Perturbation Theory

Assume we treat the interaction term $V\left(\vert{\bf r}_1 -{\bf r}_2\vert\right)$ in the Hamiltonian Eq. (III.2.17) as a perturbation,

$$\hat{H} = \hat{H}_0+ \hat{H}_1,\quad \hat{H}_0 = -\frac{\hbar^2}{2m}\Delta_1+ V({\bf r}_1) -\frac{\hbar^2}{2m}\Delta_2 + V({\bf r}_2)$$

$$\quad \hat{H}_1 = U\left(\vert{\bf r}_1 -{\bf r}_2\vert\right).$$ (2.19)

We seek the first correction to an energy level $E^{(0)}_{\alpha\beta}$ of $\hat{H}_0$,

$$\hat{H}_0 \phi_{\alpha\beta}^{\pm}({\bf r}_1,{\bf r}_2) = E^{(0)}_{\alpha\beta}\phi_{\alpha\beta}^{\pm}({\bf r}_1,{\bf r}_2),\quad E^{(0)}_{\alpha\beta} = E^{(0)}_{\alpha} + E^{(0)}_{\beta}$$

$$\phi_{\alpha\beta}^{\pm}({\bf r}_1,{\bf r}_2) = \frac{1}{\sqrt{2}}\left[\phi_{\alpha}({\bf r}_1) \phi_{\beta}({\bf r}_2) \pm \phi_{\alpha}({\bf r}_2) \phi_{\beta}({\bf r}_1) \right],$$ (2.20)

where $\phi_{\alpha}$ and $\phi_{\beta}$ are two eigenstates with eigenenergies $E^{(0)}_{\alpha}$ and $E^{(0)}_{\beta}$ of the (identical) single particle Hamiltonians $-\frac{\hbar^2}{2m}\Delta+ V({\bf r})$.

We assume the single particle levels to be non-degenerate. Still, the two-electron level $E^{(0)}_{\alpha\beta}$ is degenerate because it corresponds to the two states $\vert\phi_{\alpha\beta}^{\pm}\rangle$ ( $^+$ for the symmetric and $^-$ for the anti-symmetric state. The corresponding two-by-two matrix of $\hat{H}_1$ we need diagonalise for the degenerate first order perturbation theory in the sub-space spanned by $\vert\phi_{\alpha\beta}^{\pm}\rangle$ is however diagonal so that things become easy:

$$\underline{\underline{H}}_1 = \left( \begin{array}{cc} \langl... ...\\ 0 & A_{\alpha\beta}-J_{\alpha\beta} \\ \end{array}\right).$$ (2.21)

Inserting the definitions, we have ( $i,j,=\pm$)

$$\langle \phi_{\alpha\beta}^{i} \vert \hat{H}_1 \vert \phi_{\alpha... ...rt{\bf r}_1 -{\bf r}_2\vert\right) \phi_{\alpha\beta}^{j}({\bf r}_1,{\bf r}_2).$$ (2.22)

Exercise: Show that $\langle \phi_{\alpha\beta}^{+} \vert \hat{H}_1 \vert \phi_{... ...{\alpha\beta}^{-} \vert \hat{H}_1 \vert \phi_{\alpha\beta}^{+} \rangle=0$.

The explicit calculation of the remaining diagonal elements $\langle \phi_{\alpha\beta}^{+} \vert \hat{H}_1 \vert \phi_{\alpha\beta}^{+} \rangle$ and $\langle \phi_{\alpha\beta}^{-} \vert \hat{H}_1 \vert \phi_{\alpha\beta}^{-} \rangle$ yields

$$A_{\alpha\beta} = \int \int d{\bf r}_1 d{\bf r}_2 \vert\phi_\alpha({\bf r}_1)\vert^2 U\left(\vert{\bf r}_1 -{\bf r}_2\vert\right) \vert\phi_{\beta}({\bf r}_2)\vert^2$$

(direct term) (2.23)

$$J_{\alpha\beta} = \int \int d{\bf r}_1 d{\bf r}_2 \phi_{\alpha}^*({\bf r}_2) \phi_\... ...t{\bf r}_1 -{\bf r}_2\vert\right) \phi_{\alpha}({\bf r}_1)\phi_\beta({\bf r}_2)$$

(exchange term, exchange integral) (2.24)

Exercise: Verify these expressions.

The symmetrical orbital wave function ( $+$) belongs to the $S=0$ (singlet) spinor, whereas the anti-symmetrical orbital wave function ( $-$) belongs to the $T=0$ (triplet) spinors. Therefore, the unperturbed energy level $E^{(0)}_{\alpha\beta}$ splits into two levels

$$E^{(1)}_{\alpha\beta,S=0} = E^{(0)}_{\alpha\beta}+ A_{\alpha\beta} + J_{\alpha\beta}, \quad S=0$$ (2.25)

$$E^{(1)}_{\alpha\beta,S=1} = E^{(0)}_{\alpha\beta}+ A_{\alpha\beta} - J_{\alpha\beta}, \quad S=1 .$$ (2.26)