HW 12: Methods of Approximation

Question-1¶

For the following hamiltonians, identify the exactly solvable and the perturbation parts and compute the first order corrections to the ground state energy

• $\hat{H} = -\frac{\hbar^2}{2\mu}\frac{d^2}{dx^2}+kx^2+\gamma x^3+\eta x^4$

• $\hat{H} =-\frac{\hbar^2}{2\mu}\frac{d^2}{dx^2}+V(x)$

  • Where $V(0 \leq x\leq L/2) = 0$ and $V(L/2 \leq x\leq L)=a$

• $\hat{H} = -\frac{\hbar^2}{2\mu}\nabla^2-\frac{e^2}{4\pi \epsilon_0 r}+e\xi r cos\theta$

• $\hat{H}=-\frac{\hbar^2}{2\mu}\nabla_{\theta,\phi}^2+\mu \xi cos\theta$

Question-2¶

Complete the proof of perturbation theory following 8-43 Problem in the book

Question-3¶

Use a trial function $\phi(x)=e^{-ax^2}$ to calculate the ground state energy for a quartic oscillator:

Question-4¶

• Prove variational theorem following the hints from Problem 8-1 in the book.

• Complete the derivation of secular determinant outlined in the Problem 8-22 from the book

Question-5¶

Application of variational method: Linear combinations of trial functions

In this problem we make use of the variational method to calcuate the ground state of the particle confined in a box of size $L$ which is subject to the following piecewise potential.

As a trial function we will use a linear combination of first two eigenfunctions of particle in a box without potentials() e.g the classical exactly solved PIB model)

Here is a step by step guide to solving this problem:

• Draw the shape of the potential $V(x)$ to see how it is different from classical PIB.

• Write down matrix elements $H_{11}$, $H_{12}=H_{21}$ $S_{11}$, $S_{12}$ etc.

• Evaluate matrix elements by taking advatnage of symetry arguments when possible.

• Plug matirx elements into the determinant and solve the quadratic euqation which gives the determines the energies.

Solution key to Question-5¶

• The piecewise function $V(x)$ is symmetric with respect to L/2 on a [0,L] interval. Note the nice analogy when we compare L/2 shifted integration limits $[-L/2, +L/2]$ with harmonic oscillator eigenfunctions on $[-\infty, +\infty]$! We will take advantage of this fact below to set integrals of odd functions to zero!

• $S_{11}=\langle 1 \mid 1\rangle =\frac{2}{L} = \int^{L}_0 sin^2 \Big(\frac{\pi x}{L} \Big)dx=1$ similarly $S_{22} =1$ $S_{12}=\langle 1 \mid 2\rangle =0$ similarly $S_{21}=0$

  $$H_{12} = \langle 1 \mid \hat{H} \mid 2\rangle=\langle 1 \mid \hat{H}_0+V(x) \mid 2\rangle= \\ = E_2\langle 1 \mid 2\rangle +\langle 1 \mid V(x)\mid 2 \rangle= 0+ \frac{2}{L}\int^L_0 \Big(\frac{\pi x}{L} \Big)V(x)\Big(\frac{2\pi x}{L} \Big)dx=0+0$$

We broke down $H_{12}$ into two integrals first of which is zero because of orthogonality of PIB eigenfunctions and second is zero because we are integrating an overall odd function $\langle even\mid even \mid odd \rangle$ on a symmetric range.