Normalization of the Ground State of the Supersymmetric Harmonic

Home , Ground State (Angel), Perturbation theory

Letters in High Energy Physics LHEP-160, 2020

Normalization of the Ground State of the Supersymmetric Harmonic 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 been no unbroken supersymmetry observed in nature, and if nature is described by supersymmetry, it must be broken. Supersymmetry may be broken spontaneously at any order of perturbation theory or dynamically due to nonperturbative effects. To examine the methods of supersymmetry breaking, special attention is given to discuss the normalization of the ground state of the supersymmetric harmonic oscillator. This study explains that perturbation theory gives incorrect results for both the ground-state wave function and the energy spectrum and it fails to give an explanation to the supersymmetry breaking.

Keywords: supersymmetry breaking, perturbation theory, This work is structured as follows. After a brief intro- nonperturbative effects duction, we study the problem of a harmonic oscillator with DOI: 10.31526/LHEP.2020.160 fermionic as well as bosonic fields using the usual quantum mechanical operators method. Then, we consider supersymmetric classical mechanics and study generalized classical Pois- 1. INTRODUCTION son brackets to Dirac brackets and quantization rules in order to introduce the superalgebra. Furthermore, the formal- Supersymmetry, often abbreviated SUSY, was first introduced ism of supersymmetry in quantum mechanics is introduced. in 1971 by Gelfand and Likhtman, Raymond, and Neveu and Subsequently, we give a general background in the concepts Schwartz, and later, it was rediscovered by other groups [1, 2]. and methods of supersymmetric quantum mechanics. This fol- Supersymmetric quantum mechanics was developed by Witten lowed by providing a more specific study of the ground state in 1982 [3], as a toy model to test the breaking of supersymme- of the supersymmetric harmonic oscillator. Finally, we summa- try. In general, supersymmetry plays a main role in the mod- rize our arguments and to draw our conclusions. ern understanding of theoretical physics, in particular, quantum field theories, gravity theories, and string theories [4]. Supersymmetry is a symmetry that connects particles 2. HARMONIC OSCILLATOR of integer spins (bosons) and particles of half-integer spins (fermions) [5]. In supersymmetry, it is possible to introduce op- Supersymmetric theories necessarily include both bosons and erators that change bosons which are commuting variables into fermions, so a system that includes both bosonic and fermionic fermions which are anticommuting variables and vice versa [6]. degrees of freedom is considered here. The supersymmetry al- Supersymmetric theories are theories which are invariant un- gebra is a mathematical formalism for describing the relation der those transformations [7]. The supersymmetry algebra in- between bosons and fermions, and we will discuss this later in volves commutators as well as anticommutators [8]. Section 4. In quantum mechanics, bosons are integer spin parti- In this study, the main mathematical structure which in- cles which obey Bose-Einstein statistics, and the bosonic opera- volves supersymmetric quantum mechanics is derived starting tors satisfy the usual commutation relations, while fermions are with explaining the basic idea of supersymmetry and followed half-integer spin particles characterized by Fermi-Dirac statis- by introducing the necessary framework to make a supersym- tics, obey the Pauli exclusion principle, and can be described metric theory. We derive the basic formulation of supersym- by Grassmann variables, which are anticommuting objects; for metric quantum mechanics starting with introducing the con- review, see [9, 10, 11]. cepts of supercharges and superalgebra. We show that if there The simplest system that consists of a combination of is a supersymmetric state, it is the zero-energy ground-state. If bosonic and fermionic fields is the one-dimensional harmonic such a state exists, the supersymmetry is unbroken; otherwise, oscillator. There are many excellent reviews on this subject; we it is broken. So far, there has been no unbroken supersymmetry refer the reader to them for more and complete expositions observed in nature, and if nature is described by supersymme- [2, 12, 13, 14]. We begin by considering the Lagrangian formu- try, it must be broken. In fact, supersymmetry may be broken lation for the bosonic and fermionic harmonic oscillator: spontaneously at any order of perturbation theory or dynam- 1 2 αβ 2 ically due to nonperturbative effects. To examine the methods L = q˙ + iψαδ ψ˙ β + F + ω (iψ1ψ2 + qF) , (1) of supersymmetry breaking, we study the normalization of the 2 ground state of the supersymmetric harmonic oscillator and where q(t) and F(t) are real bosonic fields, ψα(t) (with α ∈ calculate the corrections to the ground-state energy using per- {1, 2}) are real fermionic fields, ω is a real parameter, δαβ is the turbation theory. We found out that no perturbation effect can Kronecker delta, and we used the Einstein summation conven- lead to supersymmetry breaking and it must be due to the non- tion. perturbative effects. It is straightforward, using the classical equation of motion for F + ωq = 0, to eliminate F from Eq. (1) and obtain the equiv-

1 Letters in High Energy Physics LHEP-160, 2020 alent Lagrangian: that the bosonic ladder operators aˆ and aˆ† satisfy the following commutation relations: 1 L = q˙2 + iψ δαβψ˙ − ω2q2 + iωψ ψ . (2) 2 α β 1 2 [aˆ, aˆ†] = 1, [aˆ, aˆ] = [aˆ†, aˆ†] = 0 . (11) Then, using the Legendre transformation from the variables However, on the other hand, the lowering (annihilation) and {q, q˙, ψ, ψ˙} to {q, p, ψ, π}, we obtain raising (creation) operators for fermions, bˆ† and bˆ, respectively, are defined as αβ β 1 2 2 2 H = qp˙ + ψαδ π − L = p + ω q − iωψ1ψ2 , (3) r r 2 ˆ 1 ˆ† 1 b = ψˆ1 − iψˆ2 , b = ψˆ1 + iψˆ2 . (12) α 2¯h 2¯h where π are the momenta conjugate to ψα. Later in this study, we shall see that the general proper- Also, as will be shown in Section 4, the fermionic operators ψα ties of supersymmetry have to be clear if the Hamiltonian is and ψβ satisfy the canonical quantum anticommutation con- 1 built as a larger Hamiltonian consisting of two components, dition {ψα, ψβ} = h¯ δαβ . Therefore, it can be seen that the one representing the bosonic field and the other representing fermionic ladder operators bˆ and bˆ† satisfy the following an- the fermionic field. Leaving this aside for the moment, to de- ticommutation relations: rive the supersymmetric Hamiltonian, let us define the bosonic valuable qˆ and the bosonic momentum pˆ, respectively, as fol- {bˆ, bˆ†} = 1, {bˆ, bˆ} = {bˆ†, bˆ†} = 0 . (13) lows: It is possible using Eq. (10) to write the bosonic position and I I ∂ momentum variables in terms of the bosonic ladder operators qˆ = q 2, pˆ = −ih¯ 2 . (4) ∂q as follows: r r On the other hand, the fermionic variables ψˆ and ψˆ and the 2ω h¯ 1 2 aˆ + aˆ† = qˆ ⇒ qˆ = (aˆ† + aˆ), fermionic momenta πα, respectively, have to be defined as h¯ 2ω r r r r 2 h¯ ω h¯ h¯ aˆ − aˆ† = ipˆ ⇒ pˆ = i (aˆ† − aˆ) . ψˆ1 = σ1, ψˆ2 = σ2, h¯ ω 2 2 2 (14) α ∂L i αβ π = = − δ ψβ, α, β ∈ {1, 2} . (5) ∂ψ˙ α 2 Similarly, using Eq. (12), we can write the fermionic variables ψ1 and ψ2 in terms of the fermionic ladder operators as follows: Hence, using Eqs. (4, 5), we can rewrite the previous Hamilto- r r nian (3) in the following form: 2 h¯ bˆ† + bˆ = ψˆ ⇒ ψˆ = (bˆ + bˆ†), h¯ 1 1 2 1 d2 1 r r Hˆ = −h¯ 2 + ω2q2 I + h¯ ω σ , (6) 2 h¯ 2 dq2 2 2 3 bˆ† − bˆ = i ψˆ ⇒ ψˆ = i (bˆ − bˆ†) . h¯ 2 2 2 I (15) where 2 is the 2 × 2 identity matrix and σj are the Pauli matrices; that is, Actually, we derived Eqs. (14, 15) here; however, it will be used later. At this stage, let us define the following two Hermitian I 1 0 0 1 0 −i 1 0 2 = , σ1 = , σ2 = , σ3 = . (7) 0 1 1 0 i 0 0 −1 operators, the bosonic number operator Nˆ B, and the fermionic number operator Nˆ F: We want to obtain solutions of the time-independent Schrodinger¨ H = E 1 equation Ψ Ψ; that is, Nˆ ≡ aˆ†aˆ = ω2qˆ2 − h¯ ω + pˆ2 , B 2¯hω 2 1 2 d 2 2 I 1 † −i −h¯ + ω q 2 + h¯ ω σ3 = EΨ , (8) Nˆ ≡ bˆ bˆ = ψˆ ψˆ . (16) 2 dq2 2 F h¯ 1 2 with the wave function Ψ(x) constraint to satisfy appropriate Using Eq. (16), we can rewrite the Hamiltonian (6) in the oper- boundary conditions. ator formalism As we mentioned, the supersymmetric Hamiltonian can be 1 1 Hˆ = h¯ ω Nˆ + + h¯ ω Nˆ − . (17) written as a summation of two terms representing the bosonic B 2 F 2 and the fermionic components of the Hamiltonian This means that the energy eigenstates can be labeled with the H = HB(p, q) + HF(ψ) . (9) eigenvalues of nB and nF. The action of the ladder operators, aˆ, aˆ†, bˆ, and bˆ†, upon an To this end, we define the lowering (annihilation) and raising energy eigenstate |n , n i is listed as follows: † B F (creation) operators for bosons, aˆ and aˆ, respectively, they are √ √ † aˆ|nB, nFi = nB|nB − 1, nFi, aˆ |nB, nFi = nB + 1|nB + 1, nFi, r r √ √ 1 1 ˆ ˆ† aˆ = (ωqˆ + ipˆ) , aˆ† = (ωqˆ − ipˆ) . (10) b|nB, nFi = nF|nB, nF − 1i, b |nB, nFi = nF + 1|nB, nF + 1i . 2¯hω 2¯hω (18) Later, in Section 4, we will show that the bosonic operators, qˆ and pˆ, satisfy the usual canonical quantum commutation con- 1The derivation of the fermionic anticommutators is a bit subtle and will be dition [qˆ, pˆ] = ih¯ . Taking this fact into account, we can find discussed in the next section.

2 Letters in High Energy Physics LHEP-160, 2020

It is straightforward from Eq. (18) that the associated bosonic 3. THE CONSTRAINED HAMILTONIAN ˆ ˆ number operator NB and fermionic number operator NF obey FORMALISM

Nˆ B|nB, nFi = nB|nB, nFi, Nˆ F|nB, nFi = nF|nB, nFi . (19) Before discussing the supersymmetry algebra in the next section, we have to discuss the constrained Hamiltonian formal- Under Eq. (19), the energy spectrum of the Hamiltonian ism in this section. This formalism is needed to get the cor- (17) is rect fermionic anticommutators due to the complication of con- 1 1 straints. Good discussions of the constrained Hamiltonian sys- En = h¯ ω n + + h¯ ω n − . (20) B 2 F 2 tems are presented in [15, 16, 17, 18, 19, 20]. For instance, let us recall the Hamiltonian (3) The form of Eq. (20) implies that the Hamiltonian H is symmet- † ric under the interchange of aˆ and aˆ and antisymmetric under 1 ˆ ˆ† H = p2 + ω2q2 − iωψ ψ . (25) the interchange of b and b . 2 1 2 The most important observation from Eq. (20) is that En remains invariant under a simultaneous annihilation of one bo- A constrained Hamiltonian is one in which the momenta and son (nB → nB − 1) and creation of one fermion (nF → nF + 1) or positions are related by some constraints. In general, M con- vice versa. This is one of the simplest examples of a symmetry straints between the canonical variables can be written as called ”supersymmetry” (SUSY) and the corresponding energy spectrum reads φm(q, p, ψ, π) = 0, m = 1, . . . , M. (26) En = h¯ ω (nB + nF) . (21) These are called primary constraints. Secondary constraints are Moreover, we still have two more comments before ending additional constraints relating momenta and positions which this section: can arise from the requirement that the primary constraints are time-independent; i.e., φ˙ = 0. Fortunately, we do not have to (i) Consider that the fermionic vacuum state |0i is defined m deal with such constraints in this case. In principle, there can by also be tertiary constraints arising from the equation of motion bˆ|0i ≡ 0. (22) of the secondary constraints and so on. Then, we define a fermionic one-particle state |1i by The Hamiltonian (25) is an example of constrained Hamil- tonians. In the previous section, we defined the fermionic mo- |1i ≡ bˆ†|0i . (23) menta πα in Eq. (5) using the fermionic equation of motion followed from the usual Euler-Lagrange equations with the La- It is now easy to use the anticommutation relations (13) grangian (2) which is equivalent to this Hamiltonian. In fact, to see that there are no other states, since operating on this definition gives us the relation between the fermionic mo- |1i with bˆ or bˆ† gives either the already known state |0i menta πα and the canonical coordinates ψα: or nothing: ∂L i α = = − αβ ∈ { } ˆ ˆ ˆ† ˆ† ˆ ˆ† ˆ π δ ψβ, α, β 1, 2 . (27) b|1i = bb |0i = (1 − b b)|0i = |0i − b b|0i = |0i − 0 = |0i, ∂ψ˙ α 2 ˆ†| i = ˆ† ˆ†| i = | i = b 1 b b 0 0 0 0 . (24) In light of this equation, we have the primary constraint

So, the subspace spanned by |0i and |1i is closed un- i ˆ ˆ† φα = πα + δαβψ = 0 . (28) der the action of b and b , and, therefore, it is closed 2 β under the action of any product of these operators. We can show this directly using the two facts that the op- Such constraints mean that the way we write the Hamiltonian erator bˆbˆ† can be expressed as the linear combination is ambiguous since we can exchange position and momentum 2 1 − bˆ†bˆ and bˆ2 = bˆ† = 0. So, any product of three or variables which change the equations of motion one gets. For more bˆ, bˆ† can be shortened by using either bˆ2 = 0, or example, using Eq. (27), one can rewrite the Hamiltonian (25) 2 2 in terms of the fermionic momenta as follows: bˆ† = 0. For example, bˆ†bˆbˆ† = bˆ† − bˆbˆ† = bˆ† and bˆbˆ†bˆ = ˆ − ˆ† ˆ2 = ˆ b b b b. So, it seems to be clear that all products of 1 2 2 2 1 2 ˆ ˆ† H = p + ω q + 4iωπ π . (29) b and b can always be reduced to linear combinations 2 of the following four operators 1, bˆ, bˆ†, and bˆ†bˆ. Based on this argument, there are only two possible fermionic In the next subsection, the general properties of Poisson brack- eigenstates and hence two possible eigenvalues of the ets and its relation with the Hamiltonian formalism in classi- fermionic number operator; they are nF = {0, 1}. cal mechanics are briefly discussed. Later in the same subsection, the problem caused by the constrained Hamiltonians is (ii) The ground eigenstate has a vanishing energy eigen- described. Next, Subsection 3.2 shows how to deal with this value (when nB = nF = 0). In this case, the ground eigen- problem using the constraints. state is not degenerate and we say that supersymmetry is unbroken. This zero-energy eigenvalue arises because 3.1. Poisson Brackets and First-Class Constraints of the cancellation between the bosonic and fermionic The Poisson bracket of two arbitrary functions F(qˆ, pˆ, ψˆ, πˆ ) and contributions to the supersymmetric ground-state en- G(qˆ, pˆ, ψˆ, πˆ ) is defined as ergy since the ground-state energy for the bosonic and 1 1 fermionic oscillators has the values 2 h¯ ω and − 2 h¯ ω, re- ∂F ∂G ∂F ∂G ∂F ∂G ∂F ∂G {F G} = − (−)eF − spectively. , P i i α α , (30) ∂qi ∂p ∂p ∂qi ∂ψα ∂π ∂π ∂ψα 3 Letters in High Energy Physics LHEP-160, 2020 where eF is 0 if F is Grassmann even and 1 if F is Grass- Unfortunately, this equation is inconsistent with the equation of mann odd2. Actually, in our argument here, only the case of the motion followed from the Lagrangian formalism. Even worst, odd Grassmann variables is considered. In general, the Poisson the equation of motion for πα followed from the expected Pois- brackets have the following properties: son bracket equation is also inconsistent with the equation derived from the Lagrangian formalism. For example, using the (i) A constraint F is said to be a first-class constraint if its Poisson bracket, we have Poisson bracket with all the other constraints vanishes: ∂H A π˙ 1 = {π1, H} = − = iωψ . (39) {φ , F}P = 0 , (31) P 2 ψ1

where A runs over all the constraints. If a constraint is But if we use Eq. (27) to find an expression to the equation of not a first class, it is a second class. motion for ψ˙1 and then substitute into Eq. (39), that gives us (ii) A Poisson bracket of two functions F and G is antisym- ˙ = i ˙ 1 = − metric: ψ1 2 π 2ωψ2 . (40) {F, G} = −{G, F} . (32) P P This equation is different from the Lagrangian equations of (iii) For any three functions A, B, and C, the Poisson bracket motion and inconsistent with (38) (unless everything is trivial is linear in both entries: which is also not great). Moreover, using the Poisson bracket with the Hamiltonian (29), we can find the following set of

{A, B + C}P = {A, B}P + {A, C}P . (33) equations of motion for the fermionic positions and momenta: ∂H (iv) A Poisson bracket for any three functions satisﬁes the ψ˙ = {ψ , H} = − = −4iωπ2 = −2ωψ , 1 1 P ∂π1 1 Leibniz rule: ∂H ψ˙ = {ψ , H} = − = −4iωπ1 = −2ωψ , e e 2 2 P 2 2 {A, BC}P = {A, B}PC + (−1) A B B{A, C}P, (34) ∂π α α ∂H π˙ = {π , H}P = − = 0 . (41) where eF is the Grassmann parity of F which is 0 if F is ∂ψα Grassmann even and 1 if F is Grassmann odd. The above equations are also slightly different from the La- (v) A Poisson bracket for any three functions obeys the Ja- grangian equations of motion and inconsistent with the equa- cobi identity: tions followed from the Hamiltonian (25). In brief, there is an inconsistency between the equations of { { } } + { { } } + { { } } = A, B, C P P B, C, A P P C, A, B P P 0 . motion derived from the Hamiltonian formalism and the equa- (35) tions derived from the Lagrangian formalism. The reason for (vi) The time rate of change of any arbitrary function this inconsistency is that we have a constrained Hamiltonian. i α This problem is solved in the next subsection by imposing the F(qi, p , ψα, π ), which has no explicit time dependence, is given by its Poisson bracket with the Hamiltonian constraints onto the Hamiltonian.

F˙ = {F, H}P . (36) 3.2. The Constraints and the Equations of Motion In the Lagrangian formalism, one can impose the constraints In the Hamiltonian formalism of classical mechanics, the from the beginning by introducing Lagrange multipliers which Hamilton equations of motion have equivalent expressions in enforce the constraints. So, Hamilton’s equations of motion, in terms of the Poisson bracket. Using the last property (vi) of the the presence of constraints, can be derived from the extended Poisson brackets, one can easily see that action principle:

∂H ∂H Z = = { } i = − = { i } S = q pi + ˙ α − H − A dt q˙i i qi, H P, p˙ q , H P, ˙i ψαπ φ λA , (42) ∂p ∂qi ∂H ∂H ψ˙ = − = {ψ , H} , π˙ α = − = {πα, H} . (37) where φA are our constraints, λ are Lagrange multipliers, and α ∂πα α P ∂ψ P A α A runs over the constraints. Using the extended action prin- However, the problem with the constrained system is that the ciple, we can write out the following two expressions of the equations of motion derived using Eq. (37) are not always con- extended Lagrangian and the extended Hamiltonian, respec- sistent with those followed from the Lagrangian formalism. tively: For example, if we consider the Hamiltonian (25), the equation L = q˙ pi + ψ˙ πα − H − φAλ , (43) of motion for ψα followed from the expected Poisson bracket i α A i α A equation has to be H = q˙i p + ψ˙ απ − φ λA − L . (44) ∂H ˙ = { } = − = Indeed, if the Lagrangian is independent of one coordinate, say ψα ψα, H P α 0 . (38) ∂π qi, then, we call this coordinate an ignorable coordinate. How- ever, we still need to solve the Lagrangian for this coordinate to ﬁnd the corresponding equation of motion. Since the mo- 2Grassmann-even variables refer to bosons, while Grassmann-odd variables refer to fermions. mentum corresponding to this coordinate, may still enter the

4 Letters in High Energy Physics LHEP-160, 2020

Lagrangian and affect the evolution of other coordinates, there- On the other hand, using Eqs. in (46) and considering the fore, using the Euler-Lagrange equation, we have Hamiltonian (25) and the constraints (48), we ﬁnd that

d ∂L ∂L 1 − = 0 . (45) 1 ∂H ∂φ i dt ∂q˙ ∂q π˙ = − − λ1 = iωψ2 − λ1, i i ∂ψ1 ψ1 2 2 In general, we define the generalized momentum as pi = 2 ∂H ∂φ i π˙ = − − λ2 = −iωψ1 − λ2 . (50) ∂L/∂q˙i. Now, if there is no explicit dependence of the La- ∂ψ2 ψ2 2 grangian L on generalized coordinate qi, then ∂L/∂qi = 0. Furthermore, using Eq. (27), we get Thus, Euler-Lagrange equation becomes d/dt(∂L/∂q˙i) = 0 ⇒ dpi/dt = 0. Hence, a momentum pi is conserved when the La- 1 i ˙1 i grangian L is independent of a coordinate qi. This means that π = − ψ1 ⇒ π = − ψ˙1, if the Lagrangian is independent of a certain coordinate qi, it 2 2 i 2 i ˙2 i must be also independent of its corresponding momentum p . π = − ψ2 ⇒ π = − ψ˙2 . (51) Furthermore, by extremizing the Hamiltonian (44), we ob- 2 2 tain the following generalized Hamilton’s equations of motion: Eqs. (49, 50, and 51) imply the following set of equations of A motion for the fermionic coordinates ψα: ∂H ∂φ A q˙ = + λ = {q , H + φ λ }P , i ∂pi ∂pi A i A ψ˙1 = −ωψ2, ψ˙2 = ωψ1 . (52) A i ∂H ∂φ i A p˙ = − − λA = {p , H + φ λA}P , Fortunately, these equations are the same equations of motion ∂qi ∂qi followed from the Lagrangian formalism. ∂H ∂φA ˙ = − − = { + A } For now, let us use the Poisson bracket, the Hamiltonian ψα α α λA ψα, H φ λA P , ∂π ∂π (25), and the constraints (48), to check that A α ∂H ∂φ α A π˙ = − − λA = {π , H + φ λA}P , ˙ 1 1 1 ∂ψα ∂ψα φ = {φ , H + φ λ1}P ! φA = 0 . (46) ∂φ1 ∂(H + φ1λ ) ∂φ1 ∂(H + φ1λ ) = 1 − 1 ∂q ∂pi ∂pi ∂q For example, using the Poisson brackets, we can show that i i ! Eqs. in (46) imply that for any arbitrary function F = ∂φ1 ∂(H + φ1λ ) ∂φ1 ∂(H + φ1λ ) = − 1 + 1 F(q pi α) F˙ = {F H + A } α α i, , ψα, π , we can write , φ λA P: ∂ψα ∂π ∂π ∂ψα i α ∂F ∂qi ∂F ∂p ∂F ∂ψα ∂F ∂π i i ˙ = + + + = 0 + 0 − (λ1) − (1) − iωψ2 + λ1 F i α ∂qi ∂t ∂p ∂t ∂ψα ∂t ∂π ∂t 2 2 = i (−λ1 + ωψ2) = 0 . (53) ∂F ∂F ˙i ∂F ∂F α = q˙ + p + ψ˙α + π˙ ∂q i ∂pi ∂ψ ∂πα i α The reason that the last step is zero comes from Eqs. (49, 52) ∂F ∂H ∂φA ∂F ∂H ∂φA = − ˙ = ˙ 2 = + + − − since we have λ1 ψ1 ωψ2. Similarly, φ can be calculated i i λA i i i λA ∂qi ∂p ∂p ∂p ∂q ∂q using the Poisson bracket as follows: ∂F ∂H ∂φA ∂F ∂H ∂φA + − − + − − ˙ 2 2 2 α α λA α α λA φ = {φ , H + φ λ2}P ∂ψα ∂π ∂π ∂π ∂ψ ∂ψα ∂φ2 ∂(H + φ2λ ) ∂φ2 ∂(H + φ2λ ) ∂F ∂(H + φAλ ) ∂F ∂(H + φAλ ) = 2 − 2 = A − A i i i i i ∂qi ∂p ∂p ∂qi ∂qi ∂p ∂p ∂q ∂φ2 ∂(H + φ2λ ) ∂φ2 ∂(H + φ2λ ) ∂F ∂(H + φAλ ) ∂(H + φAλ ) ∂F = − 2 + 2 − A + A α α α α α ∂ψα ∂π ∂π ∂ψα ∂ψα ∂π ∂ψ ∂π i i ≡ {F, H + φAλ } . (47) = 0 + 0 − (λ ) − (2) iωψ + λ A P 2 2 1 2 2 At this stage, we need to confirm that Eqs. in (46) give us the = i (−λ2 − ωψ1) = 0 . (54) correct equations of motion using the constrained Hamiltonian (25). Since α, β ∈ {1, 2} in the primary constraint (28), then, we Also, the last step is zero because from Eqs. (49, 52), we have ˙ have the two constraints λ2 = −ψ2 = −ωψ1. Subsequently, Eqs. (53, 54) are zero if we impose the equations of motion (52). In other words, we can say i i φ1 = π1 + ψ , φ2 = π2 + ψ . (48) that the constraints are consistent with the equations of motion. 2 1 2 2 By using Eqs. in (46) and taking into count the Hamiltonian 3.3. Dirac Brackets and Second Class Constraints (29), and the constraints (48), we find that A Constraint F is said to be a second class constraint if it has nonzero Poisson brackets, and, therefore, it requires special ∂H ∂φ1 ˙ = − − = − treatment. Given a set of second class constraints, we can de- ψ1 1 1 λ1 λ1, ∂π π fine a matrix ∂H ∂φ2 { A B} = AB ψ˙ = − − λ = −λ . (49) φ , φ P C , (55) 2 ∂π2 π2 2 2

5 Letters in High Energy Physics LHEP-160, 2020

AB where we deﬁne the inverse of C , as CAB, such that Similarly, using Eq. (62), we ﬁnd that

AB A β β ! β β ! C CBC = δ . (56) ∂ψ ∂φ ∂ψ ∂φ ∂ψ ∂φ ∂ψ ∂φ C { β} = β − β − β + β ψβ, φ P i i β β ∂qi ∂p ∂p ∂qi ∂ψβ ∂π ∂π ∂ψβ The Dirac bracket provides a modiﬁcation to the Poisson brackets to ensure that the second class constants vanish: = −1 . (66)

A {φ , F}D = 0 . (57) In addition, using the constraints, definitions (47, 62), we can define a matrix F(q p) G(q p) The Dirac bracket of two arbitrary functions , and , αβ α β is defined as C = {φ , φ }P ∂φα ∂φβ ∂φα ∂φβ ∂φα ∂φβ ∂φα ∂φβ A B = i − i − α + α {F, G}D = {F, G}P − {F, φ }PCAB{φ , G}P . (58) ∂qi ∂p ∂p ∂qi ∂ψα ∂π ∂π ∂ψα α β α β Dirac bracket has the same properties of Poisson bracket, − ∂φ ∂φ + ∂φ ∂φ ∂ψ ∂πβ ∂πβ ∂ψ which we have listed in Subsection 3.1. In addition, we have to β β notice the following. = −iδαβ . (67)

(i) If φ˙ A = 0, then, the Dirac bracket of any constraint with Moreover, from the deﬁnition (3.3), we have the extended Hamiltonian is equivalent to its Poisson αβ αβ bracket: Cαβ = −C = iδ . (68)

A A As a result of substituting Eqs. (64, 65, 66, and 68) into Eq. (63), {F, H + φ λA}D = {F, H + φ λA}P . (59) the Dirac bracket of the two variables ψα and ψβ gives us (ii) In some cases, we can return to the original Hamilto- {ψ , ψ } = iδαβ . (69) nian (25) and forget about the constraints and the mo- α β D menta πβ at the cost of replacing the Poisson brackets with Dirac brackets. For example, if we consider the case 3.4. Dirac Bracket and the Equations of Motion φ˙ A = 0, then, we have In this subsection, we show that the Dirac brackets give the correct equations of motion for the Hamiltonian (27) using the ex- A A F˙ = {F, H + φ λA}D = {F, H}D + {F, φ }DλA = {F, H}D , pression (60) ψ˙ α = {ψα, H}D , (70)

A We check that where we used Eq. (57); {φ , F}D = 0. A ψ˙ = {ψ , H} = {ψ , −iωψ ψ } = −iω{ψ , ψ ψ } (iii) If {φ , p}P = 0, then, using Eq. (58), we can ﬁnd that 1 1 D 1 1 2 D 1 1 2 D = −iω ({ψ ψ } ψ + ψ {ψ ψ } ) = −iω −iδ ψ + A B 1, 1 D 2 1 1, 2 D αβ 2 0 {q, p}D = {q, p}P − {q, φ }PCAB{φ , p} = − = − ∂q ∂p ∂q ∂p ωδαβψ2 ωψ2 , (71) = {q, p}P = − = 1 . (61) ∂q ∂p ∂p ∂q and similarly

˙ We are now in a position to calculate {ψα, ψβ}D for the ψ2 = {ψ2, H}D = {ψ2, −iωψ1ψ2}D = −iω{ψ2, ψ1ψ2}D Hamiltonian (25) with the constraints (28). That {ψα, ψβ}D in- = −iω ({ψ2, ψ1}Dψ2 + ψ1{ψ2, ψ2}D) = −iω 0 + ψ1(−iδαβ) volves a primary constraint φβ followed from the following relation which is obtained from the Lagrangian formalism: = −ωψ1δαβ = ωδαβψ1 = ωψ1 . (72)

i These equations are the same equations of motion which we φβ = πβ + δαβψ . (62) 2 α have obtained in Subsection 3.2 using the constrained Hamil- ton’s equations of motion (46). Using the deﬁnition of Dirac bracket, we get

α β 3.5. The Dirac Bracket Superalgebra {ψα, ψ } = {ψα, ψ } − {ψα, φ } C {φ , ψ } . (63) β D β P P αβ β P In this subsection, we investigate the supersymmetry of our Hamiltonian (25). For this purpose, we need to introduce some Then, since the variables ψα and ψβ are independent, their Pois- son bracket vanishes; i.e., operators Qi known as supercharges, which basically can be obtained from Noether’s¨ theorem arising from a symmetry of α the Lagrangian which exchanges bosons and fermions. In the {ψα, ψβ}P = {ψβ, ψ }P = 0 . (64) next section, we will look at the action of the supercharge oper- Furthermore, using Eq. (47), we ﬁnd that ators. For instance, the supercharge operators in the case of the system under discussion here, can be written as follows: ∂ψ ∂φα ∂ψ ∂φα ∂ψ ∂φα ∂ψ ∂φα { α} = α − α − α + α ψα, φ P i i α α Q = pψ + ωqψ ∂qi ∂p ∂p ∂qi ∂ψα ∂π ∂π ∂ψα βγ 1 1 2 Qα = pψα + ωqeαβδ ψγ ⇒ (73) = −1 . (65) Q2 = pψ2 − ωqψ1 ,

6 Letters in High Energy Physics LHEP-160, 2020 where again we suppose that {α, β} can only take the values So far, we have restricted ourselves to study classical sys- {1, 2}, and eαβ is the two index Levi-Civita symbol which de- tems. In Subsection 3.5, we have investigated the supersym- ﬁned as metry of the classical harmonic oscillator. Now, we are going  +1 if (α, β) is (1, 2), to quantize the theory to study the supersymmetry of quan-  eαβ = −1 if (α, β) is (2, 1), (74) tum systems. In a quantum mechanical supersymmetric sys-   0 if α = β . tem, the supercharges Qi together with the Hamiltonian H form a so-called superalgebra. The recipe for quantizing the Hamil- The system to be supersymmetric, the supercharges (73) to- tonian in a situation where we have second class constraints is gether with the Hamiltonian (27), and the constraints (25) must to replace the Dirac brackets with either commutator brackets satisfy the classical Dirac bracket superalgebra deﬁned by for bosonic variables or anticommutator brackets for fermionic variables multiplied with the factor ih¯ , so we have {Qα, Qβ}D = −2iδαβH,

{Qα, H}D = 0 . (75) {q, p}D = 1 ⇒ [qˆ, pˆ] = ih¯ , {ψ , ψ } = −iδ ⇒ {ψˆ , ψˆ } = h¯ δ . Now, we need to check that our system satisfies Eq. (75). To α β D αβ α β αβ do that, suppose that {α, β} have the values {1, 2}. Then, using (81) the definition of the Poisson and Dirac brackets, we find that, Taking this into account, we can see that the superalgebra for 2 2 2 2 2 N-dimensional quantum system is characterized by {Q1, Q1}D = −i p + ω q , {Q1, Q2}D = ω ψ1 + ψ2 , [ ˆ ˆ ] = = ··· 2 2 2 2 2 Qi, H 0, i 1 N, {Q2, Q1}D = −ω ψ1 + ψ2 , {Q2, Q2}D = −i p + ω q . {Qˆ , Qˆ } = h¯ δ Hˆ , i, j = 1 . . . N . (82) (76) i j ij This will be elaborated further in the next section using the ex- If we make substitution using Eq. (76), we find ample of the supersymmetrical harmonic oscillator. One more

{Qα, Qβ}D = {Q1, Q1}D + {Q1, Q2}D + {Q2, Q1}D + {Q2, Q2}D notation before ending this section is that the supercharges Qi † = 2 2 2 2 2 2 2 2 2 2 are Hermitian, i.e. Qi Qi, and this implies that = −i(p + ω q ) + ω(ψ1 + ψ2) − ω(ψ1 + ψ2) − i(p + ω q ) 2 2 2 ˆ ˆ ˆ † ˆ † = −2i(p + ω q ) ≡ −2iδαβH . (77) {Qi, Qj} = {Qi , Qj } . (83) Furthermore, using the following relations 5. SUPERSYMMETRIC HARMONIC OSCIL- {Q , H} = ωpψ − ω2qψ , {Q , H} = −ωpψ + ω2qψ , 1 P 2 1 2 D 2 1 LATOR {φ1, H}P = iωψ2, {φ2, H}P = iωψ2 , (78) Now, we turn back to the quantum harmonic oscillator prob- we ﬁnd lem as a simple example to show the supersymmetric property of a quantum mechanical system. Using Eq. (16), the har- {Qα, H}D = {Q1, H}D + {Q2, H}D monic oscillator Hamiltonian (17) can be written in terms of the = { } − { A} { B } Q1, H P Q1, φ PCAB φ , H P bosonic and fermionic laddering operators as follows: A B +{Q2, H}P − {Q2, φ }PCAB{φ , H}P † † 1 1 2 2 Hˆ = h¯ ω(aˆ aˆ + bˆ bˆ) . (84) = {Q1, H}P − {Q1, φ }PC11{φ , H}P − {Q1, φ }PC22{φ , H}P 1 1 2 2 +{Q2, H}P − {Q2, φ }PC11{φ , H}P − {Q2, φ }PC22{φ , H}P Moreover, the supercharge operators Qˆ 1 and Qˆ 2 can also be 2 2 = ωPψ2 − ω qψ1 − ωPψ2 + ω qψ1 written in terms of the bosonic and fermionic laddering opera- 2 2 tors as follows: −ωPψ1 − ω qψ2 + ω qψ2 + ωPψ1

= 0 . (79) Qˆ 1 = pψ1 + ωqψ2 r ! r ! h¯ ω h¯ It is clear from Eqs. (77, 79) that the supercharges (73) together = i (aˆ† − aˆ) (bˆ + bˆ†) with the Hamiltonian (27) and the constraints (28) satisfy the 2 2 two conditions (75) of the classical Dirac bracket superalgebra. r ! r ! h¯ h¯ + ω (aˆ)† + aˆ) i (bˆ − bˆ†) 2ω 2 √ 4. THE SUPERSYMMETRY ALGEBRA = ih¯ ω(aˆ†bˆ − aˆbˆ†) , (85) The material in this section is elaborated in detail in [21, 22, 23, and similarly, 24, 25, 26]. The supersymmetry algebra encodes a symmetry describing a relation between bosons and fermions. In general, Qˆ 2 = pψ2 − ωqψ1 the supersymmetry is constructed by introducing supersym- r ! r ! h¯ ω h¯ metric transformations which are generated by the supercharge = i (aˆ† − aˆ) i (bˆ − bˆ†) Qi operators, where the role of the supercharges Qi is to convert 2 2 a fermionic degree of freedom into a bosonic degree of freedom r ! r ! h¯ h¯ and vice versa; i.e., − ω (aˆ)† + aˆ) (bˆ + bˆ†) 2ω 2 Q|fermionici = |bosonici, Q|bosonici = |fermionici . √ † ˆ ˆ† (80) = −h¯ ω(aˆ b + aˆb ) . (86)

7 Letters in High Energy Physics LHEP-160, 2020

Using Eqs. (85, 86), one may deﬁne non-Hermitian opera- E ˆ ˆ † tors Q and Q as Q

(1) (2) 1 E E Qˆ = (Qˆ − iQˆ ) 3 2 2 1 2 √ √ 1 † † † † = ih¯ ω(aˆ bˆ − aˆbˆ ) + ih¯ ω(aˆ bˆ + aˆbˆ ) † 2 Q √ (1) (2) † E E = ih¯ ω(aˆ bˆ) , (87) 2 1 and similarly,

(1) (2) E E 1 0 Qˆ † = (Qˆ + iQˆ ) 1 2 1 2 1 √ √ = ih¯ ω(aˆ†bˆ − aˆbˆ†) − ih¯ ω(aˆ†bˆ + aˆbˆ†) (1) 2 E √ E=0 0 = −ih¯ ω(aˆbˆ†) . (88)

nF =0 nF =1 We can directly use Eq. (84, 87, 88) to check that the supercharge operators Qˆ and Qˆ † with the Hamiltonian H satisfy the quan- FIGURE 1: A schematic view showing the energy levels for tum superalgebra: a supersymmetric system consisting of only two possible fermionic states nF = {0, 1}. The supercharge operators Q and {Qˆ , Qˆ } = {Qˆ †, Qˆ †} = 0, Q† exchange bosons and fermions without affecting the energy {Qˆ , Qˆ †} = {Qˆ †, Qˆ } = H, due to the degeneracy of the energy levels of the two supersymmetric partners except for the ground state of the first partner. [H, Qˆ ] = [H, Qˆ †] = 0 . (89) where V(xˆ) is the potential, and B is a magnetic field. If those Now, it is easy using Eqs. (87, 88) to see the action of the potential and magnetic fields can be written in terms of some ˆ ˆ † operators Q, Q on the energy eigenstates: function W(x), as ˆ Q|nB, nFi ∼ |nB − 1, nF + 1i, dW(x) 2 d2W(x) ( ) = ≡ 02 ( ) = ≡ 00 † V x W , B x 2 W , Qˆ |nB, nFi ∼ |nB + 1, nF − 1i . (90) dx dx (94) Remember that there are only two possible fermionic states the Hamiltonian is supersymmetric and we shall refer to the † nF = {0, 1}, so the effect of the operators Qˆ and Qˆ can be function W(x) as a superpotential. Using Eq. (94), we can written as follows: rewrite the Hamiltonian (93) as Qˆ |n , 0i ∼ |n − 1, 1i, 1 2 02 I 1 00 B B Hˆ = pˆ + W 2 + h¯ σ3W . (95) † 2 2 Qˆ |nB, 1i ∼ |nB + 1, 0i . (91) At this stage, we define the following two Hermitian super- Also, from Eq. (89), the operators Qˆ and Qˆ † commute with the charge operators: Hamiltonian H; for that reason, we can see that 1 0 Qˆ 1 = σ1 pˆ + σ2W (xˆ) , H(Qˆ |nB, nFi) = Qˆ (H|nB, nFi) = (EB + EF)(Qˆ |nB, nFi) , (92) 2 1 Qˆ = σ pˆ − σ W0(xˆ) . (96) and this means that the whole energy of the system remains 2 2 2 1 unchanged by the action of the supercharge operators. The Hamiltonian (95) together with the supercharges (96) con- In short, the supercharge operator Qˆ acts to change one bo- stitutes a superalgebra. It is a simple matter of algebra to verify son to one fermion leaving the total energy of the system in- the conditions for the superalgebra given by Eq. (82). It may be variant. Conversely, Qˆ † changes a fermion into a boson leaving useful before verifying these conditions to check the following the energy unchanged. This is illustrated in Figure 1. commutation relation: ∂ ∂ ∂ [pˆ, f (x)] g = [−ih¯ , f (x)] g = −ih¯ ( f g) + f (g) 6. SUPERSYMMETRIC QUANTUM MECHAN- ∂x ∂x ∂x ICS ∂g ∂ f ∂g = −ih¯ f · + · g − f · = −ih¯ f 0(x) g . In this section, we study the general formalism of one- ∂x ∂x ∂x dimensional supersymmetric quantum mechanics. The ideas in (97) this section were discussed in [3, 27, 28, 29, 30, 31]. We begin by considering a Hamiltonian of the form Removing g gives us the commutator of the momentum pˆ with an arbitrary function of the position coordinate x: 1 2 I 1 0 Hˆ = pˆ + V(xˆ) 2 + hB¯ (xˆ)σ3 , (93) [pˆ, f (x)] = −ih¯ f (x). (98) 2 2 8 Letters in High Energy Physics LHEP-160, 2020

Using this relation, we find the following: it is straightforward to check that the Hamiltonian Hˆ can be expressed as [ 02] = 02 − 02 = − 0 00 pˆ, W pWˆ W pˆ 2ihW¯ W , Hˆ = {Qˆ , Qˆ †} , (107) 0 2 0 2 2 0 00 [W , pˆ ] = W pˆ − pˆ W = ih¯ {W , pˆ}, and it commutes with both the operators Qˆ and Qˆ †: {pˆ, W00} = pWˆ 00 + W00 pˆ . (99) [Qˆ , Hˆ ] = 0 & [Qˆ †, Hˆ ] = 0 . (108) Now, we can use the outcome of Eq. (99) to check the outcome of the commutator of the supercharge Qˆ 1 with the Hamiltonian Furthermore, one also finds Hˆ : {Qˆ , Qˆ } = 0 & {Qˆ †, Qˆ †} = 0 . (109) [Qˆ 1, Hˆ ] = Qˆ 1Hˆ − HQˆ 1 As we have seen in the case of the supersymmetric Har- = 1 (σ pˆ + σ W0) · pˆ2 + W02 I + hW¯ 00σ 4 1 2 2 3 monic oscillator, a supersymmetric system with two indepen- 1 2 02 I 00 0 dent supercharges, with the possible exception of the energy of − 4 pˆ + W 2 + hW¯ σ3 · (σ1 pˆ + σ2W ) the ground eigenstate, all the energy levels are split into two = 1 σ pˆ3 + σ pWˆ 02 + σ σ h¯ pWˆ 00 + σ W0 pˆ2 + σ W0W02 + σ σ hW¯ 0W00 4 1 1 1 3 2 2 2 3 eigenstates with either nF = 0 or nF = 1. For a spinor quan- 1 3 2 0 02 02 0 00 00 0 tum mechanical system, this implies that the excited energy − 4 σ1 pˆ + σ2 pˆ W + σ1W pˆ + σ2W W + σ3σ1hW¯ pˆ + σ3σ2hW¯ W eigenstates come in degenerate spin-up/spin-down pairs |En, ↑ 1 02 00 0 0 00 = 4 σ1[pˆ, W ] − σ2ih¯ {pˆ, W } + σ2[W , pˆ] + 2σ1ihW¯ W i/|En, ↓i. These degenerate spin-up/spin-down pairs are related to the acts of the supercharge operators Qˆ and Qˆ †, where = 0 . (100) the supercharge operators Qˆ convert the degenerate spin-up Similarly, one can confirm that state to the degenerate spin-down state without making any change in the energy eigenvalue of the states: [Qˆ 2, Hˆ ] = Qˆ 2Hˆ − Hˆ Qˆ 2 = 0. (101) 0 0 ↑ 0 Furthermore, we can find that Qˆ |E , ↑i ∼ = , (110) n 1 0 0 ↓ ˆ ˆ ˆ ˆ ˆ ˆ ˆ 2 {Q1, Q1} = Q1Q1 + Q1Q1 = 2Q1 while the supercharge operators Qˆ † convert the degenerate 1 02 = σ pˆ + σ2W spin-down state to the degenerate spin-up state without mak- 2 1 ing any change in the energy eigenvalue of the states: 1 2 = σ2 pˆ2 + σ2W0 + σ σ pWˆ 0 + σ σ W0 pˆ 2 1 2 1 2 2 1 ˆ † 0 1 0 ↑ 1 2 Q |En, ↓i ∼ = . (111) = (pˆ2 + W0 )I + iσ (pWˆ 0 − W0 pˆ) 0 0 ↓ 0 2 2 3 1 2 02 00 To sum up, the supersymmetric transformations occur due = (pˆ + W )I + h¯ σ W ≡ H , (102) † 2 2 3 to the supercharge operators Qˆ and Qˆ . In this case, it causes and similarly, we have transforms between the energy eigenstates spin-up/spin-down which have the same energy eigenvalues. ˆ † ˆ ˆ {Q2, Q2} = H . (103) Note that, to do the previous calculations, we used the Pauli 2 2 2 I 7. SUPERSYMMETRIC GROUND STATE matrices properties: σ1 = σ2 = σ3 = 2 and σ1σ2 = −σ2σ1 = iσ3. It is clear from Eqs. (100,101, 102, and 139) that the Hamilto- So far, we have investigated all the required information to nian (95) together with the supercharges (96) satisfies the con- describe a supersymmetric quantum mechanical system. In ditions for the superalgebra given by Eq. (82). this section, we study the supersymmetric quantum mechan- Indeed, if we define the quantities Q and Q† as ical ground state. The argument in this section follows from [32, 33, 34, 35, 36]. Let us now write the expression of the super- ˆ = ˆ − ˆ Q Q1 iQ2 symmetric Hamiltonian (95) as the summation of two separate 1 0 1 0 terms: a Hamiltonian Hˆ and a Hamiltonian Hˆ ; then, we have = σ pˆ + σ W − i σ pˆ − σ W + − 2 1 2 2 2 1 1 02 1 00 1 0 ˆ = 2 + ± = (σ − iσ )pˆ + (σ + iσ )W H± pˆ W hW¯ . (112) 2 1 2 2 1 2 2 1 In the eigenbasis of σ , the supersymmetric Hamiltonian is di- = (σ − iσ ) pˆ + iW0 3 2 1 2 agonal: 0 ˆ † = σ− pˆ + iW , (104) ˆ H+ 0 A A 0 H ≡ = † , (113) 0 Hˆ − 0 AA and † 0 † Qˆ = Qˆ 1 + iQˆ 2 = σ+ pˆ − iW , (105) and the supercharge operators Qˆ and Qˆ can be written as where σ and σ , respectively, are defined as − + 0 0 0 A† Qˆ = , Qˆ † = , (114) 1 0 0 A 0 0 0 σ− = (σ − iσ ) = , 2 1 2 1 0 where 1 0 1 σ+ = (σ + iσ ) = , (106) 0 0 2 1 2 0 0 A = pˆ + iW , A† = pˆ − iW . (115)

9 Letters in High Energy Physics LHEP-160, 2020

According to Eq. (107), we can write the supersymmetric In fact, it has no physical meaning to find the ground- Hamiltonian Hˆ in terms of the supercharges operators Qˆ and state wave function ψ0 if there are no normalizable solutions Qˆ † as follows: of such form. So, we need now to normalize the ground-state wave function (124) by determining the values of the two con- ˆ = ˆ ˆ † + ˆ † ˆ = ˆ 2 = ˆ †2 H QQ Q Q 2Q 2Q . (116) stants A and B. The ground-state wave function ψ0(x) to be normalizable must vanish at the positive and the negative infi- In the last step, we used the fact that the supercharges are Her- nite x-values. In order to satisfy this condition, we have to set mitian operators. We see from Eq. (116) that the Hamiltonian |W(x)| → ∞ as |x| → ∞. Then, we have the following three Hˆ can be written in terms of the squares of the supercharge possible cases for the supersymmetric ground-state wave func- operators. For this reason, the energy of any eigenstate of this tion: Hamiltonian must be positive or zero. Let us now consider that | i ˆ Ψ0 is the ground state of the supersymmetric Hamiltonian H. 1. The first case is when W(x) → +∞ as x → ±∞. In Based on Eq. (116), it can be seen that the ground state can have this case, the superpotential W(x) is an even function a zero energy only if it satisfies the following two conditions: and it is positive at the boundaries. Figure 2 describes 2 the behavior of the potential V(x) in this case. Since E0 = hΨ0|Hˆ |Ψ0i = hΨ0|Qˆ |Ψ0i = 0 =⇒ Qˆ |Ψ0i = 0, W(x) is positive, we cannot normalize the wave function ˆ ˆ †2 ˆ † E0 = hΨ0|H|Ψ0i = hΨ0|Q |Ψ0i = 0 =⇒ Q |Ψ0i = 0. (117) + W(x) h¯ ψ0 = Ae , but we can choose A = 0. However, we − − W(x) Therefore, if there exists a state which is annihilated by each of can normalize the wave function ϕ = Be h¯ to find ˆ ˆ † 0 the supercharge operators Q and Q which means that it is in- the value of the constant B. The complete ground-state variant under the supersymmetry transformations, such a state wave function, in this case, may be expressed in the gen- is automatically the zero-energy ground state. However, on the eral form ! other hand, any state that is not invariant under the supersym- 0 metry transformations has a positive energy. Thus, if there is a Ψ0 = − W(x) . (125) Be h¯ supersymmetric state, it is the zero-energy ground state and it is said that the supersymmetry is unbroken. Moreover, since the supersymmetry algebra (108) implies V(x) that the supercharge operators Qˆ and Qˆ † commute with the supersymmetric Hamiltonian Hˆ , so all the eigenstates of Hˆ are doubly degenerate. For that reason, it will be convenient to write the supersymmetric ground state of the system in terms ± of two components, ψ0 , as follows: x + ψ0 Ψ0 = − . (118) ψ0

One can now solve eigenvalue problem Qˆ |ψ0i = 0 to find the zero-energy ground-state wave function. If we make substitution using Eqs. (114, 118), we get FIGURE 2: The even superpotential W(x) is an even function, + where W(x) → +∞ as x → ±∞. 0 0 ψ0 Qˆ |ψ0i = − = 0 , (119) A 0 ψ0 2. The second case is when W(x) → −∞ as x → ±∞. and then, if we make substitution using Eq. (115), our problem In this case, the superpotential W(x) is an even func- reduces to solve the first-order differential equation: tion and it is negative at the boundaries. The behavior of the potential V(x) in this case is described in Figure ∂ψ+ + 0 + 0 0 + 3. Since W(x) is negative, we can normalize the wave A|ψ0 i = pˆ + iW |ψ0 i = 0 =⇒ −ih¯ + iW ψ0 = 0 . ∂x W(x) a h¯ (120) function Ψ0 = Ae and find the value of the con- It is simple and straightforward to solve the previous differen- stant A; however, we cannot normalize the wave func- − W(x) tial equation as follows: b h¯ tion Ψ0 = Be , but we can choose B = 0. The complete ground-state wave function, in this case, shall be dψ+ W0 W0 0 = dx =⇒ ln ψ+ = + ln A . (121) expressed in the general form + h 0 h ψ0 ¯ ¯ W(x) ! Thus, Ψ = Ae h¯ . (126) W 0 + h¯ 0 ψ0 = Ae , (122) and similarly, + − W ( ) → {+ − } ψ = Be h¯ . (123) 3. The other two cases correspond to W x ∞, ∞ 0 as x → {+∞, −∞}, and W(x) → {−∞, +∞} as x → Now, the general form of the ground-state wave function may {+∞, −∞}. In those two cases, the superpotential W(x) be expressed as is an odd function and it is positive at one of the bound- W ! Ae h¯ aries and negative at the other boundary. The behavior of Ψ0 = − W . (124) Be h¯ . the potential V(x) in this case is described in Figure 4. In 10 Letters in High Energy Physics LHEP-160, 2020

V(x) 8. CORRECTIONS TO THE GROUND-STATE ENERGY In this section, we consider the example of the supersymmetry harmonic oscillator which has a unique zero-energy ground state. Then, we use the conventional perturbation theory to ex- x amine whether it has nonvanishing corrections to the energy of the ground state, and accordingly, it can be responsible for the spontaneous breaking of the supersymmetry. To this end, in the next two subsections, based on [37, 38, 39, 40, 41], we are going to compute both the ﬁrst and the second corrections to the ground-state energy of the supersymmetric harmonic os- FIGURE 3: The superpotentialW(x) is an even function, where cillator. For instance, let us recall the general form of the super- W(x) → −∞ as x → ±∞. symmetric quantum mechanics Hamiltonian, which is given by Eq. (95): 1 2 02 I 1 00 those two cases, both the two constants A and B vanish Hˆ = (pˆ + W ) 2 + hW¯ σ3. (128) 2 2 and we cannot normalize the wave function. Therefore, we have Consider now that the superpotential W(x) is deﬁned as Ψ0 = 0. (127) 1 W0 = ωqˆ + 3gqˆ2 W(q) = ωqˆ2 + gqˆ3 ⇒ (129) Thus, since those two forms of the zero-energy ground- 2 W00 = ω + 6gqˆ , state wave function cannot be normalized, this means that they do not exist. where g is a perturbation. If g = 0, the previous Hamiltonian reduces to the supersymmetric harmonic oscillator Hamiltonian: V(x) 1 1 Hˆ 0 = (pˆ2 + ω2q2)I + h¯ ωσ . (130) 2 2 2 3 We have already solved this Hamiltonian and found that it has x a unique zero-energy ground state,

(0) E0 = 0 , (131)

and we verify that this state is invariant under supersymmetry. Based on our argument in the previous section and after normalization, the unbroken supersymmetric ground-state wave (a) W(x) → −∞ as x → −∞, and W(x) → ∞ as x → ∞. function can be written as follows: V(x) ! (0) 0 = 2 Ψ0 ω 1 − ωq . (132) ( ) 4 2¯h πh¯ e

Now, when g is small, the Hamiltonian (128) can be written as Hˆ = Hˆ 0 + Hˆ 0 , (133) where Hˆ 0 is the Hamiltonian of the unperturbed system and Hˆ 0 x a perturbation:

9 Hˆ 0 = ( g2qˆ4 + 3ωgqˆ3)I + 3¯hgqˆσ . (134) 2 2 3 (b) W(x) → ∞ as x → −∞, and W(x) → −∞ as x → ∞. Moreover, if we consider the matrix representation, the pertur- FIGURE 4: The superpotentialW(x) is an odd function. bative Hamiltonian Hˆ acting on the ground-state wave function (0) Ψ0 yields ! In short, the zero-energy ground-state wave function can be 9 2 4 3 0 0 (0) 2 g qˆ + 3ωgqˆ + 3¯hgqˆ 0 Hˆ = ωq2 = normalized when the superpotential W(x) has an even num- Ψ0 9 2 4 3 ω 1 − 0. 0 g qˆ + 3ωgqˆ − 3¯hgqˆ ( ) 4 2¯h 2 πh¯ e ber of zeros. In this case, there exists a zero-energy ground (135) state which is the true vacuum state and the supersymmetry In view of this last equation, it is clear that for this combination is unbroken. However on the other hand, if the superpoten- ( ) the impact of the Hamiltonian Hˆ 0 on the wave function Ψ 0 is tial W(x) has an odd number of zeros, the zero-energy ground- 0 only due to the component: state wave function cannot be normalized. Then, we immedi- ately realize that there is no zero-energy ground state in such 0 9 2 4 3 case and the supersymmetry is spontaneously broken. Hˆ = g qˆ + 3ωgqˆ − 3¯hgqˆ . (136) 2 11 Letters in High Energy Physics LHEP-160, 2020

As a result, we see that it is enough to consider the Hamiltonian Therefore, the second-order correction to ground-state energy Hˆ 0 to compute the first and second corrections to the ground- of the supersymmetric harmonic oscillator is computed as fol- state energy. lows: Furthermore, we have to mention here that, for the super- ( ) ( ) |hψ 0 | 9 g2qˆ4 + 3ωgqˆ3 − 3¯hgqˆ|ψ 0 i|2 symmetric harmonic oscillator, except for the ground state, all (0) = 0 2 1 E2 ( ) ( ) the energy levels are degenerate to two energy states. However, E 0 − E 0 because we are just interested in computing the correction to 0 1 (0) (0) the ground-state energy, which is not degenerate, we can use |hψ | 9 g2qˆ4 + 3ωgqˆ3 − 3¯hgqˆ|ψ i|2 + 0 2 2 ( ) ( ) the nondegenerate perturbation theory. E 0 − E 0 In addition, it is useful to remember from quantum me- 0 2 ( ) ( ) chanics that the wave function of degree n can be obtained by |hψ 0 | 9 g2qˆ4 + 3ωgqˆ3 − 3¯hgqˆ|ψ 0 i|2 + 0 2 3 the following recursion formula: (0) (0) E0 − E3 r r (n + 1)h¯ nh¯ (0) 9 2 4 3 0 2 |hψ0 | 2 g qˆ + 3ωgqˆ − 3¯hgqˆ|ψ4i| qˆψn = ψn+1 + ψn−1 . (137) + . (143) 2ω 2ω (0) (0) E0 − E4 As well, it is important to recall the following relation: We can simplify the previous equation by calculating the nu- hψn|ψmi = δnm . (138) merator of each term using the recurrence relation for the harmonic oscillator energy eigenfunction given by Eq. (137). We We are ready now to move forward to the next two subsections get and compute the first and the second corrections to the energy r r of the ground state. h¯ h¯ qˆψ = ψ + ψ , 1 ω 2 2ω 0 s s s 8.1. The First-Order Correction 3 3 3 3 3¯h h¯ 9¯h We know from the previous argument that the ground state of qˆ ψ1 = ψ4 + 3 ψ2 + ψ0, ω3 ω3 8ω3 the supersymmetric Hamiltonian is nondegenerate. Therefore, q 3 q 3 q 3 q 3 we can use the time-independent nondegenerate perturbation 3 = 15¯h + 72¯h + 243¯h + 3¯h qˆ ψ3 ω3 ψ6 ω3 ψ4 8ω3 ψ2 4ω3 ψ0, theory to compute the first-order correction to the ground-state q 4 q 4 q 4 q 4 q4 = 45¯h + 147¯h + h¯ + 9¯h energy of the supersymmetric harmonic oscillator as follows: ˆ ψ2 2ω4 ψ6 ω4 ψ4 12.5 ω4 ψ2 2ω4 ψ0, q 4 q 4 q 4 q 4 q 4 (1) (0) 0 (0) 4 105¯h 945¯h 2725¯h 81¯h 3¯h ˆ qˆ ψ4 = 4 ψ8 + 4 ψ6 + 4 ψ4 + 4 ψ2 + 4 ψ0 . E0 = hψ0 |H |ψ0 i ω 4ω 16ω 2ω 2ω ( ) 9 ( ) (144) = hψ 0 | g2qˆ4 + 3ωgqˆ3 − 3¯hgqˆ|ψ 0 i 0 2 0 In Eq. (144), we only list the terms which contain ψ0. The other 9 2 (0) 4 (0) (0) 3 (0) (0) (0) = 2 g hψ0 |qˆ |ψ0 i + 3ωghψ0 |qˆ |ψ0 i − 3¯hghψ0 |qˆ|ψ0 i . terms do not affect our calculations, since all of them give us (139) zero. Using Eq. (144) to calculate Eq. (143), we get

3 The terms qˆ ψ and qˆψ are linearly independent of ψ so that 2 s 2 0 0 0 2 3 r ! 4 4 both the second and the third terms of Eq. (139) are zero. How- (0) g 9¯h h¯ 81g 9¯h E2 = − 3ω × − 3¯h × −   ever, h¯ ω ω3 2ω 8¯hω 2ω4 4 4 4 4 3¯h 7¯h 9¯h 2 2 qˆ ψ0 = ψ4 + ψ2 + ψ0 , (140)  s   s  2ω4 4ω4 16ω4 g2 3¯h3 g4 9 3¯h4 − 3ω ×  −  ×  and then, substituting it into Eq. (139) gives us, 3¯hω 4ω3 4¯hω 2 2ω4

2 2 3 2 3 ( ) 9 ( ) ( ) 9 3 h¯ ( ) ( ) 9 h¯ 729 h¯ 9 h¯ 243 h¯ E 1 = g2hψ 0 |qˆ4|ψ 0 i = g2 × hψ 0 |ψ 0 i = − g2 − g4 − g2 − g4 0 2 0 0 2 4 ω2 0 0 8 ω2 16 ω5 4 ω2 32 ω5 27 h¯ 2 27 h¯ 2 1701 h¯ 3 = g2 'O(g2) . (141) = − g2 − g4 . (145) 8 ω2 8 ω2 32 ω5 In the view of the last equation, the ﬁrst-order correction to the Finally, with regard to Eqs. (141 and 145), we realize that up ground-state energy reduces to zero. to the second order the energy corrections to the ground-state energy vanish. This is concluded as 8.2. The Second-Order Correction O(g) = 0, From the nondegenerate time-independent perturbation the- 27 h¯ 2 27 h¯ 2 ory, the second-order correction to the energy is given by O(g2) = − = 0 . (146) 8 ω2 8 ω2 (0) 0 (0) 2 ( ) |hψ |Hˆ |ψ i| E 0 = n m . (142) It is worth noting that the second term in Eq. (145) should be 2 ∑ (0) (0) m6=n Em − En canceled when we extend the calculation to the fourth-order perturbation corrections. Showing this would make the calcu- Since the perturbation contains only terms of q, q3, and q4, the lations here more complicated and confusing. However, the re- numerator of Eq. (142) is zero for all m values except m = 1, 3, 4. sult can be expanded to any ﬁnite order in perturbation theory.

12 Letters in High Energy Physics LHEP-160, 2020

Hence, there are no corrections to the energy of the ground state Ψ0 = 0. This means that the wave function of the ground state and the supersymmetry remains unbroken at any finite order of is not excited, even if the perturbation theory told us that it is perturbation theory. excited and has no energy correction. Thus, the perturbation technique gives us incorrect results for both the wave function and the energy spectrum and fails 9. PROPERTIES OF SUSY QUANTUM ME- to give an explanation to the supersymmetry breaking. CHANICS 10. CONCLUSIONS W(x) In this study, we studied the basic aspects of supersymmetric quantum mechanics. We started with introducing the algebra of Grassmann variables and then looked into quantum mechanics of the supersymmetric harmonic oscillator, which includes fermionic as well as bosonic fields. Afterward, we investigated the algebraic structure of supersymmetric quantum mechanics. We started by investigating the superalgebra using Dirac brackets. Then, we introduced the concept of the su- x percharge operators Qˆ and Qˆ †. In general, the supersymmetry is constructed by introducing supersymmetric transformations which are generated by the supercharge operators, where the role of the supercharges is to change the bosonic state into the fermionic state and vice versa, while the Lagrangian remains invariant. Moreover, we have presented the basic properties of FIGURE 5: The superpotential of the harmonic oscillator supersymmetric quantum mechanics. ground state. Furthermore, we illustrated, for a supersymmetric quantum mechanical system, that the energy spectrum is degenerate A supersymmetric quantum mechanics system is said to except for the ground state, which must have a zero eigenvalue have unbroken supersymmetry if it has a zero-energy ground in order for the system to have an unbroken supersymmetry. state, which is E0 = 0, while if the system has a positive Also, we have explained that if there is a supersymmetric state, ground-state energy, E0 > 0, it is said to have a broken super- it is the zero-energy ground state. If such a zero-energy ground symmetry [1]. For example, in Section 5, we have studied the state exists, it is said that the supersymmetry is unbroken. So Hamiltonian of the supersymmetric harmonic oscillator and far, there has been no unbroken supersymmetry observed in showed that it obeys the superalgebra. Then, we calculated the nature, and if nature is described by supersymmetry, of course, ground-state energy for that system and found evidence that it it must be broken. vanishes, E0 = 0, and consequently, this supersymmetric har- In fact, supersymmetry may be broken spontaneously at monic oscillator has an unbroken supersymmetry. any order of perturbation theory or dynamically due to non- In the previous section, we applied a small perturbation g perturbative effects. To examine this statement, we studied the to the supersymmetric harmonic oscillator and calculated the normalization of the ground state of the supersymmetric har- effect on the ground-state energy. We found out that the per- monic oscillator. Then, we used perturbation theory to calcu- turbation does not affect the energy of the ground state at the late the corrections to the ground-state energy. We found out second order in perturbation theory, but this result can be ex- that the perturbation does not affect the energy of the ground panded to any finite order. This means that the supersymmetry state at second order in perturbation theory, but this result can breaking does not occur because of perturbation, and it must be expanded to any finite order. This means that the supersym- be due to the nonperturbative effects. metry breaking is not seen in perturbation theory, and it must Again, let us consider the same potential Eq. (129), which be due to the nonperturbative effects. we have used before in Section 8: 1 W(q) = ωqˆ2 + gqˆ3 . (147) ACKNOWLEDGEMENTS 2 The author acknowledges the research office of the University As we discussed in Section 7, the wave function of the ground of the Witwatersrand and the African Institute for Mathemati- state should have three zeros, Ψ0 → +∞ as x → +∞, and cal Sciences (Ghana) for financial support. Ψ0 → −∞ as x → −∞. Figure 5, shows an arbitrary diagram presenting how the wave function of the ground state should look like, and Eq. (148) gives us the nonnormalized form of the References ground-state wave function: [1] Fredrik Engbrant. Supersymmetric quantum mechanics − W ! and integrability. Master’s thesis, Uppsala University, Ae h¯ Ψ0 = − W . (148) 2012. Be h¯ [2] T. Wellman. An introduction to supersymmetry in quantum mechanical systems. Brown University Memoran- Acutely, Eq. (148) could not be normalized, and the only way dum, 2003. to solve the Schrodinger¨ equation for the ground state is taking

13 Letters in High Energy Physics LHEP-160, 2020

[3] Edward Witten. Dynamical breaking of supersymmetry. [26] Patrick Labelle. Supersymmetry Demystified. McGraw Hill Nuclear Physics B, 188(3):513–554, 1981. Professional, 2010. [4] Thomas Dumitrescu. Topics in Supersymmetric Quantum [27] Edward Witten. Constraints on supersymmetry breaking. Field Theory. PhD thesis, Princeton University, 2013. Nuclear Physics B, 202(2):253–316, 1982. [5] Matteo Bertolini. Lectures on supersymmetry. Lecture [28] R Rodrigues. The quantum mechanics susy algebra: an notes given at SISSA, 2011. introductory review. arXiv preprint hep-th/0205017, 2002. [6] Fred Cooper, Avinash Khare, and Uday Sukhatme. Su- [29] MV Ioffe, S¸Kuru, J Negro, and LM Nieto. Susy approach persymmetry and quantum mechanics. Physics Reports, to pauli hamiltonians with an axial symmetry. Journal of 251(5):267–385, 1995. Physics A: Mathematical and General, 39(22):6987, 2006. [7] Paul Batzing and Are Raklev. Lecture notes for [30] Johan Gudmundsson. Supersymmetric Quantum Mechanics. fys5190/fys9190–supersymmetry. A course given at the Lund University, 2014. University of Oslo, 2013. [31] DS Berman and Eliezer Rabinovici. Supersymmetric [8] Sven Krippendorf, Fernando Quevedo, and Oliver Schlot- gauge theories. In a NATO Advanced Study Institute, page terer. Cambridge lectures on supersymmetry and extra 141. Springer, 2002. dimensions. arXiv preprint arXiv:1011.1491, 2010. [32] Philip Argyres. Introduction to supersymmetry. Course [9] Muhammad Abdul Wasay. Supersymmetric quantum note, Cornell University, 1996. mechanics and topology. Adv. High Energy Phys., [33] Philip Argyres. An introduction to global supersymmetry. 2016:3906746, 2016. Lecture notes, 2001. [10] SO Morgan. Supersymmetric born reciprocity. PhD thesis, [34] Juan Mateos Guilarte, Marina de la Torre Mayado, and University of Tasmania, 2003. Miguel Angel Gonzalez´ Leon.´ From n= 2 supersymmetric [11] Nathan Seiberg. Noncommutative superspace, n= 1/2 su- classical to quantum mechanics and back: the susy wkb persymmetry, field theory and string theory. Journal of approximation. Monograf´ıasde la Real Academia de Ciencias High Energy Physics, 2003(06):010, 2003. Exactas, F´ısicas,Qu´ımicasy Naturales de Zaragoza, (29):113– [12] Bijan Kumar Bagchi. Supersymmetry in quantum and classi- 128, 2006. cal mechanics. CRC Press, 2000. [35] Georg Junker. Supersymmetric methods in quantum and sta- [13] Yamen Hamdouni. Supersymmetry in quantum mechan- tistical physics. Springer Science & Business Media, 2012. ics. African Institute of Mathematical Sciences, Cape Town, [36] P Salomonson and JW Van Holten. Fermionic coordi- 2005. nates and supersymmetry in quantum mechanics. Nuclear [14] Mikio Nakahara. Geometry, topology and physics. CRC Physics B, 196(3):509–531, 1982. Press, 2003. [37] Herbert Goldstein. Classical mechanics. Pearson Education [15] Donald T Greenwood. Advanced dynamics. Cambridge India, 1965. University Press, 2006. [38] David Jeffery Griffiths. Introduction to quantum mechanics. [16] Andrew Hanson, Tullio Regge, and Claudio Teitelboim. Pearson Education India, 2005. Constrained hamiltonian systems. Accademia Nazionale dei [39] Franz Schwabl. Quantum mechanics. Springer, 4th edition, Lincei, 1976. 2007. [17] Kishore B Marathe. Constrained hamiltonian systems. [40] Jun John Sakurai and Jim Napolitano. Modern quantum In Group Theoretical Methods in Physics, pages 177–182. mechanics. Addison-Wesley, 2011. Springer, 1983. [41] Ramamurti Shankar. Principles of quantum mechanics. [18] Marc Henneaux and Claudio Teitelboim. Quantization of Springer Science & Business Media, 2012. gauge systems. Princeton university press, 1992. [19] Andreas Wipf. Hamilton’s formalism for systems with constraints. In Canonical Gravity: From Classical to Quan- tum, pages 22–58. Springer, 1994. [20] Yaser Tavakoli. Lecture i: Constrained hamiltonian systems. Courses in canonical gravity given at The Federal Uni- versity of Espirito Santo, 2014. [21] Taichiro Kugo and Paul Townsend. Supersymmetry and the division algebras. Nuclear Physics B, 221(2):357–380, 1983. [22] Manuel Drees. An introduction to supersymmetry. arXiv preprint hep-ph/9611409, 1996. [23] Adel Bilal. Introduction to supersymmetry. arXiv preprint hep-th/0101055, 2001. [24] Monique Combescure, François Gieres, and Maurice Ki- bler. Are n= 1 and n= 2 supersymmetric quantum mechanics equivalent? Journal of Physics A: Mathematical and General, 37(43):10385, 2004. [25] Jonathan Bougie, Asim Gangopadhyaya, Jeffry Mallow, and Constantin Rasinariu. Supersymmetric quantum mechanics and solvable models. Symmetry, 4(3):452–473, 2012.