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In electronic structure theory, one seeks solutions of the molecular electronic Hamiltonian

$$ H_e(\vec{R}) = T_n(\vec{r}) + V_{ne}(\vec{r};\vec{R}) + V_{ee}(\vec{r}) + const, $$ where $\vec{r}$ are electronic coordinates, and $\vec{R}$ are frozen nuclear coordinates. One approach to obtaining approximate solutions is to define a set of single particle basis functions $\{\phi_p(r)\}$ (orbitals) and projecting the problem onto the space spanned by their antisymmetric products (Slater determinants). In this second-quantized representation, all the approximate solutions one obtains are antisymmetric by construction.

My question is: for the original problem, defined by the differential operator $H_e(\vec{R})$, is it true, at the level of mathematics, that the solutions are guaranteed to be anti-symmetric functions, or is this an ad hoc assumption resulting from knowledge of the anti-symmetry property for fermions.

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    $\begingroup$ When you say "at the level of mathematics", are you asking us to ignore the physics in the construction of the operator and focus solely on its mathematical form? For example, should we ignore "where r are electronic coordinates" and treat r as any 3-vector? $\endgroup$ Commented May 1 at 23:15
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    $\begingroup$ Yes, I'm interested in just the mathematical form of the operator as presented, where one can treat $\vec{r}$ as a $3N$ dimensional vector $(\vec{r}_1,\ldots,\vec{r}_N)$ (which physically would encode coordinates of $N$ electrons). From the mathematical form of the operator, could I show that all solutions are antisymmetric, or even that there exists an antisymmetric solution to begin with? If there are non-antisymmetric solutions, are these just "non-physical" solutions of a math problem I should discard, or do they have relevance to the physics? $\endgroup$ Commented May 1 at 23:33
  • $\begingroup$ @Solarflare0 please stick to just one question. If it gets answered you can ask another one in a new post. $\endgroup$ Commented May 2 at 6:55

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So, your question is "are the solutions of the Schrödinger equation guaranteed to be anti-symmetric functions, or is this an ad hoc assumption resulting from knowledge of the anti-symmetry property for fermions"?

This is a false dichotomy: to find a solution, you need to specify the symmetry. In mathematics, one can consider whatever, but in physics we have to obey nature. The solution for electrons is different from the solution for bosons, since they are different types of particles.

The mathematical proof for the existence of bosons and fermions is simple. Requiring that we have indistinguishable particles, it is clear that switching the places of particles $i$ and $j$ in the many-particle wave function $\Psi(x_1,\dots,x_i,\dots,x_j,\dots,x_N)$ once and back again has to give you the same wave function back. Thus, $ \hat{P}_{ij}^2 \Psi(x_1,\dots,x_i,\dots,x_j,\dots,x_N) = \hat{P}_{ij} \Psi(x_1,\dots,x_j,\dots,x_i,\dots,x_N) =\Psi(x_1,\dots,x_i,\dots,x_j,\dots,x_N) $. From this we find that $\Psi$ is an eigenfunction of $\hat{P}^2_{ij}$, and therefore the possible eigenvalues for $\hat{P}_{ij}$ are $\pm 1$: the upper sign for bosons and the lower for fermions.

One finds the solution by inserting a permissible Ansatz for the wave function, and then optimizing it in terms of the degrees of freedom. Typically, these are first the orbital expansion coefficients, which one finds with self-consistent field calculations, and in the next step, configuration coefficients for configuration interaction theory.

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    $\begingroup$ It's not false dichotomy. Even an asymmetric solution (e.g. an arbitrary linear combination of same-energy solutions that ignores indistinguishability of electrons) is a solution of the Schrödinger's equation. The (anti)symmetry is imposed quite separately from the differential equation, it's more like a boundary condition on the restricted $3N$-dimensional configuration space that would result if you remove all the extraneous positions obtainable by permutations of the electrons. I explore this idea in this thread on a 1D 2-electron model. $\endgroup$
    – Ruslan
    Commented May 2 at 18:45
  • $\begingroup$ @Ruslan why don't you write an answer? $\endgroup$ Commented May 3 at 2:16
  • $\begingroup$ @NikeDattani-NoFreeTime posted. $\endgroup$
    – Ruslan
    Commented May 3 at 9:12
  • $\begingroup$ @Ruslan mathematically yes, but physically no. $\endgroup$ Commented May 3 at 10:14
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    $\begingroup$ The question explicitly mentions "the level of mathematics". $\endgroup$
    – Ruslan
    Commented May 3 at 10:17
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There are several things to notice here.

First, the Hamiltonian you wrote is not a completed definition of an operator, since it lacks boundary conditions. This means that there's no eigenproblem yet. You can of course impose the natural BCs: boundedness of the wavefunction at infinity—or, for a finite but ideal crystal, Born-von Karman BCs.

But now we get to the second thing. Such description of the problem models each electron as an independent particle with its own degrees of freedom. Particularly, such electrons can be anywhere regardless of the positions of the others. This is clearly not a fermionic behavior, and there will indeed be solutions of the Schrödinger's equation for such a setup where the Pauli exclusion principle is violated. Just consider the ground state: it's non-degenerate, and it can be made non-negative on the whole $\mathbb{R}^{3N}$ (due to Courant's nodal domain theorem), which means that an exchange of a pair of electrons leaves the eigenfunction invariant, instead of yielding a factor of $-1$.

So indeed, we must impose antisymmetry by hand. One way is, as you note, to use an antisymmetric basis from which the solutions would be constructed. Another way is to reduce the configuration space, removing all the points obtainable by pairwise exchange of electrons, leaving only one instance of each such point—and then imposing boundary condition that the wavefunction vanishes on the hypersurfaces of electron-electron collision (Dirichlet BC playing the role of Pauli exclusion constraint). This approach is explored on a simple 1D two-electron model here.

And the third thing is that your Hamiltonian appears to ignore spin, and this loses some valid spatial eigenstates. Namely, if we consider a system of two electrons (e.g. a helium atom with a frozen nucleus), we should note that its ground state is a singlet (parahelium), where the spatial part of the wavefunction is symmetric with respect to an exchange of electrons. It's the spin part of the wavefunction that in this case ensures Pauli exclusion.

This means that actually, when you're solving the eigneproblem on the unreduced configuration space, you shouldn't thoughtlessly discard the symmetric spatial functions. Instead, if you start by solving the spinless Schrödinger's equation, you should post-process each of your (generally asymmetric) solutions as follows:

  1. Augment the solution chosen with spin degrees of freedom for each electron, treating these degrees of freedom as additional coordinates, e.g. $\psi(\vec r_1,\vec r_2)$ becomes $\psi(\vec r_1, s_1, \vec r_2, s_2)$ with the possible combinations of spins being

\begin{align} &\psi(\vec r_1, \uparrow, \vec r_2, \downarrow),\\ &\psi(\vec r_1, \uparrow, \vec r_2, \uparrow),\\ &\psi(\vec r_1, \downarrow, \vec r_2, \uparrow),\\ &\psi(\vec r_1, \downarrow, \vec r_2, \downarrow). \tag1 \end{align}

  1. Apply antisymmetrization to the resulting function as

$$\psi_{\text{ant}} = \sum_{P} \delta_P \hat P\psi, \tag2$$

where $P$ spans all possible permutations of the particles (i.e. of the tuples $(\vec r_i, s_i)$ passed to $\psi$) between the slots of the function $\psi,$ and $\delta_P$ is the sign of the permutation done by the operator $\hat P$ (i.e. the determinant of the corresponding permutation matrix).

  1. If the result of such operation is identical zero (e.g. for a fully symmetric spatial wavefunction for a 3-electron system), then you can discard this spatial solution. Otherwise you've obtained a fermionic solution.

  2. To make some sense from the result, try forming concrete combinations of spins (like in $(1)$) and passing them to the resulting function. For example, if you had a function $\psi(\vec r_1,s_1,\vec r_2,s_2),$ you'd have the following obtained from $(2)$:

$$\psi_{\text{ant}}(\vec r_1,s_1,\vec r_2,s_2)=\psi(\vec r_1,s_1,\vec r_2,s_2)-\psi(\vec r_2,s_2,\vec r_1,s_1). \tag3$$

If you now supply e.g. $s_1=\uparrow$ and $s_2=\downarrow,$ you can get

$$\psi_{\text{ant}\uparrow\downarrow}(\vec r_1,\vec r_2)=|\uparrow\downarrow\,\rangle\psi(\vec r_1,\vec r_2)-|\downarrow\uparrow\,\rangle\psi(\vec r_2,\vec r_1). \tag4$$

Consider the opposite case, $s_1=\downarrow, s_2=\uparrow,$ and you'll get another, also fermionic, function $\psi_{\text{ant}\downarrow\uparrow}.$ Taking linear combinations of these two you'll obtain more interesting solutions, like the above mentioned singlet state

\begin{align} \psi_{\text{singlet}}(\vec r_1, \vec r_2)&=\psi_{\text{ant}\uparrow\downarrow}(\vec r_1,\vec r_2)+\psi_{\text{ant}\downarrow\uparrow}(\vec r_1,\vec r_2)=\\ &=(|\downarrow\uparrow\,\rangle-|\uparrow\downarrow\,\rangle)(\psi(\vec r_1, \vec r_2)+(\vec r_2, \vec r_1)). \tag5 \end{align}

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It depends on what you mean. If your Hamiltonian is defined on $H_N:=\otimes^N \mathfrak h$, the $N$-fold tensor product of the single-particle Hilbert space $\mathfrak h$, then actually under quite general assumptions, a (permutation invariant) Hamiltonian of $N$ particles of the form $$\hat H_N =T+V+W \quad , $$

with kinetic energy $T$, external potential $V$ (no magnetic field) and (symmetric) interaction $W$, can be shown to have a bosonic ground state! In particular, not all eigenstates are fermionic. For a rigorous statement with proof, see e.g. Lieb and Seiringer. The stability of matter in quantum mechanics. CUP. Section 3.2.4.

If, instead, you work with the anti-symmetrized subspace $H^A_N:=\wedge^N \mathfrak h \subset H_N$, then by construction all eigenfunctions of the Hamiltonian are anti-symmetric. This of course is the case if you really consider $N$ identical fermions (such as electrons).

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  • $\begingroup$ Note: The relevant book passage should be available via Google Books. $\endgroup$
    – Jakob
    Commented May 2 at 14:12
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    $\begingroup$ Can the downvoter please explain how I can improve the answer? $\endgroup$
    – Jakob
    Commented May 3 at 11:37
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Update: I carelessly misunderstood the specific question I replied to as asking about the kinetic energy operator only. My answer isn’t applicable here, but I’m gonna leave it for reference unless someone else wants to delete it.

It seems like all the other answers are addressing a more general question than this one that you specifically asked:

My question is: for the original problem, defined by the differential operator $H_e(R)$, is it true, at the level of mathematics, that the solutions are guaranteed to be anti-symmetric functions, or is this an ad hoc assumption resulting from knowledge of the anti-symmetry property for fermions.

I'll try to answer it!

The anti-symmetric solution is included ad-hoc based on knowledge of the exchange properties of Fermions. You asked:

"is it true, at the level of mathematics, that the solutions are guaranteed to be anti-symmetric functions".

Ignoring the physics, you can prove it's false by finding a counter-example."

Note that the Hartree wave-function, which solves the equation, is just a single product of single-particle wave-functions. It is not antisymmetric.

If you insist on a linear combination of products, consider Bosons. The free particle Hamiltonian is, mathematically, exactly the same operator. The many-particle solutions are symmetric combinations of the single particle wave-functions.

Let $H \propto -\nabla^2_1 - \nabla^2_2$ be the free particle Hamiltonian for a system of two particles. $\nabla^2_i$ acts on the coordinates of $i^{th}$ particle. The many-particle solutions are $\Psi(x) \propto \psi_1(x_1) \psi_2(x_2) \pm \psi_1(x_2) \psi_2(x_1)$ where the upper/lower sign is symmetric/antisymmetric. $H\Psi = E\Psi$. The single particle eigenstates satisfy $-\nabla^2_i \psi_i = \epsilon_i \psi_i$. Plug it in: $$ H\Psi(x) = -\nabla^2_1(\psi_1(x_1) \psi_2(x_2) \pm \psi_1(x_2) \psi_2(x_1)) - \nabla^2_2(\psi_1(x_1) \psi_2(x_2) \pm \psi_1(x_2) \psi_2(x_1)) \\ = (\psi_1(x_1) \psi_2(x_2) \epsilon_1 \pm \psi_1(x_2) \psi_2(x_1) \epsilon_2 ) + (\psi_1(x_1) \psi_2(x_2) \epsilon_2 \pm \psi_1(x_2) \psi_2(x_1) \epsilon_1) \\ = (\psi_1(x_1) \psi_2(x_2) \pm \psi_1(x_2) \psi_2(x_1))(\epsilon_1 + \epsilon_2) = E \Psi . $$ Both the symmetric and antisymmetric solutions solve it with $E = \epsilon_1 + \epsilon_2$. The one that is "correct" for Fermions is determined by physics, not math!

Cheers, Ty

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    $\begingroup$ Well but here you ignore the interaction term, which is given in the question. So while the other answers may (or may not) be too general, your's is, IMHO, not general enough. In other words, you discuss a special case only. $\endgroup$
    – Jakob
    Commented May 4 at 7:04
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    $\begingroup$ The Hartree product does not solve the Schrödinger equation; it also lacks the proper symmetry, as it is neither bosonic nor fermionic. A single symmetrized configuration (e.g. antisymmetric determinant as in Hartree-Fock) is also not a solution to the Schrödinger equation, since the Coulomb operator is a two-particle operator. Instead, you need to include all possible configurations. $\endgroup$ Commented May 4 at 11:37
  • $\begingroup$ Oh yeah. When the author called $H_e$ the differential operator, I took that to mean kinetic energy operator alone. I didn’t look carefully that the same symbol above was the full interacting Hamiltonian :o $\endgroup$ Commented May 5 at 2:29

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