Explain Adjoint of Commutator Identity in Second Quantization

[tommy01]

Hi all.

I found the following identity in a textbook on second quantization:

$$([a_1^{\dagger},a_2^{\dagger}]_{\mp})^{\dagger}=[a_1,a_2]_{\mp}$$

but why?

$$([a_1^{\dagger},a_2^{\dagger}]_{\mp})^{\dagger}=(a_1^{\dagger}a_2^{\dagger}\mp a_2^{\dagger}a_1^{\dagger})^{\dagger}=a_2a_1\mp a_1a_2$$

and in the case of the commutator (and not the anticommutator) this isn't the result mentioned in the book.

i would be glad if someone can explain. thanks.

 

[javierR]

$$([a_1^{\dagger},a_2^{\dagger}]_{\mp})^{\dagger}=(a_1^{\dagger}a_2^{\dagger}\mp a_2^{\dagger}a_1^{\dagger})^{\dagger}=a_2a_1\mp a_1a_2$$

This looks fine to me

 

[haushofer]

Hi all.

I found the following identity in a textbook on second quantization:

$$([a_1^{\dagger},a_2^{\dagger}]_{\mp})^{\dagger}=[a_1,a_2]_{\mp}$$

but why?

$$([a_1^{\dagger},a_2^{\dagger}]_{\mp})^{\dagger}=(a_1^{\dagger}a_2^{\dagger}\mp a_2^{\dagger}a_1^{\dagger})^{\dagger}=a_2a_1\mp a_1a_2$$

and in the case of the commutator (and not the anticommutator) this isn't the result mentioned in the book.

i would be glad if someone can explain. thanks.

Well, my best guess is that

$$[A,B]^{\dagger} = (AB - BA)^{\dagger} = B^{\dagger}A^{\dagger}-A^{\dagger}B^{\dagger} = [B^{\dagger},A^{\dagger}] = -[A^{\dagger},B^{\dagger}]$$

So

$$[A^{\dagger},B^{\dagger}]^{\dagger} = - [A,B]$$

what you also wrote down. Which textbook are you referring to?

 

[tommy01]

Thanks for your replies.

I found the problem. The identity only holds for the special case of creation-/annihilation-operators, where the (anti-)commutator for fermions or bosons resplectively is zero.

thanks and greetings.