Creation and annihilation operators

Remarks

The map $f \mapsto a^*(f)$ is linear and $f \mapsto a(f)$ is antilinear. In the bosonic case they satisfy, on $\mathcal{F}_0$, $$[a(f), a(g)] = 0, \qquad [a(f), a^*(g)] = \ip{f}{g}\, I.$$ In the fermionic case they satisfy the canonical anticommutation relations $$\{a(f), a(g)\} = 0, \qquad \{a(f), a^*(g)\} = \ip{f}{g}\, I,$$ where $\{A, B\} = AB + BA$. Fermionic operators extend to bounded operators with $\norm{a(f)} = \norm{f}$, while bosonic ones are unbounded.

Example: let $\mathfrak{h} = \C$ and $f = 1$, in the bosonic case. Then $\mathcal{H}^{(n)}_+$ is spanned by the unit vector $e_n = 1^{\otimes n}$, and the definition gives $a^* e_n = \sqrt{n+1}\, e_{n+1}$ and $a\, e_{n+1} = \sqrt{n+1}\, e_n$. Check: $a a^* e_n = (n+1) e_n$ and $a^* a\, e_n = n\, e_n$, so $[a, a^*] = I$.

Non-example: in the fermionic case with $\mathfrak{h} = \C$, the space $\mathcal{H}^{(2)}_- = \set{0}$, so $a^*(1)^2 = 0$, and a second particle cannot be created in the same state. Hence the bosonic rule $[a, a^*] = I$ fails there: on $\C^2 = \mathcal{H}^{(0)}_- \oplus \mathcal{H}^{(1)}_-$ the operators satisfy $a a^* + a^* a = I$ instead.