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.

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


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.


1 Answer 1


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 $a^2=(a^\dagger)^2=0$.


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