# Lie bracket question

1. Feb 15, 2010

### 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:

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

$$c=\sqrt{\frac{m\omega}{\hbar}}$$

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

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

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

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

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

Can anyone tell me why this is so?

marin

2. Feb 15, 2010

### 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

$$(A B)^\dagger = B^\dagger A^\dagger$$

we have,

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

Last edited: Feb 15, 2010
3. Feb 15, 2010

### Marin

yes, of course!