Why do electrons behave as classical particles when they are far apart?

[Tio Barnabe]

If two electrons are far apart, the antisymmetrization part in the probability amplitude for position is negligible and they behave as classical particles, thus we don't need to consider antisymmetrization. My question is why is this also true when we have a large number of electrons, say $10^{20}$ electrons? Is it because electrical interactions between them keep they far away from each other?

 

[vanhees71]

What do you mean by "the antisymmetrization part" is negligible? For a two-electron system the Hilbert space is spanned by antisymmetrized two-particle product-basis states, e.g., position-spin states, $|\vec{x}_1,\sigma_{z1} \rangle \otimes |\vec{x}_2,\sigma_{z2} \rangle - |\vec{x}_2,\sigma_{z2} \rangle \otimes |\vec{x}_1,\sigma_{z1} \rangle$. You cannot neglect the antisymmetrization. It simply doesn't make sense for two indistinguishable fermions to consider other states!

 

[Tio Barnabe]

I understand what you say. If we consider two electrons far away, then they will be like classical particles, correct? So it's meaningless to consider the antisymmetrization. This is what I meant.

 

[vanhees71]

A single electron is usually not like a classical particle, because it's even (at least as far as we know today) an elementary particle. It behaves to a good approximation only classically when it is in continuous interaction with something as, e.g., in a cloud chamber where it leaves a "track" like a classical particle would do. That's well understood from quantum mechanics as has been explained in a famous paper by Neville Mott already in the early days of modern QT.

 

[Tio Barnabe]

I'd rather like to consider a two electron system with the mutual interaction between the electrons neglected and such that the Hamiltonian doesn't depend on the spin. In such case, the spatial wave function can be written as $$\phi (x_1,x_2) = \frac{1}{\sqrt{2}}[ \omega_{A}(x_1) \omega_{B}(x_2) \pm \omega_{A}(x_2) \omega_{B}(x_1)]$$ where the labels $1$ and $2$ stand for electron $1$ and $2$, respectively (even that the electrons themselves are indistinguishable). Working out the probability for finding electron $1$ in a volume around $x_1$ and electron $2$ in a volume around $x_2$ from the above equation, you will get the terms I have been talking about througout this thread.

Do you see now what I mean?

 

[PeterDonis]

I'd rather like to consider a two electron system with the mutual interaction between the electrons neglected and such that the Hamiltonian doesn't depend on the spin.

In which case you're no longer talking about electrons, but about idealized electrically neutral spin zero particles that happen to have the same mass as electrons. In this idealized model, the particles would behave like bosons, not fermions; so you would be "neglecting" antisymmetrization only in the sense that you would have to symmetrize, not antisymmetrize.

If you are really trying to model classical particles, i.e., Boltzmann statistics (instead of either Bose-Einstein or Fermi-Dirac), then you're not doing quantum mechanics any more, you're doing classical physics.

 

[vanhees71]

I'd rather like to consider a two electron system with the mutual interaction between the electrons neglected and such that the Hamiltonian doesn't depend on the spin. In such case, the spatial wave function can be written as $$\phi (x_1,x_2) = \frac{1}{\sqrt{2}}[ \omega_{A}(x_1) \omega_{B}(x_2) \pm \omega_{A}(x_2) \omega_{B}(x_1)]$$ where the labels $1$ and $2$ stand for electron $1$ and $2$, respectively (even that the electrons themselves are indistinguishable). Working out the probability for finding electron $1$ in a volume around $x_1$ and electron $2$ in a volume around $x_2$ from the above equation, you will get the terms I have been talking about througout this thread.

Do you see now what I mean?

Nevertheless, in this case you have to add the spin part to make the wave function antisymmetric, i.e.,
$$\phi (x_1,x_2) = \frac{1}{2}[ \omega_{A}(x_1) \omega_{B}(x_2) \pm \omega_{A}(x_2) \omega_{B}(x_1)] (\chi_1 \chi_2-\chi_2 \chi_1).$$

 

[blue_leaf77]

Working out the probability for finding electron 11 in a volume around x1x_1 and electron 22 in a volume around x2x_2 from the above equation, you will get the terms I have been talking about througout this thread.

So, you mean the overlap term which contains products between the $\omega_A$ and $\omega_B$. I think it should be clear to you that this overlap term will be very small if the two electrons are individually localized far away from each other. Then shouldn't it be obvious that when you generalize this to any number of electrons and if they are far away one from another, the overlap terms will be much less significant?