# Creation and annihilation operators for fermions from anticommutator

In a question, I was given that $$a^{\dagger}a + a a^{\dagger} =1$$ and asked to show what $$a|n\rangle$$ and $$a^{\dagger}|n\rangle$$ would be, given that $$H|n\rangle=(a^{\dagger}a + 1/2)$$.

I am getting the wrong answer for $$a^{\dagger}$$.

Call number operator $$N=H-1/2$$ and denote anticommutator by $$\{\}$$.

$$\{N,a^{\dagger}\}=a^{\dagger}aa^{\dagger} + a^{\dagger}a^{\dagger}a$$

$$\{N,a^{\dagger}\}=a^{\dagger}(aa^{\dagger}+ a^{\dagger}a)=a^{\dagger}$$

now for $$a$$, I get the same answer.

$$\{N,a\}=a^{\dagger}aa + aa^{\dagger}a$$

$$\{N,a\}=a$$

So then $$Na|n\rangle=(\{N,a\}-aN)|n\rangle=(a-an)|n\rangle=(1-n)a|n\rangle$$

then you can do the same thing with $$a^{\dagger}$$ .

This leads to the eigenvectors differing by a constant, I think, which is not what we want, right? in the end, per this link they have different properties.

What have I done wrong? note that I cannot use the {a,a}=0 property or the counterpart with $$a^{\dagger}$$.

Edit: with the help of Mike (thanks) now I get it. following this to the end, for satisfaction:

$$a|n\rangle = C_{a} |1-n\rangle$$

$$\langle n| a^{\dagger}a|n\rangle = |C_{a}|^{2} \langle 1-n|1-n\rangle$$

$$n = |C_{a}|^{2}$$ repeating with $$a^{\dagger}$$

$$\langle n| aa^{\dagger} |n\rangle = |C_{a^{\dagger}}|^{2} \langle 1-n|1-n\rangle$$

$$1-n= |C_{a^{\dagger}}|^{2}$$

so they're not quite the same.

I think you want to compute $$[a^\dagger, N]$$ rather than $$\{A^\dagger, N\}$$. Use $$[A,BC]= \{A,B\}C- B\{A,C\},$$ to do this.
Mind you, your algebra is correct. If $$n=1$$ then $$a$$ takes you to $$|0\rangle$$ on which $$N\to 0$$, and if $$n=0$$ then $$a|0\rangle=0$$ and $$N0=0$$. Similarly if $$n=1$$ then $$a^\dagger |1\rangle=0$$, and if $$n=0$$ we have $$a^\dagger|0\rangle =|1\rangle$$ and $$Na^\dagger|0\rangle= 1a^\dagger|0\rangle = (1-n) a^\dagger|0\rangle$$. To show that $$n=0,1$$ are the only possibilitites you need to show that $$a^2=(a^\dagger)^2=0$$.