Normalization of the Ground State of the Supersymmetric Harmonic

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arXiv:2008.02080v1 [physics.gen-ph] 24 Jul 2020 uto,w td h rbe fahroi siltrwith oscillator harmonic a of problem the study we duction, non- the to due be effects. must perturbative c it effect and breaking perturbation supersymmetry no to that lead out using found energy and We theory. ground-state oscillator turbation the to harmonic corrections supersymmetric the the calculate of th of state normalization ground the metho study the we examine breaking, To supersymmetry dynam- of effects. or nonperturbative theory to broken perturbation due be of ically may order supersymmetry any fact, at In spontaneously broken. supersymme be by must described is it nature try, if supersymmetr and unbroken nature, no been in has observed there far, otherwi So unbroken; broken. is is supersymmetry it stat the ground exists, zero-energy state the a is thersuch it if state, that supersymmetric show a We is superalgebra. con and the supercharges supersym- introducing of of with cepts starting formulation mechanics basic quantum the metric supersym- derive a We make theory. to metric framework necessary followe the and introducing supersymmetry of by idea basic the explaining starti with derived is mechanics quantum supersymmetric volves [8]. anticommutators as i well algebra as supersymmetry commutators The volves u [7]. invariant transformations are those which der theories are [ theories versa Supersymmetric vice and variables anticommuting in are variables which spi commuting fermions are half-integer which bosons introdu of change to that possible erators particles is it and supersymmetry, In [5]. 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(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 iωψ ψ , (3) − 2 − 1 2 1 1 bˆ = ψˆ iψˆ , bˆ† = ψˆ + iψˆ . (12) 2¯h 1 − 2 2¯h 1 2 where πα are the momenta conjugate to ψ . r r α 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- built as a larger Hamiltonian consisting of two components, dition ψ , ψ = h¯ δ 1. Therefore, it can be seen that the { α β} αβ 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

∂ momentum variables in terms of the bosonic ladder operators Á qˆ = q Á2, pˆ = ih¯ 2 . (4) − ∂q as follows:

On the other hand, the fermionic variables ψˆ1 and ψˆ2 and the † 2ω h¯ † α aˆ + aˆ = qˆ qˆ = (aˆ + aˆ), fermionic momenta π , respectively, have to be defined as r h¯ ⇒ r 2ω 2 h¯ ω h h aˆ aˆ† = ipˆ pˆ = i (aˆ† aˆ) . ˆ ¯ ˆ ¯ ψ1 = σ1, ψ2 = σ2, − r h¯ ω ⇒ r 2 − r 2 r 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- nian (3) in the following form: 2 h¯ bˆ† + bˆ = ψˆ ψˆ = (bˆ + bˆ†), h¯ 1 ⇒ 1 2 2 r r 1 2 d 2 2 1 Hˆ = h¯ + ω q Á2 + h¯ ω σ3 , (6) † 2 h¯ † 2 − dq2 2 bˆ bˆ = i ψˆ2 ψˆ2 = i (bˆ bˆ ) . − r h¯ ⇒ r 2 − (15) where Á is the 2 2 identity matrix and σ are the Pauli matri- 2 × j ces; that is, Actually, we derived Eqs. (14, 15) here; however, it will be used 1 0 0 1 0 i 1 0 later. At this stage, let us define the following two Hermitian Á = , σ = , σ = − , σ = . (7) 2 0 1 1 1 0 2 i 0 3 0 1 operators, the bosonic number operator Nˆ , and the fermionic − B number operator Nˆ F: We want to obtain solutions of the time-independent Schrödinger Ψ = Ψ 1 equation H E ; that is, Nˆ aˆ†aˆ = ω2qˆ2 h¯ ω + pˆ2 , B ≡ 2¯hω − 2 1 2 d 2 2 1 † i h¯ + ω q Á2 + h¯ ω σ3 = EΨ , (8) Nˆ bˆ bˆ = − ψˆ ψˆ . (16) 2 − dq2 2 F 1 2 ≡ h¯ 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 nB, nF is listed as follows: (creation) operators for bosons aˆ† and aˆ, respectively, they are | i aˆ n , n = √n n 1, n , aˆ† n , n = √n + 1 n + 1, n , | B Fi B| B − Fi | B Fi B | B Fi 1 † 1 bˆ n , n = √n n , n 1 , bˆ† n , n = √n + 1 n , n + 1 . aˆ = (ωqˆ + ipˆ) , aˆ = (ωqˆ ipˆ) . (10) | B Fi F| B F − i | B Fi F | B F i r 2¯hω r 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.

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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, nF = nB nB, nF , Nˆ F nB, nF = nF nB, nF . (19) Before discussing the supersymmetry algebra in the next sec- | i | i | i | i tion, 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- E = h¯ ω n + + h¯ ω n . (20) n B 2 F − 2 tems are presented in [16, 17, 18, 19, 20, 21]. 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 (n n 1) and creation of one fermion (n n + 1) or positions are related by some constraints. In general, M con- B → B − F → F 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 0 is defined m | i deal with such constraints in this case. In principle, there can by also be tertiary constraints arising from the equation of motion bˆ 0 0. (22) | i≡ of the secondary constraints and so on. Then, we define a fermionic one-particle state 1 by The Hamiltonian (25) is an example of constrained Hamil- | i tonians. In the previous section, we defined the fermionic mo- 1 bˆ† 0 . (23) menta πα in Eq. (5) using the fermionic equation of motion fol- | i≡ | i lowed 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- 1 with bˆ or bˆ† gives either the already known state 0 , menta πα and the canonical coordinates ψα: | i | i or nothing: ∂L i πα = = δαβψ , α, β 1, 2 . (27) bˆ 1 = bˆbˆ† 0 = (1 bˆ†bˆ) 0 = 0 bˆ†bˆ 0 = 0 0 = 0 , ∂ψ˙ − 2 β ∈{ } | i | i − | i | i− | i | i− | i α bˆ† 1 = bˆ†bˆ† 0 = 0 0 = 0 . (24) | i | i | i In light of this equation, we have the primary constraint So, the subspace spanned by 0 and 1 is closed un- i ˆ ˆ† | i | i 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) in 2 2 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 subsec- { } tion, the problem caused by the constrained Hamiltonians is (ii) The ground eigenstate has a vanishing energy eigen- described. Next, the 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 = ( )ǫF , (30) spectively. { }P ∂q ∂pi − ∂pi ∂q − ∂ψ ∂πα − ∂πα ∂ψ i i α α 3 Letters in High Energy Physics LHEP xx, xxx, 2018 where ǫF 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- ψ˙ = 2iπ˙ 1 = 2ωψ . (40) metric: 1 − 2 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 = A, B + A, C . (33) equations of motion for the fermionic positions and momenta: { }P { }P { }P ∂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 , ǫ ǫ 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 ǫF 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 + B, C, A + C, A, B = 0. { { }P}P { { }P}P { { }P}P 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 . (36) 3.2. The Constraints and the Equations of Motion { }P 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 = = i = = i = i + ˙ α A q˙i i qi, H P, p˙ q , H P, S q˙i p ψαπ H φ λA dt , (42) ∂p { } − ∂qi { } − − Z ∂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 equation has to be H = q˙ pi + ψ˙ πα φAλ L . (44) i α − A − ∂H ψ˙ = ψ , H = = 0 . (38) Indeed, if the Lagrangian is independent of one coordinate, say α α P ∂πα { } − 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

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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 ∂H ∂φ1 i = 0 . (45) π˙ 1 = λ = iωψ λ , dt ∂q˙i − ∂qi 1 2 1 − ∂ψ1 − ψ1 − 2 i 2 In general, we define the generalized momentum as p = 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. Thus, Euler-Lagrange equation becomes d/dt(∂L/∂q˙ ) = 0 Furthermore, using Eq. (27), we get i ⇒ dpi/dt = 0. Hence, a momentum pi is conserved when the La- grangian L is independent of a coordinate q . This means that 1 i ˙1 i i π = ψ1 π = ψ˙1, if the Lagrangian is independent of a certain coordinate qi, it − 2 ⇒ − 2 i must be also independent of its corresponding momentum p . 2 i ˙2 i 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 ∂H ∂φA motion for the fermionic coordinates ψα: q˙ = + λ = q , H + φAλ , i ∂pi ∂pi A { i A}P ψ˙1 = ωψ2, ψ˙2 = ωψ1 . (52) A − i ∂H ∂φ i A p˙ = λA = p , H + φ λA P , − ∂qi − ∂qi { } Fortunately, these equations are the same equations of motion A followed from the Lagrangian formalism. ∂H ∂φ A ψ˙ α = λA = ψα, H + φ λA P , For now, let us use the Poisson bracket, the Hamiltonian − ∂πα − ∂πα { } (25), and the constraints (48), to check that ∂H ∂φA π˙ α = λ = πα, H + φAλ , ∂ψ ∂ψ A A P 1 1 1 − α − α { } φ˙ = φ , H + φ λ1 P A { } φ = 0. (46) ∂φ1 ∂(H + φ1λ ) ∂φ1 ∂(H + φ1λ ) = 1 1 i i Eqs. in (46) are derived in detail in Appendix A. For example, ∂qi ∂p − ∂p ∂qi ! using the Poisson brackets, we can show that Eqs. in (46) im- 1 1 1 1 i α ∂φ ∂(H + φ λ1) ∂φ ∂(H + φ λ1) ply that for any arbitrary function F = F(q , p , ψα, π ), we can i = α + α write F˙ = F, H + φAλ : − ∂ψα ∂π ∂π ∂ψα ! { A}P i i ∂F ∂q ∂F ∂pi ∂F ∂ψ ∂F ∂πα = 0 + 0 (λ ) (1) iωψ + λ ˙ = i + + α + − 2 1 − − 2 2 1 F i α ∂qi ∂t ∂p ∂t ∂ψα ∂t ∂π ∂t = i ( λ1 + ωψ2) = 0 . (53) ∂F ∂F ∂F ∂F − = q˙ + p˙i + ψ˙ + π˙α i i α α ∂qi ∂p ∂ψα ∂π The reason that the last step is zero comes from Eqs. (49, 52) ˙ ˙ 2 ∂F ∂H ∂φA ∂F ∂H ∂φA since we have λ1 = ψ1 = ωψ2. Similarly, φ can be calculated = + + − i i λA i i i λA using the Poisson bracket as follows: ∂qi ∂p ∂p ∂p − ∂q − ∂q A A 2 2 2 ∂F ∂H ∂φ ∂F ∂H ∂φ φ˙ = φ , H + φ λ2 P + α α λA + α α λA { } ∂ψα − ∂π − ∂π ∂π − ∂ψ − ∂ψα 2 2 2 2 ∂φ ∂(H + φ λ2) ∂φ ∂(H + φ λ2) A A = ∂F ∂(H + φ λ ) ∂F ∂(H + φ λ ) i i = A A ∂qi ∂p − ∂p ∂qi i i i ∂qi ∂p − ∂p ∂q 2 2 2 2 ∂φ ∂(H + φ λ2) ∂φ ∂(H + φ λ2) A A = α + α ∂F ∂(H + φ λA) ∂(H + φ λA) ∂F − ∂ψα ∂π ∂π ∂ψα α + α α − ∂ψα ∂π ∂ψ ∂π i i = 0 + 0 (λ ) (2) iωψ + λ F,H + φAλ . (47) − 2 2 − 1 2 2 ≡{ A}P = i ( λ ωψ ) = 0 . (54) At this stage, we need to confirm that Eqs. in (46) give us the − 2 − 1 correct equations of motion using the constrained Hamiltonian Also, the last step is zero because from Eqs. (49, 52), we have (25). Since α, β 1, 2 in the primary constraint (28), then, we λ = ψ˙ = ωψ . Subsequently, Eqs. (53, 54) are zero if we ∈ { } 2 − 2 − 1 have the two constraints impose the equations of motion (52). In other words, we can say i i that the constraints are consistent with the equations of motion. φ1 = π1 + ψ , φ2 = π2 + ψ . (48) 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 xx, xxx, 2018

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 The Dirac bracket of two arbitrary functions F(q, p) and G(q, p) Cαβ = φα, φβ is defined as { }P ∂φα ∂φβ ∂φα ∂φβ ∂φα ∂φβ ∂φα ∂φβ A B F, G = F, G F, φ C φ , G . (58) = i i α + α { }D { }P −{ }P AB{ }P ∂qi ∂p − ∂p ∂qi − ∂ψα ∂π ∂π ∂ψα ∂φα ∂φβ ∂φα ∂φβ Dirac bracket has the same properties of Poisson bracket, + ∂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, and 66, 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 Hamiltonian ψ , ψ = iδαβ . (69) (25) forget about the constraints and the momenta πβ { α β}D at the cost of replacing the Poisson brackets with Dirac brackets. For example, if we consider the case φ˙ A = 0, 3.4. Dirac Bracket and the Equations of Motion 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˙ = ψ , H , (70) α { α }D We check that where we used Eq. (57); φA, F = 0. { }D ˙ (iii) If φA, p = 0, then, using Eq. (58), we can find that ψ1 = ψ1, H D = ψ1, iωψ1ψ2 D = iω ψ1, ψ1ψ2 D { }P { } { − } − { } = iω ( ψ , ψ ψ + ψ ψ , ψ ) = iω iδ ψ + 0 A B 1 1 D 2 1 1 2 D αβ 2 q, p D = q, p P q, φ PCAB φ , p − { } { } − − { } { } −{ } { } = ωδ ψ = ωψ , (71) ∂q ∂p ∂q ∂p − αβ 2 − 2 = q, p P = = 1 . (61) { } ∂q ∂p − ∂p ∂q and similarly

We are now in a position to calculate ψ , ψ for the ψ˙2 = ψ2, H D = ψ2, iωψ1ψ2 D = iω ψ2, ψ1ψ2 D { α β}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 re- − { } { } − − = ωψ δ = ωδ ψ = ωψ . (72) lation which is obtained from the Lagrangian formalism: − 1 αβ αβ 1 1 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 Nöether’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 + ωqǫαβδ ψγ (73) = 1. (65) ⇒ Q2 = pψ2 ωqψ1 , − − 6 Letters in High Energy Physics LHEP xx, xxx, 2018 where again we suppose that α, β can only take the values a fermionic degree of freedom into a bosonic degree of freedom { } 1, 2 , and ǫ is the two index Levi-Civita symbol defined as and vice versa; i.e., { } αβ if is Q fermionic = bosonic , Q bosonic = fermionic . +1 (α, β) (1, 2), | i | i | i | i if is (80) ǫαβ =  1 (α, β) (2, 1), (74)  − if So far, we have restricted ourselves to study classical sys-  0 α = β . tems. In Subsection 3.5, we have investigated the supersymmetry of the classical harmonic oscillator. Now, we are going The system to be supersymmetric, the supercharges (73) to- to quantize the theory to study the supersymmetry of quan- gether with the Hamiltonian (27) and the constraints (25) must tum systems. In a quantum mechanical supersymmetric sys- satisfy the classical Dirac bracket superalgebra which defined tem, the supercharges Q together with the Hamiltonian H form by i a so-called superalgebra. The recipe for quantizing the Hamil- tonian in a situation where we have second class constraints is Qα, Qβ D = 2iδαβH, { } − to replace the Dirac brackets with either commutator brackets Q , H = 0 . (75) { α }D for bosonic variables or anticommutator brackets for fermionic variables multiplied with the factor ih¯ , so we have Now, we need to check that our system satisfies Eq. (75). To do that, suppose that α, β have the values 1, 2 . Then, { } { } q, p D = 1 [qˆ, pˆ]= ih¯ , using the definition of the Poisson and Dirac brackets, we find { } ⇒ ψ , ψ = iδ ψˆ , ψˆ = h¯ δ . that,(see Appendix B for more details) { α β}D − αβ ⇒ { α β} αβ (81) Q , Q = i p2 + ω2q2 , Q , Q = ω ψ2 + ψ2 , { 1 1}D − { 1 2}D 1 2 Taking this into account, we can see that the superalgebra for Q , Q = ω ψ2 + ψ2 , Q , Q = i p2 + ω2q2 . N-dimensional quantum system is characterized by { 2 1}D − 1 2 { 2 2}D − (76) [Qˆ , Hˆ ]= 0, i = 1 N, i ··· Qˆ , Qˆ = h¯ δ Hˆ , i, j = 1... N . (82) If we make substitution using Eq. (76), we find { i j} ij Q , Q = Q , Q + Q , Q + Q , Q + Q , Q This will be elaborated further in the next section using the ex- { α β}D { 1 1}D { 1 2}D { 2 1}D { 2 2}D 2 2 2 2 2 2 2 2 2 2 ample of the supersymmetrical harmonic oscillator. One more = i(p + ω q )+ ω(ψ1 + ψ2) ω(ψ1 + ψ2) i(p + ω q ) − − − notation before ending this section is that the supercharges Qi = 2i(p2 + ω2q2) 2iδ H . (77) are Hermitian, i.e. Q† = Q , and this implies that − ≡− αβ i i Furthermore, using the following relations (see Appendix B) Qˆ , Qˆ = Qˆ †, Qˆ † . (83) { i j} { i j } Q , H = ωpψ ω2qψ , Q , H = ωpψ + ω2qψ , { 1 }P 2 − 1 { 2 }D − 2 1 φ , H = iωψ , φ , H = iωψ , (78) 5. SUPERSYMMETRIC HARMONIC OSCIL- { 1 }P 2 { 2 }P 2 LATOR we find Now, we turn back to the quantum harmonic oscillator prob- Qα, H D = Q1, H D + Q2, H D lem as a simple example to show the supersymmetric prop- { } { } { } A B = Q1, H P Q1, φ PCAB φ , H P erty of a quantum mechanical system. Using Eq. (16), the har- { } −{ } { } monic oscillator Hamiltonian (17) can be written in terms of the + Q , H Q , φA C φB, H { 2 }P −{ 2 }P AB{ }P bosonic and fermionic laddering operators as follows: = Q , H Q , φ1 C φ1, H Q , φ2 C φ2, H { 1 }P −{ 1 }P 11{ }P −{ 1 }P 22{ }P + Q , H Q , φ1 C φ1, H Q , φ2 C φ2, H Hˆ = h¯ ω(aˆ†aˆ + bˆ†bˆ) . (84) { 2 }P −{ 2 }P 11{ }P −{ 2 }P 22{ }P 2 2 = ωPψ2 ω qψ1 ωPψ2 + ω qψ1 − − Moreover, the supercharge operators Qˆ 1 and Qˆ 2 can also be ωPψ ω2qψ + ω2qψ + ωPψ − 1 − 2 2 1 written in terms of the bosonic and fermionic laddering opera- = 0. (79) tors as follows:

It is clear from Eqs. (77, 79) that the supercharges (73) together Qˆ 1 = pψ1 + ωqψ2 with the Hamiltonian (27) and the constraints (28) satisfy the h¯ ω h¯ two conditions (75) of the classical Dirac bracket superalgebra. = i (aˆ† aˆ) (bˆ + bˆ†) r 2 − ! r 2 ! h¯ h¯ 4. THE SUPERSYMMETRY ALGEBRA + ω (aˆ)† + aˆ) i (bˆ bˆ†) r 2ω ! r 2 − ! The material in this section is elaborated in detail in [22, 23, 24, = ih¯ √ω(aˆ†bˆ aˆbˆ†) , (85) 25, 26, 27]. The supersymmetry algebra encodes a symmetry − describing a relation between bosons and fermions. In general, the supersymmetry is constructed by introducing supersymmetric transformations which are generated by the supercharge Qi operators, where the role of the supercharges Qi is to convert 7 Letters in High Energy Physics LHEP xx, xxx, 2018 and similarly, E Q Qˆ 2 = pψ2 ωqψ1 − (1) (2) E E h¯ ω h¯ 3 2 = i (aˆ† aˆ) i (bˆ bˆ†) r 2 − ! r 2 − ! Q† h¯ h¯ (1) (2) † ˆ ˆ† E E ω (aˆ) + aˆ) (b + b ) 2 1 − r 2ω ! r 2 ! = h¯ √ω(aˆ†bˆ + aˆbˆ †) . (86) − (1) (2) E E Using Eqs. (85, 86), one may deﬁne non-Hermitian opera- 1 0 tors Qˆ and Qˆ † as

1 (1) E Qˆ = (Qˆ iQˆ ) 0 2 1 − 2 E=0 1 † † † † = ih¯ √ω(aˆ bˆ aˆbˆ )+ ih¯ √ω(aˆ bˆ + aˆbˆ ) nF =0 nF =1 2 − = ih¯ √ω(aˆ†bˆ) , (87) FIGURE 1: A schematic view showing the energy levels for a supersymmetric system consisting of only two possible and similarly, fermionic states n = 0, 1 . The supercharge operators Q and F { } Q† exchange bosons and fermions without affecting the energy † 1 due to the degeneracy of the energy levels of the two supersym- Qˆ = (Qˆ 1 + iQˆ 2) 2 metric partners except for the ground state of the first partner. 1 = ih¯ √ω(aˆ†bˆ aˆbˆ†) ih¯ √ω(aˆ†bˆ + aˆbˆ†) 2 − − = ih¯ √ω(aˆbˆ†) . (88) 6. SUPERSYMMETRIC QUANTUM MECHAN- − ICS We can directly use Eq. (84, 87, 88) to check that the supercharge operators Qˆ and Qˆ † with the Hamiltonian H satisfy the quan- In this section, we study the general formalism of one- tum superalgebra: dimensional supersymmetric quantum mechanics. The ideas in this section were discussed in [3, 28, 29, 30, 31, 32]. We begin by Qˆ , Qˆ = Qˆ †, Qˆ † = 0, considering a Hamiltonian of the form { } { } Qˆ , Qˆ † = Qˆ †, Qˆ = H, { } { } 1 2 1 Hˆ = pˆ + V(xˆ) Á2 + hB¯ (xˆ)σ3 , (93) [H, Qˆ ]= [H, Qˆ †]= 0. 2 2 (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 operators Qˆ , Qˆ † on the energy eigenstates: function W(x), as

2 2 Qˆ nB, nF nB 1, nF + 1 , dW(x) 2 d W(x) V(x)= W′ B(x)= W′′ , | i∼| − i dx ≡ dx2 ≡ Qˆ † n , n n + 1, n 1 . (90) | B Fi∼| B F − i (94) the Hamiltonian is supersymmetric and we shall refer to the Remember that there are only two possible fermionic states function W(x) as a superpotential. Using Eq. (94), we can n = 0, 1 , so the effect of the operators Qˆ and Qˆ † can be F { } rewrite the Hamiltonian (93) as written as follows:

1 2 2 1 Qˆ n , 0 n 1, 1 , Hˆ = pˆ + W′ Á2 + h¯ σ3W′′ . (95) | B i∼| B − i 2 2 † Qˆ nB, 1 nB + 1, 0 . (91) | i∼| i At this stage, we deﬁne the following two Hermitian supercharge operators: Also, from Eq. (89), the operators Qˆ and Qˆ † commute with the Hamiltonian H; for that reason, we can see that 1 Qˆ = σ pˆ + σ W′(xˆ) , 1 2 1 2 H(Qˆ n , n )= Qˆ (H n , n ) = (E + E )(Qˆ n , n ) , (92) | B Fi | B Fi B F | B Fi 1 Qˆ 2 = σ2 pˆ σ1W′(xˆ) . (96) and this means that the whole energy of the system remains 2 − unchanged by the action of the supercharge operators. The Hamiltonian (95) together with the supercharges (96) con- ˆ In short, the supercharge operator Q 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 ˆ † variant. Conversely, Q 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.

8 Letters in High Energy Physics LHEP xx, xxx, 2018

1 1 commutation relation: = σ pˆ + σ W′ i σ pˆ σ W′ 2 1 2 − 2 2 − 1 ∂ ∂ ∂ 1 [pˆ, f (x)] g =[ ih¯ , f (x)] g = ih¯ ( f g)+ f (g) = (σ iσ )pˆ + (σ + iσ )W − ∂x − ∂x ∂x 1 2 2 1 ′ 2 − ∂g ∂ f ∂g 1 = ih¯ f + g f = ihf¯ ′(x) g . = (σ1 iσ2) pˆ + iW′ − · ∂x ∂x · − · ∂x − 2 − (97) = σ pˆ + iW′ , (104) − Removing g gives us the commutator of the momentum pˆ with and † Qˆ = Qˆ + iQˆ = σ+ pˆ iW′ , (105) an arbitrary function of the position coordinate x: 1 2 − where σ and σ+, respectively, are defined as [pˆ, f (x)] = ihf¯ ′(x). (98) − − 1 0 0 Using this relation, we find the following: σ = (σ1 iσ2)= , − 2 − 1 0 2 2 2 1 0 1 [pˆ, W′ ]= pWˆ ′ W′ pˆ = 2ihW¯ ′W′′, σ+ = (σ + iσ )= , (106) − − 2 1 2 0 0 2 2 2 [W′, pˆ ]= W′ pˆ pˆ W′ = ih¯ W′′, pˆ , − { } ˆ pˆ, W′′ = pWˆ ′′ + W′′ pˆ . (99) it is straightforward to check that the Hamiltonian H can be { } expressed as Now, we can use the outcome of Eq. (99) to check the outcome Hˆ = Qˆ , Qˆ † , (107) { } of the commutator of the supercharge Qˆ with the Hamiltonian 1 and it commutes with both the operators Qˆ and Qˆ †: Hˆ : [Qˆ , Hˆ ]= 0 & [Qˆ †, Hˆ ]= 0 . (108) [Qˆ , Hˆ ]= Qˆ Hˆ HQˆ 1 1 − 1 1 2 2 = (σ pˆ + σ W ) pˆ + W Á + hW¯ σ Furthermore, one also finds 4 1 2 ′ · ′ 2 ′′ 3 1 2 2 † † pˆ + W Á + hW¯ σ (σ pˆ + σ W ) ˆ ˆ ˆ ˆ − 4 ′ 2 ′′ 3 · 1 2 ′ Q, Q = 0 & Q , Q = 0 . (109) { } { } = 1 σ pˆ3 + σ pWˆ 2 + σ σ h¯ pWˆ + σ W pˆ2 + σ W W 2 + σ σ hW¯ W 4 1 1 ′ 1 3 ′′ 2 ′ 2 ′ ′ 2 3 ′ ′′ As we have seen in the case of the supersymmetric Har- 1 σ pˆ3 + σ pˆ2W + σ W 2 pˆ + σ W 2W + σ σ hW¯ pˆ + σ σ hW¯ W monic oscillator, a supersymmetric system with two indepen- − 4 1 2 ′ 1 ′ 2 ′ ′ 3 1 ′′ 3 2 ′′ ′ dent supercharges, with the possible exception of the energy of 1 2 = σ [pˆ, W ] σ2ih¯ pˆ, W + σ2[W , pˆ]+ 2σ ihW¯ W the ground eigenstate, all the energy levels are split into two 4 1 ′ − { ′′} ′ 1 ′ ′′ eigenstates with either n = 0 or n = 1. For a spinor quan- = 0. (100) F F tum mechanical system, this implies that the excited energy eigenstates come in degenerate spin-up/spin-down pairs E , Similarly, one can confirm that | n ↑ / E , . These degenerate spin-up/spin-down pairs are re- i | n ↓i lated to the acts of the supercharge operators Qˆ and Qˆ †, where [Qˆ 2, Hˆ ]= Qˆ 2Hˆ Hˆ Qˆ 2 = 0. (101) − the supercharge operators Qˆ convert the degenerate spin-up Furthermore, we can find that state to the degenerate spin-down state without making any change in the energy eigenvalue of the states: ˆ ˆ ˆ ˆ ˆ ˆ ˆ 2 Q1, Q1 = Q1Q1 + Q1Q1 = 2Q1 { } 0 0 0 1 2 Qˆ En, ↑ = , (110) = σ pˆ + σ2W′ | ↑i ∼ 1 0 0 2 1 ↓ 1 2 2 2 2 ˆ † = σ pˆ + σ W′ + σ σ pWˆ ′ + σ σ W′ pˆ while the supercharge operators Q convert the degenerate 2 1 2 1 2 2 1 spin-down state to the degenerate spin-up state without mak- 1 2 2 = (pˆ + W′ ) Á + iσ (pWˆ ′ W′ pˆ) ing any change in the energy eigenvalue of the states: 2 2 3 − 1 2 2 † 0 1 0 = (pˆ + W′ ) Á2 + h¯ σ3W′′ H , (102) Qˆ En, = ↑ . (111) 2 ≡ | ↓i ∼ 0 0 0 ↓ and similarly, we have To sum up, the supersymmetric transformations occur due to the supercharge operators Qˆ and Qˆ †. In this case, it causes Qˆ †, Qˆ = Hˆ . (103) { 2 2} transforms between the energy eigenstates spin-up/spin-down which have the same energy eigenvalues. Note that, to do the previous calculations, we used the Pauli 2 2 2 matrices properties: σ = σ = σ = Á and σ σ = σ σ = 1 2 3 2 1 2 − 2 1 iσ3. It is clear from Eqs. (100,101, 102, and 139) that the Hamilto- 7. SUPERSYMMETRIC GROUND STATE nian (95) together with the supercharges (96) satisfies the conditions for the superalgebra given by Eq. (82). So far, we have investigated all the required information to Indeed, if we define the quantities Q and Q† as describe a supersymmetric quantum mechanical system. In this section, we study the supersymmetric quantum mechan- Qˆ = Qˆ iQˆ ical ground state. The argument in this section follows from 1 − 2 9 Letters in High Energy Physics LHEP xx, xxx, 2018

[33, 34, 35, 36, 37]. Let us now write the expression of the super- and then, if we make substitution using Eq. (115), our problem symmetric Hamiltonian (95) as the summation of two separate reduces to solve the first-order differential equation: terms: a Hamiltonian Hˆ + and a Hamiltonian Hˆ , then, we have − + + + ∂ψ0 + A ψ = pˆ + iW′ ψ = 0 = ih¯ + iW′ψ = 0. 1 2 2 1 0 0 ∂x 0 Hˆ = pˆ + W′ hW¯ ′′ . (112) | i | i ⇒ − ± 2 ± 2 (120) It is simple and straightforward to solve the previous differen- In the eigenbasis of σ3, the supersymmetric Hamiltonian is di- tial equation as follows” agonal: † + Hˆ + 0 A A 0 dψ W W Hˆ = , (113) 0 = ′ dx = ψ+ = ′ + A ≡ 0 Hˆ 0 AA† + ln 0 ln . (121) − ψ0 h¯ ⇒ h¯ ˆ ˆ † and the supercharge operators Q and Q can be written as Thus, + W ψ = Ae h¯ , (122) 0 0 0 A† 0 Qˆ = , Qˆ † = , (114) A 0 0 0 and similarly, + W ψ = Be− h¯ . (123) where 0 Now, the general form of the ground-state wave function may † A = pˆ + iW′, A = pˆ iW′ . (115) be expressed as − W Ψ Ae h¯ According to Eq. (107), we can write the supersymmetric 0 = W . (124) Be h¯ . Hamiltonian Hˆ in terms of the supercharges operators Qˆ and − ! Qˆ † as follows: In fact, it has no physical meaning to find the ground- state wave function ψ0 if there are no normalizable solutions 2 Hˆ = Qˆ Qˆ † + Qˆ †Qˆ = 2Qˆ 2 = 2Qˆ † . (116) of such form. So we need now to normalize the ground-state wave function (124) by determining the values of the two con- In the last step, we used the fact that the supercharges are Her- stants A and B. The ground-state wave function ψ0(x) to be mitian operators. We see from Eq. (116) that the Hamiltonian normalizable must vanish at the positive and the negative infi- Hˆ can be written in terms of the squares of the supercharge nite x-values. In order to satisfy this condition, we have to set operators. For this reason, the energy of any eigenstate of this W(x) ∞ as x ∞. Then, we have the following three | | → | | → Hamiltonian must be positive or zero. Let us now consider that possible cases for the supersymmetric ground-state wave func- Ψ is the ground state of the supersymmetric Hamiltonian Hˆ . tion: | 0i Based on Eq. (116), it can be seen that the ground state can have a zero energy only if it satisfies the following two conditions: 1. The first case is when W(x) +∞ as x ∞. In → → ± this case, the superpotential W(x) is an even function E = Ψ Hˆ Ψ = Ψ Qˆ 2 Ψ = 0 = Qˆ Ψ = 0, and it is positive at the boundaries. Figure 2 describes 0 h 0| | 0i h 0| | 0i ⇒ | 0i 2 the behavior of the potential V(x) in this case. Since E = Ψ Hˆ Ψ = Ψ Qˆ † Ψ = 0 = Qˆ † Ψ = 0. (117) 0 h 0| | 0i h 0| | 0i ⇒ | 0i W(x) is positive, we cannot normalize the wave function W(x) + h¯ Therefore, if there exists a state which is annihilated by each of ψ0 = Ae , but we can choose A = 0. However, we ˆ ˆ † W(x) the supercharge operators Q and Q which means that it is in- h¯ can normalize the wave function ϕ0− = Be− to find variant under the supersymmetry transformations, such a state the value of the constant B. The complete ground-state is automatically the zero-energy ground state. However, on the wave function, in this case, may be expressed in the gen- other hand, any state that is not invariant under the supersym- eral form metry transformations has a positive energy. Thus, if there is a 0 supersymmetric state, it is the zero-energy ground state and it Ψ0 = W(x) . (125) Be h¯ is said that the supersymmetry is unbroken. − ! Moreover, since the supersymmetry algebra (108) implies that the supercharge operators Qˆ and Qˆ † commute with the su- V(x) persymmetric 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:

ψ+ x Ψ = 0 . (118) 0 ψ 0− One can now solve eigenvalue problem Qˆ ψ = 0 to ﬁnd the | 0i 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, + ∞ ∞ 0 0 ψ0 where W(x) + as x . Qˆ ψ0 = = 0 , (119) → →± | i A 0 ψ− 0 10 Letters in High Energy Physics LHEP xx, xxx, 2018

2. The second case is when W(x) ∞ as x ∞. V(x) → − → ± In this case, the superpotential W(x) is an even function and it is negative at the boundaries. The behavior of the potential V(x) in this case is described in Figure 3. Since W(x) is negative, we can normalize the wave x W(x) Ψa h¯ function 0 = Ae and ﬁnd the value of the constant A; however we cannot normalize the wave func- W(x) Ψb h¯ tion 0 = Be− , but we can choose B = 0. The complete ground-state wave function, in this case, shall be expressed in the general form

W(x) Ae h¯ (a) W(x) ∞ as x ∞, and W(x) ∞ as x ∞. Ψ0 = . (126) →− →− → → 0 ! V(x)

V(x)

x x

(b) W(x) ∞ as x ∞, and W(x) ∞ as x ∞. → →− →− → FIGURE 4: The superpotentialW(x) is an odd function. FIGURE 3: The superpotentialW(x) is an even function, where W(x) ∞ as x ∞. →− →± state. Then, we use the conventional perturbation theory to examine whether it has nonvanishing corrections to the energy of the ground state, and accordingly, it can be responsible for 3. The other two cases correspond to W(x) +∞, ∞ → { − } the spontaneous breaking of the supersymmetry. To this end, as x +∞, ∞ , and W(x) ∞, +∞ as x → { − } → {− } → in the next two subsections, based on [38, 39, 40, 41, 42], we are +∞, ∞ . In those two cases, the superpotential W(x) { − } going to compute both the ﬁrst and the second corrections to is an odd function and it is positive at one of the bound- the ground-state energy of the supersymmetric harmonic os- aries and negative at the other boundary. The behavior of cillator. For instance, let us recall the general form of the super- the potential V(x) in this case is described in Figure 4. In symmetric quantum mechanics Hamiltonian, which is given by those two cases, both the two constants A and B vanish Eq. (95): and we cannot normalize the wave function. Therefore. 1 2 2 1 we have Hˆ = (pˆ + W′ ) Á + hW¯ ′′σ . (128) 2 2 2 3 Ψ0 = 0. (127) Consider now that the superpotential W(x) is deﬁned as Thus, since those two forms of the zero-energy ground- 1 W = ωqˆ + 3gqˆ2 state wave function cannot be normalized, this means W(q)= ωqˆ2 + gqˆ3 ′ (129) that they do not exist. 2 ⇒ W′′ = ω + 6gqˆ , g g = In short, the zero-energy ground-state wave function can be where is a perturbation. If 0, the previous Hamiltonian re- normalized when the superpotential W(x) has an even num- duces to the supersymmetric harmonic oscillator Hamiltonian: ber of zeros. In this case, there exists a zero-energy ground 0 1 2 2 2 1 Hˆ = (pˆ + ω q ) Á + h¯ ωσ . (130) state which is the true vacuum state and the supersymmetry 2 2 2 3 is unbroken. However on the other hand, if the superpotential W(x) has an odd number of zeros, the zero-energy ground- We have already solved this Hamiltonian and found that it has state wave function cannot be normalized. Then, we immedi- a unique zero-energy ground state, ately realize that there is no zero-energy ground state in such (0) E = 0 , (131) case and the supersymmetry is spontaneously broken. 0 and we verify that this state is invariant under supersymmetry. Based on our argument in the previous section and after nor- 8. CORRECTIONS TO THE GROUND-STATE malization, the unbroken supersymmetric ground-state wave ENERGY function can be written as follows:

In this section, we consider the example of the supersymme- (0) 0 Ψ 2 0 = 1 ωq . (132) try harmonic oscillator which has a unique zero-energy ground ω 4 2¯h ( πh¯ ) e− ! 11 Letters in High Energy Physics LHEP xx, xxx, 2018

3 Now, when g is small, the Hamiltonian (128) can be written The terms qˆ ψ0 and qˆψ0 are linearly independent of ψ0 so that as both the second and the third terms of Eq. (139) are zero. How- 0 Hˆ = Hˆ + Hˆ ′ , (133) ever, 3¯h4 7¯h4 9¯h4 where Hˆ 0 is the Hamiltonian of the unperturbed system and Hˆ ˆ4 = + + ′ q ψ0 4 ψ4 4 ψ2 4 ψ0 , (140) a perturbation: 2ω 4ω 16ω then, substituting it into Eq. (139) gives us, 9 2 4 3 ˆ = ( + ) Á + 2 H′ g qˆ 3ωgqˆ 2 3¯hgqˆσ3 . (134) ( ) 9 ( ) ( ) 9 3 h¯ ( ) ( ) 2 E 1 = g2 ψ 0 qˆ4 ψ 0 = g2 ψ 0 ψ 0 0 2 h 0 | | 0 i 2 × 4 ω2 h 0 | 0 i Moreover, if we consider the matrix representation, the pertur- 27 h¯ 2 bative Hamiltonian Hˆ acting on the ground-state wave function = g2 (g2) . (141) 8 ω2 ≃O Ψ(0) 0 yields In the view of the last equation, the first-order correction to 9 2 4 3 0 the ground-state energy reduces to zero. Another method with (0) 2 g qˆ + 3ωgqˆ + 3¯hgqˆ 0 Hˆ Ψ = ωq2 = ′ 0 9 2 4 3 ω 1 0. 0 g qˆ + 3ωgqˆ 3¯hgqˆ ( ) 4 e 2¯h more details to calculations of the first-order correction of the 2 − πh¯ − ! (135) ground-state energy is given in Appendix C. In view of this last equation, it is clear that for this combination ˆ Ψ(0) 8.2. The Second-Order Correction the impact of the Hamiltonian H′ on the wave function 0 is only due to the component From the nondegenerate time-independent perturbation theory, the second-order correction to the energy is given by 9 ˆ = g2qˆ4 + 3ωgqˆ3 3¯hgqˆ . (136) ′ (0) (0) 2 H 2 − ( ) ψ ˆ ψ E 0 = |h n |H′| m i| . (142) 2 ∑ (0) (0) As a result, we see that it is enough to consider the Hamiltonian m=n Em En 6 − ˆ ′ to compute the first and second corrections to the ground- H 3 4 state energy (see Appendix C). Since the perturbation contains only terms of q, q , and q , the Furthermore, we have to mention here that, for the super- numerator of Eq. (142) is zero for all m values except m = 1, 3, 4. symmetric harmonic oscillator, except for the ground state, all For more explanation about this point, see Appendix D. There- the energy levels are degenerate to two energy states. However, fore, the second-order correction to ground-state energy of the because we are just interested in computing the correction to supersymmetric harmonic oscillator is computed as follows: the ground-state energy, which is not degenerate, we can use (0) 9 2 4 3 (0) 2 the nondegenerate perturbation theory. For more explanation (0) ψ g qˆ + 3ωgqˆ 3¯hgqˆ ψ E = |h 0 | 2 − | 1 i| 2 ( ) ( ) about this point see Appendix C. E 0 E 0 In addition, it is useful to remember from quantum me- 0 − 1 (0) (0) chanics that the wave function of degree n can be obtained by ψ 9 g2qˆ4 + 3ωgqˆ3 3¯hgqˆ ψ 2 + |h 0 | 2 − | 2 i| ( ) ( ) the following recursion formula: E 0 E 0 0 − 2 (0) (0) (n + 1)h¯ nh¯ ψ 9 g2qˆ4 + 3ωgqˆ3 3¯hgqˆ ψ 2 qˆψ = ψ + ψ . (137) + |h 0 | 2 − | 3 i| n n+1 n 1 (0) (0) r 2ω r 2ω − E E 0 − 3 (0) As well, it is important to recall the following relation: ψ 9 g2qˆ4 + 3ωgqˆ3 3¯hgqˆ ψ0 2 + |h 0 | 2 − | 4i| . (143) (0) (0) ψ ψ = δ . (138) E0 E4 h n| mi nm − We are ready now to move forward to the next two subsections We can simplify the previous equation by calculating the nu- and compute the first and the second corrections to the energy merator of each term using the recurrence relation for the har- of the ground state. monic oscillator energy eigenfunction given by Eq. (137). We get

8.1. The First-Order Correction h¯ h¯ We know from the previous argument that the ground state of qˆψ1 = ψ2 + ψ0, r ω r 2ω the supersymmetric Hamiltonian is nondegenerate. Therefore, 3 3 3 we can use the time-independent nondegenerate perturbation 3 3¯h h¯ 9¯h qˆ ψ1 = ψ4 + 3 ψ2 + ψ0, theory to compute the ﬁrst-order correction to the ground-state s ω3 s ω3 s 8ω3 energy of the supersymmetric harmonic oscillator as follows: 3 15¯h3 72¯h3 243¯h3 3¯h3 qˆ ψ3 = ω3 ψ6 + ω3 ψ4 + 8ω3 ψ2 + 4ω3 ψ0, (1) (0) (0) q 4 q 4 q 4 q 4 ˆ 4 45¯h 147¯h h¯ 9¯h E0 = ψ0 ′ ψ0 ˆ = h |H | i q ψ2 2ω4 ψ6 + ω4 ψ4 + 12.5 ω4 ψ2 + 2ω4 ψ0, 9 (0) 2 4 3 (0) q 4 q 4 q4 q4 4 = ψ0 g qˆ + 3ωgqˆ 3¯hgqˆ ψ0 4 105¯h + 945¯h + 2725¯h + 81¯h + 3¯h h | 2 − | i qˆ ψ4 = ω4 ψ8 4ω4 ψ6 16ω4 ψ4 2ω4 ψ2 2ω4 ψ0 . (0) (0) (0) (0) (0) (0) q q q q q = 9 g2 ψ qˆ4 ψ + 3ωg ψ qˆ3 ψ 3¯hg ψ qˆ ψ . (144) 2 h 0 | | 0 i h 0 | | 0 i− h 0 | | 0 i (139) In Eq. (144), we only list the terms which contain ψ0. The other terms do not affect our calculations, since all of them give

12 Letters in High Energy Physics LHEP xx, xxx, 2018 us zero (see Appendix C). Using Eq. (144) to calculate Eq. (143), showed that it obeys the superalgebra. Then, we calculated the we get ground-state energy for that system and found evidence that it vanishes, E0 = 0, and consequently, this supersymmetric har- 2 2 monic oscillator has an unbroken supersymmetry. g2 9¯h3 h¯ 81g4 9¯h4 (0) In the previous section, we applied a small perturbation g E2 = 3ω 3 3¯h 4 − h¯ ω × ω − × r 2ω ! − 8¯hω s 2ω  to the supersymmetric harmonic oscillator and calculated the 2  2  effect on the ground-state energy. We found out that the per- g2 3¯h3 g4 9 3¯h4 turbation does not affect the energy of the ground state at the 3ω − 3¯hω  × s 4ω3  − 4¯hω  2 × s 2ω4  second order in perturbation theory, but this result can be expanded to any finite order. This means that the supersymmetry 9 h¯ 2  729 h¯ 3  9 h¯ 2 243 h¯ 3  breaking does not occur because of perturbation, and it must = g2 g4 g2 g4 − 8 ω2 − 16 ω5 − 4 ω2 − 32 ω5 be due to the nonperturbative effects. 2 3 Again, let us consider the same potential Eq. (129), which 27 h¯ 2 1701 h¯ 4 = g g . (145) we have used before in Section 8: − 8 ω2 − 32 ω5 1 For more details, another method to compute the second-order W(q)= ωqˆ2 + gqˆ3 . (147) correction of the ground-state energy is given in Appendix D. 2 Finally, with regard to Eqs. (141 and 145), we realize that up As we discussed in Section 7, the wave function of the ground to the second order the energy corrections to the ground-state state should have three zeros, Ψ +∞ as x +∞, and 0 → → energy vanish. This is concluded as Ψ ∞ as x ∞. Figure 5, shows an arbitrary diagram 0 → − → − presenting how the wave function of the ground state should (g)= 0, O look like, and Eq. (148) gives us the nonnormalized form of the 27 h¯ 2 27 h¯ 2 ground-state wave function: (g2)= = 0 . (146) O 8 ω2 − 8 ω2 W Ae− h¯ It is worth noting that the second term in Eq. (145) should be Ψ0 = W . (148) canceled when we extend the calculation to the fourth-order Be− h¯ ! perturbation corrections. Showing this would make the calculations here more complicated and confusing. However, the re- Acutely, Eq. (148) could not be normalized, and the only way sult can be expanded to any finite order in perturbation theory. to solve the Schrödinger equation for the ground state is taking Ψ Hence, there are no corrections to the energy of the ground state 0 = 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- ˆ ˆ † x percharge operators Q and Q . 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. Also, we have explained that if there is a supersymmetric state, state, which is E0 = 0, while if the system has a positive it is the zero-energy ground state. If such a zero-energy ground ground-state energy, E0 > 0, it is said to have a broken supersymmetry [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

13 Letters in High Energy Physics LHEP xx, xxx, 2018 nature, and if nature is described by supersymmetry, of course, From Eqs. (A.6, A.7), we obtain it must be broken. In fact, supersymmetry may be broken spontaneously at ∂H ∂φA q˙ = + λ = q , H + φAλ . (A.8) any order of perturbation theory or dynamically due to non- i ∂p ∂p A { i A}P perturbative effects. To examine this statement, we studied the normalization of the ground state of the supersymmetric har- (ii) monic oscillator. Then, we used perturbation theory to calculate the corrections to the ground-state energy. We found out ∂H ∂L ∂φA d ∂L ∂φA that the perturbation does not affect the energy of the ground = λA = λA ∂qi − ∂qi − ∂qi − dt ∂q˙i − ∂qi state at second order in perturbation theory, but this result can A i ∂φ be expanded to any finite order. This means that the supersym- = p˙ λA . (A.9) metry breaking is not seen in perturbation theory, and it must − − ∂qi be due to the nonperturbative effects. Thus, A i ∂H ∂φ p˙ = λA . (A.10) ACKNOWLEDGEMENTS − ∂qi − ∂qi Using the Poisson brackets, we find that The author acknowledges the research office of the University of the Witwatersrand and the African Institute for Mathemati- ∂pi ∂(H + φAλ ) ∂(H + φAλ ) ∂pi pi, H + φAλ A A cal Sciences (Ghana) for financial support. A P = i i { } ∂qi ∂p − ∂qi ∂p ! i A A i ∂p ∂(H + φ λA) ∂(H + φ λA) ∂p α α Appendix A. THE GENERALIZED HAMIL- − ∂ψα ∂π − ∂ψα ∂π ! ∂H ∂φA TON’S EQUATIONS OF MO- = . (A.11) TION − ∂qi − ∂qi Recall the Hamiltonian (25): From Eqs. (A.10, A.11), we obtain

1 2 2 2 A H = p + ω q iωψ1ψ2 . (A.1) i ∂H ∂φ i A 2 − p˙ = λA = p , H + φ λA P . (A.12) − ∂qi − ∂qi { } We showed in Subsection 3.2 that the previous Hamiltonian could be written in the form (29): (iii) 1 2 2 2 1 2 H = p + ω q + 4iωπ π . (A.2) ∂H ∂φA 2 = ψ˙ λ . (A.13) α α α A Also consider the momenta πα and the primary constraint φα, ∂π − − ∂qπ respectively, define as Thus, ∂H ∂φA α ∂L i αβ ˙ π = = δ ψβ, ψα = α α λA . (A.14) ∂ψ˙α − 2 − ∂π − ∂qπ i φα = πα + δαβψ . (A.3) Using the Poisson brackets, we find that 2 β A A i ∂ψα ∂(H + φ λ ) ∂(H + φ λ ) ∂p Downward, we explain how to use the Poisson bracket to get ψ , H + φAλ A A α A P = i the generalized Hamilton’s equations of motion, which we { } ∂qi ∂p − ∂qi ∂ψα ! have only write down in Eq. (46). A A ∂ψα ∂(H + φ λA) ∂(H + φ λA) ∂ψα α α (i) The extended Hamiltonian (44) is written as − ∂ψα ∂π − ∂ψα ∂π ! i α A ∂H ∂φA H = q˙ p + ψ˙ απ φ λ L . (A.4) = . (A.15) i − A − − ∂πα − ∂πα Therefore, From Eqs. (A.14, A.15), we obtain ∂H ∂φA = q˙i λA . (A.5) ∂pi − ∂pi ∂H ∂φA ψ˙ = λ = ψ , H + φAλ . (A.16) Thus, α − ∂πα − ∂qπα A { α A}P ∂H ∂φA q˙i = + λA . (A.6) ∂p ∂p (iv) Using Poisson brackets, we find that ∂H ∂L ∂φA A A = λA A ∂qi ∂(H + φ λA) ∂(H + φ λA) ∂qi q , H + φ λ P = ∂ψα − ∂ψα − ∂qψα { i A} ∂q ∂pi − ∂q ∂pi i i ! d ∂L ∂φA A A = λA ∂qi ∂(H + φ λA) ∂(H + φ λA) ∂qi − dt ∂ψ˙ α − ∂ψα α α − ∂ψα ∂π − ∂ψα ∂π ! A α ∂φ ∂H ∂πA = π˙ λA . (A.17) = + λA. (A.7) ∂ψ ∂pi ∂pi − − α 14 Letters in High Energy Physics LHEP xx, xxx, 2018

∂Q1 ∂Q2 ∂Q1 ∂Q2 Thus, Q1, Q2 P = i i A { } ∂qi ∂p − ∂p ∂qi α ∂H ∂φ π˙ = λA . (A.18) ∂Q1 ∂Q2 ∂Q1 ∂Q2 ∂Q1 ∂Q2 ∂Q1 ∂Q2 ∂ψ ∂ψ 1 + 1 2 + 2 − α − α − ∂ψ1 ∂π ∂π ∂ψ1 − ∂ψ2 ∂π ∂π ∂ψ2 Using the Poisson brackets, we obtain 2 2 = ω ψ1 + ψ2 , (B.27) A A i A ∂πα ∂(H + φ λA) ∂(H + φ λA) ∂p πα, H + φ λA P = { } ∂q ∂pi − ∂q ∂πα i i ! ∂Q2 ∂Q1 ∂Q2 ∂Q1 Q2, Q1 P = ∂q i i ∂q A A { } i ∂p − ∂p i ∂πα ∂(H + φ λA) ∂(H + φ λA) ∂πα α α ∂Q2 ∂Q1 ∂Q2 ∂Q1 ∂Q2 ∂Q1 ∂Q2 ∂Q1 − ∂ψα ∂π − ∂ψα ∂π ! 1 + 1 2 + 2 − ∂ψ1 ∂π ∂π ∂ψ1 − ∂ψ2 ∂π ∂π ∂ψ2 A α ∂φ = π˙ λA . (A.19) 2 2 − − ∂ψ = ω ψ + ψ , (B.28) α − 1 2 From Eqs. (A.18, A.19), we obtain

A ∂Q2 ∂Q2 ∂Q2 ∂Q2 ∂H ∂φ Q2, Q2 P = i i π˙ α = λ = π , H + φAλ . (A.20) { } ∂qi ∂p − ∂p ∂qi ∂ψ ∂ψ A α A P − α − α { } ∂Q2 ∂Q2 ∂Q2 ∂Q2 ∂Q2 ∂Q2 ∂Q2 ∂Q2 1 + 1 2 + 2 − ∂ψ1 ∂π ∂π ∂ψ1 − ∂ψ2 ∂π ∂π ∂ψ2 (v) By deﬁnition from Eq. (A.3), = 0, (B.29) A A i φ = π + ψA 2 ∂Q ∂φ1 ∂Q ∂φ1 Q , φ1 = 1 1 i i 1 P ∂q ∂pi ∂pi ∂q = ψ + ψ { } i − i A A 1 1 1 1 − 2 2 ∂Q1 ∂φ ∂Q1 ∂φ ∂Q1 ∂φ ∂Q1 ∂φ 1 + 1 2 + 2 = 0 . (A.21) − ∂ψ1 ∂π ∂π ∂ψ1 − ∂ψ2 ∂π ∂π ∂ψ2 = p , (B.30) −

Appendix B. DIRAC BRACKET AND THE 1 1 1 ∂φ ∂Q1 ∂φ ∂Q1 φ , Q1 P = i i SUPERALGEBRA { } ∂qi ∂p − ∂p ∂qi 1 1 1 1 ∂φ ∂Q1 ∂φ ∂Q1 ∂φ ∂Q1 ∂φ ∂Q1 From (Appendix A), recall the expressions of the Hamiltonian 1 + 1 2 + 2 − ∂ψ1 ∂π ∂π ∂ψ1 − ∂ψ2 ∂π ∂π ∂ψ2 (A.1) and the extended Hamiltonian (A.2): = p , (B.31) − 1 2 2 2 H = p + ω q iωψ1ψ2. (B.22) 2 − 2 2 2 ∂Q1 ∂φ ∂Q1 ∂φ Q1, φ P = i i As well, take into account the for the constraint (A.3) and sup- { } ∂qi ∂p − ∂p ∂qi 2 2 2 2 pose that α, β can only take the values 1, 2 ; then, we have ∂Q1 ∂φ ∂Q1 ∂φ ∂Q1 ∂φ ∂Q1 ∂φ { } { } 1 + 1 2 + 2 − ∂ψ1 ∂π ∂π ∂ψ1 − ∂ψ2 ∂π ∂π ∂ψ2 1 1 i φ = π + ψ1, = ωq , (B.32) 2 − 2 2 i φ = π + ψ2 . (B.23) 2 2 2 2 ∂φ ∂Q1 ∂φ ∂Q1 φ , Q1 P = i i { } ∂qi ∂p − ∂p ∂qi Furthermore, consider the supercharge formula, and once 1 2 2 2 ∂φ ∂Q1 ∂φ ∂Q1 ∂φ ∂Q1 ∂φ ∂Q1 again suppose that α, β can only take the values 1, 2 , then 1 + 1 2 + 2 { } { } − ∂ψ2 ∂π ∂π ∂ψ1 − ∂ψ2 ∂π ∂π ∂ψ2 we have = ωq , (B.33) − βγ Q1 = pψ1 + ωqψ2 Qα = pψα + ωqǫαβδ ψγ (B.24) ⇒ Q2 = pψ2 ωqψ1 , ∂Q ∂φ1 ∂Q ∂φ1 − Q , φ1 = 2 2 2 P ∂q ∂pi ∂pi ∂q αβ { } i − i where ǫαβ and δ are the Levi-Civita symbol and Kronecker 1 1 1 1 ∂Q2 ∂φ ∂Q2 ∂φ ∂Q2 ∂φ ∂Q2 ∂φ delta function, respectively. 1 + 1 2 + 2 − ∂ψ1 ∂π ∂π ∂ψ1 − ∂ψ2 ∂π ∂π ∂ψ2 To verify that the supercharges (B.24) together with the = ωq , (B.34) Hamiltonian (B.22) and the constraints (B.23) satisfy the Dirac bracket superalgebra (75), we need as a ﬁrst step to calculate 1 1 the Dirac brackets of the supercharges: 1 ∂φ ∂Q2 ∂φ ∂Q2 φ , Q2 P = i i { } ∂qi ∂p − ∂p ∂qi Qα, Qβ D = Q1, Q1 D + Q1, Q2 D + Q2, Q1 D + Q2, Q2 D . (B.25) 1 1 1 1 { } { } { } { } { } ∂φ ∂Q2 ∂φ ∂Q2 ∂φ ∂Q2 ∂φ ∂Q2 1 + 1 2 + 2 − ∂ψ1 ∂π ∂π ∂ψ1 − ∂ψ2 ∂π ∂π ∂ψ2 To simplify the calculation of Eq. (B.25), let us start with calculating the Poisson bracket of the supercharges and the con- = ωq , (B.35) straints. 2 2 ∂Q1 ∂Q1 ∂Q1 ∂Q1 2 ∂Q2 ∂φ ∂Q2 ∂φ Q1, Q1 P = i i Q2, φ P = i i { } ∂qi ∂p − ∂p ∂qi { } ∂qi ∂p − ∂p ∂qi 2 2 2 2 ∂Q1 ∂Q1 ∂Q1 ∂Q1 ∂Q1 ∂Q1 ∂Q1 ∂Q1 ∂Q2 ∂φ ∂Q2 ∂φ ∂Q2 ∂φ ∂Q2 ∂φ 1 + 1 2 + 2 1 + 1 2 + 2 − ∂ψ1 ∂π ∂π ∂ψ1 − ∂ψ2 ∂π ∂π ∂ψ2 − ∂ψ1 ∂π ∂π ∂ψ1 − ∂ψ2 ∂π ∂π ∂ψ2 = 0, (B.26) = p , (B.36) − 15 Letters in High Energy Physics LHEP xx, xxx, 2018

2 2 2 ∂φ ∂Q2 ∂φ ∂Q2 ∂Q2 ∂H ∂Q2 ∂H ∂Q2 ∂H ∂Q2 ∂H φ , Q2 P = i i Q , H = + { } ∂qi ∂p ∂p ∂qi 2 P i i α α − { } ∂qi ∂p − ∂p ∂qi − ∂ψα ∂π ∂π ∂ψα 2 2 2 2 ∂φ ∂Q2 ∂φ ∂Q2 ∂φ ∂Q2 ∂φ ∂Q2 2 ∂ψ 1 + 1 ∂ψ ∂ψ 2 + 2 ∂ψ = ωpψ ω qψ , (B.47) − 1 ∂π ∂π 1 − 2 ∂π ∂π 2 − 1 − 2 = p . (B.37) − ∂φ ∂H ∂φ ∂H ∂φ ∂H ∂φ ∂H Using Dirac bracket definition and Eq.s ( B.26-B.37), then, φ , H = 1 1 1 + 1 { 1 }P ∂q ∂pi − ∂pi ∂q − ∂ψ ∂πα ∂πα ∂ψ we get i i α α = iωψ2 , (B.48) Q , Q = Q , Q Q , φA C φB, Q { 1 1}D { 1 1}P −{ 1 }P AB{ 1}P 1 1 2 2 = Q1, Q1 P Q1, φ PC11 φ , Q1 P Q1, φ PC22 φ , Q1 P { } −{ } { } −{ } { } ∂φ ∂H ∂φ ∂H ∂φ ∂H ∂φ ∂H 2 2 2 φ , H = 2 2 2 + 2 = i(p + ω q ) , (B.38) { 2 }P ∂q ∂pi − ∂pi ∂q − ∂ψ ∂πα ∂πα ∂ψ − i i α α = iωψ . (B.49) − 1 A B Q1, Q2 D = Q1, Q2 P Q1, φ PCAB φ , Q2 P { } { } −{ } { } Therefore, using Eqs. ( B.46,B.47, B.48, and B.49), we calcu- = Q , Q Q , φ1 C φ1, Q Q , φ2 C φ2, Q { 1 2}P −{ 1 }P 11{ 2}P −{ 1 }P 22{ 2}P late the Dirac bracket of the supercharges and the Hamiltonian. 2 2 = ω(ψ1 + ψ2) , (B.39) Q , H = Q , H Q , φA C φB, H { 1 }D { 1 }P −{ 1 }P AB{ }P = Q , H Q , φ1 C φ1, H Q , φ2 C φ2, H A B { 1 }P −{ 1 }P 11{ }P −{ 1 }P 22{ }P Q2, Q1 D = Q2, Q1 P Q2, φ PCAB φ , Q1 P { } { } −{ } { } = ωpψ ω2qψ ωpψ + ω2qψ . (B.50) = Q , Q Q , φ1 C φ1, Q Q , φ2 C φ2, Q 2 − 1 − 2 1 { 2 1}P −{ 2 }P 11{ 1}P −{ 2 }P 22{ 1}P = ω(ψ2 + ψ2) , (B.40) − 1 2 Q , H = Q , H Q , φA C φB, H { 2 }D { 2 }P −{ 2 }P AB{ }P Q , H Q , φ1 C φ1, H Q , φ2 C φ2, H A B = 2 P 2 P 11 P 2 P 22 P Q , Q D = Q2, Q2 P Q2, φ PC φ , Q2 P { } −{ } { } −{ } { } { 2 2} { } −{ } AB{ } 2 2 1 1 2 2 = ωpψ1 ω qψ2 + ω qψ2 + ωpψ1 , (B.51) = Q2, Q2 P Q2, φ PC11 φ , Q2 P Q2, φ PC22 φ , Q2 P − − { } −{ } { } −{ } { } = i(p2 + ω2q2) . (B.41) Substitute Eqs. (B.50, B.51) into equation (B.45), thus, we find − Substituting Eqs. (B.38, B.39, B.40, and B.41) into Eq. (B.25), Q , H = Q , H + Q , H { α }D { 1 }D { 2 }D thus, we find = ωpψ ω2qψ ωpψ + ω2qψ 2 − 1 − 2 1 Qα, Qβ D = Q1, Q1 D + Q1, Q2 D + Q2, Q1 D + Q2, Q2 D 2 2 { } { } { } { } { } + ωpψ1 ω qψ2 + ω qψ2 + ωpψ1 2 2 2 2 2 2 2 2 2 2 − − = i(p + ω q )+ ω(ψ1 + ψ2) ω(ψ1 + ψ2) i(P + ω q ) − − − = 0. (B.52) = 2i(P2 + ω2q2) . (B.42) − It is clear from Eqs. (B.44 and B.52) that the supercharges (B.24) On the other hand, together with the Hamiltonian (B.22) and the constraints (B.23) satisfy the two conditions (75) of the Dirac bracket superalge- 2iδ H = 2iδ H 2iδ H − αβ − 11 − 22 bra. = 2i(p2 + ω2q2) 2i(P2 + ω2q2) − − = 2i(p2 + ω2q2) . (B.43) − Appendix C. THE FIRST-ORDER CORREC- Eqs. (B.42) and (B.43) imply that TION In this appendix, we use conventional perturbation theory to Q , Q = 2iδ H . (B.44) { α β}D − αβ compute the first-order correction to the ground-state energy of the supersymmetric harmonic oscillator. Good discussion of So far, as a second step to verify that the supercharges (B.24) the perturbation theory can be found in [38, 39, 40, 41, 42]. In together with the Hamiltonian (B.22) and the constraints (B.23) the nondegenerate time-independent perturbation theory, the satisfy the Dirac bracket superalgebra (75), we need to calculate first-order correction to the energy of the nth’ state is given by the Dirac brackets of the supercharges and the Hamiltonian: (1) 0 ˆ 0 Q , H = Q , H + Q , H . (B.45) En = ψn H′ ψn . (C.53) { α }D { 1 }D { 2 }D h | | i To simplify the calculation of Eq. (B.45), let us calculate the We need now to use this formula to compute the first-order cor- Poisson bracket of the supercharges and the Hamiltonian. rection to the ground-state energy of our system. As we have shown in Section 6, the supersymmetric quantum mechanics ∂Q ∂H ∂Q ∂H ∂Q ∂H ∂Q ∂H Hamiltonian is given in the general form = 1 1 1 + 1 Q1, H P i i α α { } ∂qi ∂p − ∂p ∂qi − ∂ψα ∂π ∂π ∂ψα 1 2 2 1

ˆ Á = ωpψ ω2qψ , (B.46) H = (pˆ + W′ ) 2 + hW¯ ′′σ3 . (C.54) 2 − 1 2 2

16 Letters in High Energy Physics LHEP xx, xxx, 2018

Let us here consider the superpotential W(x) deﬁned as Both the second term and the third term of the previous integration vanish since the integral functions qˆ3 exp( ωqˆ2/¯h) and − 1 W = ωqˆ + 3gqˆ2 qˆ exp( ωqˆ2/¯h) are odd functions and there integrations from W = ωqˆ2 + gqˆ3 ′ (C.55) − 2 ⇒ W′′ = ω + 6gqˆ , ∞ to +∞ are equal to zero. Therefore the previous integration − reduces to where g is a perturbation. If g is zero, the Hamiltonian Hˆ re- 2 +∞ 2 +∞ 2 (1) 9g ω ωqˆ ω ωqˆ duces to the supersymmetric harmonic oscillator Hamiltonian. 4 h¯ 2 h¯ 2 E0 = qˆ e− dq = 9g e− qˆ dq Therefore, when g is small, the Hamiltonian Hˆ can be written 2 πh¯ ∞ πh¯ 0 r Z− r Z as 5 2 0 2 ω 3 πh¯ 27 h¯ 2 2 Hˆ = Hˆ + Hˆ ′ , (C.56) = 9q = g (g ) . (C.62) πh s 5 2 r ¯ × 8 ω 8 ω ≃O where Hˆ 0 is the Hamiltonian of the unperturbed supersymmetric harmonic oscillator and Hˆ ′ is the perturbation: The last equation implies that the ﬁrst-order correction to the ground-state energy reduces to zero. Notice that to solve the 9 2 4 3 integration which appeared in the last equation, we have used Hˆ ′ = ( g qˆ + 3ωgqˆ ) Á + 3¯hgqˆσ . (C.57) 2 2 3 the integral formula

As shown in Section 5, for the supersymmetric harmonic os- ∞ 1 3 (2n 1) π cillator, except the ground energy state, all the energy levels x2n exp( ax2)dx = · ··· − . (C.63) 0 − 2n+1 a2n+1 degenerate into two energy states. Therefore, if we have an en- Z r ψa ergy level degenerates to the two energy states Ψ = n n 0 Appendix D. THE SECOND-ORDER COR- 0 RECTION and Ψn = b , we can write ψn As described in [38, 39, 40, 41, 42], the second-order correction a a to the energy in the time-independent nondegenerate pertur- ψn ψm

ˆ Á ψn H′ ψm = b f1(qˆ) 2 + f2(qˆ)σ3 b = 0. bation theory is given by the form h | | i ψn ψm D E (C.58) (0) (0) 2 0 (2) ψn ˆ ′ ψm If we interest in a ground state of the form , the previous E0 = ∑ |h |H | i| . (D.64) ψ0 (0) (0) m=n En Em equation reduces to 6 − Furthermore, the harmonic oscillator wave function is orthog- ψ Hˆ ′ ψm = ψ f (qˆ) f (qˆ) ψm = 0. (C.59) h 0| | i h 0| 1 − 2 | i onal polynomial, and the harmonic oscillator wave function of degree n can be obtained by the following recursion formula: Based on this argument, we can see for this combination that (0) ˆ Ψ 1 2 the impact of the Hamiltonian H′ on the wave function 0 is (0) 1 ω 4 ωqˆ ω ψn (qˆ)= e− 2¯h Hˆ n qˆ , (D.65) only due to the component: √2nn! πh¯ h¯ r 9 2 4 3 ˆ ˆ ′ = g q + 3ωgqˆ 3¯hgqˆ . (C.60) where Hn is the Hermite Polynomial of degree n. Also, it is H 2 − helpful to consider the recursion relation of the Hermite Poly- nomials: Actually, this allows us to compute the first and second correc- 1 tions to the ground-state energy for this system just using the qˆHˆ n = Hˆ n+1 + nHˆ n 1 . (D.66) 2 − Hamiltonian ˆ . As well, we can use the nondegenerate time- H′ independent perturbation theory to calculate the corrections to Moreover, for two wave functions ψn and ψm, we have the re- the ground-state energy of our harmonic oscillator system even lation ψ ψ = δ . (D.67) if it has degenerate energy levels. h n| mi nm Thus, taking into account the supersymmetric harmonic os- Eqs. (D.65, D.66, and D.67) imply that cillator ground-state wave function in the form (132) and the Hamiltonian (C.60), then, we can now use the formula (C.53) to 1 ψ0 q ψm = C11δ0(m+1) + C12nδ0(m 1) = C12nδ1m . (D.68) compute the first-order correction to the energy of the ground h | | i 2 − state as follows: This argument leads us to know that all the terms of the sum- (1) 0 ˆ (0) mation with m > 4 in Eq. (D.64) vanish. So, to compute the E0 = ψ0 ′ ψ0 h |H | i second-order correction to the energy of the ground state using (0) 9 2 4 3 (0) = ψ ( g qˆ + 3ωgqˆ ) Á + 3¯hqqˆσ ψ Eq. (D.64), we only need to consider the nonzero terms with h 0 | 2 2 3| 0 i m 4. +∞ 2 ω ωqˆ 9 2 4 3 ≤ = e− h¯ g qˆ + 3ωgqˆ 3¯hgqˆ dq Now, we are able to compute the first term with m = 1. ∞ πh¯ 2 − Z− r First considering the ground and the first excited state wave 2 +∞ 2 +∞ 2 9g ω 4 ωqˆ ω 3 ωqˆ functions: = qˆ e− h¯ dq + 3ωg qˆ e− h¯ dq 2 πh¯ ∞ πh¯ ∞ Z Z 2 3 2 r − r − (0) ω 1 ωq (0) 4ω 1 ωq +∞ ωqˆ2 4 2¯h and 4 2¯h ω ψ0 = ( ) e− , ψ1 = ( 3 ) qe− . (D.69) 3¯hg qeˆ − h¯ dq . (C.61) πh¯ πh¯ − πh¯ ∞ r Z− 17 Letters in High Energy Physics LHEP xx, xxx, 2018

Then, Then,

2 2 2 1 ωqˆ 1 ωqˆ 2 ω 9 ω 4ω2 12ω (0) 9 2 4 (0) 2 ( ) 4 e− 2¯h g2qˆ4 4 qˆ4 qˆ2+3 e− 2¯h (0) 3 (0) (0) (0) ψ g qˆ ψ πh¯ 2 ( 576πh¯ ) 2 h¯ (0) (0) 3ωg ψ qˆ ψ 3¯hg ψ qˆ ψ 0 2 4 h¯ − 3 2 0 1 0 1 |h | | i| = ψ0 3ωgqˆ 3¯hgqˆ ψ1 h | | i− h | | i (0) (0) D 0 4¯hω E |h | − | i| E0 E4 − (0) (0) = h¯ ω − E E 2 2 0 − 1 − 27g4 ωqˆ 2 ωqˆ 2 2¯h 4 4ω 4 12ω 2 2¯h 2 2 2 = − 2 e− qˆ 2 qˆ qˆ + 3 e− 2 ∞ ωqˆ 2 ∞ ωqˆ 128πh¯ h¯ h¯ 1 2ω 4 − h¯ 2ω 2 − h¯ − = − 6ωgˆ 2 qˆ e dq 6¯hg 2 qˆ e dq h¯ ω πh¯ 0 πh¯ 0 D 2 2 2E − 27g4 2 +∞ ωqˆ +∞ ωqˆ +∞ ωqˆ 2 4ω 8 h¯ 12ω 6 h¯ 4 h¯ q q = − 2 2 ∞ qˆ e− dq h¯ ∞ qˆ e− dq + 3 ∞ qˆ e− dq R 2 R 128πh¯ h¯ − − − − 2 3 3 2 R 2 R 2 R 2 g h h h 27g4 2 +∞ ωqˆ +∞ ωqˆ +∞ ωqˆ 2 9 2¯ 3 2¯ 9 ¯ 2 4ω 8 h¯ 12ω 6 h¯ 4 h¯ = − 2 2 qˆ e− dq qˆ e− dq + 3 qˆ e− dq = − = 2 g . (D.70) 32πh¯ h¯ 0 − h¯ 0 0 h¯ ω  4 s ω − 2 s ω  − 8 ω R R 2 R 2 27g4 105 πh¯ 5 45 πh¯ 5 9 πh¯ 5 27g4 πh¯ 5 = − + = − 3   32πh¯ 2 8 ω5 − 4 ω5 8 ω5 32πh¯ 2 ω5 q q q q Similarly, we can compute the second term with m = 2. The 3 243 h¯ second excited state wave functions is given by the form = g4 . (D.76) − 32 ω5 1 2 (0) ω 4 2ω 2 ωq Using Eqs. (D.70, D.72, D.74, and D.76), we are able to compute ψ = qˆ 1 e− 2¯h . (D.71) 2 4πh¯ h¯ − the second-order correction to the ground-state energy for the supersymmetric harmonic oscillator as follows: Then, (0) (0) 2 (0) 9 2 4 (0) 2 2 2 2 ψ 3¯hgqˆ ψ ψ g qˆ ψ 1 ωqˆ 1 ωqˆ (0) 0 1 0 2 2 ω 9 2 4 ω 4 2ω 2 E = |h |− | i| + |h | | i| (0) 9 2 4 (0) 2 ( ) 4 e− 2¯h g qˆ ( ) ( qˆ 1)e− 2¯h 2 (0) (0) (0) (0) ψ g qˆ ψ πh¯ 2 4πh¯ h¯ − |h 0 | 2 | 2 i| = E0 E1 E0 E2 E(0) E(0) D 0 2¯hω E − − 0 − 2 − (0) 3 (0) 2 (0) 9 2 4 (0) 2 2 2 ψ 3ωgqˆ ψ ψ g qˆ ψ 81g4 +∞ ωqˆ +∞ ωqˆ 2 0 3 0 2 4 4ω 6 h¯ 4 h¯ + |h | | i| + |h | | i| = − 2 qˆ e− dq 2 qˆ e− dq ( ) ( ) ( ) ( ) 16πh¯ h¯ 0 − 0 E 0 E 0 E 0 E 0 0 3 0 4 R R2 − − 4 5 5 3 2 3 2 3 81g 15 πh¯ 3 πh¯ 729 h¯ 4 9 h¯ 2 729 h¯ 4 9 h¯ 2 243 h¯ 4 = − 2 5 5 = 5 g . = g g g g 16πh¯ 4 ω − 4 ω − 16 ω 2 5 2 5 q q − 8 ω − 16 ω − 4 ω − 32 ω (D.72) 27 h¯ 2 1701 h¯ 3 = g2 g4 . (D.77) 8 2 32 5 Furthermore, the same way we can compute the second term − ω − ω with m = 3. The third excited state wave functions take the Finally, from Eqs. (C.62) and (D.77), we realize that up to the form second order the energy corrections to the ground-state energy 3 2 (0) ω 1 ω 3 ωqˆ of the supersymmetric quantum mechanical harmonic oscilla- ψ = ( ) 4 (2 qˆ 3qˆ)e− 2¯h . (D.73) 3 9πh¯ 3 h¯ − tor vanish. This is concluded as Then, (g)= 0, O 2 2 2 1 ωq 1 ωqˆ 2 2 ω 3 ω3 ω 3 27 h¯ 27 h¯ (0) (0) ( ) 4 e− 2¯h 3ωgq ( ) 4 (2 qˆ 3qˆ)e− 2¯h 2 ψ 3ωgqˆ3 ψ 2 πh¯ 9πh¯3 h¯ − (g )= = 0. (D.78) |h 0 | | 3 i| = O 8 ω2 − 8 ω2 E(0) E(0) D 0 3¯hω E 0 − 3 − 2 3 2 2 2 g ω ωqˆ 3 ω 3 ωqˆ = − e 2¯h qˆ (2 qˆ 3qˆ)e 2¯h πh¯ 3 − h¯ − − References D 2 2 E g2ω3 +∞ ωqˆ +∞ ωqˆ 2 [1] Fredrik Engbrant. Supersymmetric quantum mechanics 2 ω 6 h¯ 4 h¯ = − 3 h¯ ∞ qˆ e− dq 3 ∞ qˆ e− dq πh¯ − − − and integrability. Master’s thesis, Uppsala University, 2 3 R 2 R 2 2 g ω 4ω +∞ 6 ωqˆ +∞ 4 ωqˆ 2012. = − qˆ e h¯ dq 6 qˆ e h¯ dq πh¯ 3 h¯ 0 − − 0 − [2] T. Wellman. An introduction to supersymmetry in quan- R 2 R 2 tum mechanical systems. Brown University Memoran- g2ω3 15 πh¯ 5 9 πh¯ 5 g2ω3 3 πh¯ 5 = − = − πh¯ 3 4 ω5 − 4 ω3 πh¯ 3 2 ω3 dum, 2003. q q q [3] Edward Witten. Dynamical breaking of supersymmetry. 9 h¯ 2 2 Nuclear Physics B, 188(3):513–554, 1981. = 2 g . (D.74) − 4 ω [4] Thomas Dumitrescu. Topics in Supersymmetric Quantum Also, similarly, we can compute the term with m = 4. The Field Theory. PhD thesis, Princeton University, 2013. fourth excited state wave function is [5] Matteo Bertolini. Lectures on supersymmetry. Lecture notes given at SISSA, 2011. 1 2 2 (0) ω 4 4ω 4 12ω 2 ωqˆ [6] Fred Cooper, Avinash Khare, and Uday Sukhatme. Su- ψ = qˆ qˆ + 3 e− 2¯h . (D.75) 4 576πh¯ h¯ 2 − h¯ persymmetry and quantum mechanics. Physics Reports, 251(5):267–385, 1995. [7] Paul Batzing and Are Raklev. Lecture notes for fys5190/fys9190–supersymmetry. A course given at the University of Oslo, 2013. [8] Sven Krippendorf, Fernando Quevedo, and Oliver Schlot- terer. Cambridge lectures on supersymmetry and extra dimensions. arXiv preprint arXiv:1011.1491, 2010.

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