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My friend claimed that if we have a two-electron atom in the ground state, and somehow we get the "real" electronic wavefunction $\psi(\mathbf{x_1},\mathbf{x_2})$ of the system, the statistical probability of finding two electrons (spin-up and spin-down) at the same spot $\mathbf{x_0}$ is nonzero, and the repulsion potential between them looks like a cusp condition when an electron is near the nucleus.

I think if two electrons occupy the same place, then the repulsion must be infinitely large, which is not possible. So I believe the probability should vanish here. However, he argues that in the dynamic case, two electrons can never be in the same place, but statistically, the probability obtained by squaring the wavefunction is not zero.

I'm pretty confused now. Who is correct?

Update: The link here seems to contradict the answers below. I do not quite understand the statement "while at large distances between the electrons the conditional probability is greater than half of the corresponding unconditional one". Can someone give a hint?

Update 2: Some helpful references mentioned by DavePhD:

Fitzpatrick's Quantum Chemistry "On the other hand, in the spin-singlet state, there is an enhanced probability of finding the two electrons at the same point in space (because of the final term in the previous expression). In other words, the two electrons are attracted to one"

Advances in Quantum Chemistry volume 1, page 121: for the beryllium triplet P state two opposite spin electrons being at the same point is the most probable configuration

A statement in Atkins Physical Chemistry "The other combination [opposite spins] does not vanish when the two electrons are at the same point in space".

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Electrons do not occupy places. Other than that, you are mostly right. Indeed, $\psi(r_1,r_2)$ must vanish at $r_1=r_2$ because of the spin-statistics theorem, if not for other reason. Upd. It was brought to my attention that this reasoning is not quite right; see the comment below. Still, I would expect at least a downward-pointing cusp, much like the upward-pointing cusp found at the nucleus, only the other way around.

This, however, is not easy to visualize. We're not used to imagining multidimensional surfaces. And if we try to focus on one electron at a time, then we're back with the one-electron approximation, and the accurate multiparticle $\psi$ is lost to us.

BTW, the argument of infinite repulsion does not work. After all, $\psi$ doesn't vanish near the nucleus; wouldn't that imply infinite attraction? No, it would not. You integrate over the entire space and come up with some figure for energy, which is surely finite.

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    $\begingroup$ I think the spin part of the wavefunction has already taken care of the antisymmetry of the total wavefunction. The spatial part should be symmetric. $\endgroup$ – James Feb 14 '17 at 18:55
  • $\begingroup$ I'm very curious about the first sentence of this. What might be a good place to get more information about that? $\endgroup$ – Samy Bencherif Jan 16 at 21:48
  • $\begingroup$ Any textbook on quantum chemistry would do. $\endgroup$ – Ivan Neretin Jan 16 at 21:53
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Given that the two electrons have antisymmetric spin, your friend is right.

The phenomenon is called Fermi heaps.

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    $\begingroup$ Great link! but I think the reference here is more about how spin affects the spatial position instead of the exact coalescence condition we are interested in here. Dave's link to the Rice paper is more relevant $\endgroup$ – James Feb 21 '17 at 16:59
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    $\begingroup$ The Wikipedia about Fermi heaps and holes article also links to some nice animations quantum.bu.edu/notes/GeneralChemistry/FermniHolesAndHeaps.html $\endgroup$ – aventurin Feb 21 '17 at 21:23
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It is necessary to consider the wavefunction for the two electrons as the product of a spatial and spin part, $\psi=\psi_{space}\sigma_{spin}$. As the electrons are spin half particles (fermions) the total wavefunction must be asymmetric to exchange of coordinates. The answer to your question then depends upon whether the electrons are in a singlet or triplet spin state.

In the singlet state ('spin-paired') the spatial part of the wavefunction is symmetric and spin asymmetric; $$ \psi_S=\psi_{space}(sym)\sigma_{spin}(asym) $$ in the triplet state with 'parallel' spins it is the other way round.

The electrons can be combined into 4 possible spin states. Labelling the spins as $\alpha, \beta$ the combinations are $\alpha\alpha, \beta\beta, \alpha\beta, \beta\alpha$, of these the last two are neither symmetric nor asymmetric and so linear combinations are made to ensure proper symmetry. The four states are now $\alpha\alpha, \beta\beta, \frac{1}{\sqrt(2)}(\alpha\beta \pm \beta\alpha)$.

The arrangement is shown in more detail in the next figure, where a vector model of spin angular momentum (cones/arrows) is used to illustrate the difference between singlet and triplet spin states; ($1$) and ($2$) are used to label electrons.

spin states

As a consequence of the symmetry of the spin states the spatial states are determined to be symmetric for a singlet state and asymmetric for a triplet. This is shown in the next figure, which shows, just for illustration, the electron density plotted for particles in a box. In the singlet the electrons crowd together but in the triplet they are 'forbidden' to occupy the same region as one another by symmetry (as Pauli exclusion principle). In some texts the crowding together is called a 'fermi heap' and the lack of this a 'fermi hole' although its seems not to be common usage in chemistry.

singelt-triplet-space

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  • $\begingroup$ I'm confused. This is just for one 2D particle? $\endgroup$ – James Feb 22 '17 at 19:57
  • $\begingroup$ Isn't it a 1D box, with the two coordinates coming from the 1D position of the two particles? And how, if at all, is electrostatic repulsion include? It seems like this is really two non-electrostatically-interacting particles in a 1D box. $\endgroup$ – DavePhD Feb 23 '17 at 17:26
  • $\begingroup$ The repulsion comes from the fact that electrons are Fermions, and so from the symmetry that this imposes. $\endgroup$ – porphyrin Feb 24 '17 at 9:16
  • $\begingroup$ Have corrected error in 2D/1D in text . Wavefunctions calculated for a particles in a box using ψ(x,n)=sin(nπx/L)ψ(x,n)=sin(nπx/L) and then plotting square of Ψ=ψ(x1,1)ψ(x2,2)±ψ(x2,1)ψ(x1,2) $\endgroup$ – porphyrin Feb 24 '17 at 9:16
  • $\begingroup$ I still think you're leaving out the Coulombic electron-electron repulsion. Compare to this paper, especially section IV Coulomb Integrals. www-alavi.ch.cam.ac.uk/papers/twoelinbox.pdf $\endgroup$ – DavePhD Feb 24 '17 at 12:33
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Yes, the probability density is non-zero.

The book High-Energy Atomic Physics in the section "4.3 Kato Cusp Conditions" has a detailed explanation of why this is so for the singlet state.

("Kato" is Tosio Kato who in 1957 published a famous paper on this topic On the eigenfunctions of many-particle systems in quantum mechanics Communications in Pure and Applied Mathematics Vol. 10, pages 151-177.)

In section 4.3.2 an equation is derived from the Hamiltonian for a two electron atom.

Where $r_1$ and $r_2$ are the distances of the electrons from the nucleus and $r_{12}$ is the distance between electrons:

$\frac{\partial{\Psi}}{\partial{r_{12}}}$ (at $r_{12} = 0$) $= \frac{m\alpha}{2}\Psi $

Eventually in section "4.4.3 Approximate Wavefunctions Along Coalescence Lines" ("coalescence" meaning two or more particles being at the same point) an approximate wavefunction for two electron atoms when electrons are at the same point is derived:

$\Psi = Ne^{-2\eta R}$

where:

"R" is the distance of the two electrons from the nucleus

$\eta = mZ\alpha$

and for helium $N = 1.55(m\alpha)^3$

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  • $\begingroup$ Can I add here the references mentioned along with your bounty, like Fitzpatrick's and Atkin's? $\endgroup$ – James Feb 24 '17 at 15:57
  • $\begingroup$ @James yes, that's fine $\endgroup$ – DavePhD Feb 24 '17 at 15:58
  • $\begingroup$ @James This is the third reference I mentioned in the bounty, page 121: books.google.com/… $\endgroup$ – DavePhD Feb 24 '17 at 16:03
  • $\begingroup$ @James Fitzpatrick page 220: books.google.com/… $\endgroup$ – DavePhD Feb 24 '17 at 16:05
  • $\begingroup$ @James and various Atkins references such as page 119 here books.google.com/… and page 341 here books.google.com/… $\endgroup$ – DavePhD Feb 24 '17 at 16:09
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Your are right. If two electrons occupy the same place, the repulsion will be infinitely large in the real solution of Schrödinger equation.

But under One electron approximation framework (HF,DFT), probability of finding two electrons (spin-up and spin-down) at the same spot $x_0$ is nonzero.

As for the two electons with opposite spin , motion of electrons are uncorrelated. That means the probability of finding two electrons at $r_1$ and $r_2$ can be written as,

$$ P(r_1,r_2)= \mid \psi_1(r_1) \mid ^2 \times\mid \psi_2(r_2) \mid ^2 $$

when $r_1 = r_2$, $P(r_1,r_2) \neq 0$

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  • $\begingroup$ I understand in HF it is a different story. I am asking for the real case $\endgroup$ – James Feb 14 '17 at 20:47
  • $\begingroup$ In first paragraph I talked about the real case. Maybe I should make an implementation later. $\endgroup$ – Chenxin Feb 16 '17 at 23:33
  • $\begingroup$ Please see Ivan's analogy to the near-nucleus case. $\endgroup$ – James Feb 17 '17 at 16:54

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