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I have made a effort to recalculate the wavefunction of the electrons of the He atom. Schrödinger’s time-independent equation in natural units for the He atom is

$$\left(-\frac 12\vec\nabla^2_{r_1} - \frac 12\vec\nabla^2_{r_2} - \frac{2}{r_1} - \frac{2}{r_2} + \frac{1}{r_1 - r_2}\right)\Psi = E\Psi.\tag{1}$$

I decided to try solve the equation analytically since I am not happy with the screening of electrons to each other producing an equivalent charge. I took the initial condition

$$\Psi(r_1,0) = e^{-r_{1}} \tag{2}$$

since the electron 2 has been captured by the nucleus it produces the exact solution for the hydrogen atom. Then, I rearranged the terms to isolate the terms regarding electron 2:

$$\left(\frac 12\vec\nabla^2_{r_2} + \frac{2}{r^2}\right)\Psi = \left(-\frac 12\vec\nabla^2_{r_1} - \frac{1}{r_1} - E\right)\Psi \tag{3}$$

But since electron 2 has been captured by the nucleus, $r_2$ simply does not exist. So, I simply removed it from the equation, but kept the Laplacian term because this determines the shape of the wavefunction of electron 2. Otherwise, the solution becomes zero, which does not make sense.

So, I ended up with

$$\frac 12\vec\nabla^2_{r_2}\Psi = -\frac 12\vec\nabla^2_{r_1} - \frac{1}{r_1} - E\Psi.\tag{4}$$

Under the initial condition I calculate

$$\left(\frac 12\vec\nabla^2_{r_1} - \frac{1}{r_1} - E\right)\Psi(r_1,0) = e^{-r_1} - \frac{e^{-r_1}}{r_1} - E, \tag{5}$$

which is the solution for a hydrogen-like atom.

Now I need to start taking points, calculate the value at those points, then multiply that series of points by $r_2.$ Do I need to take points until 1, assuming that I have used natural units (because 1 is the Bohr radius in natural units)?

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    $\begingroup$ Firstly " I took the initial condition" you mean "I used the boundary condition" - initial conditions are for time dependent problems, which this ain't. I can't also see why your assumption for this boundary condition can be justified, it should be something about the wave function going to zero as the electrons get very far from the nucleus, but this isn't very important, see below. $\endgroup$ Commented yesterday
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    $\begingroup$ But most importantly you are wasting your time. This is a 3 body problem. There is no [closed form analytic solution]( en.wikipedia.org/wiki/Closed-form_expression). Take a look at Quantum Three-Body Problem if you want to see some of the fun and games that are really required here $\endgroup$ Commented yesterday
  • $\begingroup$ @IanBush it is a reduced 3 body problem since I assume that the nucleus is stationary so it is solvable. $\endgroup$ Commented yesterday
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    $\begingroup$ Change the boundary conditions for any differential equation and the results change. Specifying the boundary conditions is a fundamental part of defining the problem to be solved when dealing with differential equations. If I solve the Schrodinger equation for an atom or molecule with open boundary conditions (i.e. an isolated species) I will get a different solution from solving it with periodic boundary conditions (i.e. a crystal) $\endgroup$ Commented yesterday
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    $\begingroup$ Bethe and Salpeter wrote a whole book about solutions to one- and two-electron atoms… $\endgroup$ Commented yesterday

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(Expanding my comments into an answer)

I am afraid you are wasting your time. This is a 3 body problem. There is no closed form analytic solution. Take a look at Quantum Three-Body Problem amongst others if you want to see some of the fun and games that are required here for a good approximate, numerical solution.

More generally this is the usual problem in Chemistry - few if any problems are exactly solvable, even the solution of the Schrödinger equation for the H atom is an approximation if we dig a bit deeper than we usually do. So we make models which while not exact give us some insight to help understand what is going on. Screening is one such model. Admittedly it is a very crude one, one we learn early in our chemical career when we lack the tools to do better, but that doesn't stop it being useful so long as we understand it is a model and don't try to use it where it is not applicable.

Finally about your maths where you say "I took the initial condition" you mean "I used the boundary condition" - initial conditions are for time dependent problems, which this ain't. I also can't justify the boundary conditions you propose. They should be something along the lines of

  • When both electrons are far from the nucleus the wavefunction should approach zero
  • The solution should be regular - basically acceptable solutions should not "blow up" due to the infinities in the Coulomb operators at $r_{12}=0$

These are probably most easily written by a change of variables to $r_1+r_2$ and $r_1-r_2$, but given this is all ultimately an exercise in futility I can't be bothered to think harder.

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    $\begingroup$ Your link for "quantum 3 body problem" seems to be a duplicate of the previous link, I think. $\endgroup$ Commented yesterday
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    $\begingroup$ Yes - As I said taking my comment and turning it into an answer. Another more recent article would be Analytical solution of the Schrödinger equation for the neutral helium atom in the ground state considering the uncertainty principle, vibrational modes and quantum-electrodynamical effects - but note while this talks about an analytic solution by that it means reducing the problem to a simpler set of equations which then need to be solved on a computer (a MATLAB script is given) $\endgroup$ Commented yesterday
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    $\begingroup$ Ian: What @GlennWillen seems to be trying to say is that you had the same Wikipedia link to en.wikipedia.org/wiki/Closed-form_expression twice in your answer and left out the link to arxiv.org/abs/physics/9905051 that was in your original comment. Anyway, I just edited your answer to fix the links. $\endgroup$ Commented 14 hours ago
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    $\begingroup$ I'm really not a fan of "closed form expression" term. It references a set of "basic" functions, but this family is ill-defined. For example, trigonomatric functions are widely considered basic, but why? Why we don't add, say, gamma-function to the list? $\endgroup$ Commented 13 hours ago
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    $\begingroup$ @permeakra "For example, trigonomatric functions are widely considered basic, but why?" Mostly because of Liouville: en.wikipedia.org/wiki/Elementary_function $\endgroup$ Commented 13 hours ago
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Attempts to obtain an analytical solution for the ground state of the helium-like atoms were ongoing for the entire 20th century. Quick search obtained me a series of three papers [1-3]. The first one can be ignored and the latter two explain mostly the same procedure in slightly different wordings.

Note: I followed the solution only in key details and wrote-up the highlights here. For more rigorous treatment you should read the papers referenced.

The idea is to move from the $\psi(\vec{r}_1,\vec{r}_2)$ to $\psi(r_1,r_2,\theta)$, i.e. instead of the full set of six cartesian coordinates, one deals with only three independent variables: the two electron-nucleus distances and an angle between directions from the nucleus to each electron. This transition is an acceptable assumption because the ground state is known to be spherically symmetric. For most excited states that can be approximated as having an electron on non-s orbital this assumption breaks. But using it, we can make suitable assumptions in regard to the wavefunction: it should decasy as $\exp(-r_i)$ at large $r_i$, be symmetric in regards to swap of electons and a few more. With this, a trial wavefuction expansion is written up as

$\psi(r_1,r_2,\theta)=\displaystyle\sum_{l,i,j,p}\ C_{lijp}\ P_l(\cos \theta)\ \exp (-\lambda r_1)\ \exp (-\lambda r_2)\ \ln^p r_1$

where $P_l$ are Legendre polynomials used here for being orthogonal, making subsequent steps easier. Also, it is assumed that $r_2\leq r_1$.

Then, a few conditions can be imposed on $C_{lijp}$ due to requirements on the wavefunction, see the papers. The final requirements has form of a recurrence relation of the $C_{lijp}$ coefficients derived from application of the Schrodinger operator to the trial expansion.

From theoretical perspective, the problem is solved. From the practical perspective one should find a nice expression for the $C_{lijp}$ and write up the resulting expression. I'm too lazy to type up those formulas, so feel free to consult the papers.

  1. https://doi.org/10.1098/rspa.1982.0148
  2. https://doi.org/10.1098/rspa.1982.0149
  3. https://doi.org/10.1088/0305-4470/16/18/025
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So I am posting this as a answer since I kept going:As it turns out for small values of $r_{2}$ $\Psi(r_{1},r_{2}) = e^{-r_{1}}-e^{-r_{1}}/r_{1}-c(boundary \space condition)+r_{2}(e^{-r_{1}}-e^{-r_{1}}/r_{1}-c) = (1+r_{2})(e^{-r_{1}}-e^{-r_{1}}/r_{1}-c)$

But when we say that for small $r^{2}$ we get $1+r_{2}$ as the first term of the Maclaurin series the one naturally can assume that the general function is $e^{r_{2}}$.So the general solution becomes $\Psi(r_{1},r_{2})=e^{r_{2}}(e^{-r_{1}}-e^{-r_{1}}/r_{1}-c) =(e^{r_{2}-r_{1}}-e^{r_{2}-r_{1}}/r_{1}-ce^{r_{2}}) $

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    $\begingroup$ As @IanBush has pointed out, there is not exact solution for the Schrödinger equartion for the He atom. It is a well known fact. You can find a discussion on the approximates solutions and variational and pertubational strategies to obtain in older Quantum Chemistry textbooks. It is a long time since I studied it but sure you can find something in textbooks editions of the 80's (maybe it is a good idea checking Levine's or Szabo's Quantum Chemistry). Later editions may have given less weight to this problem since they expend more time in modern electronic structure methods. $\endgroup$ Commented 16 hours ago

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