# I Are electrons in a shared shell quantumly entangled?

1. Mar 17, 2016

### bbbl67

The reason I ask is because by definition entangled particles, like electrons, are always precisely the opposite spin of each other. At the same time it's known that for electrons to share an electronic shell with another electron, that they too must be precisely the opposite spin of each other. Am I taking a leap too far in my assumptions?

2. Mar 17, 2016

### vanhees71

It's not so clear to me what you mean. Of course, electrons are fermions and thus not more than cannot occupy the same single-particle state at a time. In the case of atoms the ground state can host two electrons with opposite spin-$z$ components only.

The full quantumtheoretical explanation is as follows: Since the orbital angular momentum of this state is $\ell=0$ the spatial part of the two-body wave function is even under exchange of the electrons and thus the spin part must be odd. That implies that the spin part is in the state
$$|\Psi_s \rangle=|1/2,-1/2 \rangle -|-1/2,1/2 \rangle.$$
This means the spin (polarization) of the electrons are entangled.

In the next shell, which is a $p$-shell, i.e., $\ell=1$, the two electrons are in a superposition of the spin-triplet states, i.e.,
$$|1/2,1/2 \rangle, \quad |-1/2,-1/2 \rangle, \quad |1/2,-1/2 \rangle + |1/2,-1/2 \rangle,$$
i.e., they can have the same spin-$z$ component or not.

3. Mar 18, 2016

### bbbl67

So in some shells they maybe entangled, and in others they may not be?

4. Mar 18, 2016

### vanhees71

exactly.

5. Mar 18, 2016

### bbbl67

But wait a minute, aren't all electronic orbitals made up of some form of binary pairing? Like for example, the 2p orbital can have 6 electrons in it, but it is subdivided into 2p_x, 2p_y, 2p_z, so wouldn't each electron be entangled to each other in each sub-orbital?

6. Mar 19, 2016

### secur

It's not the case that two entangled electrons must have opposite spins. When they share a physical state, as in the s orbital of a helium atom, then they must have opposite spins, since they can't share the same total state (including spin), by the Pauli exclusion principle. (Or, because their wave function is anti-symmetric). And, in this case, they are certainly entangled.

But in general if electrons (or any particles) are entangled it means only that one or more of their properties (could be spin, momentum, whatever) are correlated in some way. If you measure the correlated property of one, then you know that property of the other. Two entangled electrons do not have to be in the same state, of course. In a typical gedanken, they might be physically separated by a few light years. So they could be correlated to have the same spin, without offending Pauli.

Suppose we take two electrons in the same state (s-shell, say) and separate them physically without affecting spins (a very delicate operation to say the least!) They're no longer in the same state, but still, their spins must be opposite. However, with the spin axis oriented vertically, physically turn one of them upside down - now their spins must be the same. Well, actually the gyroscopic quality of the spin angular momentum would make it difficult to accomplish that precisely. So, you could use Thomas precession (just the right amount) to do it.

Another example, suppose a spin-1 particle decayed into a bunch of products, including two electrons. Suppose all the other particles' spins added up to 0. Then the two electrons are entangled and must have the same spin.

The whole discussion has a certain air of unreality since spin is dependent on the orientation of your measuring device; talking about the spin of an electron without specifying the experimental setup is, in fact, meaningless. I've been assuming that it was carefully specified, all along. Final note, just to avoid confusion: of course we're talking about the direction of spin (up, down, or a sum thereof with norm 1) not the magnitude, which is always 1/2 hbar.

As always, I could be wrong. Don't trust anything I say until someone more qualified agrees.

Last edited: Mar 19, 2016
7. Mar 19, 2016

### vanhees71

In the case of the helium you can have both. In the ground state you can have either the spins coupled antisymmetrically (para helium) to total spin 0 or symmetrically (ortho helium) to total spin 1. In the former case you can have both electrons in the 1s shell, in the other not, because it's forbidden by the Pauli principle. The ground state is para helium with the spins in the antisymmetric state and the lowest energy level of the ortho helium is with one electron in the 1s and the other in the 2s shell. Since there's no electromagnetic transition between the s states, originally one thought that there are two different elements rather than only one helium.

8. Mar 19, 2016

### secur

My phrasing "the s orbital" was meant to imply a single orbital, 1s, without getting into para/ortho details; but really it's inaccurate, so thanks for the clarification!

Come to think of it - Ortho helium makes a nice simple example of entangled electrons where the spins are the same, better than my examples (turning upside down, and decay of spin-1 particle). Measure one of an ortho helium atom's electron's spin, the (entangled) other must be the same

Last edited: Mar 19, 2016
9. Mar 22, 2016

### bbbl67

Isn't the 1s and 2s orbitals separate energy states? So why would there be no electromagnetic transition between them?

10. Mar 22, 2016

### Staff: Mentor

Selection rules.

11. Mar 22, 2016

### bbbl67

Sorry, I'm not a professional chemist or physicist, what selection rules are you talking about?

12. Mar 22, 2016

### Khashishi

I think it is more correct to say there is no electric transition between them. There is a low probability, magnetic spin-flip transition between them.

Basically, photons carry angular momentum, and 1s and 2s states both have no orbital angular momentum. So by angular momentum conservation, it's impossible for an atom to go from 2s to 1s and emit one photon, unless the spin is also flipped during the transition. The electric force doesn't interact with the spin-- you need the magnetic force, and this only happens at much lower probabilities -- so it is called "doubly forbidden".

13. Mar 22, 2016

### bbbl67

14. Mar 22, 2016

### Khashishi

Yeah. A photon carries both the electric and magnetic forces. Actually, both the electric and magnetic force are parts of the electromagnetic force, but it is useful to split them up for some calculations. And in these atomic calculations, the electric force is much stronger than the magnetic force, and we usually totally ignore the magnetic force, unless the electric force doesn't play a role in the transition.

15. Mar 23, 2016

### bbbl67

So isn't there any photons released during a magnetic transition?

16. Mar 23, 2016

### Khashishi

A photon is released.

17. Mar 24, 2016

### bbbl67

Do we know what energy level that this photon would be at?

18. Mar 24, 2016

### zonde

I would like to reiterate this question. I am interested in answer. I suppose that in covalent bonds sub-orbitals can loose symmetry (and acquire slightly different energy levels?).

19. Mar 24, 2016

### Staff: Mentor

The same as the difference in energy between the two atomic states involved in the transition.