Thank you for the fascinating quantum analogue of my "ten ball" experiment,
where you use electrons instead of steel balls. You destroy the electron in
the third "Penning trap" and then point out that electron number 3 was NOT
destroyed, because electrons are not individuals. Instead, (If I interpret
correctly) we have 9 electrons distributed among ten "Penning traps," with
equal probability in their distribution.
If we now examine each of the Penning traps for the existence of an
electron, what do we find? My guess is that we would find nine electrons in
the ten traps, and one empty trap. The identity of the empty trap would
presumably be unpredictable.
Is my guess correct?
I don't dispute this, but you are certainly correct when you say "This may
sound ridiculous. . ." This vividly demonstrates "quantum weirdness."
----- Original Message -----
"Patrick Leahy" wrote:
Quantum "uncertainty" is better thought of as "both at once" rather than
"either or". Here's a quantum analogue of your experiment.
Take ten electrons held in a row of "Penning traps" (magnetic "bottles" that
can hold single electrons) labelled 1 to 10 (the label is attached to the
trap). Introduce an anti-electron into trap number 3, causing an
annihilation, so we now have 9 electrons, held in traps 1, 2 and 4 to 10.
Does this mean that electron number 3 was destroyed?
No, because since electrons *are* genuinely identical, they are not
individuals. The wavefunction for any group of electrons is always a perfect
mixture of all possible "identity assignments", e.g. electron 1 in trap 1, 2
in trap 2 etc plus electron 2 in trap 1, 1 in trap 2 etc.
This may sound ridiculous, but without this feature matter as we know it
simply wouldn't exist, since it underlies the Pauli exclusion principle and
hence the structure of atoms and all chemical properties.