Normalization of the Ground State of the Supersymmetric Harmonic Oscillator
aa r X i v : . [ phy s i c s . g e n - ph ] J u l Letters in High Energy Physics LHEP xx, xxx, 2018
Normalization of the Ground State of the SupersymmetricHarmonic Oscillator
Ahmed Ayad
School of Physics, University of the Witwatersrand, Private Bag 3, WITS-2050, Johannesburg, South Africa
Abstract
Supersymmetry plays a main role in all current thinking about superstring theory. Indeed, many remark-able properties of string theory have been explained using supersymmetry as a tool. So far, there has beenno unbroken supersymmetry observed in nature, and if nature is described by supersymmetry, it must bebroken. Supersymmetry may be broken spontaneously at any order of perturbation theory or dynamicallydue to nonperturbative effects. To examine the methods of supersymmetry breaking, special attention isgiven to discuss the normalization of the ground state of the supersymmetric harmonic oscillator. Thisstudy explains that perturbation theory gives incorrect results for both the ground-state wave functionand the energy spectrum and it fails to give an explanation to the supersymmetry breaking.
Keywords: supersymmetry breaking, perturbation theory,nonperturbative effects
1. INTRODUCTION
Supersymmetry, often abbreviated SUSY, was first introducedin 1971 by Gelfand and Likhtman, Raymond, and Neveu andSchwartz, and later, it was rediscovered by other groups [1, 2].Supersymmetric quantum mechanics was developed by Wittenin 1982 [3], as a toy model to test the breaking of supersymme-try. In general, supersymmetry plays a main role in the mod-ern understanding of theoretical physics, in particular, quan-tum field theories, gravity theories, and string theories [4].Supersymmetry is a symmetry that connects particlesof integer spins (bosons) and particles of half-integer spins(fermions) [5]. In supersymmetry, it is possible to introduce op-erators that change bosons which are commuting variables intofermions which are anticommuting variables and vice versa [6].Supersymmetric theories are theories which are invariant un-der those transformations [7]. The supersymmetry algebra in-volves commutators as well as anticommutators [8].In this study, the main mathematical structure which in-volves supersymmetric quantum mechanics is derived startingwith explaining the basic idea of supersymmetry and followedby introducing the necessary framework to make a supersym-metric theory. We derive the basic formulation of supersym-metric quantum mechanics starting with introducing the con-cepts of supercharges and superalgebra. We show that if thereis a supersymmetric state, it is the zero-energy ground state. Ifsuch a state exists, the supersymmetry is unbroken; otherwise,it is broken. So far, there has been no unbroken supersymmetryobserved in nature, and if nature is described by supersymme-try, it must be broken. In fact, supersymmetry may be brokenspontaneously at any order of perturbation theory or dynam-ically due to nonperturbative effects. To examine the methodsof supersymmetry breaking, we study the normalization of theground state of the supersymmetric harmonic oscillator andcalculate the corrections to the ground-state energy using per-turbation theory. We found out that no perturbation effect canlead to supersymmetry breaking and it must be due to the non-perturbative effects.This work is structured as follows. After a brief intro-duction, we study the problem of a harmonic oscillator with fermionic as well as bosonic fields using the usual quantummechanical operators method. Then, we consider supersym-metric classical mechanics and study generalized classical Pois-son brackets to Dirac brackets and quantization rules in or-der to introduce the superalgebra. Furthermore, the formal-ism of supersymmetry in quantum mechanics is introduced.Subsequently, we give a general background in the conceptsand methods of supersymmetric quantum mechanics. This fol-lowed by providing a more specific study of the ground stateof the supersymmetric harmonic oscillator. Finally, our conclu-sion is given, and for completeness, some needed derivationsare broadly outlined in the appendix.
2. HARMONIC OSCILLATOR
Supersymmetric theories necessarily include both bosons andfermions, so a system that includes both bosonic and fermionicdegrees of freedom is considered here. The supersymmetry al-gebra is a mathematical formalism for describing the relationbetween bosons and fermions, and we will discuss this later inSection 4. In quantum mechanics, bosons are integer spin parti-cles which obey Bose-Einstein statistics, and the bosonic opera-tors satisfy the usual commutation relations, while fermions arehalf-integer spin particles characterized by FermiDirac statis-tics, obey the Pauli exclusion principle, and can be describedby Grassmann variables, which are anticommuting objects; forreview, see [9, 10, 11, 12].The simplest system that consists of a combination ofbosonic and fermionic fields is the one-dimensional harmonicoscillator. There are many excellent reviews on this subject; werefer the reader to them for more and complete expositions[2, 13, 14, 15]. We begin by considering the Lagrangian formu-lation for the bosonic and fermionic harmonic oscillator: L = (cid:16) ˙ q + i ψ α δ αβ ˙ ψ β + F (cid:17) + ω ( i ψ ψ + qF ) , (1)where q ( t ) and F ( t ) are real bosonic fields, ψ α ( t ) (with α ∈{
1, 2 } ) are real fermionic fields, ω is a real parameter, δ αβ is theKronecker delta, and we used the Einstein summation conven-tion.It is straightforward, using the classical equation of motionfor F + ω q =
0, to eliminate F from Eq. (1) and obtain the equiv- etters in High Energy Physics LHEP xx, xxx, 2018alent Lagrangian: L = (cid:16) ˙ q + i ψ α δ αβ ˙ ψ β − ω q (cid:17) + i ωψ ψ . (2)Then, using the Legendre transformation from the variables { q , ˙ q , ψ , ˙ ψ } to { q , p , ψ , π } , we obtain H = ˙ qp + ψ α δ αβ π β − L = (cid:16) p + ω q (cid:17) − i ωψ ψ , (3)where π α are the momenta conjugate to ψ α .Later in this study, we shall see that the general proper-ties of supersymmetry have to be clear if the Hamiltonian isbuilt as a larger Hamiltonian consisting of two components,one representing the bosonic field and the other representingthe fermionic field. Leaving this aside for the moment, to de-rive the supersymmetric Hamiltonian, let us define the bosonicvaluable ˆ q and the bosonic momentum ˆ p , respectively, as fol-lows: ˆ q = q I , ˆ p = − i ¯ h I ∂∂ q . (4)On the other hand, the fermionic variables ˆ ψ and ˆ ψ and thefermionic momenta π α , respectively, have to be defined asˆ ψ = r ¯ h σ , ˆ ψ = r ¯ h σ , π α = ∂ L ∂ ˙ ψ α = − i δ αβ ψ β , α , β ∈ {
1, 2 } . (5)Hence, using Eqs. (4, 5), we can rewrite the previous Hamilto-nian (3) in the following form: ˆH = (cid:18) − ¯ h d dq + ω q (cid:19) I + (cid:18)
12 ¯ h ω (cid:19) σ , (6)where I is the 2 × σ j are the Pauli matri-ces; that is, I = (cid:18) (cid:19) , σ = (cid:18) (cid:19) , σ = (cid:18) − ii (cid:19) , σ = (cid:18) − (cid:19) . (7)We want to obtain solutions of the time-independent Schr ¨odingerequation H Ψ = E Ψ ; that is,12 (cid:18) − ¯ h d dq + ω q (cid:19) I + (cid:18)
12 ¯ h ω (cid:19) σ = E Ψ , (8)with the wave function Ψ ( x ) constraint to satisfy appropriateboundary conditions.As we mentioned, the supersymmetric Hamiltonian can bewritten as a summation of two terms representing the bosonicand the fermionic components of the Hamiltonian H = H B ( p , q ) + H F ( ψ ) . (9)To this end, we define the lowering (annihilation) and raising(creation) operators for bosons ˆ a † and ˆ a , respectively, they areˆ a = r h ω ( ω ˆ q + i ˆ p ) , ˆ a † = r h ω ( ω ˆ q − i ˆ p ) . (10)Later, in Section 4, we will show that the bosonic operators, ˆ q and ˆ p , satisfy the usual canonical quantum commutation con-dition [ ˆ q , ˆ p ] = i ¯ h . Taking this fact into account, we can find that the bosonic ladder operators ˆ a and ˆ a † satisfy the followingcommutation relations [ ˆ a , ˆ a † ] = [ ˆ a , ˆ a ] = [ ˆ a † , ˆ a † ] = b † and ˆ b , respectivelyare defined asˆ b = r h (cid:0) ˆ ψ − i ˆ ψ (cid:1) , ˆ b † = r h (cid:0) ˆ ψ + i ˆ ψ (cid:1) . (12)Also, as will be shown in Section 4, the fermionic operators ψ α and ψ β satisfy the canonical quantum anticommutation con-dition { ψ α , ψ β } = ¯ h δ αβ . Therefore, it can be seen that thefermionic ladder operators ˆ b and ˆ b † satisfy the following an-ticommutation relations: { ˆ b , ˆ b † } = { ˆ b , ˆ b } = { ˆ b † , ˆ b † } = a + ˆ a † = r ω ¯ h ˆ q ⇒ ˆ q = r ¯ h ω ( ˆ a † + ˆ a ) ,ˆ a − ˆ a † = r h ω i ˆ p ⇒ ˆ p = i r ¯ h ω ( ˆ a † − ˆ a ) . (14)Similarly, using Eq. (12), we can write the fermionic variables ψ and ψ in terms of the fermionic ladder operators as follows:ˆ b † + ˆ b = r h ˆ ψ ⇒ ˆ ψ = r ¯ h ( ˆ b + ˆ b † ) ,ˆ b † − ˆ b = i r h ˆ ψ ⇒ ˆ ψ = i r ¯ h ( ˆ b − ˆ b † ) . (15)Actually, we derived Eqs. (14, 15) here; however, it will be usedlater. At this stage, let us define the following two Hermitianoperators, the bosonic number operator ˆ N B , and the fermionicnumber operator ˆ N F : ˆN B ≡ ˆ a † ˆ a = h ω (cid:16) ω ˆ q − ¯ h ω + ˆ p (cid:17) , ˆN F ≡ ˆ b † ˆ b = − i ¯ h (cid:0) ˆ ψ ˆ ψ (cid:1) . (16)Using Eq. (16), we can rewrite the Hamiltonian (6) in the oper-ator formalism ˆH = ¯ h ω (cid:18) ˆN B + (cid:19) + ¯ h ω (cid:18) ˆN F − (cid:19) . (17)This means that the energy eigenstates can be labeled with theeigenvalues of n B and n F .The action of the ladder operators, ˆ a , ˆ a † , ˆ b and ˆ b † , upon anenergy eigenstate | n B , n F i is listed as follows: ˆ a | n B , n F i = √ n B | n B − n F i , ˆ a † | n B , n F i = √ n B + | n B + n F i , ˆ b | n B , n F i = √ n F | n B , n F − i , ˆ b † | n B , n F i = √ n F + | n B , n F + i . (18) The derivation of the fermionic anticommutators is a bit subtle and will bediscussed in the next section. ˆN B and fermionic number operator ˆN F obey ˆN B | n B , n F i = n B | n B , n F i , ˆN F | n B , n F i = n F | n B , n F i . (19)Under Eq. (19), the energy spectrum of the Hamiltonian(17) is E n = ¯ h ω (cid:18) n B + (cid:19) + ¯ h ω (cid:18) n F − (cid:19) . (20)The form of Eq. (20) implies that the Hamiltonian H is symmet-ric under the interchange of ˆ a and ˆ a † and antisymmetric underthe interchange of ˆ b and ˆ b † .The most important observation from Eq. (20) is that E n re-mains invariant under a simultaneous annihilation of one bo-son ( n B → n B −
1) and creation of one fermion ( n F → n F +
1) orvice versa. This is one of the simplest examples of a symmetrycalled ”supersymmetry” (SUSY) and the corresponding energyspectrum reads E n = ¯ h ω ( n B + n F ) . (21)Moreover, we still have two more comments before endingthis section:(i) Consider that the fermionic vacuum state | i is definedby ˆ b | i ≡
0. (22)Then, we define a fermionic one-particle state | i by | i ≡ ˆ b † | i . (23)It is now easy to use the anticommutation relations (13)to see that there are no other states, since operating on | i with ˆ b or ˆ b † gives either the already known state | i ,or nothing: ˆ b | i = ˆ b ˆ b † | i = ( − ˆ b † ˆ b ) | i = | i − ˆ b † ˆ b | i = | i − = | i , ˆ b † | i = ˆ b † ˆ b † | i = | i = | i and | i is closed un-der the action of ˆ b and ˆ b † , and, therefore, it is closedunder the action of any product of these operators. Wecan show this directly using the two facts that the op-erator ˆ b ˆ b † can be expressed as the linear combination1 − ˆ b † ˆ b and ˆ b = ˆ b † =
0. So, any product of three ormore ˆ b , ˆ b † can be shortened by using either ˆ b =
0, orˆ b † =
0. For example ˆ b † ˆ b ˆ b † = ˆ b † − ˆ b ˆ b † = ˆ b † and ˆ b ˆ b † ˆ b = ˆ b − ˆ b † ˆ b = ˆ b . So, it seems to be clear that all products ofˆ b and ˆ b † can always be reduced to linear combinationsof the following four operators 1, ˆ b , ˆ b † , and ˆ b † ˆ b . Basedon this argument, there are only two possible fermioniceigenstates and hence two possible eigenvalues of thefermionic number operator; they are n F = {
0, 1 } .(ii) The ground eigenstate has a vanishing energy eigen-value (when n B = n F = ¯ h ω and − ¯ h ω , re-spectively.
3. THE CONSTRAINED HAMILTONIANFORMALISM
Before discussing the supersymmetry algebra in the next sec-tion, we have to discuss the constrained Hamiltonian formal-ism in this section. This formalism is needed to get the cor-rect fermionic anticommutators due to the complication of con-straints. Good discussions of the constrained Hamiltonian sys-tems are presented in [16, 17, 18, 19, 20, 21]. For instance, let usrecall the Hamiltonian (3) H = (cid:16) p + ω q (cid:17) − i ωψ ψ . (25)A constrained Hamiltonian is one in which the momenta andpositions are related by some constraints. In general, M con-straints between the canonical variables can be written as φ m ( q , p , ψ , π ) = m =
1, . . . , M . (26)These are called primary constraints. Secondary constraints areadditional constraints relating momenta and positions whichcan arise from the requirement that the primary constraints aretime-independent; i.e., ˙ φ m =
0. Fortunately, we do not have todeal with such constraints in this case. In principle, there canalso be tertiary constraints arising from the equation of motionof the secondary constraints and so on.The Hamiltonian (25) is an example of constrained Hamil-tonians. In the previous section, we defined the fermionic mo-menta π α in Eq. (5) using the fermionic equation of motion fol-lowed from the usual Euler-Lagrange equations with the La-grangian (2) which is equivalent to this Hamiltonian. In fact,this definition gives us the relation between the fermionic mo-menta π α and the canonical coordinates ψ α : π α = ∂ L ∂ ˙ ψ α = − i δ αβ ψ β , α , β ∈ {
1, 2 } . (27)In light of this equation, we have the primary constraint φ α = π α + i δ αβ ψ β = H = (cid:16) p + ω q (cid:17) + i ωπ π . (29)In the next subsection, the general properties of Poisson brack-ets and its relation with the Hamiltonian formalism in classi-cal mechanics are briefly discussed. Later in the same subsec-tion, the problem caused by the constrained Hamiltonians isdescribed. Next, the Subsection 3.2 shows how to deal with thisproblem using the constraints. The Poisson bracket of two arbitrary functions F ( ˆ q , ˆ p , ˆ ψ , ˆ π ) and G ( ˆ q , ˆ p , ˆ ψ , ˆ π ) is defined as { F , G } P = (cid:18) ∂ F ∂ q i ∂ G ∂ p i − ∂ F ∂ p i ∂ G ∂ q i (cid:19) ( − ) ǫ F (cid:18) ∂ F ∂ψ α ∂ G ∂π α − ∂ F ∂π α ∂ G ∂ψ α (cid:19) , (30)3etters in High Energy Physics LHEP xx, xxx, 2018where ǫ F is 0 if F is Grassmann even and 1 if F is Grass-mann odd . Actually, in our argument here, only the case of theodd Grassmann variables is considered. In general, the Poissonbrackets have the following properties:(i) A constraint F is said to be a first-class constraint if itsPoisson bracket with all the other constraints vanishes: { φ A , F } P = A runs over all the constraints. If a constraint isnot a first class, it is a second class.(ii) A Poisson bracket of two functions F and G is antisym-metric: { F , G } P = −{ G , F } P . (32)(iii) For any three functions A , B , and C , the Poisson bracketis linear in both entries: { A , B + C } P = { A , B } P + { A , C } P . (33)(iv) A Poisson bracket for any three functions satisfies theLeibniz rule: { A , BC } P = { A , B } P C + ( − ) ǫ A ǫ B B { A , C } P , (34)where ǫ F is the Grassmann parity of F which is 0 if F isGrassmann even and 1 if F is Grassmann odd.(v) A Poisson bracket for any three functions obeys the Ja-cobi identity: { A , { B , C } P } P + { B , { C , A } P } P + { C , { A , B } P } P = F ( q i , p i , ψ α , π α ) , which has no explicit time dependence,is given by its Poisson bracket with the Hamiltonian˙ F = { F , H } P . (36)In the Hamiltonian formalism of classical mechanics, theHamilton equations of motion have equivalent expressions interms of the Poisson bracket. Using the last property (vi) of thePoisson brackets, one can easily see that˙ q i = ∂ H ∂ p i = { q i , H } P , ˙ p i = − ∂ H ∂ q i = { q i , H } P ,˙ ψ α = − ∂ H ∂π α = { ψ α , H } P , ˙ π α = − ∂ H ∂ψ α = { π α , H } P . (37)However, the problem with the constrained system is that theequations of motion derived using Eq. (37) are not always con-sistent with those followed from the Lagrangian formalism.For example, if we consider the Hamiltonian (25), the equationof motion for ψ α followed from the expected Poisson bracketequation has to be˙ ψ α = { ψ α , H } P = − ∂ H ∂π α = Grassmann-even variables refer to bosons, while Grassmann-odd variablesrefer to fermions.
Unfortunately, this equation is inconsistent with the equation ofmotion followed from the Lagrangian formalism. Even worst,the equation of motion for π α followed from the expected Pois-son bracket equation is also inconsistent with the equation de-rived from the Lagrangian formalism. For example, using thePoisson bracket, we have˙ π = { π , H } P = − ∂ H ψ = i ωψ . (39)But if we use Eq. (27) to find an expression to the equation ofmotion for ˙ ψ and then substitute into Eq. (39), that gives us˙ ψ = i ˙ π = − ωψ . (40)This equation is different from the Lagrangian equations ofmotion and inconsistent with (38) (unless everything is trivialwhich is also not great). Moreover, using the Poisson bracketwith the Hamiltonian (29), we can find the following set ofequations of motion for the fermionic positions and momenta:˙ ψ = { ψ , H } P = − ∂ H ∂π = − i ωπ = − ωψ ,˙ ψ = { ψ , H } P = − ∂ H ∂π = − i ωπ = − ωψ ,˙ π α = { π α , H } P = − ∂ H ∂ψ α = In the Lagrangian formalism, one can impose the constraintsfrom the beginning by introducing Lagrange multipliers whichenforce the constraints. So, Hamilton’s equations of motion, inthe presence of constraints, can be derived from the extendedaction principle: S = Z (cid:16) ˙ q i p i + ˙ ψ α π α − H − φ A λ A (cid:17) dt , (42)where φ A are our constraints, λ A are Lagrange multipliers, and A runs over the constraints. Using the extended action prin-ciple, we can write out the following two expressions of theextended Lagrangian and the extended Hamiltonian, respec-tively: L = ˙ q i p i + ˙ ψ α π α − H − φ A λ A , (43) H = ˙ q i p i + ˙ ψ α π α − φ A λ A − L . (44)Indeed, if the Lagrangian is independent of one coordinate, say q i , then, we call this coordinate an ignorable coordinate. How-ever, we still need to solve the Lagrangian for this coordinateto find the corresponding equation of motion. Since the mo-mentum corresponding to this coordinate may still enter the4etters in High Energy Physics LHEP xx, xxx, 2018Lagrangian and affect the evolution of other coordinates, there-fore, using the Euler-Lagrange equation, we have ddt (cid:18) ∂ L ∂ ˙ q i (cid:19) − ∂ L ∂ q i = p i = ∂ L / ∂ ˙ q i . Now, if there is no explicit dependence of the La-grangian L on generalized coordinate q i , then ∂ L / ∂ q i = d / dt ( ∂ L / ∂ ˙ q i ) = ⇒ dp i / dt =
0. Hence, a momentum p i is conserved when the La-grangian L is independent of a coordinate q i . This means thatif the Lagrangian is independent of a certain coordinate q i , itmust be also independent of its corresponding momentum p i .Furthermore, by extremizing the Hamiltonian (44), we ob-tain the following generalized Hamilton’s equations of motion:˙ q i = ∂ H ∂ p i + ∂φ A ∂ p i λ A = { q i , H + φ A λ A } P ,˙ p i = − ∂ H ∂ q i − ∂φ A ∂ q i λ A = { p i , H + φ A λ A } P ,˙ ψ α = − ∂ H ∂π α − ∂φ A ∂π α λ A = { ψ α , H + φ A λ A } P ,˙ π α = − ∂ H ∂ψ α − ∂φ A ∂ψ α λ A = { π α , H + φ A λ A } P , φ A = F = F ( q i , p i , ψ α , π α ) , we canwrite ˙ F = { F , H + φ A λ A } P :˙ F = ∂ F ∂ q i ∂ q i ∂ t + ∂ F ∂ p i ∂ p i ∂ t + ∂ F ∂ψ α ∂ψ α ∂ t + ∂ F ∂π α ∂π α ∂ t = ∂ F ∂ q i ˙ q i + ∂ F ∂ p i ˙ p i + ∂ F ∂ψ α ˙ ψ α + ∂ F ∂π α ˙ π α = ∂ F ∂ q i (cid:16) ∂ H ∂ p i + ∂φ A ∂ p i λ A (cid:17) + ∂ F ∂ p i (cid:16) − ∂ H ∂ q i − ∂φ A ∂ q i λ A (cid:17) + ∂ F ∂ψ α (cid:16) − ∂ H ∂π α − ∂φ A ∂π α λ A (cid:17) + ∂ F ∂π α (cid:16) − ∂ H ∂ψ α − ∂φ A ∂ψ α λ A (cid:17) = (cid:16) ∂ F ∂ q i ∂ ( H + φ A λ A ) ∂ p i − ∂ F ∂ p i ∂ ( H + φ A λ A ) ∂ q i (cid:17) − (cid:16) ∂ F ∂ψ α ∂ ( H + φ A λ A ) ∂π α + ∂ ( H + φ A λ A ) ∂ψ α ∂ F ∂π α (cid:17) ≡ { F, H + φ A λ A } P . (47)At this stage, we need to confirm that Eqs. in (46) give us thecorrect equations of motion using the constrained Hamiltonian(25). Since α , β ∈ {
1, 2 } in the primary constraint (28), then, wehave the two constraints φ = π + i ψ , φ = π + i ψ . (48)By using Eqs. in (46) and taking into count the Hamiltonian(29), and the constraints (48), we find that˙ ψ = − ∂ H ∂π − ∂φ π λ = − λ ,˙ ψ = − ∂ H ∂π − ∂φ π λ = − λ . (49) On the other hand, using Eqs. in (46) and considering theHamiltonian (25) and the constraints (48), we find that˙ π = − ∂ H ∂ψ − ∂φ ψ λ = i ωψ − i λ ,˙ π = − ∂ H ∂ψ − ∂φ ψ λ = − i ωψ − i λ . (50)Furthermore, using Eq. (27), we get π = − i ψ ⇒ ˙ π = − i ψ , π = − i ψ ⇒ ˙ π = − i ψ . (51)Eqs. (49, 50, and 51) imply the following set of equations ofmotion for the fermionic coordinates ψ α :˙ ψ = − ωψ , ˙ ψ = ωψ . (52)Fortunately, these equations are the same equations of motionfollowed from the Lagrangian formalism.For now, let us use the Poisson bracket, the Hamiltonian(25), and the constraints (48), to check that˙ φ = { φ , H + φ λ } P = ∂φ ∂ q i ∂ ( H + φ λ ) ∂ p i − ∂φ ∂ p i ∂ ( H + φ λ ) ∂ q i ! = − ∂φ ∂ψ α ∂ ( H + φ λ ) ∂π α + ∂φ ∂π α ∂ ( H + φ λ ) ∂ψ α ! = + − (cid:16) i (cid:17) ( λ ) − ( ) (cid:16) − i ωψ + i λ (cid:17) = i ( − λ + ωψ ) = λ = − ˙ ψ = ωψ . Similarly, ˙ φ can be calculatedusing the Poisson bracket as follows:˙ φ = { φ , H + φ λ } P = (cid:18) ∂φ ∂ q i ∂ ( H + φ λ ) ∂ p i − ∂φ ∂ p i ∂ ( H + φ λ ) ∂ q i (cid:19) = − (cid:18) ∂φ ∂ψ α ∂ ( H + φ λ ) ∂π α + ∂φ ∂π α ∂ ( H + φ λ ) ∂ψ α (cid:19) = + − (cid:16) i (cid:17) ( λ ) − ( ) (cid:16) i ωψ + i λ (cid:17) = i ( − λ − ωψ ) = λ = − ˙ ψ = − ωψ . Subsequently, Eqs. (53, 54) are zero if weimpose the equations of motion (52). In other words, we can saythat the constraints are consistent with the equations of motion. A Constraint F is said to be a second class constraint if it hasnonzero Poisson brackets, and, therefore, it requires specialtreatment. Given a set of second class constraints, we can de-fine a matrix { φ A , φ B } P = C AB , (55)5etters in High Energy Physics LHEP xx, xxx, 2018where we define the inverse of C AB , as C AB , such that C AB C BC = δ AC . (56)The Dirac bracket provides a modification to the Poisson brack-ets to ensure that the second class constants vanish: { φ A , F } D = F ( q , p ) and G ( q , p ) is defined as { F , G } D = { F , G } P − { F , φ A } P C AB { φ B , G } P . (58)Dirac bracket has the same properties of Poisson bracket,which we have listed in Subsection 3.1. In addition, we have tonotice the following.(i) If ˙ φ A =
0, then, the Dirac bracket of any constraint withthe extended Hamiltonian is equivalent to its Poissonbracket: { F , H + φ A λ A } D = { F , H + φ A λ A } P . (59)(ii) In some cases, we can return to the original Hamiltonian(25) forget about the constraints and the momenta π β at the cost of replacing the Poisson brackets with Diracbrackets. For example, if we consider the case ˙ φ A = ˙ F = { F , H + φ A λ A } D = { F , H } D + { F , φ A } D λ A = { F , H } D , (60)where we used Eq. (57); { φ A , F } D = { φ A , p } P =
0, then, using Eq. (58), we can find that { q , p } D = { q , p } P − { q , φ A } P C AB { φ B , p } = { q , p } P = ∂ q ∂ q ∂ p ∂ p − ∂ q ∂ p ∂ p ∂ q = { ψ α , ψ β } D for theHamiltonian (25) with the constraints (28). That { ψ α , ψ β } D in-volves a primary constraint φ β followed from the following re-lation which is obtained from the Lagrangian formalism: φ β = π β + i δ αβ ψ α . (62)Using the definition of Dirac bracket, we get { ψ α , ψ β } D = { ψ α , ψ β } P − { ψ α , φ α } P C αβ { φ β , ψ β } P . (63)Then, since the variables ψ α and ψ β are independent, their Pois-son bracket vanishes; i.e., { ψ α , ψ β } P = { ψ β , ψ α } P = { ψ α , φ α } P = (cid:18) ∂ψ α ∂ q i ∂φ α ∂ p i − ∂ψ α ∂ p i ∂φ α ∂ q i (cid:19) − (cid:18) ∂ψ α ∂ψ α ∂φ α ∂π α + ∂ψ α ∂π α ∂φ α ∂ψ α (cid:19) = − { ψ β , φ β } P = ∂ψ β ∂ q i ∂φ β ∂ p i − ∂ψ β ∂ p i ∂φ β ∂ q i ! − ∂ψ β ∂ψ β ∂φ β ∂π β + ∂ψ β ∂π β ∂φ β ∂ψ β ! = − C αβ = { φ α , φ β } P = (cid:16) ∂φ α ∂ q i ∂φ β ∂ p i − ∂φ α ∂ p i ∂φ β ∂ q i (cid:17) − (cid:16) ∂φ α ∂ψ α ∂φ β ∂π α + ∂φ α ∂π α ∂φ β ∂ψ α (cid:17) − (cid:16) ∂φ α ∂ψ β ∂φ β ∂π β + ∂φ α ∂π β ∂φ β ∂ψ β (cid:17) = − i δ αβ . (67)Moreover, from the definition (3.3), we have C αβ = − C αβ = i δ αβ . (68)As a result of substituting Eqs. (64, 65, and 66, 68) into Eq. (63),the Dirac bracket of the two variables ψ α and ψ β gives us { ψ α , ψ β } D = i δ αβ . (69) In this subsection, we show that the Dirac brackets give the cor-rect equations of motion for the Hamiltonian (27) using the ex-pression ˙ ψ α = { ψ α , H } D , (70)We check that˙ ψ = { ψ , H } D = { ψ , − i ωψ ψ } D = − i ω { ψ , ψ ψ } D = − i ω ( { ψ , ψ } D ψ + ψ { ψ , ψ } D ) = − i ω (cid:16) − i δ αβ ψ + (cid:17) = − ωδ αβ ψ = − ωψ , (71)and similarly˙ ψ = { ψ , H } D = { ψ , − i ωψ ψ } D = − i ω { ψ , ψ ψ } D = − i ω ( { ψ , ψ } D ψ + ψ { ψ , ψ } D ) = − i ω (cid:16) + ψ ( − i δ αβ ) (cid:17) = − ωψ δ αβ = ωδ αβ ψ = ωψ . (72)These equations are the same equations of motion which wehave obtained in Subsection 3.2 using the constrained Hamil-ton’s equations of motion 46. In this subsection, we investigate the supersymmetry of ourHamiltonian (25). For this purpose, we need to introduce someoperators Q i known as supercharges, which basically can beobtained from N¨oether’s theorem arising from a symmetry ofthe Lagrangian which exchanges bosons and fermions. In thenext section, we will look at the action of the supercharge oper-ators. For instance, the supercharge operators in the case of thesystem under discussion here can be written as follows: Q α = p ψ α + ω q ǫ αβ δ βγ ψ γ ⇒ Q = p ψ + ω q ψ Q = p ψ − ω q ψ , (73)6etters in High Energy Physics LHEP xx, xxx, 2018where again we suppose that { α , β } can only take the values {
1, 2 } , and ǫ αβ is the two index Levi-Civita symbol defined as ǫ αβ = + if ( α , β ) is (
1, 2 ) , − if ( α , β ) is (
2, 1 ) ,0 if α = β . (74)The system to be supersymmetric, the supercharges (73) to-gether with the Hamiltonian (27) and the constraints (25) mustsatisfy the classical Dirac bracket superalgebra which definedby { Q α , Q β } D = − i δ αβ H , { Q α , H } D = { α , β } have the values {
1, 2 } . Then,using the definition of the Poisson and Dirac brackets, we findthat,(see Appendix B for more details) { Q , Q } D = − i (cid:16) p + ω q (cid:17) , { Q , Q } D = ω (cid:16) ψ + ψ (cid:17) , { Q , Q } D = − ω (cid:16) ψ + ψ (cid:17) , { Q , Q } D = − i (cid:16) p + ω q (cid:17) .(76)If we make substitution using Eq. (76), we find { Q α , Q β } D = { Q , Q } D + { Q , Q } D + { Q , Q } D + { Q , Q } D = − i ( p + ω q ) + ω ( ψ + ψ ) − ω ( ψ + ψ ) − i ( p + ω q ) = − i ( p + ω q ) ≡ − i δ αβ H . (77)Furthermore, using the following relations (see Appendix B) { Q , H } P = ω p ψ − ω q ψ , { Q , H } D = − ω p ψ + ω q ψ , { φ , H } P = i ωψ , { φ , H } P = i ωψ , (78)we find { Q α , H } D = { Q , H } D + { Q , H } D = { Q , H } P − { Q , φ A } P C AB { φ B , H } P + { Q , H } P − { Q , φ A } P C AB { φ B , H } P = { Q , H } P − { Q , φ } P C { φ , H } P − { Q , φ } P C { φ , H } P + { Q , H } P − { Q , φ } P C { φ , H } P − { Q , φ } P C { φ , H } P = ω P ψ − ω q ψ − ω P ψ + ω q ψ − ω P ψ − ω q ψ + ω q ψ + ω P ψ =
4. THE SUPERSYMMETRY ALGEBRA
The material in this section is elaborated in detail in [22, 23, 24,25, 26, 27]. The supersymmetry algebra encodes a symmetrydescribing a relation between bosons and fermions. In general,the supersymmetry is constructed by introducing supersym-metric transformations which are generated by the supercharge Q i operators, where the role of the supercharges Q i is to convert a fermionic degree of freedom into a bosonic degree of freedomand vice versa; i.e., Q | fermionic i = | bosonic i , Q | bosonic i = | fermionic i .(80)So far, we have restricted ourselves to study classical sys-tems. In Subsection 3.5, we have investigated the supersym-metry of the classical harmonic oscillator. Now, we are goingto quantize the theory to study the supersymmetry of quan-tum systems. In a quantum mechanical supersymmetric sys-tem, the supercharges Q i together with the Hamiltonian H forma so-called superalgebra. The recipe for quantizing the Hamil-tonian in a situation where we have second class constraints isto replace the Dirac brackets with either commutator bracketsfor bosonic variables or anticommutator brackets for fermionicvariables multiplied with the factor i ¯ h , so we have { q , p } D = ⇒ [ ˆ q , ˆ p ] = i ¯ h , { ψ α , ψ β } D = − i δ αβ ⇒ { ˆ ψ α , ˆ ψ β } = ¯ h δ αβ . (81)Taking this into account, we can see that the superalgebra forN-dimensional quantum system is characterized by [ ˆ Q i , ˆ H ] = i = · · · N , { ˆ Q i , ˆ Q j } = ¯ h δ ij ˆ H , i , j = N . (82)This will be elaborated further in the next section using the ex-ample of the supersymmetrical harmonic oscillator. One morenotation before ending this section is that the supercharges Q i are Hermitian, i.e. Q † i = Q i , and this implies that { ˆQ i , ˆQ j } = { ˆQ † i , ˆQ † j } . (83)
5. SUPERSYMMETRIC HARMONIC OSCIL-LATOR
Now, we turn back to the quantum harmonic oscillator prob-lem as a simple example to show the supersymmetric prop-erty of a quantum mechanical system. Using Eq. (16), the har-monic oscillator Hamiltonian (17) can be written in terms of thebosonic and fermionic laddering operators as follows: ˆH = ¯ h ω ( ˆ a † ˆ a + ˆ b † ˆ b ) . (84)Moreover, the supercharge operators ˆQ and ˆQ can also bewritten in terms of the bosonic and fermionic laddering opera-tors as follows: ˆQ = p ψ + ω q ψ = i r ¯ h ω ( ˆ a † − ˆ a ) ! r ¯ h ( ˆ b + ˆ b † ) ! + ω r ¯ h ω ( ˆ a ) † + ˆ a ) ! i r ¯ h ( ˆ b − ˆ b † ) ! = i ¯ h √ ω ( ˆ a † ˆ b − ˆ a ˆ b † ) , (85)7etters in High Energy Physics LHEP xx, xxx, 2018and similarly, ˆQ = p ψ − ω q ψ = i r ¯ h ω ( ˆ a † − ˆ a ) ! i r ¯ h ( ˆ b − ˆ b † ) ! − ω r ¯ h ω ( ˆ a ) † + ˆ a ) ! r ¯ h ( ˆ b + ˆ b † ) ! = − ¯ h √ ω ( ˆ a † ˆ b + ˆ a ˆ b † ) . (86)Using Eqs. (85, 86), one may define non-Hermitian opera-tors ˆQ and ˆQ † as ˆQ = ( ˆQ − i ˆQ )= (cid:16) i ¯ h √ ω ( ˆ a † ˆ b − ˆ a ˆ b † ) + i ¯ h √ ω ( ˆ a † ˆ b + ˆ a ˆ b † ) (cid:17) = i ¯ h √ ω ( ˆ a † ˆ b ) , (87)and similarly, ˆQ † = ( ˆQ + i ˆQ )= (cid:16) i ¯ h √ ω ( ˆ a † ˆ b − ˆ a ˆ b † ) − i ¯ h √ ω ( ˆ a † ˆ b + ˆ a ˆ b † ) (cid:17) = − i ¯ h √ ω ( ˆ a ˆ b † ) . (88)We can directly use Eq. (84, 87, 88) to check that the superchargeoperators ˆQ and ˆQ † with the Hamiltonian H satisfy the quan-tum superalgebra: { ˆQ , ˆQ } = { ˆQ † , ˆQ † } = { ˆQ , ˆQ † } = { ˆQ † , ˆQ } = H , [ H , ˆQ ] = [ H , ˆQ † ] = ˆQ , ˆQ † on the energy eigenstates: ˆQ | n B , n F i ∼ | n B − n F + i , ˆQ † | n B , n F i ∼ | n B + n F − i . (90)Remember that there are only two possible fermionic states n F = {
0, 1 } , so the effect of the operators ˆQ and ˆQ † can bewritten as follows: ˆQ | n B , 0 i ∼ | n B −
1, 1 i , ˆQ † | n B , 1 i ∼ | n B +
1, 0 i . (91)Also, from Eq. (89), the operators ˆQ and ˆQ † commute with theHamiltonian H ; for that reason, we can see that H ( ˆQ | n B , n F i ) = ˆQ ( H | n B , n F i ) = ( E B + E F )( ˆQ | n B , n F i ) , (92)and this means that the whole energy of the system remainsunchanged by the action of the supercharge operators.In short, the supercharge operator ˆQ acts to change one bo-son to one fermion leaving the total energy of the system in-variant. Conversely, ˆQ † changes a fermion into a boson leavingthe energy unchanged. This is illustrated in Figure 1. E ( ) E ( ) E ( ) E ( ) E ( ) E ( ) E ( ) E E = n F = n F = QQ † FIGURE 1: A schematic view showing the energy levels fora supersymmetric system consisting of only two possiblefermionic states n F = {
0, 1 } . The supercharge operators Q and Q † exchange bosons and fermions without affecting the energydue to the degeneracy of the energy levels of the two supersym-metric partners except for the ground state of the first partner.
6. SUPERSYMMETRIC QUANTUM MECHAN-ICS
In this section, we study the general formalism of one-dimensional supersymmetric quantum mechanics. The ideas inthis section were discussed in [3, 28, 29, 30, 31, 32]. We begin byconsidering a Hamiltonian of the form ˆH = (cid:16) ˆ p + V ( ˆ x ) (cid:17) I +
12 ¯ hB ( ˆ x ) σ , (93)where V ( ˆ x ) is the potential, and B is a magnetic field. If thosepotential and magnetic fields can be written in terms of somefunction W ( x ) , as V ( x ) = (cid:18) dW ( x ) dx (cid:19) ≡ W ′ B ( x ) = (cid:18) d W ( x ) dx (cid:19) ≡ W ′′ ,(94)the Hamiltonian is supersymmetric and we shall refer to thefunction W ( x ) as a superpotential. Using Eq. (94), we canrewrite the Hamiltonian (93) as ˆH = (cid:16) ˆ p + W ′ (cid:17) I +
12 ¯ h σ W ′′ . (95)At this stage, we define the following two Hermitian super-charge operators: ˆQ = (cid:0) σ ˆ p + σ W ′ ( ˆ x ) (cid:1) , ˆQ = (cid:0) σ ˆ p − σ W ′ ( ˆ x ) (cid:1) . (96)The Hamiltonian (95) together with the supercharges (96) con-stitutes a superalgebra. It is a simple matter of algebra to verifythe conditions for the superalgebra given by Eq. (82). It may beuseful before verifying these conditions to check the following8etters in High Energy Physics LHEP xx, xxx, 2018commutation relation: [ ˆ p , f ( x )] g = [ − i ¯ h ∂∂ x , f ( x )] g = − i ¯ h (cid:18) ∂∂ x ( f g ) + f ∂∂ x ( g ) (cid:19) = − i ¯ h (cid:18) f · ∂ g ∂ x + ∂ f ∂ x · g − f · ∂ g ∂ x (cid:19) = − i ¯ h f ′ ( x ) g .(97)Removing g gives us the commutator of the momentum ˆ p withan arbitrary function of the position coordinate x : [ ˆ p , f ( x )] = − i ¯ h f ′ ( x ) . (98)Using this relation, we find the following: [ ˆ p , W ′ ] = ˆ pW ′ − W ′ ˆ p = − i ¯ hW ′ W ′′ , [ W ′ , ˆ p ] = W ′ ˆ p − ˆ p W ′ = i ¯ h { W ′′ , ˆ p } , { ˆ p , W ′′ } = ˆ pW ′′ + W ′′ ˆ p . (99)Now, we can use the outcome of Eq. (99) to check the outcomeof the commutator of the supercharge ˆQ with the Hamiltonian ˆH : [ ˆQ , ˆH ] = ˆQ ˆH − H ˆQ = ( σ ˆ p + σ W ′ ) · (cid:16)(cid:16) ˆ p + W ′ (cid:17) I + ¯ hW ′′ σ (cid:17) − (cid:16)(cid:16) ˆ p + W ′ (cid:17) I + ¯ hW ′′ σ (cid:17) · ( σ ˆ p + σ W ′ ) = (cid:16) σ ˆ p + σ ˆ pW ′ + σ σ ¯ h ˆ pW ′′ + σ W ′ ˆ p + σ W ′ W ′ + σ σ ¯ hW ′ W ′′ (cid:17) − (cid:16) σ ˆ p + σ ˆ p W ′ + σ W ′ ˆ p + σ W ′ W ′ + σ σ ¯ hW ′′ ˆ p + σ σ ¯ hW ′′ W ′ (cid:17) = (cid:16) σ [ ˆ p , W ′ ] − σ i ¯ h { ˆ p , W ′′ } + σ [ W ′ , ˆ p ] + σ i ¯ hW ′ W ′′ (cid:17) = [ ˆQ , ˆH ] = ˆQ ˆH − ˆH ˆQ =
0. (101)Furthermore, we can find that { ˆ Q , ˆ Q } = ˆ Q ˆ Q + ˆ Q ˆ Q = Q = (cid:0) σ ˆ p + σ W ′ (cid:1) = (cid:16) σ ˆ p + σ W ′ + σ σ ˆ pW ′ + σ σ W ′ ˆ p (cid:17) = (cid:16) ( ˆ p + W ′ ) I + i σ ( ˆ pW ′ − W ′ ˆ p ) (cid:17) = (cid:16) ( ˆ p + W ′ ) I + ¯ h σ W ′′ (cid:17) ≡ H , (102)and similarly, we have { ˆQ †2 , ˆQ } = ˆH . (103)Note that, to do the previous calculations, we used the Paulimatrices properties: σ = σ = σ = I and σ σ = − σ σ = i σ . It is clear from Eqs. (100,101, 102, and 139) that the Hamilto-nian (95) together with the supercharges (96) satisfies the con-ditions for the superalgebra given by Eq. (82).Indeed, if we define the quantities Q and Q † as ˆQ = ˆQ − i ˆQ = (cid:0) σ ˆ p + σ W ′ (cid:1) − i (cid:0) σ ˆ p − σ W ′ (cid:1) = (cid:0) ( σ − i σ ) ˆ p + ( σ + i σ ) W ′ (cid:1) = ( σ − i σ ) (cid:0) ˆ p + iW ′ (cid:1) = σ − (cid:0) ˆ p + iW ′ (cid:1) , (104)and ˆ Q † = ˆ Q + i ˆ Q = σ + (cid:0) ˆ p − iW ′ (cid:1) , (105)where σ − and σ + , respectively, are defined as σ − = ( σ − i σ ) = (cid:18) (cid:19) , σ + = ( σ + i σ ) = (cid:18) (cid:19) , (106)it is straightforward to check that the Hamiltonian ˆH can beexpressed as ˆH = { ˆQ , ˆQ † } , (107)and it commutes with both the operators ˆQ and ˆQ † : [ ˆQ , ˆH ] = [ ˆQ † , ˆH ] = { ˆQ , ˆQ } = { ˆQ † , ˆQ † } = n F = n F =
1. For a spinor quan-tum mechanical system, this implies that the excited energyeigenstates come in degenerate spin-up/spin-down pairs | E n , ↑i / | E n , ↓i . These degenerate spin-up/spin-down pairs are re-lated to the acts of the supercharge operators ˆQ and ˆQ † , wherethe supercharge operators ˆQ convert the degenerate spin-upstate to the degenerate spin-down state without making anychange in the energy eigenvalue of the states:ˆ Q | E n , ↑i ∼ (cid:18) (cid:19) (cid:18) ↑ (cid:19) = (cid:18) ↓ (cid:19) , (110)while the supercharge operators ˆQ † convert the degeneratespin-down state to the degenerate spin-up state without mak-ing any change in the energy eigenvalue of the states:ˆ Q † | E n , ↓i ∼ (cid:18) (cid:19) (cid:18) ↓ (cid:19) = (cid:18) ↑ (cid:19) . (111)To sum up, the supersymmetric transformations occur dueto the supercharge operators ˆQ and ˆQ † . In this case, it causestransforms between the energy eigenstates spin-up/spin-downwhich have the same energy eigenvalues.
7. SUPERSYMMETRIC GROUND STATE
So far, we have investigated all the required information todescribe a supersymmetric quantum mechanical system. Inthis section, we study the supersymmetric quantum mechan-ical ground state. The argument in this section follows from9etters in High Energy Physics LHEP xx, xxx, 2018[33, 34, 35, 36, 37]. Let us now write the expression of the super-symmetric Hamiltonian (95) as the summation of two separateterms: a Hamiltonian ˆH + and a Hamiltonian ˆH − , then, we have ˆH ± = (cid:16) ˆ p + W ′ (cid:17) ±
12 ¯ hW ′′ . (112)In the eigenbasis of σ , the supersymmetric Hamiltonian is di-agonal: ˆH ≡ (cid:18) ˆH + ˆH − (cid:19) = (cid:18) A † A AA † (cid:19) , (113)and the supercharge operators ˆQ and ˆQ † can be written as ˆQ = (cid:18) A (cid:19) , ˆQ † = (cid:18) A † (cid:19) , (114)where A = ˆ p + iW ′ , A † = ˆ p − iW ′ . (115)According to Eq. (107), we can write the supersymmetricHamiltonian ˆH in terms of the supercharges operators ˆQ and ˆQ † as follows: ˆH = ˆQ ˆQ † + ˆQ † ˆQ = ˆQ = ˆQ † . (116)In the last step, we used the fact that the supercharges are Her-mitian operators. We see from Eq. (116) that the Hamiltonian ˆH can be written in terms of the squares of the superchargeoperators. For this reason, the energy of any eigenstate of thisHamiltonian must be positive or zero. Let us now consider that | Ψ i is the ground state of the supersymmetric Hamiltonian ˆH .Based on Eq. (116), it can be seen that the ground state can havea zero energy only if it satisfies the following two conditions: E = h Ψ | ˆH | Ψ i = h Ψ | ˆQ | Ψ i = = ⇒ ˆQ | Ψ i = E = h Ψ | ˆH | Ψ i = h Ψ | ˆQ † | Ψ i = = ⇒ ˆQ † | Ψ i = (117)Therefore, if there exists a state which is annihilated by each ofthe supercharge operators ˆQ and ˆQ † which means that it is in-variant under the supersymmetry transformations, such a stateis automatically the zero-energy ground state. However, on theother hand, any state that is not invariant under the supersym-metry transformations has a positive energy. Thus, if there is asupersymmetric state, it is the zero-energy ground state and itis said that the supersymmetry is unbroken.Moreover, since the supersymmetry algebra (108) impliesthat the supercharge operators ˆQ and ˆQ † commute with the su-persymmetric Hamiltonian ˆH , so all the eigenstates of ˆH aredoubly degenerate. For that reason, it will be convenient towrite the supersymmetric ground state of the system in termsof two components, ψ ± , as follows: Ψ = (cid:18) ψ + ψ − (cid:19) . (118)One can now solve eigenvalue problem ˆQ | ψ i = ˆQ | ψ i = (cid:18) A (cid:19) (cid:18) ψ + ψ − (cid:19) = A | ψ + i = (cid:0) ˆ p + iW ′ (cid:1) | ψ + i = = ⇒ − i ¯ h ∂ψ + ∂ x + iW ′ ψ + = d ψ + ψ + = W ′ ¯ h dx = ⇒ ln ψ + = W ′ ¯ h + ln A . (121)Thus, ψ + = Ae W ¯ h , (122)and similarly, ψ + = Be − W ¯ h . (123)Now, the general form of the ground-state wave function maybe expressed as Ψ = Ae W ¯ h Be − W ¯ h . ! . (124)In fact, it has no physical meaning to find the ground-state wave function ψ if there are no normalizable solutionsof such form. So we need now to normalize the ground-statewave function (124) by determining the values of the two con-stants A and B . The ground-state wave function ψ ( x ) to benormalizable must vanish at the positive and the negative infi-nite x -values. In order to satisfy this condition, we have to set | W ( x ) | → ∞ as | x | → ∞ . Then, we have the following threepossible cases for the supersymmetric ground-state wave func-tion:1. The first case is when W ( x ) → + ∞ as x → ± ∞ . Inthis case, the superpotential W ( x ) is an even functionand it is positive at the boundaries. Figure 2 describesthe behavior of the potential V ( x ) in this case. Since W ( x ) is positive, we cannot normalize the wave function ψ + = Ae W ( x ) ¯ h , but we can choose A =
0. However, wecan normalize the wave function ϕ − = Be − W ( x ) ¯ h to findthe value of the constant B . The complete ground-statewave function, in this case, may be expressed in the gen-eral form Ψ = Be − W ( x ) ¯ h ! . (125) xV ( x ) FIGURE 2: The even superpotential W ( x ) is an even function,where W ( x ) → + ∞ as x → ± ∞ .10etters in High Energy Physics LHEP xx, xxx, 20182. The second case is when W ( x ) → − ∞ as x → ± ∞ .In this case, the superpotential W ( x ) is an even func-tion and it is negative at the boundaries. The behaviorof the potential V ( x ) in this case is described in Figure3. Since W ( x ) is negative, we can normalize the wavefunction Ψ a = Ae W ( x ) ¯ h and find the value of the con-stant A ; however we cannot normalize the wave func-tion Ψ b = Be − W ( x ) ¯ h , but we can choose B =
0. The com-plete ground-state wave function, in this case, shall beexpressed in the general form Ψ = Ae W ( x ) ¯ h ! . (126) xV ( x ) FIGURE 3: The superpotential W ( x ) is an even function, where W ( x ) → − ∞ as x → ± ∞ .3. The other two cases correspond to W ( x ) → { + ∞ , − ∞ } as x → { + ∞ , − ∞ } , and W ( x ) → {− ∞ , + ∞ } as x →{ + ∞ , − ∞ } . In those two cases, the superpotential W ( x ) is an odd function and it is positive at one of the bound-aries and negative at the other boundary. The behavior ofthe potential V ( x ) in this case is described in Figure 4. Inthose two cases, both the two constants A and B vanishand we cannot normalize the wave function. Therefore.we have Ψ =
0. (127)Thus, since those two forms of the zero-energy ground-state wave function cannot be normalized, this meansthat they do not exist.In short, the zero-energy ground-state wave function can benormalized when the superpotential W ( x ) has an even num-ber of zeros. In this case, there exists a zero-energy groundstate which is the true vacuum state and the supersymmetryis unbroken. However on the other hand, if the superpoten-tial W ( x ) has an odd number of zeros, the zero-energy ground-state wave function cannot be normalized. Then, we immedi-ately realize that there is no zero-energy ground state in suchcase and the supersymmetry is spontaneously broken.
8. CORRECTIONS TO THE GROUND-STATEENERGY
In this section, we consider the example of the supersymme-try harmonic oscillator which has a unique zero-energy ground xV ( x ) (a) W ( x ) → − ∞ as x → − ∞ , and W ( x ) → ∞ as x → ∞ . xV ( x ) (b) W ( x ) → ∞ as x → − ∞ , and W ( x ) → − ∞ as x → ∞ . FIGURE 4: The superpotential W ( x ) is an odd function.state. Then, we use the conventional perturbation theory to ex-amine whether it has nonvanishing corrections to the energyof the ground state, and accordingly, it can be responsible forthe spontaneous breaking of the supersymmetry. To this end,in the next two subsections, based on [38, 39, 40, 41, 42], we aregoing to compute both the first and the second corrections tothe ground-state energy of the supersymmetric harmonic os-cillator. For instance, let us recall the general form of the super-symmetric quantum mechanics Hamiltonian, which is given byEq. (95): ˆH = ( ˆ p + W ′ ) I +
12 ¯ hW ′′ σ . (128)Consider now that the superpotential W ( x ) is defined as W ( q ) = ω ˆ q + g ˆ q ⇒ W ′ = ω ˆ q + g ˆ q W ′′ = ω + g ˆ q , (129)where g is a perturbation. If g =
0, the previous Hamiltonian re-duces to the supersymmetric harmonic oscillator Hamiltonian: ˆH = ( ˆ p + ω q ) I +
12 ¯ h ωσ . (130)We have already solved this Hamiltonian and found that it hasa unique zero-energy ground state, E ( ) = Ψ ( ) = ( ωπ ¯ h ) e − ω q h ! . (132)11etters in High Energy Physics LHEP xx, xxx, 2018Now, when g is small, the Hamiltonian (128) can be writtenas ˆH = ˆH + ˆH ′ , (133)where ˆH is the Hamiltonian of the unperturbed system and ˆH ′ a perturbation: ˆH ′ = ( g ˆ q + ω g ˆ q ) I + hg ˆ q σ . (134)Moreover, if we consider the matrix representation, the pertur-bative Hamiltonian ˆH acting on the ground-state wave function Ψ ( ) yields ˆH ′ Ψ ( ) = (cid:18) g ˆ q + ω g ˆ q + hg ˆ q g ˆ q + ω g ˆ q − hg ˆ q (cid:19) ( ωπ ¯ h ) e − ω q h ! = (135)In view of this last equation, it is clear that for this combinationthe impact of the Hamiltonian ˆH ′ on the wave function Ψ ( ) isonly due to the componentˆ H ′ = g ˆ q + ω g ˆ q − hg ˆ q . (136)As a result, we see that it is enough to consider the Hamiltonianˆ H ′ to compute the first and second corrections to the ground-state energy (see Appendix C).Furthermore, we have to mention here that, for the super-symmetric harmonic oscillator, except for the ground state, allthe energy levels are degenerate to two energy states. However,because we are just interested in computing the correction tothe ground-state energy, which is not degenerate, we can usethe nondegenerate perturbation theory. For more explanationabout this point see Appendix C.In addition, it is useful to remember from quantum me-chanics that the wave function of degree n can be obtained bythe following recursion formula:ˆ q ψ n = r ( n + ) ¯ h ω ψ n + + r n ¯ h ω ψ n − . (137)As well, it is important to recall the following relation: h ψ n | ψ m i = δ nm . (138)We are ready now to move forward to the next two subsectionsand compute the first and the second corrections to the energyof the ground state. We know from the previous argument that the ground state ofthe supersymmetric Hamiltonian is nondegenerate. Therefore,we can use the time-independent nondegenerate perturbationtheory to compute the first-order correction to the ground-stateenergy of the supersymmetric harmonic oscillator as follows: E ( ) = h ψ ( ) | ˆ H ′ | ψ ( ) i = h ψ ( ) | g ˆ q + ω g ˆ q − hg ˆ q | ψ ( ) i = g h ψ ( ) | ˆ q | ψ ( ) i + ω g h ψ ( ) | ˆ q | ψ ( ) i − hg h ψ ( ) | ˆ q | ψ ( ) i . (139) The terms ˆ q ψ and ˆ q ψ are linearly independent of ψ so thatboth the second and the third terms of Eq. (139) are zero. How-ever, ˆ q ψ = h ω ψ + h ω ψ + h ω ψ , (140)then, substituting it into Eq. (139) gives us, E ( ) = g h ψ ( ) | ˆ q | ψ ( ) i = g ×
34 ¯ h ω h ψ ( ) | ψ ( ) i =
278 ¯ h ω g ≃ O ( g ) . (141)In the view of the last equation, the first-order correction tothe ground-state energy reduces to zero. Another method withmore details to calculations of the first-order correction of theground-state energy is given in Appendix C. From the nondegenerate time-independent perturbation the-ory, the second-order correction to the energy is given by E ( ) = ∑ m = n |h ψ ( ) n | ˆ H ′ | ψ ( ) m i| E ( ) m − E ( ) n . (142)Since the perturbation contains only terms of q , q , and q , thenumerator of Eq. (142) is zero for all m values except m =
1, 3, 4.For more explanation about this point, see Appendix D. There-fore, the second-order correction to ground-state energy of thesupersymmetric harmonic oscillator is computed as follows: E ( ) = |h ψ ( ) | g ˆ q + ω g ˆ q − hg ˆ q | ψ ( ) i| E ( ) − E ( ) + |h ψ ( ) | g ˆ q + ω g ˆ q − hg ˆ q | ψ ( ) i| E ( ) − E ( ) + |h ψ ( ) | g ˆ q + ω g ˆ q − hg ˆ q | ψ ( ) i| E ( ) − E ( ) + |h ψ ( ) | g ˆ q + ω g ˆ q − hg ˆ q | ψ i| E ( ) − E ( ) . (143)We can simplify the previous equation by calculating the nu-merator of each term using the recurrence relation for the har-monic oscillator energy eigenfunction given by Eq. (137). Wegetˆ q ψ = r ¯ h ω ψ + r ¯ h ω ψ ,ˆ q ψ = s h ω ψ + s ¯ h ω ψ + s h ω ψ ,ˆ q ψ = q h ω ψ + q h ω ψ + q h ω ψ + q h ω ψ , ˆ q ψ = q h ω ψ + q h ω ψ + q ¯ h ω ψ + q h ω ψ , ˆ q ψ = q h ω ψ + q h ω ψ + q h ω ψ + q h ω ψ + q h ω ψ . (144)In Eq. (144), we only list the terms which contain ψ . Theother terms do not affect our calculations, since all of them give12etters in High Energy Physics LHEP xx, xxx, 2018us zero (see Appendix C). Using Eq. (144) to calculate Eq. (143),we get E ( ) = − g ¯ h ω ω × h ω − h × r ¯ h ω ! − g h ω s h ω − g h ω ω × s h ω − g h ω × s h ω = −
98 ¯ h ω g − h ω g −
94 ¯ h ω g − h ω g = −
278 ¯ h ω g − h ω g . (145)For more details, another method to compute the second-ordercorrection of the ground-state energy is given in Appendix D.Finally, with regard to Eqs. (141 and 145), we realize that upto the second order the energy corrections to the ground-stateenergy vanish. This is concluded as O ( g ) = O ( g ) =
278 ¯ h ω −
278 ¯ h ω =
9. PROPERTIES OF SUSY QUANTUM ME-CHANICS xW ( x ) FIGURE 5: The superpotential of the harmonic oscillatorground state.A supersymmetric quantum mechanics system is said tohave unbroken supersymmetry if it has a zero-energy groundstate, which is E =
0, while if the system has a positiveground-state energy, E >
0, it is said to have a broken super-symmetry [1]. For example, in Section 5, we have studied theHamiltonian of the supersymmetric harmonic oscillator and showed that it obeys the superalgebra. Then, we calculated theground-state energy for that system and found evidence that itvanishes, E =
0, and consequently, this supersymmetric har-monic oscillator has an unbroken supersymmetry.In the previous section, we applied a small perturbation g to the supersymmetric harmonic oscillator and calculated theeffect on the ground-state energy. We found out that the per-turbation does not affect the energy of the ground state at thesecond order in perturbation theory, but this result can be ex-panded to any finite order. This means that the supersymmetrybreaking does not occur because of perturbation, and it mustbe due to the nonperturbative effects.Again, let us consider the same potential Eq. (129), whichwe have used before in Section 8: W ( q ) = ω ˆ q + g ˆ q . (147)As we discussed in Section 7, the wave function of the groundstate should have three zeros, Ψ → + ∞ as x → + ∞ , and Ψ → − ∞ as x → − ∞ . Figure 5, shows an arbitrary diagrampresenting how the wave function of the ground state shouldlook like, and Eq. (148) gives us the nonnormalized form of theground-state wave function: Ψ = Ae − W ¯ h Be − W ¯ h ! . (148)Acutely, Eq. (148) could not be normalized, and the only wayto solve the Schr ¨odinger equation for the ground state is taking Ψ =
0. This means that the wave function of the ground stateis not excited, even if the perturbation theory told us that it isexcited and has no energy correction.Thus, the perturbation technique gives us incorrect resultsfor both the wave function and the energy spectrum and failsto give an explanation to the supersymmetry breaking.
10. CONCLUSIONS
In this study, we studied the basic aspects of supersymmetricquantum mechanics. We started with introducing the algebraof Grassmann variables and then looked into quantum me-chanics of the supersymmetric harmonic oscillator, which in-cludes fermionic as well as bosonic fields. Afterward, we in-vestigated the algebraic structure of supersymmetric quantummechanics. We started by investigating the superalgebra us-ing Dirac brackets. Then, we introduced the concept of the su-percharge operators ˆ Q and ˆ Q † . In general, the supersymmetryis constructed by introducing supersymmetric transformationswhich are generated by the supercharge operators, where therole of the supercharges is to change the bosonic state into thefermionic state and vice versa, while the Lagrangian remainsinvariant. Moreover, we have presented the basic properties ofsupersymmetric quantum mechanics.Furthermore, we illustrated, for a supersymmetric quan-tum mechanical system, that the energy spectrum is degenerateexcept for the ground state, which must have a zero eigenvaluein order for the system to have an unbroken supersymmetry.Also, we have explained that if there is a supersymmetric state,it is the zero-energy ground state. If such a zero-energy groundstate exists, it is said that the supersymmetry is unbroken. Sofar, there has been no unbroken supersymmetry observed in13etters in High Energy Physics LHEP xx, xxx, 2018nature, and if nature is described by supersymmetry, of course,it must be broken.In fact, supersymmetry may be broken spontaneously atany order of perturbation theory or dynamically due to non-perturbative effects. To examine this statement, we studied thenormalization of the ground state of the supersymmetric har-monic oscillator. Then, we used perturbation theory to calcu-late the corrections to the ground-state energy. We found outthat the perturbation does not affect the energy of the groundstate at second order in perturbation theory, but this result canbe expanded to any finite order. This means that the supersym-metry breaking is not seen in perturbation theory, and it mustbe due to the nonperturbative effects. ACKNOWLEDGEMENTS
The author acknowledges the research office of the Universityof the Witwatersrand and the African Institute for Mathemati-cal Sciences (Ghana) for financial support.
Appendix A. THE GENERALIZED HAMIL-TON’S EQUATIONS OF MO-TION
Recall the Hamiltonian (25): H = (cid:16) p + ω q (cid:17) − i ωψ ψ . (A.1)We showed in Subsection 3.2 that the previous Hamiltoniancould be written in the form (29): H = (cid:16) p + ω q (cid:17) + i ωπ π . (A.2)Also consider the momenta π α and the primary constraint φ α ,respectively, define as π α = ∂ L ∂ ˙ ψ α = − i δ αβ ψ β , φ α = π α + i δ αβ ψ β . (A.3)Downward, we explain how to use the Poisson bracket to getthe generalized Hamilton’s equations of motion, which wehave only write down in Eq. (46).(i) The extended Hamiltonian (44) is written as H = ˙ q i p i + ˙ ψ α π α − φ A λ A − L . (A.4)Therefore, ∂ H ∂ p i = ˙ q i − ∂φ A ∂ p i λ A . (A.5)Thus, ˙ q i = ∂ H ∂ p + ∂φ A ∂ p λ A . (A.6)Using Poisson brackets, we find that { q i , H + φ A λ A } P = ∂ q i ∂ q i ∂ ( H + φ A λ A ) ∂ p i − ∂ ( H + φ A λ A ) ∂ q i ∂ q i ∂ p i ! − ∂ q i ∂ψ α ∂ ( H + φ A λ A ) ∂π α − ∂ ( H + φ A λ A ) ∂ψ α ∂ q i ∂π α ! = ∂ H ∂ p i + ∂π A ∂ p i λ A . (A.7) From Eqs. (A.6, A.7), we obtain˙ q i = ∂ H ∂ p + ∂φ A ∂ p λ A = { q i , H + φ A λ A } P . (A.8)(ii) ∂ H ∂ q i = − ∂ L ∂ q i − ∂φ A ∂ q i λ A = − ddt ∂ L ∂ ˙ q i − ∂φ A ∂ q i λ A = − ˙ p i − ∂φ A ∂ q i λ A . (A.9)Thus, ˙ p i = − ∂ H ∂ q i − ∂φ A ∂ q i λ A . (A.10)Using the Poisson brackets, we find that { p i , H + φ A λ A } P = ∂ p i ∂ q i ∂ ( H + φ A λ A ) ∂ p i − ∂ ( H + φ A λ A ) ∂ q i ∂ p i ∂ p i ! − ∂ p i ∂ψ α ∂ ( H + φ A λ A ) ∂π α − ∂ ( H + φ A λ A ) ∂ψ α ∂ p i ∂π α ! = − ∂ H ∂ q i − ∂φ A ∂ q i . (A.11)From Eqs. (A.10, A.11), we obtain˙ p i = − ∂ H ∂ q i − ∂φ A ∂ q i λ A = { p i , H + φ A λ A } P . (A.12)(iii) ∂ H ∂π α = − ˙ ψ α − ∂φ A ∂ q π α λ A . (A.13)Thus, ˙ ψ α = − ∂ H ∂π α − ∂φ A ∂ q π α λ A . (A.14)Using the Poisson brackets, we find that { ψ α , H + φ A λ A } P = ∂ψ α ∂ q i ∂ ( H + φ A λ A ) ∂ p i − ∂ ( H + φ A λ A ) ∂ q i ∂ p i ∂ψ α ! − ∂ψ α ∂ψ α ∂ ( H + φ A λ A ) ∂π α − ∂ ( H + φ A λ A ) ∂ψ α ∂ψ α ∂π α ! = − ∂ H ∂π α − ∂φ A ∂π α . (A.15)From Eqs. (A.14, A.15), we obtain˙ ψ α = − ∂ H ∂π α − ∂φ A ∂ q π α λ A = { ψ α , H + φ A λ A } P . (A.16)(iv) ∂ H ∂ψ α = − ∂ L ∂ψ α − ∂φ A ∂ q ψ α λ A = − ddt (cid:18) ∂ L ∂ ˙ ψ α (cid:19) − ∂φ A ∂ψ α λ A = − ˙ π α − ∂φ A ∂ψ α λ A . (A.17)14etters in High Energy Physics LHEP xx, xxx, 2018Thus, ˙ π α = − ∂ H ∂ψ α − ∂φ A ∂ψ α λ A . (A.18)Using the Poisson brackets, we obtain { π α , H + φ A λ A } P = ∂π α ∂ q i ∂ ( H + φ A λ A ) ∂ p i − ∂ ( H + φ A λ A ) ∂ q i ∂ p i ∂π α ! − ∂π α ∂ψ α ∂ ( H + φ A λ A ) ∂π α − ∂ ( H + φ A λ A ) ∂ψ α ∂π α ∂π α ! = − ˙ π α − ∂φ A ∂ψ α λ A . (A.19)From Eqs. (A.18, A.19), we obtain˙ π α = − ∂ H ∂ψ α − ∂φ A ∂ψ α λ A = { π α , H + φ A λ A } P . (A.20)(v) By definition from Eq. (A.3), φ A = π A + i ψ A = − i ψ A + i ψ A = Appendix B. DIRAC BRACKET AND THESUPERALGEBRA
From (Appendix A), recall the expressions of the Hamiltonian(A.1) and the extended Hamiltonian (A.2): H = (cid:16) p + ω q (cid:17) − i ωψ ψ . (B.22)As well, take into account the for the constraint (A.3) and sup-pose that { α , β } can only take the values {
1, 2 } ; then, we have φ = π + i ψ , φ = π + i ψ . (B.23)Furthermore, consider the supercharge formula, and onceagain suppose that { α , β } can only take the values {
1, 2 } , thenwe have Q α = p ψ α + ω q ǫ αβ δ βγ ψ γ ⇒ Q = p ψ + ω q ψ Q = p ψ − ω q ψ , (B.24)where ǫ αβ and δ αβ are the Levi-Civita symbol and Kroneckerdelta function, respectively.To verify that the supercharges (B.24) together with theHamiltonian (B.22) and the constraints (B.23) satisfy the Diracbracket superalgebra (75), we need as a first step to calculatethe Dirac brackets of the supercharges: { Q α , Q β } D = { Q , Q } D + { Q , Q } D + { Q , Q } D + { Q , Q } D . (B.25)To simplify the calculation of Eq. (B.25), let us start with cal-culating the Poisson bracket of the supercharges and the con-straints. { Q , Q } P = (cid:16) ∂ Q ∂ q i ∂ Q ∂ p i − ∂ Q ∂ p i ∂ Q ∂ q i (cid:17) − (cid:16) ∂ Q ∂ψ ∂ Q ∂π + ∂ Q ∂π ∂ Q ∂ψ (cid:17) − (cid:16) ∂ Q ∂ψ ∂ Q ∂π + ∂ Q ∂π ∂ Q ∂ψ (cid:17) = { Q , Q } P = (cid:16) ∂ Q ∂ q i ∂ Q ∂ p i − ∂ Q ∂ p i ∂ Q ∂ q i (cid:17) − (cid:16) ∂ Q ∂ψ ∂ Q ∂π + ∂ Q ∂π ∂ Q ∂ψ (cid:17) − (cid:16) ∂ Q ∂ψ ∂ Q ∂π + ∂ Q ∂π ∂ Q ∂ψ (cid:17) = ω (cid:16) ψ + ψ (cid:17) , (B.27) { Q , Q } P = (cid:16) ∂ Q ∂ q i ∂ Q ∂ p i − ∂ Q ∂ p i ∂ Q ∂ q i (cid:17) − (cid:16) ∂ Q ∂ψ ∂ Q ∂π + ∂ Q ∂π ∂ Q ∂ψ (cid:17) − (cid:16) ∂ Q ∂ψ ∂ Q ∂π + ∂ Q ∂π ∂ Q ∂ψ (cid:17) = − ω (cid:16) ψ + ψ (cid:17) , (B.28) { Q , Q } P = (cid:16) ∂ Q ∂ q i ∂ Q ∂ p i − ∂ Q ∂ p i ∂ Q ∂ q i (cid:17) − (cid:16) ∂ Q ∂ψ ∂ Q ∂π + ∂ Q ∂π ∂ Q ∂ψ (cid:17) − (cid:16) ∂ Q ∂ψ ∂ Q ∂π + ∂ Q ∂π ∂ Q ∂ψ (cid:17) = { Q , φ } P = (cid:16) ∂ Q ∂ q i ∂φ ∂ p i − ∂ Q ∂ p i ∂φ ∂ q i (cid:17) − (cid:16) ∂ Q ∂ψ ∂φ ∂π + ∂ Q ∂π ∂φ ∂ψ (cid:17) − (cid:16) ∂ Q ∂ψ ∂φ ∂π + ∂ Q ∂π ∂φ ∂ψ (cid:17) = − p , (B.30) { φ , Q } P = (cid:16) ∂φ ∂ q i ∂ Q ∂ p i − ∂φ ∂ p i ∂ Q ∂ q i (cid:17) − (cid:16) ∂φ ∂ψ ∂ Q ∂π + ∂φ ∂π ∂ Q ∂ψ (cid:17) − (cid:16) ∂φ ∂ψ ∂ Q ∂π + ∂φ ∂π ∂ Q ∂ψ (cid:17) = − p , (B.31) { Q , φ } P = (cid:16) ∂ Q ∂ q i ∂φ ∂ p i − ∂ Q ∂ p i ∂φ ∂ q i (cid:17) − (cid:16) ∂ Q ∂ψ ∂φ ∂π + ∂ Q ∂π ∂φ ∂ψ (cid:17) − (cid:16) ∂ Q ∂ψ ∂φ ∂π + ∂ Q ∂π ∂φ ∂ψ (cid:17) = − ω q , (B.32) { φ , Q } P = (cid:16) ∂φ ∂ q i ∂ Q ∂ p i − ∂φ ∂ p i ∂ Q ∂ q i (cid:17) − (cid:16) ∂φ ∂ψ ∂ Q ∂π + ∂φ ∂π ∂ Q ∂ψ (cid:17) − (cid:16) ∂φ ∂ψ ∂ Q ∂π + ∂φ ∂π ∂ Q ∂ψ (cid:17) = − ω q , (B.33) { Q , φ } P = (cid:16) ∂ Q ∂ q i ∂φ ∂ p i − ∂ Q ∂ p i ∂φ ∂ q i (cid:17) − (cid:16) ∂ Q ∂ψ ∂φ ∂π + ∂ Q ∂π ∂φ ∂ψ (cid:17) − (cid:16) ∂ Q ∂ψ ∂φ ∂π + ∂ Q ∂π ∂φ ∂ψ (cid:17) = ω q , (B.34) { φ , Q } P = (cid:16) ∂φ ∂ q i ∂ Q ∂ p i − ∂φ ∂ p i ∂ Q ∂ q i (cid:17) − (cid:16) ∂φ ∂ψ ∂ Q ∂π + ∂φ ∂π ∂ Q ∂ψ (cid:17) − (cid:16) ∂φ ∂ψ ∂ Q ∂π + ∂φ ∂π ∂ Q ∂ψ (cid:17) = ω q , (B.35) { Q , φ } P = (cid:16) ∂ Q ∂ q i ∂φ ∂ p i − ∂ Q ∂ p i ∂φ ∂ q i (cid:17) − (cid:16) ∂ Q ∂ψ ∂φ ∂π + ∂ Q ∂π ∂φ ∂ψ (cid:17) − (cid:16) ∂ Q ∂ψ ∂φ ∂π + ∂ Q ∂π ∂φ ∂ψ (cid:17) = − p , (B.36)15etters in High Energy Physics LHEP xx, xxx, 2018 { φ , Q } P = (cid:16) ∂φ ∂ q i ∂ Q ∂ p i − ∂φ ∂ p i ∂ Q ∂ q i (cid:17) − (cid:16) ∂φ ∂ψ ∂ Q ∂π + ∂φ ∂π ∂ Q ∂ψ (cid:17) − (cid:16) ∂φ ∂ψ ∂ Q ∂π + ∂φ ∂π ∂ Q ∂ψ (cid:17) = − p . (B.37)Using Dirac bracket definition and Eq.s ( B.26-B.37), then,we get { Q , Q } D = { Q , Q } P − { Q , φ A } P C AB { φ B , Q } P = { Q , Q } P − { Q , φ } P C { φ , Q } P − { Q , φ } P C { φ , Q } P = − i ( p + ω q ) , (B.38) { Q , Q } D = { Q , Q } P − { Q , φ A } P C AB { φ B , Q } P = { Q , Q } P − { Q , φ } P C { φ , Q } P − { Q , φ } P C { φ , Q } P = ω ( ψ + ψ ) , (B.39) { Q , Q } D = { Q , Q } P − { Q , φ A } P C AB { φ B , Q } P = { Q , Q } P − { Q , φ } P C { φ , Q } P − { Q , φ } P C { φ , Q } P = − ω ( ψ + ψ ) , (B.40) { Q , Q } D = { Q , Q } P − { Q , φ A } P C AB { φ B , Q } P = { Q , Q } P − { Q , φ } P C { φ , Q } P − { Q , φ } P C { φ , Q } P = − i ( p + ω q ) . (B.41)Substituting Eqs. (B.38, B.39, B.40, and B.41) into Eq. (B.25),thus, we find { Q α , Q β } D = { Q , Q } D + { Q , Q } D + { Q , Q } D + { Q , Q } D = − i ( p + ω q ) + ω ( ψ + ψ ) − ω ( ψ + ψ ) − i ( P + ω q ) = − i ( P + ω q ) . (B.42)On the other hand, − i δ αβ H = − i δ H − i δ H = − i ( p + ω q ) − i ( P + ω q )= − i ( p + ω q ) . (B.43)Eqs. (B.42) and (B.43) imply that { Q α , Q β } D = − i δ αβ H . (B.44)So far, as a second step to verify that the supercharges (B.24)together with the Hamiltonian (B.22) and the constraints (B.23)satisfy the Dirac bracket superalgebra (75), we need to calculatethe Dirac brackets of the supercharges and the Hamiltonian: { Q α , H } D = { Q , H } D + { Q , H } D . (B.45)To simplify the calculation of Eq. (B.45), let us calculate thePoisson bracket of the supercharges and the Hamiltonian. { Q , H } P = (cid:18) ∂ Q ∂ q i ∂ H ∂ p i − ∂ Q ∂ p i ∂ H ∂ q i (cid:19) − (cid:18) ∂ Q ∂ψ α ∂ H ∂π α + ∂ Q ∂π α ∂ H ∂ψ α (cid:19) = ω p ψ − ω q ψ , (B.46) { Q , H } P = (cid:18) ∂ Q ∂ q i ∂ H ∂ p i − ∂ Q ∂ p i ∂ H ∂ q i (cid:19) − (cid:18) ∂ Q ∂ψ α ∂ H ∂π α + ∂ Q ∂π α ∂ H ∂ψ α (cid:19) = − ω p ψ − ω q ψ , (B.47) { φ , H } P = (cid:18) ∂φ ∂ q i ∂ H ∂ p i − ∂φ ∂ p i ∂ H ∂ q i (cid:19) − (cid:18) ∂φ ∂ψ α ∂ H ∂π α + ∂φ ∂π α ∂ H ∂ψ α (cid:19) = i ωψ , (B.48) { φ , H } P = (cid:18) ∂φ ∂ q i ∂ H ∂ p i − ∂φ ∂ p i ∂ H ∂ q i (cid:19) − (cid:18) ∂φ ∂ψ α ∂ H ∂π α + ∂φ ∂π α ∂ H ∂ψ α (cid:19) = − i ωψ . (B.49)Therefore, using Eqs. ( B.46,B.47, B.48, and B.49), we calcu-late the Dirac bracket of the supercharges and the Hamiltonian. { Q , H } D = { Q , H } P − { Q , φ A } P C AB { φ B , H } P = { Q , H } P − { Q , φ } P C { φ , H } P − { Q , φ } P C { φ , H } P = ω p ψ − ω q ψ − ω p ψ + ω q ψ . (B.50) { Q , H } D = { Q , H } P − { Q , φ A } P C AB { φ B , H } P = { Q , H } P − { Q , φ } P C { φ , H } P − { Q , φ } P C { φ , H } P = − ω p ψ − ω q ψ + ω q ψ + ω p ψ , (B.51)Substitute Eqs. (B.50, B.51) into equation (B.45), thus, we find { Q α , H } D = { Q , H } D + { Q , H } D = (cid:16) ω p ψ − ω q ψ − ω p ψ + ω q ψ (cid:17) + (cid:16) − ω p ψ − ω q ψ + ω q ψ + ω p ψ (cid:17) = Appendix C. THE FIRST-ORDER CORREC-TION
In this appendix, we use conventional perturbation theory tocompute the first-order correction to the ground-state energyof the supersymmetric harmonic oscillator. Good discussion ofthe perturbation theory can be found in [38, 39, 40, 41, 42]. Inthe nondegenerate time-independent perturbation theory, thefirst-order correction to the energy of the n th ’ state is given by E ( ) n = h ψ n | ˆH ′ | ψ n i . (C.53)We need now to use this formula to compute the first-order cor-rection to the ground-state energy of our system. As we haveshown in Section 6, the supersymmetric quantum mechanicsHamiltonian is given in the general form ˆH = ( ˆ p + W ′ ) I +
12 ¯ hW ′′ σ . (C.54)16etters in High Energy Physics LHEP xx, xxx, 2018Let us here consider the superpotential W ( x ) defined as W = ω ˆ q + g ˆ q ⇒ W ′ = ω ˆ q + g ˆ q W ′′ = ω + g ˆ q , (C.55)where g is a perturbation. If g is zero, the Hamiltonian ˆH re-duces to the supersymmetric harmonic oscillator Hamiltonian.Therefore, when g is small, the Hamiltonian ˆH can be writtenas ˆH = ˆH + ˆH ′ , (C.56)where ˆH is the Hamiltonian of the unperturbed supersymmet-ric harmonic oscillator and ˆH ′ is the perturbation: ˆH ′ = ( g ˆ q + ω g ˆ q ) I + hg ˆ q σ . (C.57)As shown in Section 5, for the supersymmetric harmonic os-cillator, except the ground energy state, all the energy levelsdegenerate into two energy states. Therefore, if we have an en-ergy level degenerates to the two energy states Ψ n = (cid:18) ψ an (cid:19) and Ψ n = (cid:18) ψ bn (cid:19) , we can write h ψ n | ˆH ′ | ψ m i = D (cid:18) ψ an ψ bn (cid:19) (cid:12)(cid:12)(cid:12) f ( ˆ q ) I + f ( ˆ q ) σ (cid:12)(cid:12)(cid:12) (cid:18) ψ am ψ bm (cid:19) E = (cid:18) ψ (cid:19) , the previousequation reduces to h ψ | ˆH ′ | ψ m i = h ψ | f ( ˆ q ) − f ( ˆ q ) | ψ m i = ˆH ′ on the wave function Ψ ( ) isonly due to the component:ˆ H ′ = g q + ω g ˆ q − hg ˆ q . (C.60)Actually, this allows us to compute the first and second correc-tions to the ground-state energy for this system just using theHamiltonian ˆ H ′ . As well, we can use the nondegenerate time-independent perturbation theory to calculate the corrections tothe ground-state energy of our harmonic oscillator system evenif it has degenerate energy levels.Thus, taking into account the supersymmetric harmonic os-cillator ground-state wave function in the form (132) and theHamiltonian (C.60), then, we can now use the formula (C.53) tocompute the first-order correction to the energy of the groundstate as follows: E ( ) = h ψ | ˆ H ′ | ψ ( ) i = h ψ ( ) | ( g ˆ q + ω g ˆ q ) I + hq ˆ q σ | ψ ( ) i = Z + ∞ − ∞ r ωπ ¯ h e − ω ˆ q h (cid:18) g ˆ q + ω g ˆ q − hg ˆ q (cid:19) dq = g r ωπ ¯ h Z + ∞ − ∞ ˆ q e − ω ˆ q h dq + ω g r ωπ ¯ h Z + ∞ − ∞ ˆ q e − ω ˆ q h dq − hg r ωπ ¯ h Z + ∞ − ∞ ˆ qe − ω ˆ q h dq . (C.61) Both the second term and the third term of the previous inte-gration vanish since the integral functions ˆ q exp ( − ω ˆ q /¯ h ) andˆ q exp ( − ω ˆ q /¯ h ) are odd functions and there integrations from − ∞ to + ∞ are equal to zero. Therefore the previous integrationreduces to E ( ) = g r ωπ ¯ h Z + ∞ − ∞ ˆ q e − ω ˆ q h dq = g r ωπ ¯ h Z + ∞ e − ω ˆ q h ˆ q dq = q r ωπ ¯ h × s π ¯ h ω =
278 ¯ h ω g ≃ O ( g ) . (C.62)The last equation implies that the first-order correction to theground-state energy reduces to zero. Notice that to solve theintegration which appeared in the last equation, we have usedthe integral formula Z ∞ x n exp ( − ax ) dx = · · · · ( n − ) n + r π a n + . (C.63) Appendix D. THE SECOND-ORDER COR-RECTION
As described in [38, 39, 40, 41, 42], the second-order correctionto the energy in the time-independent nondegenerate pertur-bation theory is given by the form E ( ) = ∑ m = n |h ψ ( ) n | ˆ H ′ | ψ ( ) m i| E ( ) n − E ( ) m . (D.64)Furthermore, the harmonic oscillator wave function is orthog-onal polynomial, and the harmonic oscillator wave function ofdegree n can be obtained by the following recursion formula: ψ ( ) n ( ˆ q ) = √ n n ! (cid:16) ωπ ¯ h (cid:17) e − ω ˆ q h ˆH n (cid:18)r ω ¯ h ˆ q (cid:19) , (D.65)where ˆH n is the Hermite Polynomial of degree n . Also, it ishelpful to consider the recursion relation of the Hermite Poly-nomials: ˆ q ˆH n = ˆH n + + n ˆH n − . (D.66)Moreover, for two wave functions ψ n and ψ m , we have the re-lation h ψ n | ψ m i = δ nm . (D.67)Eqs. (D.65, D.66, and D.67) imply that h ψ | q | ψ m i = C δ ( m + ) + C n δ ( m − ) = C n δ m . (D.68)This argument leads us to know that all the terms of the sum-mation with m > m ≤ m = ψ ( ) = ( ωπ ¯ h ) e − ω q h , and ψ ( ) = ( ω π ¯ h ) qe − ω q h . (D.69)17etters in High Energy Physics LHEP xx, xxx, 2018Then, |h ψ ( ) | ω g ˆ q − hg ˆ q | ψ ( ) i| E ( ) − E ( ) = (cid:12)(cid:12)(cid:12) ω g h ψ ( ) | ˆ q | ψ ( ) i− hg h ψ ( ) | ˆ q | ψ ( ) i (cid:12)(cid:12)(cid:12) − ¯ h ω = − h ω (cid:20) ω ˆ g q ω π ¯ h R ∞ ˆ q e − ω ˆ q h dq − hg q ω π ¯ h R ∞ ˆ q e − ω ˆ q h dq (cid:21) = − g ¯ h ω s h ω − s h ω = −
98 ¯ h ω g . (D.70)Similarly, we can compute the second term with m =
2. Thesecond excited state wave functions is given by the form ψ ( ) = (cid:16) ω π ¯ h (cid:17) (cid:18) ω ¯ h ˆ q − (cid:19) e − ω q h . (D.71)Then, |h ψ ( ) | g ˆ q | ψ ( ) i| E ( ) − E ( ) = (cid:12)(cid:12)(cid:12)D ( ωπ ¯ h ) e − ω ˆ q h (cid:12)(cid:12)(cid:12) g ˆ q (cid:12)(cid:12)(cid:12) ( ω π ¯ h ) ( ω ¯ h ˆ q − ) e − ω ˆ q h E(cid:12)(cid:12)(cid:12) − h ω = − g π ¯ h (cid:20)(cid:12)(cid:12)(cid:12) ω ¯ h R + ∞ ˆ q e − ω ˆ q h dq − R + ∞ ˆ q e − ω ˆ q h dq (cid:12)(cid:12)(cid:12) (cid:21) = − g π ¯ h (cid:18) q π ¯ h ω − q π ¯ h ω (cid:19) = − h ω g . (D.72)Furthermore, the same way we can compute the second termwith m =
3. The third excited state wave functions take theform ψ ( ) = ( ω π ¯ h ) ( ω ¯ h ˆ q − q ) e − ω ˆ q h . (D.73)Then, |h ψ ( ) | ω g ˆ q | ψ ( ) i| E ( ) − E ( ) = (cid:12)(cid:12)(cid:12)D ( ωπ ¯ h ) e − ω q h (cid:12)(cid:12)(cid:12) ω gq (cid:12)(cid:12)(cid:12) ( ω π ¯ h ) ( ω ¯ h ˆ q − q ) e − ω ˆ q h E(cid:12)(cid:12)(cid:12) − h ω = − g ω π ¯ h (cid:20)(cid:12)(cid:12)(cid:12)D e − ω ˆ q h (cid:12)(cid:12)(cid:12) ˆ q (cid:12)(cid:12)(cid:12) ( ω ¯ h ˆ q − q ) e − ω ˆ q h E(cid:12)(cid:12)(cid:12) (cid:21) = − g ω π ¯ h (cid:20)(cid:12)(cid:12)(cid:12) ω ¯ h R + ∞ − ∞ ˆ q e − ω ˆ q h dq − R + ∞ − ∞ ˆ q e − ω ˆ q h dq (cid:12)(cid:12)(cid:12) (cid:21) = − g ω π ¯ h (cid:20)(cid:12)(cid:12)(cid:12) ω ¯ h R + ∞ ˆ q e − ω ˆ q h dq − R + ∞ ˆ q e − ω ˆ q h dq (cid:12)(cid:12)(cid:12) (cid:21) = − g ω π ¯ h (cid:20)(cid:12)(cid:12)(cid:12) q π ¯ h ω − q π ¯ h ω (cid:12)(cid:12)(cid:12) (cid:21) = − g ω π ¯ h (cid:18) q π ¯ h ω (cid:19) = −
94 ¯ h ω g . (D.74)Also, similarly, we can compute the term with m =
4. Thefourth excited state wave function is ψ ( ) = (cid:16) ω π ¯ h (cid:17) (cid:18) ω ¯ h ˆ q − ω ¯ h ˆ q + (cid:19) e − ω ˆ q h . (D.75) Then, |h ψ ( ) | g ˆ q | ψ ( ) i| E ( ) − E ( ) = (cid:12)(cid:12)(cid:12)D ( ωπ ¯ h ) e − ω ˆ q h (cid:12)(cid:12)(cid:12) g ˆ q (cid:12)(cid:12)(cid:12) ( ω π ¯ h ) (cid:16) ω h ˆ q − ω ¯ h ˆ q + (cid:17) e − ω ˆ q h E(cid:12)(cid:12)(cid:12) − h ω = − g π ¯ h (cid:20)(cid:12)(cid:12)(cid:12)D e − ω ˆ q h (cid:12)(cid:12)(cid:12) ˆ q (cid:12)(cid:12)(cid:12) (cid:16) ω ¯ h ˆ q − ω ¯ h ˆ q + (cid:17) e − ω ˆ q h E(cid:12)(cid:12)(cid:12) (cid:21) = − g π ¯ h (cid:20)(cid:12)(cid:12)(cid:12) ω ¯ h R + ∞ − ∞ ˆ q e − ω ˆ q h dq − ω ¯ h R + ∞ − ∞ ˆ q e − ω ˆ q h dq + R + ∞ − ∞ ˆ q e − ω ˆ q h dq (cid:12)(cid:12)(cid:12) (cid:21) = − g π ¯ h (cid:20)(cid:12)(cid:12)(cid:12) ω ¯ h R + ∞ ˆ q e − ω ˆ q h dq − ω ¯ h R + ∞ ˆ q e − ω ˆ q h dq + R + ∞ ˆ q e − ω ˆ q h dq (cid:12)(cid:12)(cid:12) (cid:21) = − g π ¯ h (cid:20)(cid:12)(cid:12)(cid:12) q π ¯ h ω − q π ¯ h ω + q π ¯ h ω (cid:12)(cid:12)(cid:12) (cid:21) = − g π ¯ h (cid:20)(cid:12)(cid:12)(cid:12) q π ¯ h ω (cid:12)(cid:12)(cid:12) (cid:21) = − h ω g . (D.76)Using Eqs. (D.70, D.72, D.74, and D.76), we are able to computethe second-order correction to the ground-state energy for thesupersymmetric harmonic oscillator as follows: E ( ) = |h ψ ( ) | − hg ˆ q | ψ ( ) i| E ( ) − E ( ) + |h ψ ( ) | g ˆ q | ψ ( ) i| E ( ) − E ( ) + |h ψ ( ) | ω g ˆ q | ψ ( ) i| E ( ) − E ( ) + |h ψ ( ) | g ˆ q | ψ ( ) i| E ( ) − E ( ) = −
98 ¯ h ω g − h ω g −
94 ¯ h ω g − h ω g = −
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278 ¯ h ω −
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