Discussion
The Hamiltonian for a quantum particle of mass \(m\) is given below, where \(\omega \lt \Omega\). The Schrödinger equation for this system can be solved exactly using the orthogonal transformations: \(x = \frac{x_{1} - x_{2}}{\sqrt{2}}\) and \(y = \frac{x_{1} + x_{2}}{\sqrt{2}}\).
\[ \begin{aligned} H = &-\frac{\hbar^{2}}{2m}\left[\frac{\partial^{2}}{\partial x^{2}} + \frac{\partial^{2}}{\partial y^{2}}\right] \\ &+ \frac{1}{2}m\Omega^{2}(x^{2} + y^{2}) + m\omega^{2}xy \end{aligned} \]
The ground state energy of this system is
Answer: A
The ground state energy is \(\frac{\hbar}{2}[\sqrt{\Omega^{2}-\omega^{2}}+\sqrt{\Omega^{2}+\omega^{2}}]\).
The rotation keeps the kinetic term and \(x^{2} + y^{2} = x_{1}^{2} + x_{2}^{2}\), while \(xy = \frac{1}{2}(x_{1}^{2} - x_{2}^{2})\). The potential separates into two independent oscillators:
\[ \begin{aligned} V &= \tfrac{1}{2}m\Omega^{2}(x_{1}^{2} + x_{2}^{2}) + \tfrac{1}{2}m\omega^{2}(x_{1}^{2} - x_{2}^{2}) \\ &= \tfrac{1}{2}m(\Omega^{2} + \omega^{2})x_{1}^{2} + \tfrac{1}{2}m(\Omega^{2} - \omega^{2})x_{2}^{2} \end{aligned} \]
The normal-mode frequencies are \(\sqrt{\Omega^{2} \pm \omega^{2}}\) (real because \(\omega \lt \Omega\)), and the ground state has \(\frac{1}{2}\hbar\) times each. The option without the factor \(\frac{1}{2}\) counts a full quantum per mode, and the options with \(\Omega^{2} \pm \Omega\omega\) under the roots do not match the normal-mode frequencies, since the coupling term is \(m\omega^{2}xy\), with no factor of \(\Omega\).
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