Phy5646 PerturbationExample1

Posted by student team #5 (Chelsey Morien, Anthony Kuchera, Jeff Klatsky)

Adapted from Zettili Quantum Mechanics - Concepts and Application; Solved Problem 9.6

Consider a system whose Hamiltonian is given by $E_{0}{\begin{pmatrix}1+\lambda &0&0&0\\0&8&0&0\\0&0&3&-2\lambda \\0&0&-2\lambda &7\end{pmatrix}}$ where $\lambda \ll 1$

(a) By decomposing the Hamiltonian into ${\mathcal {H}}={\mathcal {H}}_{0}+{\mathcal {H}}'$, find the eigenvalues and eigenvectors of the unperturbed Hamiltonian.

(b) Diagonalize ${\mathcal {H}}$ to find the exact eigenvalues of ${\mathcal {H}}$; expand each eigenvalue to the second power of $\lambda$

(c) Using first and second-order non-degenerate perturbation theory, find the approximate eigenenergies of ${\mathcal {H}}$ and the eigenstates to the first order. Compare these with the exact values obtained in (b).

Solution:

(a) The matrix of ${\mathcal {H}}$ can be separated:

${\mathcal {H}}={\mathcal {H}}_{0}+{\mathcal {H}}'=E_{0}{\begin{pmatrix}1&0&0&0\\0&8&0&0\\0&0&3&0\\0&0&0&7\end{pmatrix}}+E_{0}{\begin{pmatrix}\lambda &0&0&0\\0&0&0&0\\0&0&0&-2\lambda \\0&0&-2\lambda &0\end{pmatrix}}$

${\mathcal {H}}_{0}$ is already diagonalized, so reading off its eigenvalues and eigenstates are trivial:

$E_{1}^{(0)}=E_{0},$ $E_{2}^{(0)}=8E_{0},$ $E_{3}^{(0)}=3E_{0},$ $E_{4}^{(0)}=7E_{0}$

$|\psi _{1}\rangle ={\begin{pmatrix}1\\0\\0\\0\end{pmatrix}},$ $|\psi _{2}\rangle ={\begin{pmatrix}0\\0\\1\\0\end{pmatrix}},$ $|\psi _{3}\rangle ={\begin{pmatrix}0\\0\\1\\0\end{pmatrix}},$ $|\psi _{4}\rangle ={\begin{pmatrix}0\\0\\0\\1\end{pmatrix}}$

(b) The diagonalization of ${\mathcal {H}}$ leads to the following equation:

$det{\begin{pmatrix}(1+\lambda )E_{0}-E&0&0&0\\0&8E_{0}-E&0&0\\0&0&3E_{0}-E&-2\lambda E_{0}\\0&0&-2E_{0}\lambda &7E_{0}-E\end{pmatrix}}=0$

which is equivalent to:

$(E_{0}+\lambda E_{0}-E)(8E_{0}-E)\left[(3E_{0}-E)(7E_{0}-E)-4\lambda ^{2}E_{0}^{2}\right]=0$

Solving the above equation for E yields the following exact eigenenergies:

$E_{1}^{}=(1+\lambda )E_{0}$

$E_{2}^{}=8E_{0}$

$E_{3}=(5-2{\sqrt {1+\lambda ^{2}}})E_{0}$

$E_{4}=(5-2{\sqrt {1+\lambda ^{2}}})E_{0}$

Since we have defined $\lambda <<1$, we can expand ${\sqrt {1+\lambda ^{2}}}$, keeping only terms up to second order in $\lambda _{}^{}$:

${\sqrt {1+\lambda ^{2}}}\simeq 1+{\frac {\lambda ^{2}}{2}}$, which leads to:

$E_{3}\simeq (3-\lambda ^{2})E_{0}$

$E_{4}\simeq (7+\lambda ^{2})E_{0}$

(c)