Supersymmetric Quantum Mechanics Damon Binder PHYS3001 Project 2 September 29, 2014 Abstract In Section 5, repeated factorisation is used to create a hierachy of Hamiltonians from a given initial Hamil- In this paper, the application of supersymmetry to the tonian, and this leads to the concept of shape invariant solving of quantum mechanical Hamiltonians is explored. potentials (SIP) in Section 6, a concept first introduced A method for factorising Hamiltonians is developed and by Gendenshtein in 1983 [4]. For this class of potentials, it is shown that factorisation allows one to construct a the energy eigenvalues can easily be calculated using al- second Hamiltonian with an almost identical eigenspec- gebraic methods. Curiously, all standard solvable poten- tra. This is then applied to solving Shape Invariant Po- tials, such as the harmonic oscillator, the free particle, tentials (SIPs), and the close relationship between SIPs and the infinite square well, are shape invariant poten- and analytically solvable potentials is discussed. tials, demonstrating an intimate connection between an- alytically solvable problems in QM, supersymmetry, and the factorisation method. 1 Introduction One of the major goals of modern physics has been to 2 Supersymmetric QM unify the various symmetries found in nature. Initially symmetries which were generated by a Lie algebra were Historically, SUSY QM arose out of the attempts of considered; these are sets of symmetries which are closed physicists to create toy models of SUSY which they could only under commutation. However, a no-go theorem use as a testing ground for the theory. We shall take a discovered by Coleman and Mandula in 1967[1] proved spin 1/2 model where: that it would not be possible to achieve this unification through Lie algebras. In order to circumvent the theo- H 0 H = 1 rem, superalgebras were developed, which, unlike Lie al- 0 H2 gebras, are closed under both commutation and anticom- mutation. The type of symmetries generated by these is the Hamiltonian, and where: superalgebras was known as supersymmetries (SUSY). 0 A In order to study supersymmetric quantum field the- Q = ories, supersymmetric quatum mechanics (SUSY QM) 0 0 was developed. This was initial used as a toy model in is known as the supercharge operator. These shall satisfy order to test mathematical methods that could be ap- the anti-commutator relations: plied to the more complicated field theories. In Section 2 we begin by introducing one such model, originally stud- fQ; Qg = 0; fQ; QT g = H (1) ied by Witten in [2]. It was soon discovered that these models were of interest in their own right, as they could This algebra was first introduced by Nicolai[5]. The first be used to create Hamiltonians pairs with closely related anticommutator relationship is equivalent to stating that eigenspectras. Q is nilpotent. By expanding the second anticommutator There is a close relationship between SUSY QM and relation, we find: Hamiltonian factorisation, which we shall explore in Sec- tion 3. The latter method was explored by Infeld and AT A 0 H = QQT + QT Q = Hull [3] in the 1950s; however, the inimate connection 0 AAT between factorisation and supersymmetry would only be discovered thirty years later. The results derived in Sec- While SUSY QM originated in order to understand the tion 3 are applied to scattering states in Section 4. properties of H, it was soon found that H1 and H2 were 1 closely related, and that this could be used to solve one- Let 0 be the ground state of H1. Then: dimensional problems in QM. To see this, let: h 0jH1j 0i = hA 0jA 0i = 0 1 1 1 H1j ni = Enj ni This implies that Aj 0i = 0, and thus: be the eigenvectors of H1. For n > 0: ~ d 0 1 T 1 1 1 1 p + W (x) 0(x) = 0 H2Aj ni = AA Aj ni = AH1j ni = EnAj ni (2) 2m dx 2 2 2 Likewise, if H2j ni = Enj ni, then: We can then rearrange this to find the superpotential: T 2 T 2 2 T 2 0 (x) d ln H1A j ni = A H2j ni = EnA j ni (3) W (x) = −p~ 0 = −p~ 0 (4) 2m 0(x) 2m dx These relationships tell us that, except for states of zero energy, the two Hamiltonians have an identical energy We can now calculate the SUSY partner Hamiltonian: spectra, and closely related eigenvectors. These degen- 2 2 T ~ d 2 ~ 0 eracies are in fact a result of the supersymmetry of H: H2 = AA = − + W (x) − p W (x) 2m dx2 2m [H; Q] = HQ − QH The potentials: = QQT Q + QT Q2 − Q2QT − QQT Q = 0 2 ~ 0 V1;2(x) = W (x) ± p W (x) (5) In other words, as H and Q commute, Q can be seen 2m as a symmetry of H, and this symmetry results in the degenarate eigenvalues of H. are known as SUSY partner potentials. From (2) and (3) we know that the nonzero spectra of H1 and H2 are identical. We can, however, derive an even stronger rela- 3 Factorisation of a Hamiltonian tionship between H1 and H2 by proving that H2 has no zero eigenvalues. In the previous section we found that if we have Hamil- 2 Assume that this is false, and that H2 (x) = 0. T T 0 tonians of the form A A and AA , then they have an Then: (almost) identical spectra and closely related eigenvec- tors as a result of supersymmetry. In order to apply 2 d2 2 0 = − ~ 2(x) + W (x)2 − ~ W 0(x) 2 our results to quantum mechanical problems, we need a 2m dx2 0 2m 0 method of factorising Hamiltonians into the form AT A. Our treatment of factorisation loosely follows the treat- which, by rearranging and applying (4), simplifies to: ment in [6]. d d Let us begin with the Hamiltonian of a single particle: −p~ ln 1(x) = p~ ln 2(x) 2m dx 0 2m dx 0 ~2 d2 H1 = − 2 + V1(x) Therefore: 2m dx 2 1 −1 0(x) = C( 0(x)) which has a ground state of zero energy. This require- which is not normalisable. Thus, H has no zero eigen- ment does not restrict us, as we can always add a con- 2 values. stant to the Hamiltonian in order to achieve this. As E1 = 0 and E2 6= 0, we can use (2) and (3) to We wish to factorise the Hamiltonian as H = AT A. 0 0 1 deduce that: In order to do so we can use the ansatz: 1 2 1 E0 = 0;En = En+1 (6) ~ d T ~ d A = p + W (x);A = −p + W (x) j 2 i = (E1)−1=2Aj 1 i (7) 2m dx 2m dx n−1 n n j 1 i = (E2)−1=2AT j 2 i (8) where W (x) is known as the superpotential [6]. Then: n+1 n n where the normalisation constants in (7) and (8) can be ~ d ~ d found using: H1 = −p + W (x) p + W (x) 2m dx 2m dx 1 1 1 T 1 1 1 1 hA njA ni = h njA Aj ni = h njH1j ni = En 2 2 ~ d 2 ~ 0 = + W (x) − p W (x) 2 2 2 T 2 2 2 2 2m dx2 2m hA njA ni = h njAA j ni = h njH2j ni = En 2 1;2 ikx −ik−x As an example, we will find the partner potential of lim (x) = e + R1;2e x→−∞ E the infinite square well. Modifying the potential in the centre so that the ground energy is 0, we find that: where: 2 1=2 k± = (E ± W−) (n + 1)2π2 2 π2 2 n(n + 2)π2 2 E1 = ~ − ~ = ~ We can now use relationships (2) and (3) to derive: n 2mL2 2mL2 2mL2 2 1 1=2 lim E(x) = lim NA E(x) 2 (n + 1)πx x→±∞ x→±∞ 1 (x) = sin n L L where N is an arbitary normalisation constant. Thus: The superpotential of the infinite square well is [6]: ik+x ik+x T1e = NT2(−ik+ + W+)E π cos(πx=L) π πx p~ p~ W (x) = − = − cot ik−x −ik−x ik−x −k−x L 2m sin(πx=L) L 2m L e +R1e = N[(−ik−+W−)e +(ik−+W−)e R2] and the partner potential is: Equating the coefficients of the exponential terms: 2 2 W− + ik− ~ π 2 πx R (E) = R (E) V2(x) = csc − 1 1 2 2mL2 L W− − ik− For both W (x) and V2(x), the functions are defined on W+ − ik+ T1(E) = T2(E) (9) the interval [−L; L], and are infinite otherwise. W− − ik− 2 2 ~ π By shifting V2(x) by 2 , we can deduce the eigen- 2 2 2mL From these relationships we can deduce jR1j = jR2j values of: 2 2 and jT1j = jT2j ; in other words, the reflection and 2 d2 2π2 πx transmission probabilities of parnter potentials are iden- H = − ~ + ~ csc2 2m dx2 2mL2 L tical. In the special case of W+ = W−, we have T1(E) = T2(E), and thus in this special case the phase shift of are the same as those of the infinite square well excluding transmission are also equal. the ground state, and that the eigenvectors can be found From the previous discussion, we can conclude that a by multiplying the eigenvectors of the infinite square well potential is reflectionless, then so is its partner poten- by A. Thus, for H, the first two eigenvalues and eigen- tial. As an example, let us consider the superpotential states are: W (x) = −A tanh(αx).