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-

© 2001 by Chapman & Hall/CRC

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

© 2001 by Chapman & Hall/CRC

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)

2β

1

F 2 (t)dt

’

where A’1 = D[F ]e 2β .

The Fokker-Planck eqution for the probability distribution P is