Consider an anisotropic two-dimensional oscillator given by the Hamiltonian
\[ \begin{aligned} H = &-\frac{\hbar^{2}}{2m}\left(\frac{\partial^{2}}{\partial x^{2}} + \frac{\partial^{2}}{\partial y^{2}}\right) \\ &+ \frac{1}{2}m\omega^{2}x^{2} + 2m\omega^{2}y^{2} \end{aligned} \]
The eigen energy of the oscillator is given by :
MH SET - 2023, SET Exam - 2023,
Answer: A
The energy is \(E_{n} = (n + 3/2)\hbar\omega\).
Write the y potential as \(2m\omega^{2}y^{2} = \frac{1}{2}m(2\omega)^{2}y^{2}\). The Hamiltonian separates into an x oscillator of frequency \(\omega\) and a y oscillator of frequency \(2\omega\):
\[ \begin{aligned} E &= \left(n_{x} + \tfrac{1}{2}\right)\hbar\omega + \left(n_{y} + \tfrac{1}{2}\right)2\hbar\omega \\ &= \left(n_{x} + 2n_{y} + \tfrac{3}{2}\right)\hbar\omega \end{aligned} \]
With \(n = n_{x} + 2n_{y}\) this is \((n + 3/2)\hbar\omega\); the ground state is \(\frac{3}{2}\hbar\omega\), the sum of the two zero-point energies \(\frac{1}{2}\hbar\omega + \hbar\omega\).
\((n + 1)\hbar\omega\) is the isotropic two-dimensional oscillator with both frequencies equal to \(\omega\). \((2n + 1)\hbar\omega\) and \((n + 2/3)\hbar\omega\) have the wrong zero-point energy.
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