Levi-Civita symbol and Hermitian conjugate When we take the Hermitian conjugate/dagger of an operator expression which contains a Levi-Civita symbol, do we need to transpose the Levi-Civita symbol? E.g., for the crossproduct
$$(\textbf{a}\times \textbf{b})_i~=~\epsilon_{ijk}\textbf{a}_{j}\textbf{b}_{k},$$
do we have
$$(\textbf{a}\times \textbf{b})_i^{\dagger}~=~\textbf{b}_{k}^{\dagger}\textbf{a}_{j}^{\dagger}\epsilon_{ikj}~?$$
 A: As ACuriousMind said your expression is not correct:
$$(\mathbf{a}\times \mathbf{b})_i = \epsilon_{ijk} a_j b_k$$
So taking the dagger:
$$(\mathbf{a}\times \mathbf{b})_i^{\dagger} = (\epsilon_{ijk})^\dagger a_j^\dagger b_k^\dagger\\
=\epsilon_{ijk} a_j^* b_k^*$$
because $\epsilon_{ijk}$ is real scalar, and $a_j$ and $b_k$ are complex scalars. (for scalars the transpose conjugate is just the conjugate, and for real scalars it changes nothing at all).

Since you objected to levi civita symbol being a scalar I will show you in another way that the symbol indeed doesn't change.
$$\mathbf{a}\times \mathbf{b} = \left[\begin{matrix} a_2b_3-a_3b_2 \\
-a_1b_3+a_3b_1\\
a_1b_2-a_2b_1 \end{matrix}\right] \implies 
\left[\begin{matrix}\epsilon_{123}=1 & \epsilon_{132}=-1\\
\epsilon_{213}=-1 & \epsilon_{231}=1\\
\epsilon_{312}=1 & \epsilon_{321}=-1
\end{matrix}\right]$$
which are the correct values of the Levi Civita symbol. 
$$(\mathbf{a}\times \mathbf{b})^\dagger = \left[\begin{matrix} a_2^*b_3^*-a_3^*b_2^* &
-a_1^*b_3^*+a_3^*b_1^*&
a_1^*b_2^*-a_2^*b_1^* \end{matrix}\right] \implies 
\left[\begin{matrix}\epsilon_{123}=1 & \epsilon_{132}=-1\\
\epsilon_{213}=-1 & \epsilon_{231}=1\\
\epsilon_{312}=1 & \epsilon_{321}=-1
\end{matrix}\right]$$
Which is exactly what the first part before the line break will give you. 
A: I want to add, that the situation changes if the elements of these vectors are operators. Since this is what your notation might imply. Taking the definition of what a vector product for these quantities should mean, i.e.
$$(\mathbf{a}\times\mathbf{b})_k = \epsilon_{ijk} a_i b_j \hspace{1cm} (1)$$
then gives by definition (note that $(AB)^\dagger = B^\dagger A^\dagger$)
$$(\mathbf{a}\times\mathbf{b})_k^\dagger = \epsilon_{ijk} b_j^\dagger a_i^\dagger = -\epsilon_{jik} b_j^\dagger a_i^\dagger=-(\mathbf{b}^\dagger\times\mathbf{a}^\dagger)_k$$
As pointed out by gautam1168, $\epsilon_{ijk}$ appears as a real scalar in the right hand side of (1). You see however that, taking the transpose conjugate of the initial statement is equivalent to perform a contraction with a certain kind of transpose of the Levi-Cita tensor with respect to the first two indices in such a way, that again a definition of a vector product is recovered. And, very important: the statement differs from yours by a minus sign (due to the anti-symmetry of $\epsilon_{ijk}$ swapping $i\leftrightarrow j$).
A: Are you curious about the case where the $a_j, b_k$ are matrices or operators?
As in gautam1168's answer.
$$(\mathbf{a}\times \mathbf{b})_i = \epsilon_{ijk} a_j b_k$$
So taking the dagger:
$$(\mathbf{a}\times \mathbf{b})_i^{\dagger} = (\epsilon_{ijk})^\dagger b_k^\dagger a_j^\dagger,$$
because $(\epsilon_{ijk})^\dagger$ is just a number which commutes with whatever. For the same reason, if $a_j, b_k$ are numbers, then one can change the order of $b_k^\dagger, a_j^\dagger$ if he likes.
