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I recently set up a numeric solver of the Schrödinger equation and can receive solutions for single-particle quantum mechanical problems. I became interested in simulating atoms, since there is a closed-form solution only for hydrogen. The coulomb potential for hydrogen is simple to model, since we can apply the Schrödinger equation to the electron that has is at a distance r with respect to a proton. In the case of helium however, the formulations of the potential/hamiltonian I see online contain a term that depends on the distance between electrons (for example: https://quantummechanics.ucsd.edu/ph130a/130_notes/node35.html). This poses a problem to me since the position of the two electrons is uncertain. Unlike a 3-body problem in classical mechanics, I cannot simply compute the distance between the two particles. So how would I approach this problem numerically?

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    $\begingroup$ a term that depends on the distance between electrons … This poses a problem to me since the position of the two electrons is uncertain. In the Schrodinger equation for hydrogen, there is a term that depends on the distance between the proton and the electron. But the position of the electron (and the proton, actually) is uncertain. $\endgroup$
    – Ghoster
    Commented Mar 5 at 20:55
  • $\begingroup$ @Ghoster What confuses me is that we can just say that the position of the electron is with respect to the proton and therefore apply the usual schroedinger equation with a single (particle) uncertain position. So in the case of helium instead of expecting to apply the schroedinger equation separately for each particle (and reach two wavefunctions), we should solve a single schroedinger equation for both particles to get a single wavefunction that predicts both of the particle's positions? $\endgroup$
    – Teo Tyrov
    Commented Mar 5 at 21:18
  • $\begingroup$ Yes, this is called a many-body wave function. The two electrons together set up the energy levels of helium. $\endgroup$ Commented Mar 5 at 22:06
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    $\begingroup$ The Schrodinger equation for helium is a 6-dimensional PDE with 3 coordinates for each electron. The wavefunction is $\psi(x_1,y_1,z_1,x_2,y_2,z_2)$ in Cartesian coords. And the PDE doesn’t separate as far as I know. $\endgroup$
    – Ghoster
    Commented Mar 5 at 23:33
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    $\begingroup$ You should read about quantum chemistry. $\endgroup$
    – my2cts
    Commented Mar 6 at 5:15

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The Hamilton operator of the helium atom is given by $$ H=-\frac{\hbar^2}{2m}\Delta_1 - \frac{\hbar^2}{2m} \Delta_2-\frac{2e^2}{r_1}-\frac{2e^2}{r_2}+\frac{e^2}{r_{12}},\qquad r_{1}= |\vec x_{1} |, \; r_2= |\vec{x}_2|,\; r_{12}= |\vec{x}_1-\vec{x}_2|,$$ acting on wave functions $\psi(\vec{x}_1,\sigma_1; \vec{x}_2,\sigma_2)$. According to the Pauli exclusion principle for fermions, the wave function has to be antisymmetric under the interchange $(\vec{x}_1,\sigma_1) \leftrightarrow (\vec{x}_2, \sigma_2) $ of the position and spin coordinates of the two particles, i.e. $$\psi(\vec{x}_1,\sigma_1; \vec{x}_2,\sigma_2) =- \psi(\vec{x}_2,\sigma_2; \vec{x}_1, \sigma_1).$$ As the Hamilton operator does not depend on the spins $\vec{S}_{1,2}$, the operators $H$, $\vec{S}^2$ and $S_z$ (with $\vec{S}=\vec{S}_1+\vec{S}_2$) can be diagonalized simultaneously. The addition of two spin $1/2$ angular momenta can either result in a total spin $s=0$ (spin singlet) or a total spin $s=1$ (spin triplet). In the first case (parahelium), the wave function assumes the form $$\begin{pmatrix} \psi(\vec{x}_1, \uparrow; \vec{x}_2, \uparrow)\\ \psi(\vec{x}_1, \uparrow; \vec{x}_2, \downarrow) \\\psi(\vec{x}_1, \downarrow; \vec{x}_2 \uparrow) \\ \psi(\vec{x}_1 \downarrow; \vec{x}_2, \downarrow)\end{pmatrix}=\frac{1}{\sqrt{2}}\begin{pmatrix} 0 \\1 \\ -1\\0 \end{pmatrix} \phi_s(\vec{x}_1, \vec{x}_2),$$ whith a symmetric spatial wave function $\phi_s(\vec{x}_1, \vec{x}_2)=\phi_s(\vec{x}_2,\vec{x}_1).$ The triplet states (orthohelium) with total spin $s=1$ and $s_z= +1,0,-1$ are described by $$\begin{pmatrix} 1\\ 0 \\0\\0 \end{pmatrix} \phi_a(\vec{x}_1, \vec{x}_2), \quad \frac{1}{\sqrt{2}}\begin{pmatrix} 0\\ 1 \\ 1 \\ 0 \end{pmatrix}\phi_a(\vec{x}_1,\vec{x}_2), \quad \begin{pmatrix}0\\0\\0\\1 \end{pmatrix}\phi_a(\vec{x}_1,\vec{x}_2)$$ with an antisymmetric spatial wave function $\phi_a(\vec{x}_1,\vec{x}_2) =-\phi_a(\vec{x}_2, \vec{x}_1)$.

Thus, in principle, finding the energy eigenvalues of the He atom boils down to solving the eigenvalue problem $H \phi = E\phi$ for functions $\phi(\vec{x}_1, \vec{x}_2) \in L^2(\mathbb R^6)$ being either symmetric (parahelium) or antisymmetric (orthohelium) under $\vec{x}_1 \leftrightarrow \vec{x}_2$. However, in contrast to the simple case of the hydrogen atom, the energy eigenfunctions of the He atom cannot be found in closed form and one has to appeal to suitable approximation methods. Note that an excellent upper bound for the He ground state energy can be found by a variational ansatz for the ground state wave function, whereas a lower bound can be derived from a certain operator inequality (see e.g. Walter Thirring, Quantum Mathematical Physics - Atoms, Molecules and Large Systems, Springer, chapter 4.3).

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You don't need to know the position of the electron. Numerically, you calculate the wave function for EVERY position (x,y,z) on the grid. To simplify, you assume every involved particle to be everywhere. Many combinations fall out, because most of the time, not the total position, but the relative position of the particles are important.

Does this make it clear?

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  • $\begingroup$ Yes, the previous comments helped a lot. I'm not a science student so I guess I was too used to applying motion equations on each object separately and combining everything on some sort of system of equations. Annoyingly this doubles the amount of variables which explodes the number of points in the numeric integration grid. Is there literature for numerically solving helium? I'll probably start with a "1D helium atom" and see how far I can go with my current compute $\endgroup$
    – Teo Tyrov
    Commented Mar 6 at 11:20
  • $\begingroup$ I don't know about numerical methods, but I learned that the best approach to this problem is the Born-Oppenheimer approximation: en.wikipedia.org/wiki/Born%E2%80%93Oppenheimer_approximation ... take this as a starting point $\endgroup$
    – kai90
    Commented Mar 6 at 12:57

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