# 16.2: Exercises - Perturbations


An asterisk denotes a harder problem, which you are nevertheless encouraged to try!

The following trigonometric identities may prove useful in the first two questions:

$\cos^2 A \equiv 1 − \sin^2 A \nonumber$

$\sin 2A \equiv 2 \sin A \cos A \nonumber$

$\sin A \sin B \equiv \frac{1}{2} [\cos (A − B) − \cos (A + B)] \nonumber$

$\cos A \cos B \equiv \frac{1}{2} [\cos (A − B) + \cos (A + B)] \nonumber$

1. A quantum dot is a self assembled nanoparticle in which a single electron state can be confined. A model for such an object is a particle moving in one dimension in the potential

$V (x) = \infty , \quad |x| > a, \quad V (x) = V_0 \cos (\pi x/2a), \quad |x| \leq a \nonumber$

Identify an appropriate unperturbed system and perturbation term.

Calculate the energies of the two lowest states to first order in perturbation theory.

What is the sign of $$V_0$$?

State two ways in which the colour of a material containing dots can be shifted towards the red.

2. A particle moves in one dimension in the potential

$V (x) = \infty , \quad |x| > a, \quad V (x) = V_0 \sin(\pi x/a), |x| \leq a \nonumber$

• show that the first order energy shift is zero;
• *obtain the expression for the second order correction to the energy of the ground state,

$\boxed{\Delta E_{1}^{(2)}=-\left(\frac{32 V_{0}}{15 \pi}\right)^{2} \frac{8 m a^{2}}{3 \pi^{2} \hbar^{2}}-\left(\frac{64 V_{0}}{105 \pi}\right)^{2} \frac{8 m a^{2}}{15 \pi^{2} \hbar^{2}}-\ldots} \nonumber$

3. The 1-d anharmonic oscillator: a particle of mass m is described by the Hamiltonian

$\hat{H} = \frac{\hat{p}^2}{2m} + \frac{1}{2}m\omega^2 \hat{x}^2 + \gamma \hat{x}^4 \nonumber$

• Assuming that $$\gamma$$ is small, use first-order perturbation theory to calculate the ground state energy;

$E_n \simeq (n + \frac{1}{2} )\hbar \omega + 3\gamma \left( \frac{\hbar}{2m\omega} \right)^2 (2n^2 + 2n + 1) \nonumber$

Hint: to evaluate matrix elements of powers of $$\hat{x}$$, write $$\hat{x}$$ in terms of the harmonic oscillator raising and lowering operators $$\hat{a}$$ and $$\hat{a}^{\dagger}$$. Recall that the raising and lowering operators are defined by

$\hat{a} \equiv \sqrt{\frac{m\omega}{2\hbar}} \hat{x} + \frac{i}{\sqrt{2m\omega \hbar}} \hat{p} \quad \text{ and } \quad \hat{a}^{\dagger} \equiv \sqrt{\frac{m\omega}{2\hbar}} \hat{x} − \frac{i}{\sqrt{2m\omega \hbar}} \hat{p} \nonumber$

with the properties that

$\hat{a}|n \rangle = \sqrt{n} |n − 1\rangle \quad \text{ and } \quad \hat{a}^{\dagger} |n \rangle = \sqrt{n + 1} |n + 1 \rangle \nonumber$

4. A 1-dimensional harmonic oscillator of mass $$m$$ carries an electric charge, $$q$$. A weak, uniform, static electric field of magnitude $$\mathcal{E}$$ is applied in the $$x$$-direction. Write down an expression for the classical electrostatic potential energy for a point particle at x.

The quantum operator is given by the same expression with $$x \rightarrow \hat{x}$$. By considering the symmetry of the integrals, or otherwise, how that, to first order in perturbation theory, the oscillator energy levels are unchanged, and calculate the second-order shift. Can you show that the second-order result is in fact exact?

Hint: to evaluate matrix elements of $$\hat{x}$$, write $$\hat{x}$$ in terms of the harmonic oscillator raising and lowering operators $$\hat{a}$$ and $$\hat{a}^{\dagger}$$ and use the results $$\hat{a}|n \rangle = \sqrt{ n}|n − 1 \rangle$$ and $$\hat{a}^{\dagger} |n \rangle = \sqrt{n + 1}|n + 1 \rangle$$. To obtain an exact solution, change variables to complete the square in the potential and show it remains a harmonic oscillator.

5. Starting from the relativistic expression for the total energy of a single particle, $$E = (m^2 c^4 + p^2 c^2 )^{1/2}$$, and expanding in powers of $$p^2$$, obtain the leading relativistic correction to the kinetic energy, for a plane wavefunction $$\Phi (x) = A \cos(kx)$$, and determine whether $$\Phi (x)$$ is an eigenstate for a relativistic free particle. Hint For normalisation of the wavefunction, it helps to keep the integral form for $$|A|^{−2} = \int \cos^2 kxdx$$.

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