Question

a) Use the variational principle to prove that first-order non-degenerate perturbation theory always overestimates (or at any rate never underestimates) the ground state energy.

(b) In view of (a), you would expect that the second-order correction to the ground state is always negative. Confirm that this is indeed the case, by examining Equation 6.15.

Short answer

(a) As, ${{\mathrm{\psi }}_{\mathrm{gs}}}^0\left|\mathrm{H}\right|{{\mathrm{\psi }}_{\mathrm{gs}}}^0⟩={\mathrm{E}}^0+{\mathrm{E}}^1\ge {\mathrm{E}}_{\mathrm{gs}}$this proves thatfirst-order non-degenerate perturbation theory overestimates the ground state energy.

(b) The value of $E_{gs}^2$is definitely negative.

Step-by-step solution

Define the variational principle

The variational principle states that the ground-state energy is always smaller than or equal to the calculated with the trial wavefunction expectation value. We can approximate the wave function and energy of the ground-state by changing until the expectation value of is minimized.

${\mathbf{E}}_{\mathbf{gs}}\mathbf{\le }\mathbf{⟨}{{\mathbf{\psi }}_{\mathbf{gs}}}^{\mathbf{o}}\mathbf{\left|}\mathbf{H}\mathbf{\right|}{{\mathbf{\psi }}_{\mathbf{gs}}}^{\mathbf{o}}\mathbf{⟩}$

Where$E_{gs}$ is the energy in the ground state.

Prove the first-order perturbation

(a)

Solve the problem by using ${{\psi }_{gs}}^o$ as our trial wave function as:

The variational principle tells us that:

${\mathrm{E}}_{\mathrm{gs}}\le ⟨{{\mathrm{\psi }}_{\mathrm{gs}}}^{\mathrm{o}}\left|\mathrm{H}\right|{{\mathrm{\psi }}_{\mathrm{gs}}}^{\mathrm{o}}⟩$

Using the perturbation theory as well;

$H=H^0+H^1$

Where ${\mathrm{H}}^1$ is the first order perturbation, and $H^{\circ }$is the unperturbed Hamiltonian.

Thus,

$$\begin{gathered} ⟨{{\mathrm{\psi }}_{\mathrm{gs}}}^0\left|\mathrm{H}\right|{{\mathrm{\psi }}_{\mathrm{gs}}}^0⟩=⟨{{\mathrm{\psi }}_{\mathrm{gs}}}^0\left|{\mathrm{H}}^{\circ }\right|{{\mathrm{\psi }}_{\mathrm{gs}}}^0⟩+⟨{{\mathrm{\psi }}_{\mathrm{gs}}}^0\left|{\mathrm{H}}^1\right|{{\mathrm{\psi }}_{\mathrm{gs}}}^0⟩ \\ ⟨{{\mathrm{\psi }}_{\mathrm{gs}}}^0\left|{\mathrm{H}}^{\circ }\right|{{\mathrm{\psi }}_{\mathrm{gs}}}^0⟩=E_{gs}^0 \\ ⟨{{\mathrm{\psi }}_{\mathrm{gs}}}^0\left|{\mathrm{H}}^1\right|{{\mathrm{\psi }}_{\mathrm{gs}}}^0⟩=E_{gs}^1 \end{gathered}$$

Then, $⟨{{\mathrm{\psi }}_{\mathrm{gs}}}^0\left|\mathrm{H}\right|{{\mathrm{\psi }}_{\mathrm{gs}}}^0⟩={\mathrm{E}}^0+{\mathrm{E}}^1\ge {\mathrm{E}}_{\mathrm{gs}}$and this proves the statement.

Define the value of second order correction.

(b)

The ground state's second order correction is denoted by $E_{gs}^2$ and from the second order perturbation theory,

${\mathrm{E}}_{\mathrm{gs}}^0=\sum _{\mathrm{m}\ne n}\frac{|⟨{\mathrm{\psi }}_{\mathrm{m}}^0|{\mathrm{H}}^1|{{\mathrm{\psi }}_{\mathrm{gs}}}^0⟩{|}^2}{{\mathrm{E}}_n^0-{\mathrm{E}}_{\mathrm{m}}^0}$

Therefore;

role="math" localid="1658317167106" ${\mathrm{E}}_{\mathrm{gs}}^0=\sum _{\mathrm{m}\ne \mathrm{gs}}\frac{|⟨{\mathrm{\psi }}_{\mathrm{m}}^0|{\mathrm{H}}^1|{{\mathrm{\psi }}_{\mathrm{gs}}}^0⟩{|}^2}{{\mathrm{E}}_{\mathrm{gs}}^0-{\mathrm{E}}_{\mathrm{m}}^0}$

Since role="math" localid="1658317017097" $E_{gs}^0$ is the ground state,

Then numerator is positive, but the denominator is negative as:

${{\mathrm{E}}_{\mathrm{gs}}}^0-{\mathrm{E}}_{\mathrm{m}}^0<0$ for all m.

Thus, role="math" localid="1658316906095" $|⟨{\mathrm{\psi }}_{\mathrm{m}}^0|{\mathrm{H}}^1|{{\mathrm{\psi }}_{\mathrm{gs}}}^0⟩{|}^2$, ${\mathrm{E}}_{\mathrm{gs}}^2$is definitely negative.