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Electrons are indistinguishable particles, however, when I set up two independent experiments (at two positions), I can talk about "the electrons in Experiment x". What's going on here?

I can model each electron by $|x, s \rangle$ such a state, so one electron can be in a 4-dimensional space spanned by: $|position1, +1/2 \rangle$, $|position2, +1/2 \rangle$, $|position1, -1/2 \rangle$, $|position2, -1/2 \rangle$.

The system of two electrons in there is 16 dimensional, but because of the anti symmetry we only have 6 basis vectors:

$|p1,+1/2, p1 -1/2 \rangle$, $|p2,+1/2, p2 -1/2 \rangle$, $|p1,-1/2, p2 -1/2 \rangle$, $|p1,-1/2, p2 +1/2 \rangle$, $|p1,+1/2, p2 -1/2 \rangle$ and $|p1,+1/2, p2 +1/2 \rangle$

In what way can we now talk about "the electron in experiment 1"? Or do we just mean the electron that happens to have position 1? What is the right way to think about this?

It seems that the question is being missunderstood: We already know that Electrons (ALL the electrons) are not distinguishable. The standard model describes electrons as excitations of an anticommuting quantum field, and this is a established model for electons. I don't ask wether this is true, and I don't ask why this is true.

YET when I set up an Ion trap or a quantum-computer that realizes q-bits via quantum states of electrons, it is an experimental fact that I talk about the "quantum states of the experiment". I don't talk about "all the electrons in the universe", I'm talking about the electrons in the specific machine that I have build.

And when I build another machine beneath the first one, then those two machines will (hopefully) work independently from one another.

This means although electrons are not distinguishable, we at least have a way of talking about electrons whose state will have a measurable effect in either Machine A or Machine B.

This two facts are seemingly contradictory. My question is how this contradiction can be resolved.

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  • $\begingroup$ Just wait 'til you read about the guys who claim there's only one electron in the universe which shows up everywhere in different phases. $\endgroup$ Commented May 5, 2022 at 13:56
  • $\begingroup$ Electrons cannot be distinguished. $\endgroup$
    – my2cts
    Commented May 5, 2022 at 14:06
  • $\begingroup$ @my2cts I know that they can't be distinguished. This is QFT and Many-Body QM. Yet when you set up two Ion traps (in two labs), you make statements about the electrons IN the trap, and only about those. This is experimental fact. I try to reconcile those two (seemingly contradictory) statements. $\endgroup$ Commented May 5, 2022 at 14:23
  • $\begingroup$ My answer there is related to this. $\endgroup$ Commented May 5, 2022 at 16:24
  • $\begingroup$ Perhaps the answer can be found from the many-electron density matrix formalism $\endgroup$
    – my2cts
    Commented May 5, 2022 at 18:58

1 Answer 1

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As you say, this is an experimental fact. It's the way the universe (sometimes) works. This fact constrains the theory.

The way the theory deals with this is that you may effectively distinguish the electrons if the two-electron wave function can be factored into two independent one-electron wave functions.

Note that the theory accommodates the fact, but it does not explain "why", since the theory is constructed from the fact. That's physics.

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  • $\begingroup$ one could think that independence follows from being able to write the pdf as a product $|\psi|^2 = |\phi(1)|^2 |\phi(2)|^2$ but this argument still leaves the phases arbitrary. How do you justify that the phases must also add? $\endgroup$
    – hyportnex
    Commented May 5, 2022 at 12:30
  • $\begingroup$ @hyportnex I don't think I have to. What demands that the phases must add when the electrons are experimentally separable? What observation can you make to test the assertion? That the phases add is a property of the theory. $\endgroup$
    – John Doty
    Commented May 5, 2022 at 12:35
  • $\begingroup$ How would you factor any of the given 6 basis vectors that the state must live in into a product of two states? These states are constructed in the way that this factorization won't work. $\endgroup$ Commented May 5, 2022 at 12:36
  • $\begingroup$ @Quantumwhisp You cannot in general treat the electrons as separate. Thus, you cannot in general perform the factorization. $\endgroup$
    – John Doty
    Commented May 5, 2022 at 12:46
  • $\begingroup$ of course, you do not have to justify anything, it is what it is; still would be nice to hear some handwaving argument as why it should be so besides our experiments say that the theory needs it. I wish I knew one. $\endgroup$
    – hyportnex
    Commented May 5, 2022 at 12:50

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