You are obviously wrong.

When you have operators with anti-commutation relations, you have :

$${a^+}_\lambda {a^+}_{\lambda'} + {a^+}_{\lambda'} {a^+}_{\lambda} = 0 \tag{1}$$

Taking $\lambda = \lambda'$, you get :

$${a^+}_\lambda {a^+}_{\lambda} = 0 \tag{2}$$

If you have operators with commutation relations, you have :

$${a^+}_\lambda {a^+}_{\lambda'} - {a^+}_{\lambda'} {a^+}_{\lambda} = 0 \tag{3}$$

Taking $\lambda = \lambda'$, you get :

$${a^+}_\lambda {a^+}_{\lambda} - {a^+}_{\lambda} {a^+}_{\lambda} = 0 \tag{4}$$
which is a trivial equation ($x=x$)

So, it is not true, with commutation relations, that you have : ${a^+}_\lambda {a^+}_{\lambda} = 0$, so your equation $n_\lambda ^2=n_\lambda$ is obviously false for operators with commutations relations.

This post imported from StackExchange Physics at 2014-05-04 11:38 (UCT), posted by SE-user Trimok