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H+ ψ 0 = (2.76)
2
corresponding to the solution given in (2.73).
All this can be generalized by rewriting the previous steps as
follows. We search for a potential V1 that satis¬es the Schroedinger
equation
1 d2
+ V1 ψ1 = ’χ2 ψ1
’ (2.77)
1
2
2 dx
with V1 +χ2 signifying a zero-energy bound state. The relation (2.74)
1
is re-expressed as
W1 ’ W1 = V! + χ2
2
(2.78)
1

with χ1 obtained from W1 +W1 = χ2 and V1 identi¬ed with ’2χ2 sech2
2
1 1
χ1 (x’x0 ). Note that V1 (±∞) = 0. Further requring V f 1 to be sym-
metric means that W1 must be an odd function.
For an arbitrary n levels, we look for a chain of connections

Wn + Wn = Vn’1 + χ2
2
(2.79)
n

with Vn’1 assumed to be known (notethat V0 = 0). Then Vn is
obtained from
Wn ’ Wn = Vn + χ2
2
(2.80)
n

where Wn (0) is taken to be vanishing.
Linearization of (2.79) is accomplished by the substition Wn =
gn /gn yielding
’ gn + Vn’1 gn = ’χ2 gn (2.81)
n



© 2001 by Chapman & Hall/CRC
1
As with (2.75), has also Un = satis¬es
gn

’ Un + (Wn ’ Wn )un = ’un + (Vn + χ2 )Un = 0
2
(2.82)
n

That is
’ Un + Vn Un = ’χ2 Un (2.83)
n

which may be looked upon as a generalization of (2.76) to n-levels.
In this way one arrves at a form of the Schroedinger equation which
has n distinct eigenvalues. Evidently V1 = ’2χ2 sech2 χ1 (x ’ x0 ) is
1
re¬‚ectionless.
In the study of nonlinear systems, V1 can be regarded as an in-
stantaneous frozen one-soliton solution of the KdV equation ut =
’uxxx + 6uux . The n-soliton solution of the KdV, similarly, also
emerges [79-85] as families of re¬‚ectionless potentials. It may be
remarked that if we solve (2.81) and use gn (x) = gn (’x) then we
uniquely determine Wn (x). For a further discussion of the construc-
tion of re¬‚ectionless potentials supporting a prescribed spectra of
bound states we refer to the work of Schonfeld et al. [85].

(c) SUSY and derivation of a hierarchy of Hamiltonians

The ideas of SUSYQM can also be used to derive a chain of
Hamiltonians having the properties that the adjacent members of
the hierarchy are SUSY partners. To look into this we ¬rst note that
an important consequence of the representations (2.29) is that the
partner potentials V± are related through

d2 +
V+ (x) = V’ (x) + 2 log ψ0 (x) (2.84)
dx
where we have used (2.56). The above equation implies that once
the properties of V’ (x) are given, those of V+ (x) become immedi-
ately known. Actually in our discussion of re¬‚ectionless potentials
we exploited this feature.
We now proceed to generate a sequence of Hamiltonians employ-
ing the preceding results of SUSY. Sukumar [29] pointed out that if a
certain one-dimensional Hamiltonian having a potential V1 (x) allows
for M bound states and has the ground-state eigenvalue and eigen-
(i) (i)
function as E0 and ψ0 , respectively, one can express this Hamilto-


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nian in a similar form as H+
1 d2
H1 =’ + V1 (x)
2 dx2
1+ (i)
= A1 A1 + E0 (2.85)
2
(1)
where A1 and A+ are de¬ned in terms of ψ0 . Using (2.34) and
1
+
(2.56) A1 and A1 can be expressed as
d (1) (1)
A1 = ’ ψ0 /ψ0
dx
d (1) (1)
A+ = ’ ’ ψ0 /ψ0 (2.86)
1
dx
where a prime denotes a derivative with respect to x.
The supersymmetric partner to H1 is obtained simply by inter-
changing the operators A1 and A+ 1

1 d2 1 (1)
+ V2 (x) = A1 A+ + E0
H2 = ’ (2.87)
1
2 dx2 2
where the correlation between V1 and V2 is provided by (2.84)
d2 (1)
V2 (x) = V1 (x) ’ 2 lnψ0 = V1 (x) + A1 , A+ (2.88)
1
dx
From (2.59) we can relate the eigenvalues and eigenfunctions of
H1 and H2 as
(1) (2)
En+1 = En
(1) ’1/2
(1) (1)
(2)
ψn = 2En+1 ’ 2E0 A1 ψn+1 (2.89)

To generate a hierarchy of Hamiltonians we put H2 in place of
H1 and carry out a similar set of operations as we have just now
done. It turns out that H2 can be represented as
1 d2 1 (2)
+ V2 (x) = A+ A2 + E0
H2 = ’ (2.90)
2
2
2 dx 2
with
d (2) (2)
A2 = ’ ψ0 /ψ0
dx
d (2) (2)
A+ =’ ’ ψ0 /ψ0 (2.91)
2
dx


© 2001 by Chapman & Hall/CRC
H2 induces for itself a supersymmetric partner H3 which can be
obtained by reversing the order of the operators A+ and A2 . In
2
this way we run into H4 and build up a sequence of Hamiltonian
H4 , H5 , . . . etc. A typical Hn in this family reads
1 d2 1 (n) (n’1)
+Vn (x) = A+ An +E0 = An’1 A+ +E0
Hn = ’ (2.92)
n n’1
2
2 dx 2
with
d (n) (n)
An = ’ ψ0 /ψ0
dx
d (n) (n)
A+ = ’ ’ ψ0 /ψ0 (2.93)
n
dx
and having the potential Vn
d2 (n’1)
Vn (x) = Vn’1 (x) ’ 1nψ0 , n = 2, 3, . . . M (2.94)
dx2
Further, the eigenvalues and eigenfunctions of Hn are given by
(n’1) (1)
(n)
Em = Em+1 = . . . = E(m+n’1) ,
m = 0, 1, 2, . . . M ’ n, n = 2, 3, . . . M (2.95)
(1) (1) (1) (1)
(n)
ψm = 2Em+n’1 ’ 2En’2 2Em+n’1 ’ 2En’3 . . .
’1
(1) (1) (n+m’1)
2
2Em+n’1 ’ 2E0 — An’1 An’2 . . . A1 ψ1 (2.96)
The following two illustrations will make clear the generation of
Hamiltonian hierarchy.

(1) Harmonic oscillator

Take V1 = 1 ω 2 x2 . The ground-state wave function is known to
2
(0) 2
be ψ1 ∼ e’ωx /2 . It follows from (2.84) that V2 (x) = V1 (x) + ω,
V3 (x) = V2 (x) + ω = V1 (x) + 2ω etc. leading to Vk (x) = V1 (x) + (k ’
1)ω. This amounts to a shifting of the potential in units of ω.

(2) Particle in a box problem

Here the relevant potential is given by
V1 = 0 |x| < a
= ∞ |x| = a


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The energy spectrum and ground-state wave function are well known

π2
(1)
(m + 1)2 , m = 0, 1, 2 . . .
Em = 2
8a πx
(1)
ψ0 = A cos
2a
where A is a constant. From (2.89) we ¬nd for Vn the result
π πx
n(n ’ 1) sec2
Vn (x) = V1 (x) + n = 1, 2, 3 . . .
8a 2 2a
π
(1)
(n)
= Em+n’1 = 2 (n + m)2 m = 0, 1, 2 . . .
Em
8a
We thus see that the “particle in the box” problem generates a se-
ries of sec2 πx potentials. The latter is a well-studied potential in
2a
quantum mechanics and represents an exactly solvable system.

(d) SUSY and the Fokker-Planck equation

As another example of SUSY in physical systems let us examine
its subtle role [18] on the evaluation of the small eigenvalue associated
with the “approach to equilibrium” problem in a bistable system. For
a dissipative system under a random force F (t) we have the Langevin
equation
‚U
x=’
™ + F (t) (2.97)
‚x
where U is an arbitrary function of x and F (t) depicts the noise term.
1
Assuming F (t) to have the “white-noise” correlation (β = T )

F (t) = 0, F (t)F (t ) = 2βδ(t ’ t ) (2.98)

the probability of ¬nding F (t) becomes Gaussian

1
F 2 (t)dt
P [F (t)] = A exp ’ (2.99)

1
F 2 (t)dt

where A’1 = D[F ]e 2β .
The Fokker-Planck eqution for the probability distribution P is

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