How to write the output state of a beam splitter? I refer to this pdf.
On Page 41 (Quantum-state transformation of number states) the output state on a beam splitter is derived, based on 
$$b_1^{+} = T^{*} a_1^{+} + R^{*} a_2^{+}$$
$$b_2^{+} = -R a_1^{+} + T a_2^{+}$$
The matrix, making up this transformation ist unitary:
$$U_{T ^*} =\begin{pmatrix} T^* & R^*\\ -R & T \end{pmatrix}$$

Let us assume that the radiation field was initially prepared in a
  product state of two single-photon Fock states,

$$| \Psi_{in}\rangle = | 1\rangle |1 \rangle$$
For my opinion this must be
$$| 1\rangle |1 \rangle = a_1^{+}a_2^{+} | 0\rangle |0 \rangle$$
since 
$a_{1,2}^{+}$ are the creation operators for the input fields, and $b_{1,2}^{+}$ are for the output fields.
Why do they write $| \Psi_{out}\rangle = a_1^{+} a_2^{+}| 0\rangle |0 \rangle$? Shouldn't it be $| \Psi_{out}\rangle = b_1^{+} b_2^{+}| 0\rangle |0 \rangle$ instead?
EDIT: 
$b_1^{+} b_2^{+}| 0\rangle |0 \rangle = | 1\rangle  | 1\rangle$, 
but this cannot be the output state...
So it seems, the first variant is ok. But I still do not understand the formalism...it seems, that in order to get the output state, $a$ has to replaced by $b$, according to the inverse of $U_{T^*}$.
 A: $\newcommand{\ket}[1]{\left|#1\right>}$
For reference, this approach is called second quantization.
Basically, your confusion comes from a confusion between the Schrödinger picture — where the state evolves, $\ket{Ψ_{\text{out}}}≠\ket{Ψ_{\text{in}}}$) and the operators are constant — and the Heisenberg Picture — where the state is constant $\ket{Ψ_{\text{out}}}=\ket{Ψ_{\text{in}}}$) and the operators evolve. In practice, one often switch from one picture to the other (and sometimes in intermediate pictures), depending on which is more practical.
The second quantization approach is in Heisenberg picture, where formally $\ket{Ψ_{\text{out}}}=\ket{Ψ_{\text{in}}}$, but the observables evolve. The input operators, written as products of $a_x^\dagger$ and $a_x$, become output operators, written as products of $b_x^\dagger$ and $b_x$, the transformation being given by the unitary matrix $U_{T^*}^{-1}$.
More specifically, $\ket{Ψ_{\text{out}}}=\ket{Ψ_{\text{in}}}=a_1^\dagger a_2^\dagger\ket{0}=\ket{1,1}_{\text{in}}$, but $b_1^\dagger b_2^\dagger\ket{0}=\ket{1,1}_{\text{out}}≠\ket{1,1}_{\text{in}}$: one photon in each of the input modes is not the same state as one photon in the each of the output modes. The matrix $U_{T^*}$ gives you a relation between the input operators and the output operators:
\begin{align}
 \begin{bmatrix} b_1^\dagger\\b_2^\dagger \end{bmatrix}
  &= U_{T^*} \begin{bmatrix} a_1^\dagger\\a_2^\dagger \end{bmatrix}; &
 \text{therefore }\begin{bmatrix} a_1^\dagger\\a_2^\dagger \end{bmatrix}
  &= U_{T^*}^{-1} \begin{bmatrix} b_1^\dagger\\b_2^\dagger \end{bmatrix}
   = U_{T^*}^{\dagger} \begin{bmatrix} b_1^\dagger\\b_2^\dagger \end{bmatrix}, 
\end{align}
and you can rewrite 
$a_1^\dagger a_2^\dagger$ as $(T b_1^\dagger-R^*b_2^\dagger)(R b_1^\dagger+T^*b_2^\dagger)$ in the expression of $\ket{Ψ_{\text{out}}}$ to get the expression of the state in the “output basis”.
