QM: Spin States of Electrons & Positrons

[cragar]

If I have two electrons in the ground state of a helium atom, then one is spin up and the other is spin down . And there is no way of knowing which spin state each electron is until I measure it. So when I measure one electron to be spin up does this instantaneously make the other one spin down? Or if this doesn't work with the helium atom let's use a neutral pi-meson that decays into and electron and positron .
I have studied QM a little bit and any input will be much appreciated.

 

[khemist]

I believe it does.

Once one is measured, the other is known.

 

[jtbell]

To be picky about the wording, I would say that when we measure one electron to be spin up, we know that the other electron must now be spin down. Whether that means we have actually "done" something to the other electron to "make" it spin down, I think there is no generally accepted answer (it depends on interpretations of QM).

 

[cragar]

ok i see, interesting

 

[tom.stoer]

In addition I would like to stress the meaning of the two-electron state. |up, down> is not fully correct, it must read |up, down> - |down, up>. What does that mean?

You can't say that the first electron has spin=up and the second electron has spin=down; you can't even "label" the two electrons, they do not exist as separate entities, that's why you have this entangled quantum state. Instead you have to say that one electron has spin=up, the other one has spin=down w/o referring to "this electron" or to "the first electron". Trying to distinguish the two electrons will results in classical statistics which is ruled out for quantum systems experimentally.

 

[cragar]

ok but if i know the state of one of them then i know he state of the other one

 

[tom.stoer]

Yes, you are right, but you can only say "one electron has spin=up, the other one has spin=down". One should be careful and try to avoid to say "this electron" or "that electron"; as long as they are inside the Helium atom (and not separated by some experiment) they are not individual electrons (like classical coins).

 

[apolokph]

great answer
i think tom

 

[maverick_starstrider]

Yes, you are right, but you can only say "one electron has spin=up, the other one has spin=down". One should be careful and try to avoid to say "this electron" or "that electron"; as long as they are inside the Helium atom (and not separated by some experiment) they are not individual electrons (like classical coins).

I don't believe this is so, since the two electrons have opposite spins they are no longer indistinguishable. One could easily separate the two by applying an external field and exploiting the Zeeman effect to break the degeneracy of the ground state. Thus you would say that there is a distinct and distinguishable electron with spin up and a distinct and distinguishable electron with spin down. The N! gibbs factor should not be applied here and thus the system WOULD obey classical statistics and not quantum ones would it not? Keep in mind that the derivation of distinguishability is firmly linked to the parity of an electron ground-state and thus if they WERE distinguishable they would be Pauli excluded but they are not (since they have different spin) and thus they can't be indistinguishable.

 

[dextercioby]

I don't believe this is so, since the two electrons have opposite spins they are no longer indistinguishable. One could easily separate the two by applying an external field and exploiting the Zeeman effect to break the degeneracy of the ground state. [...]

Then I'm afraid you don't understand what <indistinguishable> really means. The notion applies to the state of the compound system and to the values of the measurable observables prior_to_measurement.

 

[maverick_starstrider]

Then I'm afraid you don't understand what <indistinguishable> really means. The notion applies to the state of the compound system and to the values of the measurable observables prior_to_measurement.

So you're saying that a system of two electrons for which one is spin up and the other is spin down behaves as a system of size N=2 obeying Fermi-Dirac statistics and not Maxwell-Boltzmann statistics? I would say, provided there exists no mechanism in our Hamiltonian which allows spins to flip, that this system would obey classical statistics (Maxwell-Boltzmann). There exists no mechanism through which they compete for states.

 

[cragar]

The electrons are not in a spin state until you measure them.

 

[tom.stoer]

The electrons are not in a spin state until you measure them.

You can prepare a quantum state, e.g. S=0 which means you have a "spin state" |up, down> - |down, up>; you do not have a definitre spin for one electron, but as a quantum state it's perfectly valid.