Two-Level System Consideration: Electron Indistinguishability

[aaaa202]

Consider a two level system, for example the spin 1/2 system. For a single electron there are infinitely many possible states for this system since any state:
ls> = a lup> + bldown> with lal^2 + lbl^2 = 1
is an allowed state.
Now if we again consider the same system but with two electrons in it, it seems there is not the same freedom for how the state of the system can look. Because by the symmetrization requirement we have:
ls> = 1/sqrt(2)(lup,1>ldown,2)-lup,2>ldown,1>)
And that is the only possible state that meets this requirement. Is this true? If so, I guess it just seems weird to me that introducing a second electron takes away the freedom for the single electron. And how does this change physically happen. Imagine we have an electron in the spin up state and bring it close to another electron. The symmetrization requirement now means there is a 50% chance that the first electron is instead in the spin-down state. I think the answer to the last question is that I'm thinking this too classically and that electrons are indistinguishable on a very fundamental level. But it is nevertheless still weird to me.

 

[Simon Bridge]

The spin is a magnetic moment - i.e. a magnet.
When you put one magnet on a table, you can orient it however you like - but put two on the table and they will assume a limited range of relationships.

 

[stevendaryl]

Consider a two level system, for example the spin 1/2 system. For a single electron there are infinitely many possible states for this system since any state:
ls> = a lup> + bldown> with lal^2 + lbl^2 = 1
is an allowed state.
Now if we again consider the same system but with two electrons in it, it seems there is not the same freedom for how the state of the system can look. Because by the symmetrization requirement we have:
ls> = 1/sqrt(2)(lup,1>ldown,2)-lup,2>ldown,1>)
And that is the only possible state that meets this requirement. Is this true?

Yes, but that's more general than it looks. Consider a different basis,

$|U\rangle = a |up\rangle + b |down\rangle$
$|D\rangle = -b^* |up\rangle + a^* |down\rangle$

with $|a|^2 + |b|^2 = 1$

Then we could just as well form $\frac{q}{\sqrt{2}}(|U\rangle |D\rangle - |D\rangle |U\rangle)$
but that happens to be equal to $\frac{q}{\sqrt{2}}(|up\rangle |down\rangle - |down\rangle |up \rangle)$ (because if you multiply it out, the other two possibilities cancel). So this single state doesn't preclude finding the component particles to have spin-up in any arbitrary direction.

But you're right, that there is something a little peculiar about this. Normally, you expect that the number of degrees of freedom for a composite system has to be greater than or equal to the number of degrees of freedom of each subsystem, but that's not the case, quantum mechanically.