The non-relativistic Schrödinger Equation is:


Where $\widehat{H}$ is the Molecular Hamiltonian in Atomic Units and has the following terms:

$$ \widehat{H} = -\sum_{A=1}^M\frac{1}{2M_A}\nabla^2_A -\sum_{i=1}^N\frac{1}{2}\nabla^2_i +\sum_{A=1}^M\sum_{B>A}^M\frac{Z_AZ_B}{R_{AB}} +\sum_{i=1}^N\sum_{j>i}^N\frac{1}{r_{ij}} -\sum_{i=1}^N\sum_{A=1}^M\frac{Z_A}{r_{iA}} $$

Where labels A,B,.. denote nuclei and labels i,j,.. denote electrons. $\nabla^2_A$ and $\nabla^2_i$ are the corresponding Laplacians. $M_A$ is the mass of nuclei. $R_{AB}$ and $r_{ij}$ are the distances between corresponding Nuclear pairs and Electron pairs, respectively. $r_{iA}$ is the inter-nuclear-electron distance. $Z_A$ is the corresponding Nuclear Charge.

After applying the Born-Oppenheimer Approximation and treating the nuclei as stationary, and treating the Electronic Hamiltonian separately, i.e.

$$ \widehat{H}_{ele} = -\sum_{i=1}^N\frac{1}{2}\nabla^2_i -\sum_{i=1}^N\sum_{A=1}^M\frac{Z_A}{r_{iA}} +\sum_{i=1}^N\sum_{j>i}^N\frac{1}{r_{ij}} $$

I would like to know a mathematical explaination of why we cannot analytically solve the non-relativistic Schrödinger equation using the above electronic hamiltonian? Is it specifically the 3rd term that causes the problem?

In general many-body problems cannot be solved analytically, that is true. I would like to know the explanation for this specific case I mentioned above- for a Molecular System. I haven't found any resources to explain this point in more mathematical detail.

I have been reading Szabo and Ostlund's book. I would like to read more material regarding this specific point. Whether someone has worked this out for simple cases like say H2+ or H2?

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    $\begingroup$ This may be better asked in the Maths group $\endgroup$
    – Ian Bush
    Commented Sep 7, 2022 at 5:03
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    $\begingroup$ Even a classical, gravitationally bound system of 3 bodies cannot be generally solved analytically, Neither a helium atom can be. Why do you think molecular systems could be? $\endgroup$
    – Poutnik
    Commented Sep 7, 2022 at 5:30
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    $\begingroup$ No, of course, I don't think it can be solved for a molecular system. I get your point regarding the 3-body gravitational system. I was hoping for whether someone has rigorously worked it out for a molecular system. I will probably take Ian Bush's suggestion and ask it in the Maths group. I will also look at the 3-body gravitational problem in mathematical detail. Thanks. $\endgroup$
    – Guri
    Commented Sep 7, 2022 at 6:25
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    $\begingroup$ For the 3-body classical problem, I am not sure if the general solution just is not known, or if it was formally proved it does not exist. That is mathematical magic unavailable to merely mortal chemists. $\endgroup$
    – Poutnik
    Commented Sep 7, 2022 at 7:31
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    $\begingroup$ I wonder if there isn't in fact an analytical solution, but it's an infinite series, of which numerical solutions are only a truncated approximation. The answer probably has to do with the universality of the convergence of the sums in the series (eg in a 2 electron system, because of the three "distances" involved: r1, r2 and r12). $\endgroup$
    – Buck Thorn
    Commented Sep 9, 2022 at 10:37

2 Answers 2


The fact that the Schrödinger equation for a multi-body case cannot be solved, i.e. has no closed form solution, has nothing to do with quantum mechanics itself but is a more general issue with dynamical systems governed by invert square interaction (Coulomb, gravitational force, etc.). This is usually refer to as the three-body problem (or more generally the $n$-body problem).


I will show specifically the case for a Helium atom, but it would explain in general the difficulty of solving multi-electronic systems. The Hamiltonian for Helium can be represented as given below:

$$\hat{H} = -\frac{1}{2}\nabla^2_1 - \frac{1}{2}\nabla^2_2 - \frac{Z}{r_1} - \frac{Z}{r_2} + \frac{1}{r_{12}}$$

This can be re-written as

$$\hat{H} = \hat{H_1} + \hat{H_2} + \frac{1}{r_{12}}$$

where $\hat{H_i} = -\frac{1}{2}\nabla^2_i - \frac{Z}{r_i}; i = 1,2$

If it were not for the repulsion term in the hamiltonian, it would be completely separable. Now, you also have to consider the fact that the wavefunction, $\Psi(r_1,r_2)$, essentially is a function of two electrons. If we ignored the electron-electron repulsion term, we could separate the wavefunction into the product of two functions for each electron: $\Psi(r_1,r_2) = \psi(r_1)\cdot\psi(r_2)$ and solve the differential equation operating $\hat{H_i}$ on the corresponding $\psi(r_1)$ or $\psi(r_2)$. However, this would be like solving for two separate hydrogen atoms but with Z = 2.

The electron repulsion term makes this separation impossible as the term depends on $r_1$ and $r_2$. This is why there is a slight burden in solving the non-relativistic Schrödinger equation for a multi-electron system, I believe.

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    $\begingroup$ That's true, but your answer only explains why the Schrödinger equation can't be solved through separation of variables, it doesn't explain why the Schrödinger equation can't be solved, full stop. I expect OP is looking for the latter. Of course, proving that something cannot be solved is rather non-trivial, and well beyond my capabilities too... $\endgroup$ Commented Sep 7, 2022 at 8:44
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    $\begingroup$ @orthocresol Proving something cannot be solved always looked to me as an advanced math magic, approching proving that something does not exist. :-) (Nobody can prove a unicorn does not exist, by principle absence of evidence is not evidence of absence. ) $\endgroup$
    – Poutnik
    Commented Sep 7, 2022 at 9:40
  • $\begingroup$ @orthocresol yes I am looking for the latter. So suppose a person would look at the H atom case and say it's solvable analytically. Then they would look at the H2+ molecule and say well you can't solve it analytically because of the following reasons and therefore we go to approximate methods like Hartree-Fock Method. I am looking specifically, for those mathematical reasons. So for eg what @ M.L answered is right. I want to look at the deeper mathematical reason for it. And this is why I have also posted in Mathematics SE. $\endgroup$
    – Guri
    Commented Sep 7, 2022 at 10:10
  • $\begingroup$ Maybe take a look at this paper: arxiv.org/pdf/0704.0383.pdf and specifically the conclusion if that helps. I kind of got lost a quarter of the way through. $\endgroup$
    – M.L
    Commented Sep 7, 2022 at 15:49
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    $\begingroup$ Part of this may be related to the mathematical tools we have available. Algebra has been known for millenia, but it was impossible to solve cubic equations such as $x^3-3x-1=0$ for three real roots until the concept of imaginary numbers (whose cube roots appear in the solution) was developed. By extension, maybe someday we will find tools to solve at least simpler systems such as the helium atom. $\endgroup$ Commented Sep 9, 2022 at 10:36

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