# Supersymmetry and Singular Potentials

###### Abstract

The breaking of supersymmetry due to singular potentials in supersymmetric quantum mechanics is critically analyzed. It is shown that, when properly regularized, these potentials respect supersymmetry, even when the regularization parameter is removed.

## 1 Introduction

Supersymmetry is a beautiful and, simultaneously, a tantalizing symmetry [1-7]. On the one hand, supersymmetry leads to field theories and string theories with exceptional properties [8-9]. The improved ultraviolet behavior, the natural solution of the hierarchy problem are just a few of the nice features of supersymmetric theories. On the other hand, supersymmetry also predicts degenerate superpartner states (superpartner states with degenerate mass) corresponding to every physical particle state of the theory. Unfortunately, such superpartners with degenerate masses are not observed experimentally and, consequently, one expects that supersymmetry must be spontaneously (dynamically) broken much like the spontaneous breaking of ordinary symmetries in physical theories. However, unlike ordinary symmetries, spontaneous breaking of supersymmetry has so far proved extremely difficult in the conventional framework. Consequently, in the context of supersymmetry, one constantly looks for alternate, unconventional methods of breaking of this symmetry [6-7]. There is, of course, the breaking of supersymmetry due to instanton effects which is well understood. However, several authors, in recent years have suggested that supersymmetry may be broken in the presence of singular potentials or boundaries in a nonstandard manner [10-12]. We will explain the details of the proposed mechanism later, but the gist of the argument is that in such systems, the superpartner states may not belong to the physical Hilbert space thereby leading to a breaking of supersymmetry. This would, of course, explain why the superpartner states would not be observable. Even more interesting is the possibility that since such a breaking is nonstandard, the usual theorems of supersymmetry breaking may not apply and the ground state may continue to have zero energy thereby leading to a solution of the cosmological constant problem also.

The examples, where such a breaking has been discussed, are simple quantum mechanical models which nonetheless arise from the non-relativistic limit of some field theories. It is for this reason that, in an earlier paper, we had examined [13] a candidate relativistic dimensional field theory to see if the manifestation of such a mechanism was possible in a field theory. However, a careful examination of the theory revealed that supersymmetry prevails at the end although it might appear naively, in the beginning, that supersymmetry would be broken in the nonstandard manner. This prompted us to re-analyze the quantum mechanical models, where this mechanism was demonstrated, more carefully in order to understand if supersymmetry is truly broken in these models and if so what may be the distinguishing features in these models from a relativistic field theory. A systematic and critical examination, once again, reveals that when carefully done, supersymmetry is manifest even in such singular quantum mechanical models which is the main result of this paper.

Since our discussion would be entirely within the context of one dimensional supersymmetric quantum mechanics, let us establish the essential notations here. Given a superpotential, , we can define a pair of supersymmetric potentials as

(1) |

where “prime” denotes differentiation with respect to . With and , we can, then, define a pair of Hamiltonians which describe a supersymmetric system as

(2) |

In fact, defining the supercharges as

(3) |

we recognize that we can write the pair of Hamiltonians in eq. (2) also as

(4) |

It is clear now that if represents an eigenstate of with a nonzero energy , then, would be an eigenstate of with the same energy, namely,

(5) |

In other words, the states and (alternately, we can also denote them as and respectively) would correspond to the degenerate superpartner states. All the eigenstates of the two Hamiltonians and would be degenerate except for the ground state with vanishing energy which would correspond to the state satisfying

(6) |

For a given superpotential, at most one of the two conditions in eq. (6) can be satisfied (that is, at most, only one of the two conditions in (6) would give a normalizable state). Namely, the ground state with vanishing energy is unpaired and can belong to the spectrum of either or depending on which of the conditions leads to a normalizable state. This corresponds to the case of unbroken supersymmetry. For example, with , the pair of Hamiltonians

(7) |

describe the supersymmetric harmonic oscillator and it can be easily checked that the normalizable ground state belongs to the spectrum of (In fact, we will choose to have the ground state throughout this paper.).

If, on the other hand, the superpotential is such that neither of the states in eq. (6) is normalizable, then, supersymmetry is known to be broken by instanton effects [6]. For example, this can happen when the superpotential is an even polynomial. In contrast, the new mechanism described, in the presence of a boundary or a singular potential [10-12], corresponds to the case where the action of the supercharges takes a state out of the physical Hilbert space (although and belong to the Hilbert space) so that, say, if is in the Hilbert space, would not belong to the Hilbert space leading to the breaking of the degeneracy of states and, therefore, supersymmetry.

In this paper, we carefully analyze the models [10-12] where this mechanism is thought to be operative. We note that a quantum mechanical potential with a singular structure is best studied with a regularization because this brings out the correct boundary conditions naturally. Furthermore, the regularization must be chosen carefully preserving supersymmetry when one is dealing with a supersymmetric potential with a singular structure. Keeping this in mind, we examine a supersymmetric model with a boundary, namely, the supersymmetric oscillator on the half line [10] in section 2. We briefly recapitulate the results of a harmonic oscillator on the half line and then show through a careful analysis that supersymmetry is, in fact, manifest in such a model in spite of the boundary. In section 3, we analyze a model with a singular potential, namely, the oscillator with a potential. This potential is quite interesting and, in spite of several studies in the literature [14-16], the analysis is not quite complete and we present a systematic and complete analysis of this problem. In section 4, we take up the study of the supersymmetric oscillator with a potential [11] and show that supersymmetry is, in fact, manifest in this system as well, in spite of the singular nature of the potential. There are several interesting aspects of this model which emerge from a careful analysis which we bring out. In section 5, we solve this problem algebraically as well which supports the analysis of section 4. In section 6, we describe the solution to a puzzle raised in the literature [11] in the context of the supersymmetric oscillator with a potential and present our conclusions in section 7.

## 2 Super “Half” Oscillator

To understand the super “half” oscillator, it is useful to recapitulate briefly the results of the “half” oscillator. Let us consider a particle moving in the potential

(8) |

The spectrum of this potential is quite clear intuitively. Namely, because of the infinite barrier, we expect the wave function to vanish at the origin leading to the conclusion that, of all the solutions of the oscillator on the full line, only the odd solutions (of course, on the “half” line there is no notion of even and odd) would survive in this case. While this is quite obvious, let us analyze the problem systematically for later purpose.

First, let us note that singular potentials are best studied in a regularized manner because this is the only way that appropriate boundary conditions can be determined correctly. Therefore, let us consider the particle moving in the regularized potential

(9) |

with the understanding that the limit is to be taken at the end. The Schrödinger equation

can now be solved in the two regions. Since at the end, for any finite energy solution, we have the asymptotically damped solution, for ,

(10) |

Since the system no longer has reflection symmetry, the solutions, in the region , cannot be classified into even and odd solutions. Rather, the normalizable (physical) solution would correspond to one which vanishes asymptotically. The solutions of the Schrödinger equation, in the region , are known as the parabolic cylinder functions [17] and the asymptotically damped physical solution is given by

(11) |

The parabolic cylinder function, , of course, vanishes for large values of . For small values of , it satisfies

(12) |

It is now straightforward to match the solutions in eqs. (10, 11) and their first derivatives across the boundary at and their ratio gives

(13) |

It is clear, then, that as , this can be satisfied only if

(14) |

In other words, when the regularization is removed, the energy levels that survive are the odd ones, namely, (remember that the zero point energy is already subtracted out in (8) or (9))

(15) |

The corresponding physical wave functions are nontrivial only on the half line and have the form

(16) |

Namely, only the odd Hermite polynomials survive leading to the fact that the wave function vanishes at . Thus, we see that the correct boundary condition naturally arises from regularizing the singular potential and studying the problem systematically.

We now turn to the analysis of the supersymmetric oscillator on the half line. One can define a superpotential [10]

(17) |

which would, naively, lead to the pair of potentials

(18) |

Since, this involves singular potentials, we can study it, as before, by regularizing the singular potentials as

(19) |

with the understanding that at the end.

The earlier analysis can now be repeated for the pair of potentials in eq. (19). It is straightforward and without going into details, let us simply note the results, namely, that, in this case, we obtain

(20) |

Here . There are several things to note from this analysis. First, only the odd Hermite polynomials survive as physical solutions since the wave function has to vanish at the origin. This boundary condition arises from a systematic study involving a regularized potential. Second, the energy levels for the supersymmetric pair of Hamiltonians are no longer degenerate. Furthermore, the state with no longer belongs to the Hilbert space (since it corresponds to an even Hermite polynomial solution). This leads to the conventional conclusion that supersymmetry is broken in such a case and let us note, in particular, that in such a case, it would appear that the superpartner states do not belong to the physical Hilbert space (Namely, in this case, the supercharge is an odd operator and hence connects even and odd Hermite polynomials. However, the boundary condition selects out only odd Hermite polynomials as belonging to the physical Hilbert space.).

There is absolutely no doubt that supersymmetry is broken in this case. The question that needs to be addressed is whether it is a dynamical property of the system or an artifact of the regularization (and, hence the boundary condition) used. The answer is quite obvious, namely, that supersymmetry is broken mainly because the regularization (and, therefore, the boundary condition) breaks supersymmetry. In other words, for any value of the regularizing parameters, (even if ), the pair of potentials in eq. (19) do not define a supersymmetric system and hence the regularization itself breaks supersymmetry. Consequently, the breaking of supersymmetry that results when the regularization is removed cannot be trusted as a dynamical effect.

### Regularized Superpotential

Another way to understand this is to note that for a supersymmetric system, it is not the potential that is fundamental. Rather, it is the superpotential which gives the pair of supersymmetric potentials through Riccati type relations. It is natural, therefore, to regularize the superpotential which would automatically lead to a pair of regularized potentials which would be supersymmetric for any value of the regularization parameter. Namely, such a regularization will respect supersymmetry and, with such a regularization, it is, then, meaningful to ask if supersymmetry is broken when the regularization parameter is removed at the end. With this in mind, let us look at the regularized superpotential

(21) |

Here is the regularization parameter and we are supposed to take at the end. Note that although, at this level, both the signs of the regularization parameter are allowed, existence of a normalizable ground state (see eq. (6)) selects out (otherwise, the regularization would have broken supersymmetry through instanton effects as we have mentioned earlier).

The regularized superpotential now leads to the pair of regularized supersymmetric potentials

(22) |

which are supersymmetric for any . Let us note that the difference here from the earlier case where the potentials were directly regularized (see eq. (19)) lies only in the presence of the terms in the potentials. Consequently, the earlier solutions in the regions and continue to hold. However, the matching conditions are now different because of the delta function terms. Carefully matching the wave function and the discontinuity of the first derivative across for each of the wavefunctions and taking their ratio, we obtain the two conditions

(24) |

It is now clear that, as , (2) and (24) give respectively

(25) |

The corresponding wave functions, in this case, have the forms

(26) |

This is indeed quite interesting for it shows that the spectrum of contains the ground state with vanishing energy. Furthermore, all the other states of and are degenerate in energy corresponding to even and odd Hermite polynomials as one would expect from superpartner states. Consequently, it is quite clear that if the supersymmetric “half” oscillator is defined carefully by regularizing the superpotential, then, supersymmetry is manifest in the limit of removing the regularization. This should be contrasted with the general belief that supersymmetry is broken in this system (which is a consequence of using boundary conditions or, equivalently, of regularizing the potentials in a manner which violates supersymmetry).

### Alternate Regularization

Of course, we should worry at this point as to how regularization independent our conclusion really is. Namely, our results appear to follow from the matching conditions in the presence of singular delta potential terms and, consequently, it is worth investigating whether our conclusions would continue to hold with an alternate regularization of the superpotential which would not introduce such singular terms to the potentials. With this in mind, let us choose a regularized superpotential of the form

(27) |

Here is the regularization parameter and we are to take the limit at the end. Once again, we note that, although both signs of appear to be allowed, existence of a normalizable ground state (see eq. (6)) would select .

This regularized superpotential would now lead to the pair of supersymmetric potentials of the form

(28) |

There are no singular delta potential terms with this regularization. In fact, the regularization merely introduces a supersymmetric pair of oscillators for whose frequency is to be taken to infinity at the end.

Since there is a harmonic oscillator potential for both and , the solutions are straightforward. They are the parabolic cylinder functions which we have mentioned earlier. Now matching the wave function and its first derivative at for each of the Hamiltonians and taking the ratio, we obtain

(30) |

It is clear now that, as , eqs. (2) and (30) give respectively

(31) |

The corresponding wave functions are given by

(32) |

These are, of course, the same energy levels and wave functions as obtained in eqs. (25) and (26) respectively showing again that supersymmetry is manifest. Furthermore, this shows that this conclusion is independent of the regularization used as long as the regularization preserves supersymmetry which can be achieved by properly regularizing the superpotential.

## 3 Oscillator with Potential

In the last section, we showed that, in the presence of one kind of singularity, namely, a boundary, supersymmetry is unbroken. In what follows, we will study another class of supersymmetric models, namely, the supersymmetric oscillator with a potential, where there is a genuine singularity in the potential not necessarily arising from a boundary. A naive analysis of this model [11] also shows that supersymmetry is broken by such a singular potential (for certain parameter ranges). However, this conclusion can be understood, again, as a consequence of regularizing the potential which, as we have seen before, does not respect supersymmetry. In stead, we will show through a careful analysis that, when the superpotential is regularized, supersymmetry is manifest in this model as well (with a lot of interesting features). In this section, however, we will systematically analyze only the quantum mechanical system corresponding to an oscillator in the presence of a potential (postponing the discussion of the supersymmetric case to the next section). This system has been analyzed by several people [14-16] and the most complete analysis appears to be in ref. [16]. However, we feel that, while the energy levels derived in [16] are correct, the wave functions are not (namely, the extensions of the solutions from the positive to the negative axis are incomplete and the wave functions, of course, become quite crucial when one wants to extend the analysis to a supersymmetric system) and, consequently, we present a careful analysis of this system regularizing the singular potential in a systematic manner. With the supersymmetric system in mind (to follow in the next section), we write the potential for the system as (with )

(33) |

Consequently, the Schrödinger equation that we want to study has the form

(34) |

The singular potential is repulsive for or while it is attractive for . Furthermore, for ease of comparison, let us make the identifications with the notations of ref. [16] (note that our energies are shifted since we have in mind the supersymmetric system to study later)

(35) |

It is also worth noting here that the Schrödinger equation in (34) is invariant under

(36) |

This symmetry, of course, would also be reflected in the solutions. Furthermore, the fixed point of this symmetry, namely, separates the two branches (namely, for every value of there exist two distinct values of corresponding to two distinct branches separated at the branch point) in the parameter space.

### Regularized Potential

The Schrödinger equation in (34) can be solved quite easily for as was also done in [16]. However, to determine correctly how this wavefunction should be extended to the negative axis, it is more suitable to regularize the potential near the origin and study the problem carefully. Let us consider a potential of the form

(37) |

Namely, we have regularized the potential in a continuous manner preserving the symmetry in eq. (36) with the understanding that the regularization parameter at the end. With this regularization, the Schrödinger equation has to be analyzed in three distinct regions. However, since the potential has reflection symmetry, we need to analyze the solutions only in the regions and .

The potential is a constant in the region and hence the Schrödinger equation is quite simple here. The solutions can be classified into even and odd ones and take the forms

(38) |

where we have defined

(39) |

Since is small (and we are to take the vanishing limit at the end), the last equality holds only if which we will assume. The special values of corresponding to the absence of a singular potential have to be treated separately and we will come back to this at the end of this section. We note here that the normalization constants, and , can, in principle depend on the regularization parameter which we have allowed for in writing down the form of the solutions in eq. (38).

The potential is much more complicated in the region . However, if we make the definitions

(40) |

where

(41) |

the Schrödinger equation of (34) takes the form

(42) |

An equation of the form

(43) |

is known as the confluent hypergeometric equation [17] and the only solution of this equation which is damped for large values of the coordinate has the form

(44) |

Here are known as the confluent hypergeometric functions which have the series expansion

and satisfy

(45) |

It is clear now that eq. (42) simply is the confluent hypergeometric equation and the asymptotically damped physical solutions are nothing other than the functions with appropriate parameters. Since takes two possible values (see eq. (41)), it would seem that there would be two independent solutions of eq. (42). However, it can be easily checked that the two solutions corresponding to the two values of are really proportional to each other and not independent. Thus, we can write the general solution of the Schrödinger equation, for , as

Once again, we have allowed for a dependence of the normalization constant, , on the regularization parameter, . However, for a nontrivial solution to exist, we require that

So far, we have the general solutions, in the two regions, where energy is not quantized and which should arise from the matching conditions. Furthermore, we have not bothered to evaluate the solution in the region which clearly would be the same as in the region . However, the matching conditions would determine how we should extend the solutions in the region to the region . Therefore, let us now examine the matching conditions systematically since there are two possible cases.

Even Solution

We can match the even solution of the region and its derivative with those of the region at . Taking the ratio and remembering that is small (which is to be taken to zero at the end), we obtain to the leading order in

(47) |

Since the left hand side is independent of , for consistency, the right hand side must also be and this can happen in two different ways.

First, for , it is clear that relation (47) can be satisfied if (we assume from now on that .)

(48) |

with a suitable choice of .

On the other hand, for , if

(49) |

relation (47) can be satisfied with a suitable choice of . It is clear that the two possible branches of the solution simply reflect the symmetry in eq. (36).

This analysis shows that when the regularization is removed (namely, ), we have an even extension of the solution of the forms

(50) |

with

(51) |

(52) |

with

(53) |

Odd Solution

We can also match the odd solution of the region and its derivative with those of the region at and taking the ratio, we obtain to leading order

(54) |

Clearly, the analysis following from eq. (47) goes through identically so that we conclude that in the limit , we have an odd extension of the solution of the forms

(55) |

with

(56) |

(57) |

with

(58) |

### Understanding of the Result

The conclusion following from this analysis, therefore, is that every energy level of this system is doubly degenerate. Both even and odd extensions of the solution are possible for every value of the energy level. The energy levels, as given in eqs. (50) and (52) (or, alternately, (55) and (57)) are, of course, identical to those obtained in [16]. The crucial difference is in the structure of the wave functions, namely, that both even and odd extensions of the solution are possible for every value of the energy (Incidentally, the solutions we have obtained in terms of confluent hypergeometric functions also coincide with generalized Laguerre polynomials as was obtained in ref. [16].). It is crucial, therefore, to ask if such a conclusion is physically plausible. To understand this question, let us recapitulate the results from a simple quantum mechanical model which is well studied. Namely, let us look at a particle moving in a potential of the form

It is well known that the solutions of this system can be classified into even and odd ones with energy levels ()

The even and the odd solutions, of course, have distinct energy values for any finite strength of the delta potential. However, when , both the even and the odd solutions become degenerate in energy. Namely, a delta potential with an infinite strength leads to a double degeneracy of every energy level corresponding to both even and odd solutions. The connection of this example with the problem we are studying is intuitively clear. Namely, we can think of

It is clear that for , the singular potential behaves like a delta potential with an infinite strength and it is quite natural, therefore, that this system has both even and odd solutions degenerate in energy.

It is also clear from this analysis that it is meaningless to take the limit from the results obtained so far simply because the characters of the two problems are quite different. As we have argued, for any finite value of not coinciding with those special values, the potential behaves, at the origin, like a delta potential of infinite strength while for the special values, there is no such potential. The two cases are related in a drastically discontinuous manner. As a result, one cannot treat the as a perturbation and obtain the full, correct solution simply because there is nothing perturbative (small) about this potential for any “nontrivial” value of . Another way of saying this is to re-emphasize what we have already observed following eq. (39), namely, the character of and, therefore, the matching conditions change depending on whether or not differs from the special values .

To see how the standard results of the harmonic oscillator would emerge from this analysis, let us work out the case only for . In this case, the matching conditions for the even solution would lead to the relation for the ratios

(59) |

which can be satisfied only if

(60) |

We recognize these to be the even levels of the oscillator (remember the shifted zero point energy of the system in eq. (33) for ). Similarly, matching the odd solution and its derivative leads to

(61) |

which can be satisfied only if

(62) |

These are, of course, the odd energy levels of the harmonic oscillator. (The corresponding wavefunctions are the even and odd Hermite polynomials respectively.) There is no longer any degeneracy of these levels. In other words, the matching conditions change drastically and so do the solutions of the problem depending on whether or not equals one of the special values and the results for the special values cannot be (and, in fact, should not be) obtained from the general result in a limiting manner.

## 4 Supersymmetric Oscillator with Potential

In this section, we will analyze the supersymmetric version of the case studied in the last section. Let us consider a superpotential of the form [11]

(63) |

so that the pair of supersymmetric potentials would have the form

(64) |

To analyze this problem, we should, of course, regularize the superpotential. However, even before introducing the regularization, let us observe some general features associated with this system, namely, that this supersymmetric system would have a ground state satisfying

(65) |

which would be damped for asymptotically large values of the coordinate. On the other hand, from the behavior of this wave function for small values of , it is clear that normalizability of the wave function would require that . We also note that since the supersymmetric system involves two Hamiltonians with different dependence, the symmetry observed in the previous section, namely, (or, alternately, ) cannot be a symmetry of the whole system (Another way of saying this is to note that the superpotential has no such symmetry. We will come back to this question later in this section.). Furthermore, we cannot naively take over the results from the previous section since, as we have seen earlier, regularizing the potential directly may not respect supersymmetry.

Therefore, to study this problem systematically, we regularize the superpotential as

(66) |

Here, as before, is the regularization parameter which should be taken to zero at the end and we have regularized the superpotential such that it is continuous across the boundary. This has the nice feature that there are no delta potential terms in the potentials. In fact, the regularized superpotential leads to the pair of supersymmetric potentials of the forms

(67) |

It is worth noting here that for , the singular potential at the origin is not present in both the Hamiltonians (namely, it is truly not there). On the other hand, for , although the singular potential disappears from only one of the Hamiltonians, the complete system remembers about it through supersymmetry as is clear from the structure of the regularized potentials.

### Spectrum of

Let us now analyze the spectrum of the two different Hamiltonians systematically. First, let us note that, for the Hamiltonian , the potential is that of a harmonic oscillator in the region . We can write down the even and the odd solutions in this region as [17] (We will assume throughout that )

(68) |

The solution in the region can also be obtained from an analysis as given in the earlier section and leads to

Once again, we can match the solutions in the two regions and their derivatives across to obtain the relevant quantization conditions. As before, there are two possibilities.

Even Solution

If we match the even solution in eq. (68) and its derivative with those of eq. (4) at and take the ratio, we obtain to leading order in (remember is small and is to be taken to zero at the end),

(70) | |||||

Clearly, this can be satisfied only if ()

(71) |

Thus, we obtain that there exists an even extension of the solution of the form (when the regularization is removed)