1

√ (bx ’ iby )

B= (2.53)

2

2

pz

we may diagonilize (2.50a) to obtain 1 H ’ 2m = ω B + B + 1

2 2

• + ω σ which is a look-alike of (2.28). Summarizing, the two-

23

dimensional Pauli equation (2.50) gives a simple illustration of how

SUSY can be realized in physical systems.

2.4 Properties of the Partner Hamiltonians

As interesting property of the supersymmetric Hamiltonian Hs is

that the partner components H+ and H’ are almost isopectral. In-

deed if we set

+ ++

H+ ψ n = En ψ n (2.54)

it is a simple exercise to work out

1

+

AA+ Aψn +

H’ Aψn =

2

1+ +

=A A Aψn

2

+ +

= En Aψn (2.55)

+

This clearly shows En to be the energy spectra of H’ also. However,

+ +

Aψ0 is trivially zero since ψ0 being the ground-state solution of H+

satis¬es

+ +

’(ψ0 ) + (W 2 ’ W )ψ0 = 0 (2.56a)

and so is constrained to be of the form

x

+

ψ0 = C exp ’ W (y)dy (2.56b)

C is a constant.

We conclude that the spectra of H+ and H’ are identical except

for the ground state (n = 0) which is nondegenerate and, in the

present setup, is with the H+ component of Hs . This is the case of

© 2001 by Chapman & Hall/CRC

unbroken SUSY (nondegenerate vacuum). However, if SUSY were

to be broken (spontaneously) then H+ along with H’ can not posses

any normalizable ground-state wave function and the spectra of H+

and H’ would be similar. In other words the nondegeneracy of the

ground-state will be lost.

For square-integrability of ψ0 in one-dimension we may require

from (2.56) that W (y)dy ’ ∞ as |x| ’ ∞. One way to realize

this condition is to have W (x) di¬ering in sign at x ’ ±∞. In other

words, W (x) should be an odd function. As an example we may

keep in mind the case W (x) = ωx. On the other hand, if W (x) is

an even function, that is it keeps the same sign at x ’ ±∞, the

square-integrability condition cannot be ful¬lled. A typical example

is W (x) = x2 .

From (2.54) and (2.55) we also see for the following general eigen-

value problems of H±

(+) (+) (+)

H+ ψn+1 = En+1 ψn+1

(’) (’) (’)

H’ ψ n = En ψ n (2.57a, b)

+ +

that if Aψ0 = 0 holds for a normalizable eigenstate ψ0 of H+ , then

+ +

since H+ ψ0 ≡ 1 A+ Aψ0 = 0, it follows that such a normalizable

2

+

eigenstate is also the ground-state of H+ with the eigenvalue E0 = 0.

Of course, because of the arguments presented earlier, H’ does not

possess any normalHzed eigenstate with zero-energy value.

To inquire how the spectra and wave functions of H+ and H’

are related we use the decompositions (2.36) to infer from (2.57) the

eigenvalue equations

1

H+ A+ ψn = A+ A A+ ψn = A+ H’ ψn = En A+ ψn (2.58a)

’ ’ ’ ’ ’

2

1

H’ Aψn = AA+ Aψn = AH+ ψn = En Aψn

+ + + + +

(2.58b)

2

It is now transparent that the spectra and wave functions of H+ and

H’ are related a la [52]

+ +

’

En = En+1 , n = 0, 1, 2, . . . ; E0 = 0 (2.59a)

’1

+ +

’

2En+1 2

ψn = Aψn+1 (2.59b)

1

+

ψn+1 = (2En )’ 2 A+ ψn

’ ’

(2.59c)

We now turn to some applications of the results obtained so far.

© 2001 by Chapman & Hall/CRC

2.5 Applications

(a) SUSY and the Dirac equation

One of the important aspects of SUSY is that it appears natu-

rally in the ¬rst quantized massless Dirac operator in even dimen-

sions. To examine this feature [47, 54-74] we consider the Dirac equa-

tion in (1+2) dimensions with minimal electromagnetic coupling

(iγ µ Dµ ’ m) ψ = 0 (2.60)

where Dµ = Dµ +iqAµ with q = ’|e|. The γ matrices may be realized

in terms of the Pauli matrices since in (1+2) dimensions (2.60) can

be expressed in a 2 — 2 matrix form: γ0 = σ3 , γ1 = iσ1 and γ2 = iσ2 .

Introducing covariant derivatives

‚

D1 = ’ ieA1 ,

‚x

‚

D2 = ’ ieA2 (2.61)

‚y

Then (2.60) translates, in the massless case, to

’ (σ1 D1 + σ2 D2 ) ψ = σ3 Eψ (2.60a)

The above equation is also representative of

0 A

ψ = ’σ3 Eψ (2.62)

A+ 0

where A = D1 ’ iD2 and A+ = D1 + iD2 . From (2.35) and (2.36)

we therefore conclude

2Hs ψ = E 2 ψ (2.63)

The supersymmetric Hamiltonian thus gives the same eimen-

function and square of the energy of the original massless equation.

This also makes clear the original curiosity [75] of SUSY which was

to consider the “square root” of the Dirac operator in much the same

manner as the “square-root” of the Klein-Gordon operator was uti-

lized to arrive at the Dirac equation. In the case of a massive fermion

the eigenvalue in (2.63) gets replaced as E 2 ’ E 2 ’ m2 .

© 2001 by Chapman & Hall/CRC

In connection with the relation between chiral anomaly and

fermionic zero-modes, Jackiw [68] observed some years ago that the

Dirac Hamiltonian for (2.60), namely

’

’

’

H = ± . p + eA (2.64)

’

where ± = (’σ 2 , σ 1 ), displays a conjugation-symmetric spectrum

with zero-modes under certain conditions for the background ¬eld.

The symmetry, however, is broken by the appearance of a mass term.

Actually, in a uniform magnetic ¬eld the square of H coincides with

the Pauli Hamiltonian. As already noted by us the latter exhibits

SUSY which when exact possesses a zero-value nondegenerate vac-

uum.

Hughes, Kostelecky, and Nieto [69] have studied SUSY of mass-

less Dirac operator in some detail by focussing upon the role of

Foldy-Wouthusen (FW) transformations and have demonstrated the

relevance of SUSY in the ¬rst-order Dirac equation. To bring out

Dirac-FW equivalence let us follow the approach of Beckers and De-

bergh [71]. These authors have pointed out that since SUSYQM is

characterized by the algebra (2.32) and (2.33) involving odd super-

charges, it is logical to represent the Dirac Hamiltonian as a sum of

odd and even parts

HD = Q1 + βm (2.65)

where Q1 is odd and the mass term being even has an attached

multiplicative coe¬cient β that anticommutes with Q1

{Q1 , β} = 0 (2.66)

Squaring (2.65) at once yields

HD = Q2 + m2

2

1

= Hs + m2 (2.67)

from (2.44).

We an interpret (2.67) from the point of view of FW transfor-

mation which works as

= U HD U ’1

HF W

= β(Hs + m2 )1/2 (2.68)

© 2001 by Chapman & Hall/CRC

implying that the square of HF W is just proportional to the right-

hand side of (2.67). Note that U , which is unitary, is given by

S = S+ : U = exp(iS)

i

S = ’ βQ1 K ’1 θ

2

K

tanθ =

m

[θ, β] = 0

{HD , S} = 0 (2.69)

√

where K is even and stands for Hs with the positive sign. We can

also write

E + βQ1 + m

U= (2.70)

1/2

[2E(E + m)]

with E = (Hs + m2 )1/2 .

The SUSY of he massless Dirac operator links directly to two

very important ¬elds in quantum theory, namely index theorems and

anomalies. Indeed it is just the asymmetry of the Dirac ground state

that leads to these phenomena.

(b) SUSY and the construction of re¬‚ectionless potentials

In quantum mechanics it is well known that symmetric, re¬‚ec-

tionless potentials provide good approximations to con¬nement and

their constructions have always been welcome [48,76-78]. In the fol-

lowing we demonstrate [76-86] how the ideas of SUSYQM can be

exploited to derive the forms of such potentials.

Of the two potentials V± , let us impose upon V’ the criterion

that it possesses no bound state. So we take it to be a constant 1 χ2

2

1 1

W2 + W = χ2 > 0

V’ ≡ (2.71)

2 2

Equation (2.71) can be linearized by a substitution W = g /g which

converts it to the form

g

= χ2 (2.72)

g

The solution of (2.72) can be used to determine W (x) as

W (x) = χ tanh χ(x ’ x0 ) (2.73)

© 2001 by Chapman & Hall/CRC

Knowing W (x), V+ can be ascertained to be

1

W2 ’ W

V+ ≡

2

12

χ 1 ’ 2sech2 χ (x ’ x0 )

= (2.74)

2

One can check that H+ possesses a zero-energy bound state wave

function given by

1

ψ0 ∼ ∼ sechχ(x ’ x0 ) (2.75)

g

becausc

1

’ψ0 + W 2 ’ W ψ0 = 0