Chapter4problem2

(Submitted by group 9, based on 7.5 Griffiths)

(a) Use the variational principle to prove that first order non-degenerate perturbation theory always overestimates the ground state energy.

solution

Let the unperturbed ground state be the trial wave function. $\Psi _{gs}^{0}$

According to the variational principle$<\Psi _{gs}^{0}|H|\Psi _{gs}^{0}>\geq E_{gs}^{0}$. But $H=H^{0}+H'$, so $<\Psi _{gs}^{0}|H|\Psi _{gs}^{0}>=<\Psi _{gs}^{0}|H^{0}|\Psi _{gs}^{0}>+<\Psi _{gs}^{0}|H'|\Psi _{gs}^{0}>$

But $<\Psi _{gs}^{0}|H^{0}|\Psi _{gs}^{0}>=E_{gs}^{0}$(the unperturbed ground state energy), and $<\Psi _{gs}^{0}|H'|\Psi _{gs}^{0}>$ is the first order correction to the ground state energy, so $E_{gs}^{0}+E_{gs}^{1}\geq E_{gs}.$

(b) Show that the second-order correction to the ground state is always negative.

solution

The second order correction $(E_{gs}^{2})$ is $E_{gs}^{2}=\sum _{m\neq gs}{\frac {|<\Psi _{m}^{0}|H|\Psi _{gs}>|^{2}}{E_{gs}^{0}-E_{m}^{0}}}$.

The numerator is clearly positive, and the denominator is always negative (since $E_{gs}^{0}<E_{m}^{0}$ for all m), so $E_{gs}^{2}$ is negative.