QED on electromagnetic interactions in atoms

[WWCY]

TL;DR

2 electrons with opposite spin can occupy same orbital in atom - says Pauli
2 electrons repel - says QED
What does QED say about 2 electrons in an atom?

I think my main question is pretty much summarised in the TL;DR.

I have another related question: Is it possible for one to "create" two fermionic particles of the same charge but a different spin (using creation operators $\hat{a}_{ \downarrow , + }(x,t)\hat{a}_{ \uparrow , + }(x,t) |0 \rangle$ ) at the same spacetime location?

My knowledge in QED is extremely limited so broad-brush replies would be nice, cheers.

PS I haven't been using PF for a while now, and I have noticed that my Latex isn't rendering in my Preview, is this normal?

 

[Meir Achuz]

Yes.

 

[PeterDonis]

2 electrons repel - says QED

What do you mean by "repel"? The usual meaning of that would just be that both electrons have negative charge so there is a repulsive electromagnetic force between them. Is that what you mean? If so, it's by no means specific to QED; classical EM says the same thing. If you mean something else, what?

What does QED say about 2 electrons in an atom?

An awful lot. But as far as what Pauli said, QED says the same thing.

I have another related question: Is it possible for one to "create" two fermionic particles of the same charge but a different spin (using creation operators $\hat{a}_{ \downarrow , + }(x,t)\hat{a}_{ \uparrow , + }(x,t) |0 \rangle$ ) at the same spacetime location?

Mathematically, I don't see why this wouldn't be possible, but I don't know of any actual examples; AFAIK all known fermions have spin 1/2.

 

[vanhees71]

If you count baryon resonances then there are fermions with spin 3/2. It's of course a composite particle. All elementary fermions within the Standard Model, i.e., quarks and leptons have of course spin 1/2.

 

[WWCY]

Thanks for all the responses

What do you mean by "repel"?

I meant EM repulsion. Allow me to rephrase my questions, the original ones were poorly phrased.

What I intended to ask was: Pauli says fermions cannot occupy the same state. This means that we could find situations in which two fermions of differing spin have the exact same wavefunction, since they already differ in spin. Would this mean that these two fermions could be found at the same spacetime position?

I meant to substitute "fermions" for "electrons". But I totally forgot to consider the EM repulsive interaction between them, which I guess means that the probability of finding them decreases as their separation distance decreases - is this correct? And could non-interacting/oppositely-charged fermions be found "stacked" on the same spacetime location? As for this question:

I have another related question: Is it possible for one to "create" two fermionic particles of the same charge but a different spin (using creation operators $\hat{a}_{ \downarrow , + }(x,t)\hat{a}_{ \uparrow , + }(x,t) |0 \rangle$ ) at the same spacetime location?

Why wouldn't EM repulsion prevent two fermions of the same charge from being created at the same spacetime location? And if nothing prevents this from happening, how do these particles interact?

Hope this makes my questions clearer, cheers for helping.

 

[PeterDonis]

This means that we could find situations in which two fermions of differing spin have the exact same wavefunction, since they already differ in spin.

This doesn't make sense. If they have different spin, they can't have the exact same wave function.

the EM repulsive interaction between them, which I guess means that the probability of finding them decreases as their separation distance decreases - is this correct?

The repulsive potential will, in general, lower the amplitude of the wave function for smaller separations, yes.

Why wouldn't EM repulsion prevent two fermions of the same charge from being created at the same spacetime location?

I haven't looked at the details of the math, but I don't think the wave function amplitude goes to zero for zero separation for this case. Remember that we're talking about quantum objects, not little billiard balls with charge. So we're talking about wave function amplitudes, not definite positions.

 

[A. Neumaier]

Pauli says fermions cannot occupy the same state. This means that we could find situations in which two fermions of differing spin have the exact same wavefunction, since they already differ in spin.

No. Differing in spin implies differing in wave function!

 

[DrClaude]

Consider a two-electron atom, with orbitals indicated by $\phi_i(x)$. I will distinguish here "orbitals," which are spatial wave functions, from "spinorbitals," including spin.

The two electrons can be in one of the symmetric spin states (the "triplet state")
$$
| \uparrow \uparrow \rangle \\
\frac{1}{\sqrt{2}} \left( | \uparrow \downarrow \rangle + | \downarrow \uparrow \rangle \right) \\
| \downarrow \downarrow \rangle
$$
in which case the spatial part of the wave function must be antisymmetric (Pauli principle),
$$
\frac{1}{\sqrt{2}} \left[ \phi_i (x_1) \phi_j (x_2) - \phi_j (x_1) \phi_i (x_2) \right]
$$
with $x_i$ the coordinates of electron $i$. In this case, you can't have $i=j$, as the wave function would be 0, which is the Pauli exclusion principle. You also get that if you calculate the probability of the two electrons being at the same place, $x_i = x_j$, you also get 0, so the electrons avoid each other.

The two electrons can also be in an anti-symmetric spin states (the "singlet state"), which is was people usually refer to as "having opposite spin",
$$
\frac{1}{\sqrt{2}} \left( | \uparrow \downarrow \rangle - | \downarrow \uparrow \rangle \right)
$$
in which case the spatial part of the wave function must be symmetric (Pauli principle),
$$
\frac{1}{\sqrt{2}} \left[ \phi_i (x_1) \phi_j (x_2) + \phi_j (x_1) \phi_i (x_2) \right] \quad \textrm{or} \quad
\phi_i (x_1) \phi_i (x_2)
$$
so there both electrons can occupy the same orbital. The probability of the two electrons being close to each other is no longer 0.

This leads to what are called the Fermi heap and Fermi hole. For the same electronic configuration, it makes the triplet be lower in energy than the singlet.

 

[vanhees71]

Isn't it the other way around, and the spin-singlet state should have lower energies? Take helium. There in the spin-singlet state you can occupy with both electrons the $\ell=0$ groundstate, and indeed looking at the term scheme of ortho- and para-helium you see that the ground state is a para-helium ($S=0$) state:

https://de.wikipedia.org/wiki/Heliumatom#Ortho-_und_Parahelium

 

[DrClaude]

Isn't it the other way around, and the spin-singlet state should have lower energies? Take helium. There in the spin-singlet state you can occupy with both electrons the $\ell=0$ groundstate, and indeed looking at the term scheme of ortho- and para-helium you see that the ground state is a para-helium ($S=0$) state:

https://de.wikipedia.org/wiki/Heliumatom#Ortho-_und_Parahelium

I mentioned explicitly "for the same electronic configuration." In that diagram you linked to, you see that, e.g., for the 1s2s configuration, 2³S₁ is lower in energy than 2¹S₀.

For 1s², the fact that both electrons can be in a lower energy orbital more than compensates for the increase in e-e repulsion compared to 1s2s.

It is also the basis for the first Hund rule: for multi-electron atoms with many terms corresponding to the ground electronic configuration, you start by considering the terms with the highest multiplicity.