Introduction to Quantum Mechanics · Problem 3.47

Formalism — David J. Griffiths

Physics — Quantum Mechanics · supersymmetry

Problem

Supersymmetry. Consider the two operators A^=ip^2m+W(x)andA^†=−ip^2m+W(x),(3.116)\hat{A} = \frac{i\hat{p}}{\sqrt{2m}} + W(x) \quad \text{and} \quad \hat{A}^\dagger = \frac{-i\hat{p}}{\sqrt{2m}} + W(x), \tag{3.116} for some function W(x)W(x). These may be multiplied in either order to construct two Hamiltonians: H^1≡A^†A^=p^22m+V1(x)andH^2≡A^A^†=p^22m+V2(x);(3.117)\hat{H}_1 \equiv \hat{A}^\dagger \hat{A} = \frac{\hat{p}^2}{2m} + V_1(x) \quad \text{and} \quad \hat{H}_2 \equiv \hat{A}\hat{A}^\dagger = \frac{\hat{p}^2}{2m} + V_2(x); \tag{3.117} V1V_1 and V2V_2 are called supersymmetric partner potentials. The energies and eigenstates of H^1\hat{H}_1 and H^2\hat{H}_2 are related in interesting ways. (a) Find the potentials V1(x)V_1(x) and V2(x)V_2(x), in terms of the superpotential, W(x)W(x). (b) Show that if ψn(1)\psi_n^{(1)} is an eigenstate of H^1\hat{H}_1 with eigenvalue En(1)E_n^{(1)}, then A^ψn(1)\hat{A}\psi_n^{(1)} is an eigenstate of H^2\hat{H}_2 with the same eigenvalue. Similarly, show that if ψn(2)(x)\psi_n^{(2)}(x) is an eigenstate of H^2\hat{H}_2 with eigenvalue En(2)E_n^{(2)}, then A^†ψn(2)\hat{A}^\dagger \psi_n^{(2)} is an eigenstate of H^1\hat{H}_1 with the same eigenvalue. The two Hamiltonians therefore have essentially identical spectra. (c) One ordinarily chooses W(x)W(x) such that the ground state of H^1\hat{H}_1 satisfies A^ψ0(1)(x)=0,(3.118)\hat{A}\psi_0^{(1)}(x) = 0, \tag{3.118} and hence E0(1)=0E_0^{(1)} = 0. Use this to find the superpotential W(x)W(x), in terms of the ground state wave function, ψ0(1)(x)\psi_0^{(1)}(x). (The fact that A^\hat{A} annihilates ψ0(1)\psi_0^{(1)} means that H^2\hat{H}_2 actually has one less eigenstate than H^1\hat{H}_1, and is missing the eigenvalue E0(1)E_0^{(1)}.) (d) Consider the Dirac delta function well, V1(x)=mα22ℏ2−αδ(x),(3.119)V_1(x) = \frac{m\alpha^2}{2\hbar^2} - \alpha\delta(x), \tag{3.119} (the constant term, mα2/2ℏ2m\alpha^2/2\hbar^2, is included so that E0(1)=0E_0^{(1)} = 0). It has a single bound state (Equation 2.132) ψ0(1)(x)=mαℏexp⁡[−mαℏ2∣x∣].(3.120)\psi_0^{(1)}(x) = \frac{\sqrt{m\alpha}}{\hbar} \exp\left[-\frac{m\alpha}{\hbar^2}|x|\right]. \tag{3.120} Use the results of parts (a) and (c), and Problem 2.23(b), to determine the superpotential W(x)W(x) and the partner potential V2(x)V_2(x). This partner potential is one that you will likely recognize, and while it has no bound states, the supersymmetry between these two systems explains the fact that their reflection and transmission coefficients are identical (see the last paragraph of Section 2.5.2).

Browse the problem library