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."

Norman Samish

----- 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.

Paddy Leahy 

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