Loyola eCommons Loyola eCommons Supersymmetric Quantum Mechanics and Solvable Models Supersymmetric Quantum Mechanics and Solvable Models

: We review solvable models within the framework of supersymmetric quantum mechanics (SUSYQM). In SUSYQM, the shape invariance condition insures solvability of quantum mechanical problems. We review shape invariance and its connection to a consequent potential algebra. The additive shape invariance condition is speciﬁed by a difference-differential equation; we show that this equation is equivalent to an inﬁnite set of partial differential equations. Solving these equations, we show that the known list of (cid:126) -independent superpotentials is complete. We then describe how these equations could be extended to include superpotentials that do depend on (cid:126)


Introduction
Supersymmetric quantum mechanics (SUSYQM) is a generalization of the factorization method commonly used for the harmonic oscillator.The factorization technique begun by Darboux [1] about one hundred years ago, and used by Schrödinger [2][3][4] in the 1940's and Infeld and Hull in the 1950's [5], could be considered a precursor of SUSYQM.
The current form of SUSYQM appeared in 1981 [6] as a model of dynamical symmetry breaking.It was developed further by the authors of [7,8] among others.This simplified model turned out to have important applications in quantum mechanics.
In the next section, we will describe the general formalism of SUSYQM and in Section 3 we introduce the shape invariance condition that makes a potential solvable.Section 4 is dedicated to a description of potential algebra and its connection to shape invariance.In Section 5 we will describe a method for determining solutions of the translational shape invariance condition.We will then conclude with the analysis of recently discovered shape invariant potentials that are inherently functions of .

Supersymmetric Quantum Mechanics
Throughout this paper, we use units such that 2m = 1.
For the harmonic oscillator, the Hamiltonian [9] Similarly, in the SUSYQM formalism [10,11] a general Hamiltonian H − = − 2 d 2 dx 2 + V − (x, a) is written as a product of A + ≡ − d dx + W (x, a) and A − ≡ d dx + W (x, a), where the function W (x, a) is known as the superpotential.The product A + A − is then given by Thus, the product A + A − indeed reproduces the Hamiltonian H − above, provided the potential To see the underlying supersymmetry of this formalism, let us construct a generator Q − and its adjont Q + by: Operators Q − and Q + generate the following supersymmetry algebra: The groundstate energy of this Hamiltonian is then given by Thus, the non-vanishing of the groundstate energy implies that either Q + |Ψ 0 = 0 or Q − |Ψ 0 = 0, and hence signals the spontaneous breaking of the supersymmetry.We therefore require that Ψ 0 |H|Ψ 0 = 0 to preserve unbroken supersymmetry.
The Hamiltonians H + and H − are intertwined; i.e., A + H + = H − A + and A − H − = H + A − .This leads to the following relationships among their eigenvalues and eigenfunctions [12] Since H ± are products of the operator A − and its adjoint A + , their eigenvalues are either zero or positive [13].The ground state eigenvalue of one of these Hamiltonians must be zero in order to have unbroken supersymmetry.Without loss of generality, we choose that Hamiltonian to be H − .Thus, we have x 0 W (y,a) dy (6) where x 0 is an arbitrary point in the domain and N is the normalization constant that depends on the choice of x 0 .Thus, if ψ is a normalizable groundstate, we have a system with unbroken supersymmetry.Its groundstate E − 0 is zero and the operator A − annihilates the corresponding eigenstate ψ (−) 0 .For all higher states of H − , as indicated in Equation ( 5), there is an one-to-one correspondence with the states of H + .

Example
Consider a system described by the superpotential W (x, b) = −b cot x, defined over the domain (0, π).Corresponding partner potentials are given by This is a rather complicated potential, and rarely analyzed in quantum mechanics courses.However, for b = , the potential V − (x, ) reduces to the infinite square well, with the bottom of the potential at − 2 and zero groundstate energy.We know that the corresponding eigenstates are given by ψ . Thus, using the familiarity with the relatively simpler potential V − , we are able to derive all of the eigenvalues and eigenfunctions of V + .Then the inter-relations expressed through Equation ( 5) enable us to determine eigenvalues and eigenfunctions of Although we assume that our Hamiltonians are hermitian, hermiticity is not necessary to generate real eigenvalues.Replacing the sufficient but not necessary condition of hermiticity with the weaker condition of PT symmetry, has led to the discovery of new potentials with real energy eigenvalues [14][15][16].Considerable work has been done on the study of SI potentials with PT symmetry.In reference [17], the author examined the shape invariant hyperbolic Rosen-Morse potential as a case of exact solvability with PT invariance.In reference [18], the Scarf II potential was shown as an example of spontaneous PT-symmetry-breaking.The case of a square well with discrete PT symmetry but with intervals of non-hermiticity was shown in reference [19] to produce real eigenvalues.Authors of [20] constructed the spectrum of a square well of imaginary strength, to obtain the hierarchy of SI potentials, while authors of [21] constructed a set of solvable rational potentials, and related the PT symmetry condition to the condition that they be free of singularities.However, in this paper we limit ourselves to the case of hermitian Hamiltonians, i.e., real W (x, a).
The remainder of this manuscript is devoted to the study of the shape invariance condition and its solutions.

Shape Invariance in Supersymmetric Quantum Mechanics
If the superpotential W (x, a i ) of a system obeys the condition where a i+1 = η(a i ), the system is called shape invariant [5,[22][23][24].Various forms of the function η define classes of shape invariance.The most commonly discussed classes are: 1. Translational or Additive Shape Invariance: a i+1 = η(a i ) = a i + ; 2. Multiplicative or Scaling Shape Invariance [25,26]: a i+1 = η(a i ) = r a i where 0 < r < 1 ; 3. Cyclic Shape Invariance [27]: From Equation (7) it follows that for a shape invariant system, the partner potentials V + (x, a i ) and V − (x, a i+1 ) differ only by values of parameter a i and additive constants g(a i ).In particular, In terms of operators A ± , the shape invariance condition becomes As we will see in Section 4, Equation (9) implies that for every shape invariant system, there is always an underlying potential algebra [28][29][30][31][32] that guarantees its solvability.In the rest of this section, we will show how shape invariance enables us to find the eigenvalues and eigenfunctions of the system.

Determination of Eigenfunctions
Again from Equation (9), we see that since H + (x, a 0 ) and H − (x, a 1 ) only differ by the constant g(a 1 ) − g(a 0 ), they must have common eigenfunctions.Hence, from Equation (6), we have ψ W (y,a 1 ) dy .Then the isospectrality condition (5), requires ψ (−) 1 (x, a 0 ) ∼ A + (a 0 ) e − 1 x W (y,a 1 ) dy .Iterating this procedure, we obtain Thus, for a system with a given shape invariant superpotential, the eigenvalues and eigenfunctions can be determined analytically.This result makes it very important to find all such potentials.In the past, researchers had found a list of additive shape invariant potentials [10,11], mostly by trial and error [34,35].In Section 5, we will discuss how to find solutions of Equation ( 9) for the additive case.Before that, however, in the next section we will show why the shape invariance condition leads to solvability.

Shape Invariance and Potential Algebra
We will now show that the symmetry behind the shape invariance is essential in building the algebraic structures known as potential algebras.As we will show below, a potential algebra is in general a deformation of a three-dimensional Lie algebra, whose representations yield the spectrum of the corresponding shape invariant system.

Building the Algebra
The starting point of our construct is the shape invariance condition given by Equation ( 9), which we rewrite as where f (a 0 ) = g(a 0 )−g(a 1 ) ≡ g(a 0 )−g(η(a 0 )), and η is a function that models the change of parameter a 1 = η(a 0 ).For example, if the change of parameter is a translation, a 1 ≡ η(a 0 ) = a 0 + s.
The left hand side of Equation ( 12) resembles a commutation relation.This suggests that we use A − and A + to build the generators of the potential algebra.To transform the above shape invariance condition into an exact commutator, we replace operators A − (x, a 0 ) and A + (x, a 0 ) by where s is a constant parameter, φ is an auxiliary variable, and ∂ φ ≡ ∂ ∂φ .The function χ will be appropriately chosen to emulate the relationship between parameters a 0 and a 1 .Thus, to generate J − , we multiplied the operator A − (x, a 0 ) from right by e −isφ and replaced the parameter a 0 by the differential operator χ (i ∂ φ ).If we now compute the commutator between operators J + and J − , we find [36] [ Observe now that the right hand side of Equation ( 14) matches the left hand side of Equation ( 12) provided that we make the following mappings Since we know that a 1 = η(a 0 ), these mappings require that the function χ(i∂ φ ) satisfy Let us look at some examples to illustrate this procedure.
• Translation: a 1 ≡ η(a 0 ) = a 0 + s If the change of parameters is a translation, then the function χ that models it is the identity function We have χ(z + s) = z + s = χ(z) + s, which gives the desired change of parameters.
• Scaling: a 1 ≡ η(a 0 ) = ra 0 For shape invariant potentials characterized by a scaling change of parameters, the corresponding function χ is the exponential Indeed χ(z + s) = e z+s = e s e z = r χ(z) where we denoted e s = r.
• Other choices of parameters follow from more complicated choices for χ(z).For example taking we obtain the change of parameters: Now, let us get back to the building of the potential algebra.In terms of operators J + and J − , the shape invariance condition (12) becomes: Thus, the commutation relation of operators J + (x, φ) and J − (x, φ) generates an operator f (χ (i ∂ φ )) that has no x-dependence.If we now define a third operator J 3 in terms of the operator i ∂ φ , the shape invariance condition becomes simply one of the commutation relations of the potential algebra.In particular, if we define where k is an arbitrary constant, the three commutators among these generators are given by [39] Putting together the above results, we arrive at the following Lemma: To any shape invariant system characterized by for which we can find a function χ such that χ(z + s) = η (χ(z)) for arbitrary parameters z and s, we can associate an algebra [40] generated by satisfying the commutation relations where The function f in Equation ( 29) is given by the shape invariance condition (24), while the function χ satisfies the compatibility equation: , where η models the change of parameter a 1 = η(a 0 ) of Equation (24).
As an example let us build the potential algebra corresponding to the Pöschl-Teller II potential.The potential The shape invariance is now evident if we observe that Therefore, in terms of A + and A − operators, the shape invariance (24) for the Pöschl-Teller II potential reads Now we can identify the main objects of our model and build the corresponding algebra: 1.The parameters of the model are a 0 = α − l and a 1 = α − l + , where α is an arbitrary positive constant greater than l so that a 0 is a positive quantity; 2. The change of parameter a 1 = η(a 0 ) is thus given by a 1 = a 0 + .This is a translational change of parameter a 1 = a 0 + s with the translation parameter s = .Translation implies that the function χ satisfying ( 16) is the identity function χ(z) = z; 3. From the concrete shape invariance condition (32) of this potential we get f (a 0 ) = − 2 (α − a 0 ) = − 2α + 2a 0 .Then, the function F (J 3 ) is given by 4. Defining J ± and J 3 as prescribed by Equations ( 25) and (27), we obtain the differential realization of the algebra's generators: satisfying the commutation relations [J 3 , J ± ] = ±J ± , and [J + , J − ] = 2J 3 .Thus, shape invariance of this system implies that the system has a potential algebra given by su(2) [41,42].

Obtaining the Energy Spectrum from Algebra Representations
Once we know the potential algebra for a given potential, we can use its representations to obtain the energy spectrum for the Hamiltonian.Using Equations ( 25) and ( 26), we observe that From the reciprocal of the mapping Equation (13), we obtain Â+ (x, χ(i∂ φ + s)) Â− (z, χ(i∂ φ + s)) → A + (x, a 1 ) A − (x, a 1 ) = H − (x, a 1 ).Consequently, the spectrum of the operator J + J − gives the spectrum of the Hamiltonian.To find the concrete values for the energy, we need to know the action of individual operators J + and J − respectively on a set of eigenvectors of the operator J 3 .In this general case, J + , J − and J 3 commute with the Casimir operator given by with the function G (defined up to an additive constant) such that It can be explicitly checked that C does indeed commute with J + , J − and J 3 [41].In a basis in which J 3 and C are diagonal, J ± play the role of raising and lowering operators, respectively.Operating on an arbitrary eigenstate | q we have where we have used J + = (J − ) † .Keeping in mind that H − (x, a 1 ) → J + J − , and observing that J + J − |q = |a(q)| 2 |q , we see that in order to find the energies of the system one must find the coefficients a(q).If we apply the third commutation relation of Equation (23) to | q , we obtain using Equation (36) Next, we will determine the allowed values of q and the corresponding values a(q).Let us say q = q min corresponds to the lowest state in a given representation of the algebra.This implies that J − | q min = a (q min ) | q − 1 = 0, which means a(q min ) = 0. From Equation (38) we get Iterating this procedure we can generate a general formula for a(q) The profile of G(q) determines the dimension of the representation.For example, let us consider the two cases presented in Figure 1.
Figure 1.Generic behaviors of G(q).Case (a) corresponds to a finite, and (b) to an infinite representation of the potential algebra.
min min max q min q + n min . . .

G(q)
One obtains the finite dimensional representations of Figure 1a, by starting from a point on the G(q) vs. q graph corresponding to q = q min , and moving in integer steps parallel to the q-axis to the point corresponding to q = q max .At the end points, a(q min ) = a(q max + 1) = 0, and we get a finite representation.If G(q) is decreasing monotonically, as in Figure 1b, there exists only one end point at q = q min .Starting from q min the value of q can be increased in integer steps up to infinity.In this case we have an infinite dimensional representation.As in the finite case, q min labels the representation.The difference is that here q min takes continuous values.Similar arguments apply for a monotonically increasing function G(q).Having the representation of the algebra associated with a characteristic model, we obtain, using Equation (41), the complete spectrum of the system.
For example, let us consider the scaling change of parameters a 1 = ra 0 .Consider the simple choice f (a 0 ) = −r 1 a 0 , where r 1 is a constant.This choice generates self-similar potentials studied in references [25,26,44].In this case, combining Equation (18) with Equation (28) yields: which is a deformation of the standard so(2, 1) Lie algebra.For this case, from Equations ( 42) and (36) one gets Note that for scaling problems [25], one requires 0 < r < 1, which leads to s < 0. From the monotonically decreasing profile of the function G(q), it follows that the unitary representations of this algebra are infinite dimensional.If we label the lowest weight state of the operator J 3 by q min , then a(q min ) = 0. Without loss of generality we can choose the coefficients a(q) to be real.Then one obtains from (38) for an arbitrary q = q min + n, n = 0, 1, 2, . . .
The spectrum of the Hamiltonian H − (x, a 1 ) is given by Therefore, the eigenenergies are in agreement with the known results [25].

How Do We Find Additive Shape Invariant Superpotentials?
Since we have demonstrated the value of shape-invariant superpotentials, the question now becomes how to find such superpotentials.This question is equivalent to asking how to solve the difference-differential Equation ( 9) to find the list of desired superpotentials W .For this section, we will restrict ourselves to considering cases of translational shape invariance.Before we embark on solving this equation, let us first note that quantum mechanical potentials generally have terms of two very different orders: One "large" and another "small".For example, the classical and quantum potentials for the radial oscillator system are 1  4 ωr 2 + L 2 r 2 and 1 4 ωr 2 + ( +1) 2 r 2 respectively.To make the transition from the quantum to the classical system, one takes the limit lim { →∞ , →0} with the constraint that • → L. Thus, the quantum Hamiltonian can be written as 1 4 ωr 2 + L 2 r 2 + L r 2 .This shows that in quantum mechanics, the potential generally has one small term that depends on [43].In SUSYQM, since the potential is given by W 2 (x, a) − d W (x,a) dx , the derivative term always brings in a factor of , even if the superpotential is independent of .In the following analysis, as we determine how to solve Equation ( 9) to find shape invariant superpotentials, we will consider the cases of -independent and -dependent superpotentials separately.

Known -Independent Shape Invariant Superpotentials
We begin our discussion of known shape invariant systems by considering only superpotentials that do not depend explicitly on , which we call "conventional" superpotentials.In Table 1 we list the known "conventional" superpotentials that meet this criterion.
Previous work [45][46][47] has proven that this list of conventional shape-invariant superpotentials is complete.We now show a new proof of this completeness which has the advantage of being significantly more straightforward and elegant than it predecessors.

New Proof of Completeness of the Conventional Shape-Invariant Superpotentials
Because of additive shape invariance, the dependence of W on a and is through the linear combination a + ; therefore, the derivatives of W with respect to a and are related by: ∂a .Since Equation ( 7) must hold for an arbitrary value of , if we assume that W does not depend explicitly on , we can expand in powers of , and the coefficient of each power must separately vanish.Expanding the right hand side in powers of yields Thus, all conventional additive shape invariant superpotentials are solutions of the above set of non-linear partial differential equations [46,47].Although this represents an infinite set, note that if equations of O( ) and O( 3) are satisfied, all others automatically follow.Therefore, we proceed to find all possible solutions to the two partial differential equations: and In doing so, we derive a new proof that the superpotentials shown in Table 1 are the only possible solutions.The general solution to Equation ( 51) is Therefore, to generate all shape invariant superpotentials, we need to determine all possible combinations of u(a), X 1 (x), and X 2 (x) that satisfy Equation (50).We will ignore the case when both X 1 (x), and X 2 (x) are constants, as this corresponds to a flat potential with no x-dependence.We will also ignore the case in which X 1 (x), and X 2 (x) are linearly dependent on each other; i.e., X 2 (x) = αX 1 (x) + β.In this case, W (x) = aX 1 (x) + αX 1 (x) + β + u(a) = (a + α)X 1 (x) + β + u(a).If we redefine ã ≡ a + α, this case becomes equivalent to a superpotential with a shifted parameter and constant X 2 which will be considered shortly.We can therefore assume that X 1 and X 2 are linearly independent of each other without loss of generality.Note that from here onward, we will use lower case Greek letters to denote constants that are independent of both a and x.
To determine W (x, a), we first focus on determining u(a).To do so, we take two derivatives of (50) with respect to a.This leads to the following differential equation: where dots and primes represent derivatives taken with respect to a and x respectively.Since ∂a 2 ∂x W (x, a) = 0, this simplifies to: Inserting the form of the general solution ( 52) into (53) yields where H(a) = − 3 uü + u ... u + 1 2 ... g is a function of a, and is independent of x.Since X 1 and X 2 are linearly independent, we find that there are only three possible ways for the left-hand-side of Equation ( 54) to be independent of x: • Case 1: X 1 is a constant and ... u = 0; • Case 2: X 2 is a constant and 3ü + a ... u = 0; • Case 3: Neither X 1 nor X 2 are not constants, but 3ü + a ... u = 0 and ...
For each of these cases we can determine the form of u(a).Then we can determine X 1 (x) and X 2 (x) for these three cases.This we do by taking two derivatives of (50), this time one with respect to a and another with respect to x.This yields: Inserting the form of the general solution ( 52) into (55) yields Now, we will analyze each of the three cases in detail.
We now find X 2 by inserting the above W into Equation (50).This yields dg da or equivalently, where h(a) = − 1 2 dg da .Since X 2 (x) is independent of a, and the left side of (57) is a sum of four linearly independent functions of a (a 0 , a 1 , a 2 , and a 3 ), and the term h(a) on the right-hand-side is independent of x, the coefficient of each power of a must separately be independent of x.The linear term in a therefore requires that 2X 2 γ + 2νγ + η 2 be independent of x.Since a constant X 2 leads to a trivial solution, we must have γ = 0.The remaining x-dependent terms on the left side of (57), X 2 η−X 2 must be a constant: The solution depends on the value of the constants η and β.

Case 2: X 2 Is Constant
In this case, let X 2 = α; then Equation ( 54) requires 3ü + a ... u = 0.This yields u(a) = µa + ν + γ a .We now insert this form of u and X 2 = α into (56) to get an ordinary differential equation in x for X 1 : Integrating it once, we get This equation can be simplified by setting X1 = X 1 + µ.This leads to The solutions for X1 depend on the constant θ.
5.2.3.Case 3: X 1 and X 2 Are not Constant, but 3ü + a ... u = 0 and ... u = 0 In this case, since ... u = 0, and 3ü + a ... u = 0, we have ü = 0. Therefore u(a) = µa + ν.In this case, Equation (56) yields Thus, again we have Note that this is the same differential equation as (60) and will therefore give the same solutions for X 1 as Case 2. However, in this case, u = µa + ν (this is equivalent to choosing γ = 0 in Case 2) and X 2 is not constant.Instead, in each case we must plug our solutions for u(a) and X 1 (x) into Equation (50), which yields This equation is again simplified by setting X1 = X 1 + µ, which yields We now extend our formalism to include "extended" superpotentials that contain explicitly.To do so, we expand the superpotentials in powers of .Hence, we define We will now substitute Equation (68) into the shape invariance condition given in Equation ( 7), for which we compute ∂W ∂x a=a 0 and W 2 (x, a 0 , ).We obtain Similarly, We substitute these into Equation (7) and stipulate that the equation hold for any value of .After some significant algebraic manupulation we find that the following equation must be true separately for each positive integer value of n: For n = 1, we obtain This equation is identical to Equation (50) for -independent W 's. We have already found a set of solutions for Equation (70) that includes all known -independent superpotentials.The extended cases [50,51] are solutions to (69) as well, as shown in [46].Note that Equation (69) provides a consistency condition for all -dependent potentials; however, these are not easy to solve to determine new potentials.
Additionally, Equation (70) provides a constraint for the possible energy spectra of the extended potentials.from Equation (70), the function g(a) is given by Thus, the function g(a), and hence the energy of the system, is given entirely in terms of the -independent part of the superpotential.Hence, the eigenvalues are not affected by the -dependent extension of the superpotential.
Thus far, each of the known extended potentials contains a solution from Table 1 as the -independent term of the superpotential W 0 .Therefore, each of the expanded potentials is isospectral with its corresponding conventional potential.Future possibilities for finding new shape-invariant superpotentials fall into one of two categories: 1. Further extended superpotentials may be found based on the conventional superpotentials.In this case, the potentials derived from the extended superpotential will be isospectral with the potentials derived from the corresponding conventional superpotential; 2. While W 0 is required to satisfy Equation (70), which is equivalent to Equation (50) for -independent W 's, W 0 is not required to satisfy Equation (48).Rather, the additional constraints for an extended W are supplied by Equation (69).It therefore may be possible to find an -dependent superpotential whose -independent term W 0 is not equivalent to a conventional superpotential.Intriguingly, it therefore may still be possible to discover shape-invariant systems with new energy spectra.

Summary and Conclusions
While supersymmetric quantum mechanics began as a simplified model to account for dynamical symmetry breaking, the application of this formalism to quantum mechanics has become an important field in its own right.In this manuscript we have reviewed research on supersymmetric quantum mechanics with a particular emphasis on the property of shape invariance.As we have shown, shape invariance is a sufficient condition for exact solvability of quantum mechanical problems; i.e., given a superpotential with shape invariance, all its eigenvalues and eigenfunctions can be determined analytically.
However, in its traditional form, the shape invariance condition Equation ( 7) is a difference-differential equation and is difficult to solve.It has recently been established that for additive shape-invariant superpotentials that do not explicitly depend on , this condition can be written as a set of local partial differential equations [45][46][47].The solution to these equations showed that the list of such superpotentials was indeed complete.In this manuscript, we have presented a more straightforward proof of this result.
Since 2008, new sets of additive shape invariant potentials have been discovered [50][51][52][53][54].We have reviewed the development of these "extended" shape-invariant systems and have provided an infinite set of partial differential equations that all extended potentials (where superpotentials are inherently functions of ) must obey [46,47].We have also discussed the constraints placed on the energy spectra of these extended potentials as well as possibilities for finding additional as-yet-undiscovered cases of additive shape invariance.
It may also be possible to extend this method to other forms of shape invariance such as multiplicative or cyclic.For these, the potentials are generally not available in terms of known functions, except in very special cases (N = 2 for cyclic and limiting cases for multiplicative).It remains to be shown whether the shape invariance condition for these classes can be transformed from a difference-differential equation into a set of partial differential equations and be subjected to similar analysis.Introduction; World Scientific: Singapore, 2010. 12. .

Table 1 .
The complete family of -independent additive shape-invariant superpotentials.