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)

+