Fermionic anti-commutation relations

Question

For Pauli's exclusion principle to be followed by fermions, we need these anti-commutators $$[a_{\lambda},a_{\lambda}]_+=0 $$ and $$[a_{\lambda}^{\dagger},a_{\lambda}^{\dagger}]_+=0 $$ Then $$n_{\lambda}^{2}=a_{\lambda}^{\dagger}a_{\lambda}a_{\lambda}^{\dagger}a_{\lambda}=a_{\lambda}^{\dagger}\left(1-a_{\lambda}^{\dagger}a_{\lambda}\right)a_{\lambda}=a_{\lambda}^{\dagger}a_{\lambda}=n_{\lambda}.$$ which gives $ n_{\lambda}=0,1 $. Here we used the anti-commutator $$[a_{\lambda},a_{\lambda^{\prime}}^{\dagger}]_+= \delta_{\lambda,\lambda^{\prime}} $$ But we could have used even a commutator instead of the anti-commutator and still got the same result i.e. if we choose $[a_{\lambda},a_{\lambda^{\prime}}^{\dagger}]_{-}=\delta_{\lambda,\lambda^{\prime}} $ then $n_{\lambda}^{2}=a_{\lambda}^{\dagger}a_{\lambda}a_{\lambda}^{\dagger}a_{\lambda}=a_{\lambda}^{\dagger}\left(1+a_{\lambda}^{\dagger}a_{\lambda}\right)a_{\lambda}=a_{\lambda}^{\dagger}a_{\lambda}=n_{\lambda} $ which also gives $n_{\lambda}=0,1 $
What conditions make us impose the last anti-commutation relation $$[a_{\lambda},a_{\lambda^{\prime}}^{\dagger}]_+= \delta_{\lambda,\lambda^{\prime}} $$ instead of $[a_{\lambda},a_{\lambda^{\prime}}^{\dagger}]_{-}=\delta_{\lambda,\lambda^{\prime}} $ ?

I mean, we do not need all relations to be anti-commuting. I can take 2 of them to be anti-commuting but the third one i.e. relation between creation and annihilation operator to be commuting and still maintain the Pauli's exclusion

Answer 1

Suppose operators $a$ and $a^\dagger$ satisfy $$\left\{ a,a\right\} =0\mbox{ and }\left[a,a^\dagger\right]=1.$$

Therefore $a^{2}=0$ and $aa^\dagger=a^\dagger a+1$.

Now consider $aaa^\dagger$: $$0=aaa^\dagger =a\left(a^\dagger a+1\right)=aa^\dagger a+a=a^\dagger aa+2a=2a.$$

So we get $a=0$.

Answer 2

We need it because we want the "occupied" state to have $n_\lambda=1$; I will omit the $\lambda$ argument everywhere. In other words, we need $$ n a^\dagger |0\rangle \equiv a^\dagger a a^\dagger |0\rangle = 1\cdot a^\dagger $$ But the left hand side has the operator that is $$a^\dagger a a^\dagger |0\rangle = a^\dagger ([a,a^\dagger]_+ - a^\dagger a) = a^\dagger [a,a^\dagger]_+ $$ where the last term was dropped in the last term because $(a^\dagger)^2=0$. So we demand $$a^\dagger[a,a^\dagger]_+ |0\rangle = a^\dagger|0\rangle.$$ In combination with your other conditions, this is only possible if the anticommutator is one – we may "cancel" the $|0\rangle$ ket vector because a similar condition may be derived for $|1\rangle$ as the ket vector.

Answer 3

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.