Operator of annihilation

In Second quantification an operator of annihilation is an operator who makes pass of a state of the Space of Fock container N  ≥  1 particles in a state containing (N-1) particles. The action of an operator of annihilation on the vacuum ( N=0 ) gives zero. In the case of the Boson S, the action of the operator of annihilation a_i which destroys a particle in the state I writes:

$a\_i\; \backslash \; mid\; N\_1,\; N\_2,\; \backslash \; ldots,\; N\_i,\; \backslash \; ldots\; \backslash \; rangle\; =\; \backslash \; sqrt\; \{N\_i\}\; \backslash \; mid\; N\_1,\; N\_2,\; \backslash \; ldots,\; N\_i-1,\; \backslash \; ldots\; \backslash \; rangle$

In addition, in the case of the bosons, the operators of annhilation commutate between them:

, a_j '' = 0

In the case of the Fermion S, as (Principle of exclusion of Pauli) a state can be occupied only by one particle, the action of the operator of annihilation c_i is defined by:

$c\_i\; \backslash \; mid\; N\_1,\; N\_2,\; \backslash \; ldots,\; 1\_i,\; \backslash \; ldots\; \backslash \; rangle=\; \backslash \; mid\; N\_1,\; N\_2,\; \backslash \; ldots\; 0\_i,\; \backslash \; ldots\; \backslash \; rangle$

In addition, the operators of annhilation of the fermions anticommutent between them:

{ c_i, c_j } = c_i c_j + c_j c_i = 0

The combined square one of an operator of annihilation is a Opérateur of creation.

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