Indistinguishable particles when wavefunctions don't overlap

Question

My question is the following :

Imagine that we study two electrons, one has spin up and the other down.

If the two wavefunctions overlap, then I have the symetrisation postulate that occurs, the state will be :

$$ \frac{|+-\rangle - |-+\rangle}{\sqrt{2}} $$

But do you agree with me that if the wavefunctions don't overlap, I can distinguish my two electrons, so I can write in that case $$ |\Psi\rangle=|+-\rangle$$

And the reason behind is because in fact there will be other quantum numbers acting here : my state will be in fact something like :

$$ |\Psi\rangle=|+\rangle|\phi_1\rangle\otimes|-\rangle|\phi_2\rangle$$

And because $\phi_1$ and $\phi_2$ don't overlap, then it allows me to distinguish my particles (they act like other quantum numbers : if my particles were point-like i could write $|\phi_1\rangle=|r_1\rangle$ and $|\phi_1\rangle=|r_2\rangle$ for example ).

I just want to be sure that I understood well the mathematical reason behind "when it doesn't overlap I don't have to symmetrise my ket".

[Edit]

I have read : Pauli principle for particles very far apart from each other, the first answer for example when he says "$|\Psi(\vec{r}_2,\vec{r}_1)| = |\Psi(\vec{r}_1,\vec{r}_2)|$".

But I don't understand this if the particles don't overlap.

Indeed without an overlap, I can label the identicals particles "by mind". If the particle $1$ is on earth and the particle $2$ is on the moon without any overlapping, and if I take $r_1$ as a cooordinate on earth and $r_2$ as a coordinate on the moon, then $|\Psi(\vec{r}_1,\vec{r}_2)| \neq 0$ but $|\Psi(\vec{r}_2,\vec{r}_1)| = 0$ because the first particle can't be on moon because it is on earth.

Answer 1

By Symmetry's answer is correct, but I don't think it answers quite the question that you're getting at.

You are making a very common conceptual mistake for students first learning about quantum statistics, which is that quantum particles "need" to be indistinguishable and that (anti)symmetrization is the way that they "satisfy that need." But this isn't the case - one could certainly imagine a quantum system of multiple particles with completely identical mass, charge, spin, etc., but where the particles are "distinguishable" in the sense that their joint wavefunction is neither symmetric nor antisymmetric under particle exchange. Such particles would be neither bosons nor fermions. (In fact, condensed matter physicists do this all the time when they consider magnetic spin systems, and less commonly, systems of "anyons" which are neither boson nor fermionic.)

So why do intro QM textbooks (almost) always only consider bosonic or fermionic particles? Because every single fundamental particle in the Standard Model has been experimentally shown to be either a boson or a fermion. Recall that any physical system has an associated "Hilbert space," which is the set of quantum states in which it is physically possible to find the system. The set of totally symmetric wavefunctions forms the "bosonic Hilbert space" and the set of totally antisymmetric wavefunctions forms the "fermionic Hilbert space," and all known species of elementary particles are described by one or the other space. But a more general Hilbert space would certainly be logically and mathematically consistent.

(When you study QM at a more advanced level, you'll learn that there are deeper reasons why bosonic and fermionic theories are more "natural." In these two special cases, we can use a very elegant mathematical formalism called "second quantization" that saves us a lot of work. Spin and anyonic systems are in some ways more difficult to deal with than systems of bosons and fermions, because you can't use second quantization - or sometimes you can use it, but in a much, much more complicated way, by mapping the system to a bosonic or fermionic "gauge theory." Also, there is no known way to make a particle that is neither bosonic nor fermionic compatible with special relativity.)

Anyway, this is a long-winded way of saying that all multi-electron wavefunctions are always antisymmetrized, no matter how far away they are. (Although it turns out that you can't use this antisymmetrization to send information faster than light, so special relativity is safe.) But recall that in quantum mechanics, only inner products are physically observable, not the wavefunctions themselves. (Actually, only the norm-squared of an inner products is physically observable.) If the particles don't have any spatial overlap, then it turns out that if we want to evaluate $\langle \hat{X} \rangle$, we'll get the same answer whether or not we use the correct, antisymmetrized wavefunction, or a hypothetical unsymmetrized wavefunction. So even though the unsymmetrized wavefunction doesn't even lie in the Hilbert space and doesn't make any sense physically, we can get away with using it even though it's "wrong." Try it! Write down two non-overlapping wavefunctions and calculate $\langle \hat{X} \rangle$ with respect to both the antisymmetrized and non-antisymmetrized states - you'll get the same answer either way. So really, strictly speaking you still "have to" antisymmetrize, but you can get away with "cheating" and neglecting to do so if the particle are far away and have negligible spatial overlap.

Finally, you might wonder "if all known fundamental particles are either bosons or fermions, then why do condensed matter physicists ever bother to consider these weird magnetic spin and anyonic systems?" The answer is actually different in the two cases. Magnetic spin systems are actually electron systems where every electron's wavefuncton falls off so fast that we can treat them as "far away," even if they're only separated by an interatomic spacing! So we can get away with ignoring the antisymmetrization, even though it's "really" there. Anyons are even stranger and don't correspond to any individual fundamental particles at all - they are "collective excitations" that only emerge when you take ginormous numbers of electrons and couple them together in very special ways.

Answer 2

The total wavefunction of a pair of electrons must always be antisymmetric. So the general wavefunction has the form. $$ |\Psi\rangle = \frac{1}{2}\left(|\sigma_1\phi_1\rangle|\sigma_2\phi_2\rangle - |\sigma_2\phi_2\rangle|\sigma_1\phi_1\rangle\right) $$ If the spacial wavefunctions of the two electrons are identical, i.e. the two electrons are in the same "spacial state", then the spacial wavefunction must be symmetric and so, in order for the total wavefunction to be antisymmetric the spin part of the wavefunction must be antisymmetric. $$ |\Psi\rangle = \frac{1}{\sqrt{2}}\left(|+-\rangle - |-+\rangle\right)|\phi_1\,\phi_1\rangle $$

If, however, the spacial wavefunctions are not identical, then the spacial part of the wavefunction can be either symmetric or antisymmetric and so the spin wavefunction can then be antisymmetric or symmetric respectively \begin{align} |\Psi\rangle = \frac{1}{2}\left(|+-\rangle - |-+\rangle\right)\left(|\phi_1\,\phi_2\rangle + |\phi_2\,\phi_1\rangle\right)\\ |\Psi\rangle = \frac{1}{2}\left(|\sigma_1\sigma_2\rangle + |\sigma_2\sigma_1\rangle\right)\left(|\phi_1\,\phi_2\rangle - |\phi_2\,\phi_1\rangle\right) \end{align} where $\sigma_1$ and $\sigma_2$ can be either $+$ or $-$. The overlap of the wavefunctions does not matter. For that matter the wavefunctions generally decay exponentially at large distances so the overlap is never quite $0$. The point where overlap integrals do come into play is when there is an interaction between the electrons so there is an energy associated with having the electrons close together, which will be normally higher if the electrons are in the same state, resulting in a symmetric spacial configuration configuration being either favoured or avoided.