Root of number operator?

[KFC]

I seen in some paper, there is an matrix whose element has a square root of number operator, e.g.

$$A = \left( \begin{matrix} \alpha & \gamma \sqrt{\hat{a}\hat{a}^\dagger} \\ -\gamma \sqrt{\hat{a}^\dagger\hat{a} & \beta \end{matrix} \right)$$
where $$\alpha, \beta, \gamma$$ are real number.

What is $$A^\dagger$$? Can I write it as the following?
$$A^\dagger = \left( \begin{matrix} \alpha & -\gamma \sqrt{\hat{a}^\dagger\hat{a}} \\ \gamma \sqrt{\hat{a}\hat{a}^\dagger & \beta \end{matrix} \right)$$

By the way, if I have it operate on any Fock state, how could the operators in the matrix operating those states?

 

[jensa]

The square root of the number operator is probably defined by it's taylor expansion. This means that letting the the root of the number operator act on a Fock-state gives you $\sqrt{N}$ times the state. Also since the number operator is self-adjoint this implies that any function of it will be self-adjoint, in particular $(\sqrt{a^\dag a})^\dag=\sqrt{a^\dag a}$.

One can think of the number operator as matrices themselves (and thus also the square root of the number operator) so what you have is a 2x2 block-matrix. The hermitian adjoint of which is given by what you wrote.

I don't think that this matrix can act on the Fock-space generated by the algebra of a^dag and a. I could imagine that the matrix never acts on the Fock-space by itself only in the form of some "scalar product". What I mean with this is basically some new operator C:

$$C=(a_1, a_2)A\begin{pmatrix}b_1\\ b_2\end{pmatrix}$$

where a_1/2 and b_1/2 can be numbers or operators. Then C can act on the Fock-space.

Compare it with how one sometimes uses "vectors" of operators like

$$c=\begin{pmatrix} a_1 \\ a_2 \end{pmatrix}, \quad c^\dag=(a_1^\dag, a_2^\dag)$$

but they never really act on Fock-space in this vector form but only in a "scalar product" form. For example the Hamiltonian may look something like

$$\mathcal{H}=c^\dag H c$$

usually the elements of the matrix H are real numbers but in principle they could also be operators.