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

  • $\begingroup$ Perhaps my eyes are deceiving me, but it seems that you actually used the anti-commutator you were wondering about in your second equality after "because." $\endgroup$ Sep 15, 2013 at 7:53
  • $\begingroup$ @joshphysics made the change. I mean, I could have used even a commutator instead the anti-commutator and still got the same result $\endgroup$
    – cleanplay
    Sep 15, 2013 at 9:39
  • $\begingroup$ Related: physics.stackexchange.com/q/17893/2451 $\endgroup$
    – Qmechanic
    Sep 15, 2013 at 14:41
  • $\begingroup$ @Qmechanic : thanks, though it addresses the question but it is a bit vague for me. $\endgroup$
    – cleanplay
    Sep 15, 2013 at 17:12

3 Answers 3


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 $a=0$.


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.

  • $\begingroup$ @Motl I mean we could also use the commutator $ [a_{\lambda},a_{\lambda^{\prime}}^{\dagger}]_{-}=\delta_{\lambda,\lambda^{\prime}} $ and still get the same result. Why should I use only the anti-commutator ? I can use the anti-commutators $$[a_{\lambda},a_{\lambda}]_+=0 $$ and $$[a_{\lambda}^{\dagger},a_{\lambda}^{\dagger}]_+=0, $$ along with $$ [a_{\lambda},a_{\lambda^{\prime}}^{\dagger}]_{-}=\delta_{\lambda,\lambda^{\prime}} $$ $\endgroup$
    – cleanplay
    Sep 15, 2013 at 9:50

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.

  • $\begingroup$ My question is only about the last anti-commutation relation which you did not use in your proof. I understand that you need the two anti-commutation relations that you have used, in order to prove the Pauli's exclusion principle. My question is on what basis we choose the third relation i.e the relation of the creation and annihilation operator to be of anti-commutator type and not commutator type. Read my comment in response to Lubos Motl's answer. $\endgroup$
    – cleanplay
    Sep 15, 2013 at 17:05
  • $\begingroup$ @cleanplay : This is an other question... In short, this has to do with the spin-statistics theorem. For instance, in quantum field theory, if you write the hamiltonian for a fermionic field (for instance a Dirac field), you will find something like $H = \sum_k (b^+_kb_k - d_kd^+_k)$ ($b$ concerns particles and $d$ concerns anti-particles). But this hamiltonian has to be bounded below, and you have to choose anti-commutation relations, to have $H = \sum_k (b^+_kb_k + d^+_k d_k)$, up to a (infinite) constant. $\endgroup$
    – Trimok
    Sep 15, 2013 at 17:21

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