Franz Schwabl Quantum Mechanics 4th Edit · Problem 11.2

Relativistic Corrections

Physics — Quantum Mechanics · coupled-rotators-perturbation-theory

Problem

Two coupled, identical rotators are described by the Hamiltonian H=A(pΘ12+pΘ22)−Bcos⁡(Θ1−Θ2),H = A\left(p_{\Theta_1}^2 + p_{\Theta_2}^2\right) - B\cos(\Theta_1 - \Theta_2) \quad , where AA and B>0B > 0, pΘi=(ℏ/i)∂/∂Θip_{\Theta_i} = (\hbar/\mathrm{i})\partial/\partial\Theta_i, and Θi+2π\Theta_i + 2\pi is equivalent to Θi\Theta_i. For the case B≪Aℏ2B \ll A\hbar^2, calculate the energy eigenvalues using perturbation theory. In the opposite case B≫Aℏ2B \gg A\hbar^2, assume that only small oscillations occur, and construct the solution in analogy to the harmonic oscillator. Hint: Introduce new variables x=Θ1+Θ2x = \Theta_1 + \Theta_2 and y=Θ1−Θ2y = \Theta_1 - \Theta_2. In terms of these variables, the Hamiltonian becomes H=−2Aℏ2(∂2∂x2+∂2∂y2)−Bcos⁡y.H = -2A\hbar^2\left(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}\right) - B\cos y \quad .

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