Perturbation Theory Questions and Answers

Answer: D

The correct option is \(E_{1} \ge E_{0},\ E_{2} \ge E_{0},\ E_{1} \ge E_{2}\).

The variational principle says that the expectation value of the Hamiltonian in any trial state is never below the true ground state energy, so both estimates satisfy E₁ ≥ E₀ and E₂ ≥ E₀.

The trial function φ₂ contains φ₁ as the special case g = 0, and adds a factor depending on the electron separation that lets the wavefunction keep the electrons apart. Minimizing over a larger family can only lower (or keep) the minimum, so E₂ ≤ E₁: the Hylleraas-type function gives a better estimate.

Every option with E₁ ≤ E₀ or E₂ ≤ E₀ contradicts the variational principle.

Answer: D

The cross section is independent of both \(\theta\) and \(\phi\).

In the first Born approximation the scattering amplitude is the Fourier transform of the potential at the momentum transfer \(\vec{q}\):

\[ \begin{aligned} f(\theta) &= -\frac{m}{2\pi\hbar^{2}}\int e^{-i\vec{q}\cdot\vec{r}}\, g\delta^{3}(\vec{r})\, d^{3}r \\ &= -\frac{mg}{2\pi\hbar^{2}} \end{aligned} \]

The delta function picks out \(\vec{r} = 0\), where the exponential is 1, so \(f\) is a constant and \(\frac{d\sigma}{d\Omega} = |f|^{2} = \frac{m^{2}g^{2}}{4\pi^{2}\hbar^{4}}\) is isotropic.

Any \(\phi\) dependence would need a potential that is not spherically symmetric, and \(\theta\) dependence enters only through \(q = 2k\sin(\theta/2)\), which drops out for a point interaction.

Answer: A

The corrections are 0 and \(\frac{\lambda^{2}\Delta^{2}}{|E_{1} - E_{2}|}\).

In the unperturbed basis \(\sigma_{x}\) has zeros on the diagonal, so the first-order correction \(\langle 1|H'|1\rangle\) vanishes. The second-order correction comes from the off-diagonal element \(\langle 2|H'|1\rangle = \lambda\Delta\):

\[ \begin{aligned} E_{1}^{(2)} &= \frac{|\langle 2|H'|1\rangle|^{2}}{E_{1} - E_{2}} \\ &= \frac{\lambda^{2}\Delta^{2}}{E_{1} - E_{2}} \end{aligned} \]

Its magnitude is \(\frac{\lambda^{2}\Delta^{2}}{|E_{1} - E_{2}|}\). The exact eigenvalues expand to the same result:

\[ \begin{aligned} E_{\pm} &= \frac{E_{1} + E_{2}}{2} \\ &\pm \sqrt{\left(\frac{E_{1} - E_{2}}{2}\right)^{2} + \lambda^{2}\Delta^{2}} \end{aligned} \]

A non-zero first-order term would need diagonal elements, which \(\sigma_{x}\) does not have, and the extra \(\frac{1}{2}\) in the last option is not in the formula.

Answer: C

The cross-section depends on p sin θ/2.

In the first Born approximation the scattering amplitude is the Fourier transform of the potential with respect to the momentum transfer \(\vec{q} = \vec{p}' - \vec{p}\). For elastic scattering \(|\vec{p}'| = |\vec{p}| = p\), so

\[ q = 2p\sin\frac{\theta}{2} \]

For a central potential the amplitude, and so the cross-section \(|f|^{2}\), is a function of q alone, that is of \(p\sin(\theta/2)\). It is therefore not linear in \(\theta\), not a function of \(p\sin^{2}\theta\), and not of the momentum alone, since it varies with angle.

Answer: C

The probability is zero.

In first-order time-dependent perturbation theory, the amplitude to reach state \( |n\rangle \) is proportional to the matrix element \( \langle n|x|0\rangle \). Since \( x = \sqrt{\frac{\hbar}{2m\omega}}(a + a^{+}) \), the perturbation connects the ground state only to \( |1\rangle \):

\[ \begin{aligned} \langle n|x|0\rangle &= \sqrt{\frac{\hbar}{2m\omega}}\, \langle n|1\rangle \\ &= \sqrt{\frac{\hbar}{2m\omega}}\, \delta_{n,1} \end{aligned} \]

So the system can be found in the first excited state, but the transition to the third excited state (\( n = 3 \)) is forbidden in this order; this selection rule is the reading the key relies on. The time dependence \( e^{-at} \) affects only the size of the transition to \( n = 1 \), so 1/3, 1 and \( e^{-a/3} \) have no basis.

Answer: C

A pair of eigenvalues is 3, 7 + 2\(\varepsilon\).

The unperturbed H has a 2 by 2 block with eigenvalues 7 and 3 and eigenvectors \((1, 1, 0)/\sqrt{2}\) and \((1, -1, 0)/\sqrt{2}\), plus the eigenvalue 2 with \((0, 0, 1)\). First-order shifts are the diagonal elements of H' in this basis:

\[ \begin{aligned} E_{7} &: \tfrac{1}{2}(1 + 1 + 1 + 1)\varepsilon = 2\varepsilon \\ E_{3} &: \tfrac{1}{2}(1 - 1 - 1 + 1)\varepsilon = 0 \\ E_{2} &: H'_{33} = \varepsilon \end{aligned} \]

So the levels become 7 + 2\(\varepsilon\), 3 and 2 + \(\varepsilon\). Only the third option lists two of them; 3 + 2\(\varepsilon\) gives the shift of the 7 level to the 3 level, and 2 + 2\(\varepsilon\) doubles the shift of the lowest level.

Answer: A

The first-order correction is \(\frac{bL}{2}\).

The first excited state of the well is \(\psi_{2} = \sqrt{\frac{2}{L}}\sin\frac{2\pi x}{L}\). The first-order correction is the expectation value of the perturbation:

\[ \begin{aligned} E^{(1)} &= \langle \psi_{2}|b(L - x)|\psi_{2} \rangle \\ &= bL - b\langle x \rangle \\ &= bL - b\frac{L}{2} = \frac{bL}{2} \end{aligned} \]

Every stationary state of the infinite well has \(|\psi|^{2}\) symmetric about the centre, so \(\langle x \rangle = L/2\) and the answer is the same for any level. Options bL and 2bL ignore the averaging of x, and \(bL/\sqrt{2}\) would come from mixing up the normalisation constant.

Answer: A

The Born amplitude is \(-\frac{2m\beta}{\hbar^{2}(\mu^{2} + b^{2})}\).

For a central potential the first Born approximation is \(f(\theta) = -\frac{2m}{\hbar^{2}b}\int_{0}^{\infty} rV(r)\,\text{sin}(br)\,dr\), where b is the momentum transfer \(2k\,\text{sin}(\theta/2)\). With \(V = \beta e^{-\mu r}/r\):

\[ \begin{aligned} \int_{0}^{\infty} e^{-\mu r}\,\text{sin}(br)\,dr \\ &= \frac{b}{\mu^{2} + b^{2}} \\ f(\theta) &= -\frac{2m\beta}{\hbar^{2}(\mu^{2} + b^{2})} \end{aligned} \]

Letting \(\mu \to 0\) recovers the Rutherford amplitude, proportional to \(1/b^{2}\), a useful check. The cross-section \(|f|^{2}\) has the square of the bracket; the options with the bracket squared, cubed or square-rooted in the amplitude itself are wrong.

Answer: C

The upper bound is \(\frac{6\hbar^{2}}{ma^{2}}\).

Normalisation: \(2A^{2}\int_{0}^{a/2}x^{2}dx = A^{2}a^{3}/12 = 1\), so \(A^{2} = 12/a^{3}\). The slope of \(\psi\) is +A on the left half and -A on the right, so \(\psi''\) is a delta function of strength -2A at the peak x = a/2:

\[ \begin{aligned} \langle H\rangle &= -\frac{\hbar^{2}}{2m}\int\psi\psi''\,dx \\ &= -\frac{\hbar^{2}}{2m}\psi\!\left(\tfrac{a}{2}\right)(-2A) \\ &= \frac{\hbar^{2}}{2m}\cdot\frac{Aa}{2}\cdot 2A = \frac{\hbar^{2}A^{2}a}{2m} \end{aligned} \]

\[ \begin{aligned} \langle H\rangle &= \frac{\hbar^{2}}{2m}\cdot\frac{12}{a^{2}} \\ &= \frac{6\hbar^{2}}{ma^{2}} \end{aligned} \]

This is above the exact ground state \(\pi^{2}\hbar^{2}/2ma^{2} \approx 4.93\hbar^{2}/ma^{2}\), as the variational principle requires. \(5\hbar^{2}/ma^{2}\) is close to the exact value but is not what this trial function gives, and 7 and 8 come from arithmetic slips in the normalisation.

Answer: A

The first-order correction is \(ba_{0}^{2}\).

First-order perturbation theory gives \(E^{(1)} = \langle 100|bx^{2}|100\rangle = b\langle x^{2}\rangle\). The hydrogen ground state is spherically symmetric, so \(\langle x^{2}\rangle\), \(\langle y^{2}\rangle\) and \(\langle z^{2}\rangle\) are equal, each \(\frac{1}{3}\langle r^{2}\rangle\), and

\[ \begin{aligned} \langle r^{2}\rangle &= \frac{4}{a_{0}^{3}}\int_{0}^{\infty} r^{4}e^{-2r/a_{0}}dr = 3a_{0}^{2} \\ \langle x^{2}\rangle &= a_{0}^{2} \end{aligned} \]

so \(E^{(1)} = ba_{0}^{2}\). Using \(\langle r^{2}\rangle\) directly would give \(3ba_{0}^{2}\), not offered, and the halved or quartered values come from mistakes in the radial integral.