l**********n
发帖数: 72 | 1 Lately fisherdad is clearing away the obscure and ambiguous
concepts with spin, so I have one question about spin too.
We know that fermi distribution and bose distribution
(without interaction between the particles) differs only in
the +&- in the formula, and it's due to the antisymmetric or
symmetric of the wavefunction, or commutivity relations
between the particle state. So that means
[a,b]=delta(a,b)
{a,b}=delta(a,b)
where b is a dagger. While ab-ba or ab+ba differs only in
the signs there. | f*******d
发帖数: 339 | 2
oops, I don't know much about anyons and fractional statistics.
Maybe someone could answer your question.
However, I remember that for anyons, the phase factor after particle
exchange is path-dependent, not just a fixed number. The reason is,
if the phase factor is path independent, it could be proved that it must
equal to either +1 or -1.
【在 l**********n 的大作中提到】
: Lately fisherdad is clearing away the obscure and ambiguous
: concepts with spin, so I have one question about spin too.
: We know that fermi distribution and bose distribution
: (without interaction between the particles) differs only in
: the +&- in the formula, and it's due to the antisymmetric or
: symmetric of the wavefunction, or commutivity relations
: between the particle state. So that means
: [a,b]=delta(a,b)
: {a,b}=delta(a,b)
: where b is a dagger. While ab-ba or ab+ba differs only in
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