Q102 · CSIR-NET Chemistry, December 2013

Paper: CSIR-NET December 2013 · Subject: Physical Chemistry · Chapter: Quantum Chemistry · Topic: Harmonic Oscillator · Marks: 2 · Difficulty: Medium

Consider a two-dimensional harmonic oscillator with potential energy $\mathrm{V}(\mathrm{x}, \mathrm{y})=\frac{1}{2} \mathrm{k}_{\mathrm{x}} \mathrm{x}^{2}+\frac{1}{2} \mathrm{k}_{\mathrm{y}} \mathrm{y}^{2}$. If $\psi_{\mathrm{n}_{\mathrm{x}}}(\mathrm{x})$ and $\psi_{\mathrm{n}_{\mathrm{y}}}(\mathrm{y})$ are the eigensolutions and $\mathrm{E}_{\mathrm{n}_{\mathrm{x}}}$ and $\mathrm{E}_{\mathrm{n}_{\mathrm{y}}}$ are the eigenvalues of harmonic oscillator problem in x and y direction with potential $\frac{1}{2} \mathrm{k}_{\mathrm{x}} \mathrm{x}^{2}$ and $\frac{1}{2} \mathrm{k}_{\mathrm{y}} \mathrm{y}^{2}$ respectively, the wave function and eigenvalues of the above two-dimensional harmonic oscillator problem are
(a)$\begin{array}{l}\psi_{\mathrm{n}_{\mathrm{x}}, \mathrm{n}_{\mathrm{y}}}=\psi_{\mathrm{n}_{\mathrm{x}}}(\mathrm{x})+\psi_{\mathrm{n}_{\mathrm{y}}}(\mathrm{y}) \\ \mathrm{E}_{\mathrm{n}_{\mathrm{x}}, \mathrm{n}_{\mathrm{y}}}=\mathrm{E}_{\mathrm{n}_{\mathrm{x}}}+\mathrm{E}_{\mathrm{n}_{\mathrm{y}}}\end{array}$
(b)$\begin{array}{l}\psi_{\mathrm{n}_{\mathrm{x}}, \mathrm{n}_{\mathrm{y}}}=\psi_{\mathrm{n}_{\mathrm{x}}}(\mathrm{x}) \cdot \psi_{\mathrm{n}_{\mathrm{y}}}(\mathrm{y}) \\ \mathrm{E}_{\mathrm{n}_{\mathrm{x}}, \mathrm{n}_{\mathrm{y}}}=\mathrm{E}_{\mathrm{n}_{\mathrm{x}}} \cdot \mathrm{E}_{\mathrm{n}_{\mathrm{y}}}\end{array}$
(c)$\begin{array}{l}\psi_{\mathrm{n}_{\mathrm{x}}, \mathrm{n}_{\mathrm{y}}}=\psi_{\mathrm{n}_{\mathrm{x}}}(\mathrm{x}) \cdot \psi_{\mathrm{n}_{\mathrm{y}}}(\mathrm{y}) \\ \mathrm{E}_{\mathrm{n}_{\mathrm{x}}, \mathrm{n}_{\mathrm{y}}}=\mathrm{E}_{\mathrm{n}_{\mathrm{x}}}+\mathrm{E}_{\mathrm{n}_{\mathrm{y}}}\end{array}$
(d)$\begin{array}{l}\psi_{\mathrm{n}_{\mathrm{x}}, \mathrm{n}_{\mathrm{y}}}=\psi_{\mathrm{n}_{\mathrm{x}}}(\mathrm{x}) \cdot \psi_{\mathrm{n}_{\mathrm{y}}}(\mathrm{y}) \\ \mathrm{E}_{\mathrm{n}_{\mathrm{x}}, \mathrm{n}_{\mathrm{y}}}=\mathrm{E}_{\mathrm{n}_{\mathrm{x}}} \cdot \mathrm{E}_{\mathrm{n}_{\mathrm{y}}}\end{array}$
Answer
Answer: C ✓ checked by 4AB · confidence high

The source book printed no answer; this one was worked out and checked — see the explanation.

Explanation
For a separable Hamiltonian H = Hx + Hy, the wavefunction is the product ψnx(x)·ψny(y) and the energy is the sum Enx + Eny (c).

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