Harmonic Oscillator Questions and Answers

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.

Answer: D

\(a^{+}a|n + 1\rangle = (n + 1)|n + 1\rangle\).

The product \(a^{+}a\) is the number operator N, and the energy eigenstates are its eigenstates: \(N|m\rangle = m|m\rangle\). Step by step:

\[ \begin{aligned} a|n + 1\rangle &= \sqrt{n + 1}\,|n\rangle \\ a^{+}\sqrt{n + 1}\,|n\rangle &= \sqrt{n + 1}\sqrt{n + 1}\,|n + 1\rangle \end{aligned} \]

giving \((n + 1)|n + 1\rangle\). The state is unchanged, so options ending in \(|n\rangle\) are wrong. \(n|n + 1\rangle\) applies the eigenvalue of the wrong state, and \(\sqrt{n(n + 1)}\) mixes the factors from two different steps.

Answer: D

The first-order shift is 0.

The unperturbed ground state is a product of two one-dimensional ground states, \(\psi_{0}(x)\psi_{0}(y)\), each a Gaussian centred on the origin. The first-order shift is the expectation value of the perturbation:

\[ \begin{aligned} \Delta E^{(1)} &= \lambda mw^{2}\langle x\rangle\langle y\rangle \\ &= 0 \end{aligned} \]

because the ground-state Gaussian is even, so \(\langle x \rangle = \langle y \rangle = 0\).

The perturbation does change the energy at second order, and the exact problem (rotating to \(x \pm y\) coordinates) gives frequencies \(w\sqrt{1 \pm \lambda}\); expanding the ground-state energy shows the linear terms in \(\lambda\) cancel. The options \(\lambda\hbar w\), \(\lambda\hbar w/2\) and \(\lambda\hbar w/4\) would be shifts linear in \(\lambda\), which do not occur here.

Answer: C

The integer can be \(n = m + 3\).

At first order the transition amplitude is proportional to the matrix element \(\langle n|(a + a^{\dagger})^{3}|m\rangle\). Each factor of \(a + a^{\dagger}\) raises or lowers the level by one, so three factors change the level by an odd amount from \(-3\) to \(+3\):

\[ n \in \{m - 3,\ m - 1,\ m + 1,\ m + 3\} \]

(with \(m - 3\) and \(m - 1\) allowed only if non-negative). Of the options, only \(n = m + 3\) is in this set; it is reached by applying \(a^{\dagger}\) three times.

\(n = m - 2\) and \(n = m + 4\) are even changes, which an odd power of \(a + a^{\dagger}\) cannot produce: the cubic perturbation has odd parity and connects only states of opposite parity.