From Wikipedia "Spin-Statistics Theorem" General section:(adsbygoogle = window.adsbygoogle || []).push({});

Two indistinguishable particles, occupying two separate points, have only one state, not two. This means that if we exchange the positions of the particles, we do not get a new state, but rather the same physical state. In fact, one cannot tell which particle is in which position.

A physical state is described by a wavefunction, or – more generally – by a vector, which is also called a "state"; if interactions with other particles are ignored, then two different wavefunctions are physically equivalent if their absolute value is equal. So, while the physical state does not change under the exchange of the particles' positions, the wavefunction may get a minus sign.

Bosons are particles whose wavefunction is symmetric under such an exchange, so if we swap the particles the wavefunction does not change. Fermions are particles whose wavefunction is antisymmetric, so under such a swap the wavefunction gets a minus sign, meaning that the amplitude for two identical fermions to occupy the same state must be zero. This is the Pauli exclusion principle: two identical fermions cannot occupy the same state. This rule does not hold for bosons.

So,

Two indistinguishable particles... therefore one cannot tell which particle is is in which position.

Two particles have one state, not two.

Physical state described by a wavefunction (therefore the two particles are described by one wavefunction)

Two different wavefunctions are physically equivalent if absolute value is equal

(so the change in sign is ignored, I'm assuming this is a change in sign of the value of the amplitude)

Bosons; the swap does not change the wavefunction

Two indistinguishable particles... one wavefunction... the swap is the same as rotating the wave function "180 degrees" to realign with the two particles' original positions? Where the center of rotation is the midpoint between them?

There is no mention here of adding two things together, only a comparison of two things...

Fermions; the swap changes the wavefunction sign

So it would take two "180 degree" rotations to realign the points back to their original positions.

But how does change in sign of the wavefunction yield the amplitude zero? Are two wavefunctions or two wavefunction amplitudes from two locations being summed to zero? What happened to "one state, one wavefunction"? Is a single wavefunction amplitude value being derived from two individual wavefunction locations with opposite signs?

In the example given, we start with two identical fermions in the same state, no? Then swapping them gives a null result... how is this swap, or rotation, any different from shifting/rotating the position from which they are observed? If the amplitudes were of opposite sign before the swap, how are they in the same state before, but not after, with only an exchange of sign?

Ultimately, if two fermions are antisymmetric because of sign, what about a third fermion? Will it be antisymmetric with either or both of the others? If the state is redefined for three fermions, how can all three be antisymmetric with each other? What kind of rotations are implied? How are an arbitrary number of fermions all antisymmetric with each other?

Any help on clarifying this appreciated. As simple as possible, also appreciated.

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# Spin-Statistics Theorem

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