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The two-electron wavefunction of the ground state of helium is $$ \psi(r_1,r_2)=\phi_{1s}(r_1)\phi_{1s}(r_2)\otimes (|\uparrow_1\downarrow_2-\downarrow_2\uparrow_1\rangle)/\sqrt{2} $$ where $\phi_{1s}(r_1)$ and $\phi_{1s}(r_2)$ are both Gaussians peaked about $r_1$ and $r_2$ respectively. If one puts $r_1=r_2$, the Gaussians overlap. However, this overlapping of spatial part of the wavefunction seems to be a perfectly valid possibility. As all quantum numbers, namely $n,l,m_l,m_s$ are not the same, in the limit $r_1=r_2$ (which is not a quantum number) the total wavefunction $\psi(r_1,r_2)$ remains nonzero.

But is it possible that somehow the case $r_1=r_2$ is less favoured than $|r_1-r_2|\neq 0$? I'm assuming the electrons to be uncharged.

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    $\begingroup$ I would just like to point out that the limit $|r_1-r_2| \gg 0$ is nonsensical, since any positive number is much much larger than zero. IN order to make a comparison like that you need finite length scales on both sides of the equation. Maybe you are thinking of $|r_1-r_2|\sim a_0$, there a_0 is the Bhor radius? $\endgroup$ – Mikael Fremling Jan 18 at 9:11
  • $\begingroup$ @MikaelFremling I am asking whether $r_1-r_2=0$ can somehow be less favoured than the case $|r_1-r_2|\neq 0$. I have changed the notation. $\endgroup$ – mithusengupta123 Jan 18 at 9:15
  • $\begingroup$ I am confused by the spin wave function. The "1" is always up and the "2" is always down, but they switch positions. If it interchange 1,2 it doesn't flip sign, it flips spins. $\endgroup$ – JEB Jan 18 at 15:58
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There's a couple of deeply-wrong claims in your question

The two-electron wavefunction of the ground state of helium is $$ \psi(r_1,r_2)=\phi_{1s}(r_1)\phi_{1s}(r_2)\otimes (|\uparrow_1\downarrow_2-\downarrow_2\uparrow_1\rangle)/\sqrt{2} $$ where $\phi_{1s}(r_1)$ and $\phi_{1s}(r_2)$ are both Gaussians peaked about $r_1$ and $r_2$ respectively.

No, that's not what that equation says.

  • $\phi_{1s}(r)$ is not a Gaussian - it is reasonably well approximated by a Slater-type orbital with an exponential dependence, $\phi_{1s}(\vec r) \sim e^{-\kappa r}$, and a Gaussian gets it wrong at the low-$r$ region (it's smooth at $\vec r=0$, instead of having a cusp at the nuclear position) and at the high-$r$ region (it decays too fast, so its 'tail' is much smaller than it needs to be).
  • $\phi_{1s}(r_1)$ and $\phi_{1s}(r_2)$ are both [...] peaked about $r_1$ and $r_2$ respectively

    That's not what that notation means. It means that you have two orbitals $\phi_{1s}(r_1)$ and $\phi_{1s}(r_2)$ which are both peaked at the origin, and which are being evaluated at $r_1$ and $r_2$.

  • I'm assuming the electrons to be uncharged.

    Great! That's a very interesting model system in which one can play with some features of multi-electron systems and what they imply. But it isn't helium.


As for your actual question,

is it possible that somehow the case $r_1=r_2$ is less favoured than $|r_1-r_2|\neq 0$?

... that still doesn't make sense.

You're approximating your state as a single Slater determinant. For helium, this is a reasonable approximation, though it is never fully true in any atom. (For a full description of any atom, you need post-Hartree-Fock methods, where multiple different Slater determinants can contribute to any state.)

However, since you're

assuming the electrons to be uncharged.

then the electrons are fully uncoupled, and the uncorrelated single-Slater-determinant state you wrote down is actually the rigorous ground state of the system.

This means that the probability densities for the two electrons are fully uncoupled and uncorrelated, and asking about correlated observables like the probability of the electrons being at any given distance from each other is counter-productive and essentially meaningless. The probability that an electron will be at some point $\vec r$ is completely independent of where the other electron is, and the probability distribution of their relative distance is just whatever ends up following from that uncorrelated distribution.

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  • $\begingroup$ Thanks for the answer. I don't know what I was thinking when I said Gaussians :-( :-( I was trying to understand whether Pauli exclusion will play any role if two such electrons approach each other (apart from Coulomb repulsion). Since you have written an answer there is no point in editing it now. :-( $\endgroup$ – mithusengupta123 Jan 21 at 11:48
  • $\begingroup$ So if my wavefunction describes two noninteracting fermions there is nothing which prevents them to come arbitrarily close? One effective doesn't see other? $\endgroup$ – mithusengupta123 Jan 21 at 11:53
  • $\begingroup$ The Pauli exclusion principle is already factored in by the symmetry of your state, particularly in the antisymmetry of the spin state. And yes, since you've got noninteracting fermions in an uncorrelated spatial state, they really don't see each other. $\endgroup$ – Emilio Pisanty Jan 21 at 11:55
  • $\begingroup$ Can Pauli exclusion ever forbid two electrons (say, uncharged) to come arbitrarily close? Actually I found in C. Kittel that when electron orbitals overlap due to Pauli exclusion principle comes into play and that gives rise to a repulsion. $\endgroup$ – mithusengupta123 Jan 21 at 12:04
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If you look at the structure of your wave functions you will see that the probability to find a specific $r_1$ and $r_2$ are completely uncorrelated. This is since the two probability densities factorize.

All the asymmetric components (i.e. fermionic content) is in the spin-singlet part of the wave-function, and the special dependents is thus symmetric.

Thus the probability to find the electrons at exactly the same place is a measure of order zero (i.e very very unlikely) compared with all other possibilities.

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  • $\begingroup$ To really establish your last claim do you not need to compute the $\langle r_1-r_2\rangle$ first in the state $\psi(r_1,r_2)$? But that gives 0. $\endgroup$ – mithusengupta123 Jan 18 at 9:45
  • $\begingroup$ @mithusengupta123 it is true that to find the exact number on needs to compute $\left<r_1-r_2\right>$ (or something simlar). However to see that this is vanishingly small one only needs to consider the dimentionallity of the problem. The full space of $r_1,r_2$ pairs is 6-dimensional. The space of $r_1=r_2$ pairs is only 3-dimensional. Since it has lower dimentionality it will always be of measure zero. $\endgroup$ – Mikael Fremling Jan 18 at 9:51
  • $\begingroup$ Maybe a better question to ask is "whether a $|r_1-r_2|=a_0$ is less probable/favoured than $|r_1-r_2|=2a_0$"? $\endgroup$ – mithusengupta123 Jan 18 at 9:56
  • $\begingroup$ @mithusengupta123 I think you may be missing the point. You can ask for the probability density to find e.g. $r_1=r_2$ but the probability will always be zero, since the probability is an integral over the probability density (which is the squared probability amplitude). The probability can only be nonzero if you are integrating over a finite volume element, but the the probability depends on the size of this volume element to. In any case, since the the positions of $r_1$ and $r_2$ are independent for this wave function i'm not sure what you want to get out of your modified question... $\endgroup$ – Mikael Fremling Jan 18 at 10:14

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