A|N, l >
= A(ElN + F )|N, l > (6.26)

which implies that

Hl+j [A|N, l >] = ElN + F ’ G [A|N, l >] (6.27)


© 2001 by Chapman & Hall/CRC
(6.27) shows A|N, l > to be an eigenket of Hl+j as well. But since
the eigenvalues of Hl are already known to be ElN from (6.18), we
have for N = N ’ i and l = l + j the relation
N ’i
El+j = ElN + F ’ G (6.28)

Now by repetitive applications of the operator A on |N, l > a
sequence of eigenkets can be created in a manner
N ’(k’1)i
N N ’i
Ak |N, l >= ±l ±l+j . . . ±l+(k’1)j |N ’ ki, l + kj > (6.29)

where k is a positive integer and indicates how many times A is
applied to |N, l >. Since this cannot go inde¬nitely, N being ¬nite,
the sequence has to terminate. So we set A|0, l >= 0 for N = 0
which amounts to restricting ±l = 0. The latter ¬xes F = ’El0
0

from (6.25). It is also consistent to choose i = 1 since k is a positive
integer ≥ 1 and N can take values 0, 1, 2, etc. As such from (6.28)
we have
N ’1
G = ElN ’ El+j ’ El0 (6.30)
So we are led to a scheme [9,10] of SUSY in which the operators
A+ A and AA+ yield the same spectrum of eigenvalues

|±l |2 = ElN ’ El0
N
(6.31)

corresponding to the eigenkets |N, l > and |N ’ i, l + j > except for
the ground state which satis¬es

Hl |0, l >= A+ A|0, l >= 0 (6.32)

We can thus de¬ne a supersymmetric Hamiltonian analogous to
(2.36)
1
diag AA+ , A+ A ≡ diag (H’ , H+ )
Hs = (6.33)
2
with the arguments of (2.55) holding well. In other words, the ground
state is nondegenerate (SUSY exact) and is associated with H+ only.
The H± in (6.33) have the representations.

H+ = Hl ’ El0 , H’ = Hl+j ’ G (6.34a, b)

with G given by (6.30).


© 2001 by Chapman & Hall/CRC
We now turn to some applications of the results (6.34).

(a) Coulomb problem

Here V (r) and ElN are

1
V (r) = ’ (6.35)
r
1
ElN =’ (6.36)
2(N + l + 1)2

From (6.30) we determine G to be
1
G=’
2(N + l + 1)2
1
+
2(N ’ 1 + l + j + 1)2
1
+ (6.37)
2(l + 1)2

For G to be N independent it is necessary that j = 1. The super-
symmetric partner Hamiltonians then read

1 d2
H+ =’
2 dr2
l(l + 1) 1
+ ’
2r2 r
1
+ (6.38)
2(l + 1)2
1 d2
H’ =’
2 dr2
(l + 1)(l + 2) 1
+ ’
2r2 r
1
+ (6.39)
2(l + 1)2

The partner potentials corresponding to the above Hamiltonians are
identical to those in (6.5) and (6.7), respectively, and so similar con-
clusions hold.




© 2001 by Chapman & Hall/CRC
(b) Isotropic oscillator problem

In this case the potential and energy levels are given by
12
V (r) = r (6.40)
2
3
ElN = 2N + l + (6.41)
2
From (6.30) we ¬nd G to be
3
G=2’ l+j+ (6.42)
2
which being free of N does not impose any restriction upon the pa-
rameter j. The supersymmetric partner Hamiltonians can be derived
from (6.34) to be

1 d2
H+ =’
2 dr2
l(l + 1)
+
2r2
r2 3
+ ’ l+ (6.43)
2 2
1 d2
H’ =’
2 dr2
(l + j)(l + j + 1)
+

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