It's useful to see how our quantum perturbation theory works in a case that we can solve
Question:
It's useful to see how our quantum perturbation theory works in a case that we can solve exactly. Let's consider a two-state system in which the Hamiltonian is
\[\hat{H}_{0}=\left(\begin{array}{cc}E_{0} & 0 \tag{10.113}\\0 & E_{1}\end{array}\right)\]
for energies \(E_{0} \leq E_{1}\). Now, let's consider adding a small perturbation to this Hamiltonian:
\[\hat{H}^{\prime}=\left(\begin{array}{ll}\epsilon & \epsilon \tag{10.114}\\\epsilon & \epsilon\end{array}\right)\]
for some \(\epsilon>0\). The complete Hamiltonian of our system we are considering is \(\hat{H}=\hat{H}_{0}+\hat{H}^{\prime}\).
(a) First, calculate the exact energy eigenstates and eigenvalues of the complete Hamiltonian. From your complete result, Taylor expand it to first order in \(\epsilon\).
(b) Now, use our formulation of quantum perturbation theory to calculate the first corrections to the energy eigenstates and eigenvalues. Do these corrections agree with the explicit expansion from part (a)?
(c) From the Taylor expansion in part (a), for what range of \(\epsilon\) does the Taylor expansion converge?
Step by Step Answer:
Quantum Mechanics A Mathematical Introduction
ISBN: 9781009100502
1st Edition
Authors: Andrew J. Larkoski