Singly Ionized Li, Perturbation and Integral Help

[logic smogic]

Homework Statement

An approximate value for the ground state energy of a two electron atom can be found by starting with the energy of two noninteracting electrons in 1s states and calculating the first order correction due to the perturbation [itex]V= \frac{ {e}^{2} }{ 4 \pi {\epsilon}_{0} } \frac{ 1 }{ \left |{r}_{1}-{r}_{2} \right |}[/itex]. Perform the necessary integration and find an estimate for the ground state energy of singly ionized Li.2. Homework Equations and attempt at solution

First, I recognize the Hamiltonian of the unperturbed system as that of two Hydrogenic Hamiltonians with modified nucleonic charges:
[itex] {H}^{0} = \left[ - \frac{ {\hbar}^{2} }{ 2m } { {\nabla}_{1} }^{2} - \frac{1}{4 \pi {\epsilon}_{0}} \frac{3{e}^{2}}{{r}_{1}} \right ] + \left[ - \frac{ {\hbar}^{2} }{2m} {{\nabla}_{2}}^{2} - \frac{1}{4 \pi {\epsilon}_{0}} \frac{3{e}^{2}}{{r}_{2}} \right ][/itex]

And for the perturbation, we consider:
[itex]H= {H}^{0} + \frac{ {e}^{2} }{ 4 \pi {\epsilon}_{0} } \frac{ 1 }{ \left |{r}_{1}-{r}_{2} \right |}[/itex]

The Schrodinger Equation for [itex]{H}^{0}[/itex] separates:
[itex] \psi ({r}_{1},{r}_{2}) = {\psi}_{nlm}({r}_{1}){\psi}_{n'l'm'}({r}_{2})[/itex]

The ground state, of course (propagating the nucleonic charge through) will be:
[itex] {\psi}_{0} = \frac{27}{ \pi {a}^{3} } {e}^{-3({r}_{1} + {r}_{2})/a} [/itex] where a is the Bohr radius.

Question: I'm a little hazy on perturbations, and I'm not sure how to go about finishing this problem, and where *exactly* the integration comes in. Could someone just lay out the steps I need to take? (and of course, I'll go through with the actual calculations, just seeing how the structure would be very enlightening)

My best guess is to integrate [itex] \int_{- \infty }^{ \infty } V dr [/itex], but a friend mentioned that it should be a six-fold integral. Any ideas? Thanks!

 

[nrqed]

Homework Statement

An approximate value for the ground state energy of a two electron atom can be found by starting with the energy of two noninteracting electrons in 1s states and calculating the first order correction due to the perturbation [itex]V= \frac{ {e}^{2} }{ 4 \pi {\epsilon}_{0} } \frac{ 1 }{ \left |{r}_{1}-{r}_{2} \right |}[/itex]. Perform the necessary integration and find an estimate for the ground state energy of singly ionized Li.


2. Homework Equations and attempt at solution

First, I recognize the Hamiltonian of the unperturbed system as that of two Hydrogenic Hamiltonians with modified nucleonic charges:
[itex] {H}^{0} = \left[ - \frac{ {\hbar}^{2} }{ 2m } { {\nabla}_{1} }^{2} - \frac{1}{4 \pi {\epsilon}_{0}} \frac{3{e}^{2}}{{r}_{1}} \right ] + \left[ - \frac{ {\hbar}^{2} }{2m} {{\nabla}_{2}}^{2} - \frac{1}{4 \pi {\epsilon}_{0}} \frac{3{e}^{2}}{{r}_{2}} \right ][/itex]

And for the perturbation, we consider:
[itex]H= {H}^{0} + \frac{ {e}^{2} }{ 4 \pi {\epsilon}_{0} } \frac{ 1 }{ \left |{r}_{1}-{r}_{2} \right |}[/itex]

The Schrodinger Equation for [itex]{H}^{0}[/itex] separates:
[itex] \psi ({r}_{1},{r}_{2}) = {\psi}_{nlm}({r}_{1}){\psi}_{n'l'm'}({r}_{2})[/itex]

The ground state, of course (propagating the nucleonic charge through) will be:
[itex] {\psi}_{0} = \frac{27}{ \pi {a}^{3} } {e}^{-3({r}_{1} + {r}_{2})/a} [/itex] where a is the Bohr radius.

Question: I'm a little hazy on perturbations, and I'm not sure how to go about finishing this problem, and where *exactly* the integration comes in. Could someone just lay out the steps I need to take? (and of course, I'll go through with the actual calculations, just seeing how the structure would be very enlightening)

My best guess is to integrate [itex] \int_{- \infty }^{ \infty } V dr [/itex], but a friend mentioned that it should be a six-fold integral. Any ideas? Thanks!

The equation for first order perturbation theory (for non-degenerate states) is simply
[itex] \int \psi_0^* H_{pert} \psi_0 [/tex]
which here means an integral over the six coordinates.