Why don't electrons in the same orbital repel

[Maisara-WD]

Hi

I think it's all descriped in the title:

Why don't electrons in the same orbital repel..

I don't want the expected answer: because the direction of the magnetic field arising from the electron spin motion of each is opposite to the other... I don't get it.. how does a magnetic field arise and how is the direction opposite.. I mean.. I can't get it in my head, I can't imagine it.. if u got a diagram or any illustrating material please enhance your explanation with it. it seems too be too easy but I don't understand it... I'm waiting for you

 

[weejee]

It what context are you saying that two electrons in the same orbital don't repel?

They actually do repel. The simplest example is the helium atom, in which two electrons in the 1s orbital repel each other. Therefore the ground state energy is higher than what would be expected if the Coulomb repulsion were not present.

 

[alxm]

Specifically (since I happen to know the numbers offhand), non-repulsive Helium would have a ground state of -4.0 Hartrees. In reality it's -2.903.

 

[Tipi]

I'm not sure, but his post make me think he wonder about something like the difference between Hubbard model and general band structure theory.

Could you clarify your question, and give a source of the affirmation you are giving here? It will help us to answer...

TP

 

[inempty]

They do repel. We regard them as independent when we just want to an approximation answer, as it is said by alxm.

 

[Tipi]

In the band theory of metals, two electron in the same energy level (orbital) don't interact with each other (at all). It's the Fermi gaz... This model cannot explain the transition from conducting to insulating when we increase the nearest neighbor spacing (by heating for example).

The Hubbard model "correct it" adding an on-site interaction, U, between electron in the same orbital. This model is closer to reality. It successfully explains the transition stated above.

And be careful. Saying that electrons are in the same orbital (in a metal) doesn't mean they are at the same position in space. They have a definite momentum so they are completely delocalized in space (plane waves).

TP

 

[ZapperZ]

We should hold off on offering any more responses to this thread until the OP comes back and explain more. It is obvious that the original premise of the question is faulty (i.e. we DO consider repulsion effects between multiple electrons atom, for example). Otherwise, we might be offering answers beyond what the OP is asking for, or answering a moot question that is faulty in the first place.

Zz.

 

[nomadreid]

electric repulsion and electrons sharing the same orbital

Although most people concentrate on the negative part of the Pauli exclusion principle, I am puzzled by its positive part, which allows, for example, two electrons with opposite spin to be in the same energy level. True, Coulomb's law applies to point charges, and the two electrons in the same orbital are spread out over a region of space, so strictly the two principles (Pauli and Coulomb) do not conflict. So my question is whether the electric repulsion between two electrons in a single orbital need to be taken into consideration in calculating the properties of the orbital. If so, how does one calculate electric repulsion between two electrons in indeterminate states? Thanks.

 

[alxm]

True, Coulomb's law applies to point charges, and the two electrons in the same orbital are spread out over a region of space, so strictly the two principles (Pauli and Coulomb) do not conflict.

Electrons are point charges in the Schrödinger equation as well. (so is the nucleus to a very good approximation). So you still have a zero probability of two electrons being in the same spot at the same time due to the Coulomb repulsion - the potential energy goes to infinity and the wave function to zero at that point. Since the wave function is continuous, this also means you have an area around one electron with a low probability of finding the other electron. This is called the he 'Coulomb hole'. Similarly you have the 'Fermi hole' between electrons of the same spin due to the Exclusion Principle. (Note you don't see any 'holes' in the orbitals, because they get spread out over space, since the electron's position is)

So my question is whether the electric repulsion between two electrons in a single orbital need to be taken into consideration in calculating the properties of the orbital.

Yes of course. But it's worth noting that although the energy changes quite a bit due to the electron-electron repulsion in an orbital, the shape of the orbital does not change so much. (due to how it's 'spread out' as mentioned)

If so, how does one calculate electric repulsion between two electrons in indeterminate states? Thanks.

I'm not sure what you mean by an 'indeterminate state'. But if you're asking how to calculate the orbitals and their respective energies; well that's the central problem of quantum chemistry. You have to solve the time-independent Schrödinger equation for the electrons (even if you want a time-dependent solution you'd still typically start with solving the time independent one). This is very difficult, precisely because there's electron-electron repulsion, which makes the thing impossible to solve analytically.

(BTW, mods: Move this to quantum or atomic, since it surely doesn't belong under 'classical physics')

 

[Bob S]

In the neutral helium atom with two electrons in the 1s ground state, the binding energy of the first electron in the 1s state is 4 x 13.6 eV = 54.4 eV.

The ionization potential of the neutral helium atom is 24.59 eV, so the total binding energy of both electrons is 54.4 + 24.6 = 79 eV = 5.81 Rydbergs.

This excerpt from the book uses variational principles to get 5.69 Rydbergs. See pgs 393-395 and Eq 17.16 in

http://books.google.com/books?id=qH...ge&q=neutral helium atom ground state&f=false

[Exerpt from Quantum Mechanics: from Basic Principles to Numerical Methods and Applications by Louis Marchildon]

Bob S

 

[nomadreid]

Thank you very much, alxm and Bob S.

 

[ZapperZ]

Two different threads asking practically the same question have been merged.

Zz.

 

[Jesvant]

Electron don't repel becuse they are orbiting, hence their attraction to the proton in the nucleus is more than their repulsion

 

[ZapperZ]

Electron don't repel becuse they are orbiting, hence their attraction to the proton in the nucleus is more than their repulsion

This is not correct. Please read the previous responses to this thread.

Zz.

 

[cgk]

Electrons are point charges in the Schrödinger equation as well. (so is the nucleus to a very good approximation). So you still have a zero probability of two electrons being in the same spot at the same time due to the Coulomb repulsion - the potential energy goes to infinity and the wave function to zero at that point.

I'm sorry for reviving this thread at this point in time, but I wanted to point out that this statement is not actually true. There is a Coulomb hole, but the probability of finding two opposite-spin electrons at the same point in space is not zero in general. In fact, in the high-density limit of the electron gas, there is not even a Coulomb hole. You can read this up, for example, in Perdew and Yang, ``pair-distribution function and its coupling-constant average for the spin-polarized electron gas'', Phys Rev B. 46, 12947 (1992)

The singularity in the Coulomb interaction can be canceled by another singularity in the kinetic energy term, which is caused by the wave function not being differentiable with respect the the interelectronic distance at electron coalescence.

Since the wave function is continuous, this also means you have an area around one electron with a low probability of finding the other electron. This is called the he 'Coulomb hole'. Similarly you have the 'Fermi hole' between electrons of the same spin due to the Exclusion Principle. (Note you don't see any 'holes' in the orbitals, because they get spread out over space, since the electron's position is)

Well, the hole would not be in the orbitals (orbitals are one-particle functions!), but in the pair distribution function of the electrons.

 

[nomadreid]

cgk, thank you very much for this added input. A small correction:

You can read this up, for example, in Perdew and Yang, ``pair-distribution function and its coupling-constant average for the spin-polarized electron gas'', Phys Rev B. 46, 12947 (1992)

You mean Perdew and Wang (it is Y. Wang, from whence probably the confusion).

I will try to follow up on your indications, since I do not have access to that journal except for the abstracts of the articles.

 

[weejee]

From the following integral,
$$\int d\Omega dr r^2 |\psi(\vec{r})|^2 V(\vec{r})$$

I think a repulsive potential should be at least as strong as 1/r^3 around the origin for two particles to have zero probability of spatial overlap in principle. If it is weaker than 1/r^3, the integral is still finite around r=0.

 

[alxm]

I'm sorry for reviving this thread at this point in time, but I wanted to point out that this statement is not actually true. There is a Coulomb hole, but the probability of finding two opposite-spin electrons at the same point in space is not zero in general.

Yeah, I got it wrong. It's true that it's not zero at high density/momentum. You do always have a cusp in the density/wave function though.

Well, the hole would not be in the orbitals (orbitals are one-particle functions!), but in the pair distribution function of the electrons.

Orbitals can still be somewhat correlated (e.g. HF orbitals are correlated w.r.t. spin), my point was just that orbitals do take into account most of the interelectronic interaction.

You mean Perdew and Wang (it is Y. Wang, from whence probably the confusion).

Actually I suspect he confused Y. Wang with another important DFT theorist, W. Yang. Confusing eh? :) There's Parr & Yang and Perdew & Wang.