# Fermion annihilation operators from position and momentum

1. May 30, 2010

### haael

Is it possible to express fermion annihilation operator as a function of position and momentum?

I've seen on Wikipedia the formula for boson annihilation operator:
$$\begin{matrix} a &=& \sqrt{m\omega \over 2\hbar} \left(x + {i \over m \omega} p \right) \\ a^{\dagger} &=& \sqrt{m \omega \over 2\hbar} \left( x - {i \over m \omega} p \right) \end{matrix}$$

But what about fermions? Is it possible to get anticommutation relations from canonical relations alone, or is it necessary to postulate something else?

2. May 30, 2010

### tom.stoer

You have to use Grassmann numbers or something like that. This can be seen by inspection of the (fermionic) anti-commutators

$$\{b, b\} = 2b^2 = 0$$
$$\{b^\dagger, b^\dagger\} = 2{b^\dagger}^2 = 0$$

which cannot be derived from commuting objects.