Are two electrons in an hydrogen orbital positionally dependent ?

[xortdsc]

Hi,

I was wondering if 2 electrons in an 1s-orbital of a hydrogen anion (or alternatively neutral helium) are positionally dependent. By that I mean if some knowledge about one electron's position would somewhat give knowledge about the other ? Or is it completely nonsense to talk about positions in an orbital as the electrons are distributed and never have a specific position at all (even theoretically) ? And if that is the case, are there other correlations besides position (similar to their spin which has to be opposite for the two electrons in the same orbital, so knowing one determines the other) ?

Thanks and cheers

 

[Nugatory]

Yes, no, maybe, depends on what you mean by "positionally dependent"

The interactions between the two electrons are part of the Hamiltonian of the system; this Hamiltonian appears in Schrodinger's equation which we solve for the wave function of the system; this wave function gives us probability amplitudes for various observables, including the positions of the electrons.

Whether these positions are interesting (in the sense that they provide any useful insight into the behavior of the atom) is a different question, one that we can also ask about the ordinary single-electron hydrogen atom.

 

[DrClaude]

We know that the electrons can't be exactly at the same place at the same time (see Fermi hole), but apart from that you can only say that there is a higher probability of them being far apart than close together

 

[cgk]

To the OP: Yes, the instantaneous position of the second electron depends on the position of the first electron. For a given N-electron wave function $\Phi$, this property is described in the so called pair distribution function:
$$\rho(\vec x_1, \vec x_2)=\frac{\langle\Phi|\Psi^\dagger(\vec x_1)\Psi^\dagger(\vec x_2)\Psi(\vec x_2)\Psi(\vec x_1)|\Phi\rangle}{\langle\Phi|\Psi^\dagger(\vec x_1)\Psi(\vec x_1)|\Phi\rangle\langle\Phi|\Psi^\dagger(\vec x_2)\Psi(\vec x_2)|\Phi\rangle}$$
where $\Psi(\vec x)$ are the 'wave operators' destroying a particle at a specific place, and $\vec x_i=(\vec r_i, s_i)$ are the electron's combined space/spin coordinates. This quantity, effectively, describes the probability of finding electrons at the places $\vec x_1$ and $\vec x_2$ simultaneously, divided by the probability of finding them there individually (the denominator is the product of the electron densities at x1/x2).

This quantity, the pair distribution function, can be calculated for real wave functions in real molecules, and it is most definitely not uniformly 1. However, it does approach 1.0 as the places x1 and x2 become farther separated in space, or as x1 and x2 belong to different atomic shells (e.g., core electrons are hardly displaced by nearby valence electrons).

To DrClaude: Electrons *CAN* be right on top of each other, as long as they have different spin coordinates[1]. The Fermi hole (here often called exchange-correlation hole) does not necessarily go down to 0.0. Indeed, in the high-density limit of the uniform electron gas (=low correlation limit), the correlation hole vanishes and only an exchange hole remains. That means that in this limit opposite spin electrons are effectively uncoupled.

[1] (they can't if they have the same spin coordinate; for same-spin electrons the pair distribution function is always 0 at zero inter-electronic distance, due to the anti-symmetry of the wave function. But this does not apply to different-spin-electrons).

 

[DrClaude]

To DrClaude: Electrons *CAN* be right on top of each other, as long as they have different spin coordinates[1]. The Fermi hole (here often called exchange-correlation hole) does not necessarily go down to 0.0. Indeed, in the high-density limit of the uniform electron gas (=low correlation limit), the correlation hole vanishes and only an exchange hole remains. That means that in this limit opposite spin electrons are effectively uncoupled.

The OP was asking about two electrons in the 1s orbital of a hydrogenic atom. In that case, the electrons can be found in a strictly anti-symmetric spatial wave function, for which the probability of finding the electrons at the same place is exactly 0.

My answer was too short and lacked some important details about what I was considering. Your answer was more complete.

 

[cgk]

The OP was asking about two electrons in the 1s orbital of a hydrogenic atom. In that case, the electrons can be found in a strictly anti-symmetric spatial wave function, for which the probability of finding the electrons at the same place is exactly 0.

If they are both in the 1s hydrogen orbital, then they are in a symmetric spatial wave function (they are both in the same spatial orbital). It is then the spin wave function which is anti-symmetric. So the chance of finding them at the same place is not zero for this case.

In the dominant single determinant (Hartree-Fock approximation), the positions of the two electrons are even entirely uncorrelated in this case.

 

[xortdsc]

thanks guys. that helps. :)