# Root of number operator?

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?

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.