Harmonic Oscillator Questions and Answers

Answer: C

The matrix is \(\sqrt{\frac{\hbar}{2m\omega}}\) times the matrix with 1 and \(\sqrt{2}\) next to the diagonal:

\[ \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & \sqrt{2} \\ 0 & \sqrt{2} & 0 \end{pmatrix} \]

Using \(a^{\dagger}|\psi_{n}\rangle = \sqrt{n+1}|\psi_{n+1}\rangle\) and \(a|\psi_{n}\rangle = \sqrt{n}|\psi_{n-1}\rangle\), the only non-zero elements of \(a + a^{\dagger}\) are

\[ \begin{aligned} \langle n+1|(a + a^{\dagger})|n\rangle &= \sqrt{n+1} \\ \langle n|(a + a^{\dagger})|n+1\rangle &= \sqrt{n+1} \end{aligned} \]

So the elements next to the diagonal are 1 (between \(n = 0\) and 1) and \(\sqrt{2}\) (between 1 and 2), and the diagonal is zero, because \(x\) changes \(n\) by one. The matrix is symmetric, as a Hermitian operator with real elements must be.

The diagonal matrix with 1, \(\sqrt{2}\), \(\sqrt{3}\) is not how \(x\) acts; the option with 1, 1 misses the \(\sqrt{2}\); and the last option couples states 0 and 2, which \(x\) cannot do.

Answer: A

The ratio is \(\frac{3}{7}\).

The infinite wall at \(x = 0\) forces the wavefunction to vanish there. Of the full oscillator's eigenfunctions, only the odd ones (\(n = 1, 3, 5, \ldots\)) vanish at the origin, and on \(x \gt 0\) they still satisfy the Schrodinger equation. So the allowed energies are

\[ \begin{aligned} E &= \left(n + \tfrac{1}{2}\right)\hbar\omega,\ n = 1, 3, \ldots \\ \frac{E_{1}}{E_{3}} &= \frac{\frac{3}{2}\hbar\omega}{\frac{7}{2}\hbar\omega} = \frac{3}{7} \end{aligned} \]

\(\frac{1}{3}\) is the ratio for the full oscillator (\(\frac{1}{2}\) to \(\frac{3}{2}\)), and \(\frac{3}{5}\) uses consecutive levels \(n = 1, 2\), ignoring that even states are forbidden.

Answer: B

The matrix element is non-zero when \(m = n\) or \(m = n \pm 2\).

Write \(\hat{X} \propto (a + a^{\dagger})\) and \(\hat{P} \propto i(a^{\dagger} - a)\). Then

\[ \begin{aligned} \hat{P}\hat{X} &\propto (a^{\dagger} - a)(a + a^{\dagger}) \\ &= a^{\dagger 2} - a^{2} + a^{\dagger}a - aa^{\dagger} \\ &= a^{\dagger 2} - a^{2} - 1 \end{aligned} \]

\(a^{\dagger 2}\) raises \(n\) by 2, \(a^{2}\) lowers it by 2, and the constant \(-1\) (from \([a, a^{\dagger}] = 1\)) gives a diagonal element. Indeed \(\langle n|\hat{P}\hat{X}|n\rangle = -\frac{i\hbar}{2}\), not zero.

'\(m = n \pm 2\) only' misses that diagonal term; \(n \pm 1\) would come from a single operator, and \(n \pm 3\) from three.

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 \(\Omega\).

Answer: A

The second excited state has \(E = \frac{7}{2}\hbar\omega\) and \(d = 6\).

For the 3D isotropic oscillator \(E = (n_{x} + n_{y} + n_{z} + \frac{3}{2})\hbar\omega\). The ground state has n = 0, the first excited state n = 1, and the second excited state n = 2, so \(E = \frac{7}{2}\hbar\omega\).

Its degeneracy is the number of ways to write 2 as \(n_{x} + n_{y} + n_{z}\): (2,0,0), (0,2,0), (0,0,2), (1,1,0), (1,0,1), (0,1,1), six in all, matching \((n + 1)(n + 2)/2 = 6\). d = 3 counts only one of the two kinds of combination. \(\frac{5}{2}\hbar\omega\) is the first excited state, whose degeneracy is 3, not the second.

Answer: D

The ground state energy changes by \(-18\).

Completing the square shows the new potential is the same oscillator, shifted sideways and downwards:

\[ \begin{aligned} \tfrac{1}{2}kx^{2} + 3ax &= \tfrac{1}{2}k\left(x + \tfrac{3a}{k}\right)^{2} \\ &\quad - \tfrac{9a^{2}}{2k} \end{aligned} \]

The shift in \(x\) does not change the spectrum, since \(k\) and so \(\omega\) are unchanged; every level, including the ground state \(\frac{1}{2}\hbar\omega\), drops by the constant \(\frac{9a^2}{2k}\).

\[ \begin{aligned} \Delta E &= -\frac{9a^{2}}{2k} \\ &= -\frac{9 \times 4}{2 \times 1} = -18 \end{aligned} \]

The shift grows with the square of the linear coefficient \(3a\), so the smaller options each leave out part of \(\frac{9a^2}{2k}\). The result is exact, not a first-order estimate, because a linear term in a harmonic potential can be removed completely.

Answer: D

The correct figure is (D): a Gaussian ground-state density, peaked at the centre and leaking past the turning points, drawn with the classical U-shaped density.

Classically the particle moves slowest near the turning points, so it spends most time there: \(P_{cl}(x) \propto \frac{1}{v(x)} \propto \frac{1}{\sqrt{x_{0}^{2} - x^{2}}}\). This is smallest at the centre and grows without limit at \(x = \pm x_{0}\), the dashed U shape.

The ground state is \(\psi_{0} \propto e^{-m\omega x^{2}/2\hbar}\), so \(P(x) \propto e^{-m\omega x^{2}/\hbar}\). It is largest at \(x = 0\). Its turning points are at \(x_{0}^{2} = \frac{\hbar}{m\omega}\), where \(P\) has only fallen to \(\frac{1}{e}\) of its peak, so it extends well beyond them into the classically forbidden region.

(A) wrongly draws a flat-topped classical curve. (B) draws a quantum curve that has died out before the turning points, so no penetration. (C) draws a quantum curve shaped like a parabola that ends exactly at the turning points, which a wavefunction never does.

Answer: D

The operator that commutes with the Hamiltonian is \( a^{+}a \), the number operator.

For the oscillator \( H = \hbar\omega\left(a^{+}a + \frac{1}{2}\right) \), so \( H \) is a function of \( a^{+}a \) alone and \( [H, a^{+}a] = 0 \). The energy eigenstates are the number states \( |n\rangle \).

The ladder operators do not commute with it: \( [H, a] = -\hbar\omega a \) and \( [H, a^{+}] = \hbar\omega a^{+} \), which is why they lower and raise the energy. The momentum does not either: \( [H, P_{x}] = i\hbar m\omega^{2}x \), since the potential depends on \( x \).