I was wondering if someone could confirm this for me. I'm attempting to convert a paper into matrix notation but I keep getting confused.
I first have a unitary matrix that makes a transformation on some set of vectors (the unitary matrix is made of only real values) such that
$$b_{n} = \sum_mU_{nm}a_{m}.$$
If I then take the transpose on either side such that
$$U^{-1}b = (U^{-1}U)a = Ia,$$ I have in index notation that \begin{align*} \sum_{n} U^{\dagger}_{kn}b_{n} &= \sum_{m} \sum_{n} U^{\dagger}_{kn}U_{nm}a_{m} \\ \sum_{n} U^{\dagger}_{kn}b_{n} &= \sum_{m}(U^{\dagger}U)_{km}a_{m} \end{align*} where $$ (U^{\dagger}U)_{km} = δ_{km} $$
therefore we end up with;
$$ \sum_{n} U^{\dagger}_{kn}b_{n} = a_{k} $$
I know I can re-label any of the indices as long as I am consistent on both sides, however, what I need to know here is since $U$ is real then the Hermitian conjugate just simplifies to the transpose and transposing a matrix with index notation means just flipping the indices (or changing the sum), since when you do the transpose explicitly you're really just summing over the other index (i.e., when you have a vector it's the difference between it being a row or a column vector). Thus, the following should be equivalent:
$$ \sum_{n} U^{\dagger}_{kn}b_{n} = \sum_{n} U_{nk}b_{k} $$
Does this seem reasonable? It looks to me from my beginning point it is as the original U is summing over the second index, m. Where as the final expression is summing over the first index, n.
thanks for taking a look guys, if you want any information on the paper and such that's not a problem
As you note, $U$ is unitary so $U^\dagger$ = $U^{T}$, and so \begin{align*} a_k &= \sum_n \left[U^\dagger\right]_{kn} b_n \\ &= \sum_n \left[U^T\right]_{kn} b_n \\ &= \sum_n U_{nk} b_n. \end{align*} That is to say, $b_k$ on the right-hand side of your expression should be $b_n$.