lie bracket question

Marin

Hi!

I was doing an assignment in quantum mechanics and came upon the following fact I cannot explain to me.

I hope someone of you can and will be willing to :)


Consider the creation and annihilation operators: a^+ and a and also the momentum and position operators p and x:

[tex]x=\frac{1}{\sqrt 2 c}(a+a^{\dagger})[/tex]
[tex]p=\frac{\hbar c}{\sqrt 2 i}(a-a^{\dagger})[/tex]
[tex]a=\frac{1}{\sqrt 2}(cx+\frac{i}{c\hbar}p)[/tex]
[tex]a^{\dagger}=\frac{1}{\sqrt 2}(cx-\frac{i}{c\hbar}p)[/tex]

[tex]c=\sqrt{\frac{m\omega}{\hbar}}[/tex]


and the canonical commucator relation: [tex][x,p]=i\hbar 1[/tex], where 1 is the identity operator


It follows immediately from the canonical commutator relation between x and p that

[tex][a,a^{\dagger}]=1}[/tex]
Now, observe what happens when I take the adjoint of this equation:

[tex]([a,a^{\dagger}])^{\dagger}=(aa^{\dagger}-a^{\dagger}a)^{\dagger}=a^{\dagger}a-aa^{\dagger}=-[a,a^{\dagger}]=-1[/tex]

which is peculiar since I thought that the Identity is hermitian: [tex]1^{\dagger}=1[/tex], which apperantly doesn't hold here..

Can anyone tell me why this is so?

thanks in advance,

marin

solitonion

I think I found your mistake here
Hi Martin,

Using the fact that transposing the product of two operator swaps them and gives the product of their transposes

[tex] (A B)^\dagger = B^\dagger A^\dagger [/tex]

we have,

[tex] ([a,a^{\dagger}])^{\dagger}=(aa^{\dagger}-a^{\dagger}a)^{\dagger}=a^{\dagger\dagger}a^\dagger - a^\dagger a^{\dagger\dagger} = a a^\dagger - a^\dagger a = 1 [/tex]

Marin

yes, of course!

sorry for asking...

and thanks once again!