A. Das

S. Pernice

English

Research Papers in Economics

Response to “Supersymmetry in the
Half-Oscillator - Revisited”
A. Das* and S. Pernice**
*Department of Physics and Astronomy,
University of Rochester,
Rochester, New York, 14627
** Univ. del CEMA,
Buenos Aires, Argentina, 1054
Abstract
We point out the ﬂaw in the analysis of Gangopadhyaya and Mallow, hep-
th/0206133, where it is claimed that supersymmetry is broken in the SUSY
half-oscillator, even with a regularization respecting supersymmetry.
In an earlier paper [1], we had shown that supersymmetry, in quantum mechanical
theories with singular potentials or nontrivial boundaries, is preserved if the system is
regularized in a manner respecting supersymmetry. This is in contrast to the earlier
claims [2, 3, 4] that supersymmetry is broken in such systems. We had shown, in
particular, that if the superpotential is regularized, as opposed to the conventional
1wisdom of regularizing the potential, supersymmetry is maintained in such systems
in a natural manner. The reason for this is quite clear. A regularized superpotential
leads to a pair of supersymmetric Hamiltonians for every value of the regularization
parameter whereas a conventionally regularized potential does not lead to a pair of
supersymmetric Hamiltonians for any value of the regularization parameter.
In a recent paper, Gangopadhyaya and Mallow [5] claim that of the examples
treated in [1], the supersymmetric half-oscillator does not possess supersymmetry
in spite of using the supersymmetric regularizations proposed in [1]. In fact, their
analysis is exactly the same as given in [1]. However, their ﬁnal solution of the
boundary condition and, therefore, their conclusion is incorrect for a very simple
reason that we wish to point out.
Let us recall that the SUSY half-oscillator was ﬁrst studied in [3] where the authors
concluded that supersymmetry is broken. Physically, their result can be understood
as follows. If we require the wavefunctions to vanish at the origin (as these authors do
and as would be the case when one conventionally regularizes the potential), only odd
solutions of the harmonic oscillator are allowed. On the other hand, supersymmetry
requires the partner wavefunctions to have opposite parity and since this is not allowed
by the boundary conditions (which follows from the particular regularization used),
supersymmetry must be broken. The problem with this conclusion, as pointed out in
[1], is that such a treatment of the system corresponds to regularizing the potential
in a manner which itself breaks supersymmetry and, therefore, the result of such an
analysis can be an artifact of the regularization used. In stead, we had proposed to
2regularize the superpotential directly as
W(x) = !x(x) + c(x) (1)
where c is the regularization parameter and we take c ! +1 at the end. This
superpotential leads to the regularized supersymmetric potentials:
V+(x) =
1
2
[(!
2x
2  !)(x) + c
2(x)  c(x)]
V(x) =
1
2
[(!
2x
2 + !)(x) + c
2(x) + c(x)] (2)
The crucial diﬀerence between this and the earlier treatment of this problem [3] is the
appearance of Dirac delta functions in the potentials. The delta function is attractive
for one of the Hamiltonians while it is repulsive for the other. It is clear, therefore,
that, in the presence of an attractive delta function potential, the usual argument
for the non existence of even solutions is not automatic. In fact, we had shown
explicitly that supersymmetry is maintained in the limit c ! +1 in this theory. We
had also proposed an alternate regularization in [1] that avoids the appearance of
delta functions, and had shown that the end result is the same if the regularization
maintains supersymmetry.
It is surprising, therefore, that Gangopadhyaya and Mallow [5], who carry out ex-
actly our analysis would arrive at a diﬀerent conclusion, namely, that supersymmetry
is broken in this system. Let us note that the boundary condition (eq. (23) in [1] and
eq. (10) in [5]) is given by (we choose ! = 1
2 for simplicity)

p
2
Γ(1
2  )
Γ()
=
p
c2  2  c (3)
where  represents the energy of the eigenstates. Gangopadhyaya and Mallow claim
that  = 0 is not a solution of the above equation, in which case supersymmetry will
3be broken. It is, of course, obvious that  = 0 is a solution of (3) for any value of c
and, in particular, for c ! +1, but let us look at this more carefully to understand
the ﬂaw in the analysis of [5].
Let us deﬁne, for simplicity,
F()  
p
2
Γ(1
2  )
Γ()
and G(;c) 
p
c2  2  c ; (4)
so that we can represent the boundary condition (3) as
F() = G(;c) (5)
In a quantum theory, the boundary condition (5) may not hold for all values of .
The fact that energy is quantized means that it can be satisﬁed only for some discrete
values of  = n. Let us, in particular, investigate whether  = 0 is a solution of the
equation, and if it is, whether it remains a solution in the limit c ! +1.
Let us note that both F() and G(;c), as complex functions of , are analytic
around  = 0. For F(), we can see this from the fact that the function 1
Γ() is
entire in the whole  plane with simple zeroes at the points  = n, (n = 0;1;2;:::),
and the function Γ(1
2  ) is also analytic around  = 0, with Γ(1
2) =
p
. For
G(;c) =
p
c2  2  c, the same is also true.
Since both these functions are analytic around  = 0, a Taylor expansion around
this value leads to,
F() ' 0 +
p
2  + O(
2) (6)
G(e;c) ' 0 
1
c
 + O(
2) (7)
The fact that  = 0 is a solution of equation (5) is a consequence of the fact that the
zeroth order terms in the Taylor expansion in equations (6) and (7) coincide for that
4value. Note that this result is independent of c and, therefore, also holds in the limit
c ! +1.
The mistake in the analysis of Gangopadhyaya and Mallow lies in the fact that,
for whatever reason, they impose that the coeﬃcient of the linear terms in the Taylor
expansion should match, in addition to the zeroth order terms. Indeed, the equation
below equation (11) (for ! = 1
2) in [5] is precisely the requirement of the equality for
the linear terms in equations (6) and (7). This imposition, however, is unjustiﬁed,
since that would tantamount to requiring
F(0) = G(0;c); and F
0(0) = G
0(0;c) (8)
The boundary condition (5) does not have to hold for all values of , not even in a
neighborhood of  = 0. It should hold only for discrete values of . For the case
 = 0, this translates into the fact that only the zeroth order terms in the Taylor
expansion of F and G should be equal, but no extra conditions must be imposed
on higher order terms. The incorrect conclusion in [5] results from the imposition of
this additional condition. Supersymmetry is, in fact, unbroken in this theory, when
analyzed properly.
Acknowledgment:
This work was supported in part by US DOE Grant number DE-FG 02-
91ER40685.
5References
[1] A. Das and S. Pernice, Nucl. Phys. B561 (1999) 357.
[2] A. Jevicki and J. Rodrigues, Phys. Lett. B146 (1984) 55.
[3] P. Roy and R. Roychoudhury, Phys. Rev. D32 (1985), 1597.
[4] F. Cooper, A. Khare and U. Sukhatme, Phys. Rept. 251 (1995) 267.
[5] A. Gangopadhyaya and J. Mallow, “Supersymmetry in the Half-Oscillator - Re-
visited”, hep-th/0206133.
6

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