Given the operator $$\hat{p} = \sqrt{2mt}\sum_j[i\left|j-1\right\rangle \left\langle j|-i|j+1\right\rangle\left\langle j \right|]$$ I want to show that $\hat{p} = \hat{p}^{\dagger}$
My thoughts were the following
$$\begin{align} \hat{p}^{\dagger} &= (\sqrt{2mt}\sum_j[i\left|j-1\right\rangle \left\langle j|-i|j+1\right\rangle\left\langle j \right|])^{\dagger}\\ &= \sqrt{2mt}\sum_j ([i\left|j-1\right\rangle \left\langle j|-i|j+1\right\rangle\left\langle j \right|])^{\dagger} \end{align} $$
Now I'm stuck. (I realize I didn't came that far ;-)). My question is the following, how does the complex conjugate work on bra's and kets?
Update Is it correct to say the following?
$$\begin{align} \hat{p}^{\dagger} &= (\sqrt{2mt}\sum_j[i\left|j-1\right\rangle \left\langle j|-i|j+1\right\rangle\left\langle j \right|])^{\dagger}\\ &= \sqrt{2mt}\sum_j ([i\left|j-1\right\rangle \left\langle j|-i|j+1\right\rangle\left\langle j \right|])^{\dagger} \\ &= \sqrt{2mt}\sum_j([i\left\langle j \right|)^{\dagger}(\left\langle j \right| -i\left| j+1\right\rangle)^{\dagger}(\left|j-1\right\rangle)^{\dagger}\ \ \ ? \end{align} $$
Recall $(|n\rangle\langle k| )^\dagger =|k\rangle\langle n| $. So, then, complex conjugate and transpose a purely imaginary antisymmetric infinite matrix, $$ \hat{p}^{\dagger} = \sqrt{2mt}\sum_j \Bigl (i\left|j-1\right\rangle \left\langle j|~~-i~|j+1\right\rangle\left\langle j \right|\Bigr )^{\dagger}\\ = \sqrt{2mt}\sum_j \Bigl (i\left|j\right\rangle \left\langle j+1|~~-i~|j \right\rangle\left\langle j-1 \right|\Bigr ) = \hat{p}.$$ The identification with your original expression follows relabelling your dummy indices summed over, $j\mapsto j-1$ and $j\mapsto j+1$ in the first and second terms inside the sum, respectively.