Perturbation Theory Questions and Answers

Answer: B

The upper bound is \(\frac{1}{2}\hbar\omega\).

The variational energy for this Gaussian trial function is:

\[ \begin{aligned} \langle H \rangle &= \frac{\hbar^{2}b}{2m} \\ &\quad + \frac{m\omega^{2}}{8b} \end{aligned} \]

Minimising with respect to \(b\) gives \(b = \frac{m\omega}{2\hbar}\) and \(\langle H \rangle = \frac{1}{2}\hbar\omega\). The bound is exact here because a Gaussian of this width is the true ground state of the harmonic oscillator.

\(\frac{3}{2}\hbar\omega\) is the first excited state energy, and \(\frac{3}{4}\hbar\omega\) and \(\hbar\omega\) are values the variational energy takes for non-optimal widths, not the best bound.

Answer: C

The first order correction is \(\frac{2\alpha}{L}\).

The ground state of the box is \(\psi_{1} = \sqrt{2/L}\sin(\pi x/L)\). First-order perturbation theory gives the expectation value of the perturbation, and the delta function picks out the value at \(L/2\):

\[ \begin{aligned} E^{(1)} &= \alpha|\psi_{1}(L/2)|^{2} \\ &= \alpha\cdot\frac{2}{L}\sin^{2}\frac{\pi}{2} \\ &= \frac{2\alpha}{L} \end{aligned} \]

The correction would be 0 for the first excited state, which has a node at the centre. \(\frac{\pi^{2}\hbar^{2}}{2mL^{2}}\alpha\) multiplies the unperturbed energy by \(\alpha\), and \(L/\alpha\) has the wrong dimensions.

Answer: C

The first Born approximation is valid for large incident energy and weak scattering potential.

The Born approximation replaces the true wave inside the potential by the incident plane wave. That is accurate only if the potential distorts the wave little, which requires a weak potential, and if the particle passes quickly, which a high energy ensures. A common criterion is \(|V|a/\hbar v \ll 1\), where \(a\) is the range and \(v\) the speed.

A strong potential at low energy is the opposite regime, where partial-wave methods are used. The approximation is not universal, so 'any central potential' and 'independent of energy and potential' are wrong.

Answer: C

The dependence is \((A^{2} + \sin^{2}\theta/2)^{-2}\).

For the Yukawa potential, the first Born amplitude is proportional to the Fourier transform of \(\frac{e^{-\mu r}}{r}\):

\[ \begin{aligned} f(\theta) &\propto \frac{1}{\mu^{2} + q^{2}}, \quad q = 2k\sin\frac{\theta}{2} \\ \frac{d\sigma}{d\Omega} &= |f|^{2} \propto \frac{1}{(\mu^{2} + 4k^{2}\sin^{2}\theta/2)^{2}} \end{aligned} \]

Dividing through by \(4k^{2}\) gives the form \((A^{2} + \sin^{2}\theta/2)^{-2}\) with \(A = \frac{\mu}{2k}\). The power \(-1\) is the amplitude, not the cross-section, and positive powers would make scattering grow at large angles. As \(\mu \to 0\) this reduces to the Rutherford \(\sin^{-4}(\theta/2)\).

Answer: B

The shift is \(\lambda a_{o}^{2}\).

First-order perturbation theory gives the shift as \(\lambda\langle x^{2} \rangle\) in the ground state. The ground state is spherically symmetric, so \(\langle x^{2}\rangle = \langle y^{2}\rangle = \langle z^{2}\rangle\) \(= \langle r^{2}\rangle/3\). For the 1s state:

\[ \begin{aligned} \langle r^{2} \rangle &= \frac{4}{a_{o}^{3}}\int_{0}^{\infty} r^{4}e^{-2r/a_{o}}\,dr \\ &= 3a_{o}^{2} \\ \Delta E &= \lambda\cdot\frac{3a_{o}^{2}}{3} = \lambda a_{o}^{2} \end{aligned} \]

(The integral uses \(\int_{0}^{\infty} r^{4}e^{-br}dr = 24/b^{5}\) with \(b = 2/a_{o}\).)

Forgetting to divide by 3 would give \(3\lambda a_{o}^{2}\), and the options \(2\lambda a_{o}^{2}\), \(\lambda a_{o}^{2}/2\) and \(\sqrt{2}\lambda a_{o}^{2}\) do not follow from \(\langle r^{2} \rangle = 3a_{o}^{2}\).