Physical meaning of Transpose of an Operator in Quantum Mechanics? What's the physical meaning of transpose of a matrix in Quantum Mechanics? 
Although for Unitary or Orthogonal operators, I know that transpose of that operator would reverse the action and that's because, $A^T=A^{-1}$.
But if the operator is neither orthogonal nor unitary, then in that context what does transpose of an operator means?
Apart of this interpretation, that Transpose of an operator is equivalent to same operation but in dual space! .
 A: Suppose  a linear map $A:V\to V$ is represented in a basis ${\bf e}_n$ by the  matrix ${A^n}_m$. This means that $A$ maps ${\bf e}_n$ to
$$
A[{\bf e}_n] = {\bf e}_n{A^m}_n.
$$
The map $A$ automatically  induces a "conjugate"  map that is  usually called $A^*$ by functional analysts---  a rather bad notation from the viewpoint of physics   as  no complex conjugation is needed nor implied.   This induced map acts  the dual space $V^*$ consisting of linear functions from ${\bf f}:V\to {\mathbb C}$
by
$$ 
(A^*{\bf f})[{\bf x}] = {\bf f}(A[{\bf x}])
$$
If ${\bf e}^{*n}$ is the set of functions constituting the   dual basis such that ${\bf e}^{*n}[{\bf e}_m]= \delta^n_m$,
then
$$ 
A^* [{\bf e}^{*n}] =   {\bf e}^{*m}{A^n}_m,
$$
so the indices $n$ and $m$ have been interchanged. In other words the matrix representing $A^*$ in the basis ${\bf e}^{*n}$ is the transpose of the matrix representing $A$. As a result of this some linear algebraists call $A^*$ the transpose of $A$ and may use the notation $A^T:V^*\to V^*$.
In Dirac notation the original $A$ acts on the space of kets $|n\rangle$ by
$$
A|n\rangle= |m\rangle\langle m|A|n\rangle
$$
and the conjugate map acts on the dual space of bra vectors by
$$
\langle n|A= \langle n|A|m\rangle\langle  m|
$$
In Dirac  notation ${A^n}_m\to  \langle n|A|m\rangle$ so we lose the distiction between upstairs/downstairs indices and so are restricted to orthonormal bases. The equation
$$ 
(A^*{\bf e}^{*m})[{\bf e}_n] = {\bf e}^{*m}(A[{\bf e}_n])
$$
now reads
$$
(\langle m|A)|n\rangle= \langle m|(A|n\rangle)
$$
so Dirac does not need the confusing notation $A^*$, nor does he need an explicit  transpose because the operator is acting to its left rather than to its right.  The operation to the left on the bra is nonetheless the action of the transposed matrix.
As far as I know this story is only consistent intepretation of the transpose in QM.
A: From a physical point of view, the only relevant operators are those that are "symmetric" (in fact, self-adjoint). These operators have the properties of remaining unchanged under a "transposition". Any other operator that does not satisfy this property is not observable.
Having said this, there are some interesting cases, like the one mentioned in the OP, like unitary operators. These can be given the physical meaning of transformations, in a rather broad sense. More generally, one can consider more general objects, like partial isometries. A partial isometry $V$ maps a subspace of a Hilbert space into another subspace of the same dimension. Hence, every unitary is a partial isometry, but not every partial isometry is a unitary. If $V$ maps $K$ to $K'$, then it is easy to see that $V^T$ maps $K'$ back to $K$. For any bounded operator $A$ on a Hilbert space, one can find a positive operator $|A|$ and a partial isometry $V$ such that
$$A = V|A|$$
This is known as the polar decomposition of $A$. Using this, one then finds
$$A^T = |A|V^T$$
which can help to get an idea of what the transposition is doing on the operator $A$.
