Quantum entropy in term of density matrix Why in von Neumann expression of quantum entropy we have trace of density matrix expression? Why don't off diagonal term play a role?
 A: There are a number of compelling reasons to adopt the von-Neumann expression
\begin{align}
  S_\mathrm{vn}(\rho) = -k_B\mathrm {tr} \rho\ln\rho
\end{align}
as the definition of entropy for a quantum mechanical system with density matrix $\rho$.
Agreement with Gibbs entropy
First, note that following work by Boltzmann, Gibbs introduced a certain statistical notion of entropy, what is now called the Gibbs entropy, in the late 19th century via the expression
\begin{align}
  S_\mathrm{Gibbs}(p_1, \dots, p_n) =-k_B\sum_{i=1}^n p_i\ln p_i
\end{align}
for a statistical mechanical system with microstates $1, \dots, n$ and a probability $p_i$ that the system will access microstate $i$ in its evolution.  I'll leave it to you to read about the relevance of the Gibbs entropy formula to statistical mechanics and thermodynamics.
Now, consider an $n$-dimensional quantum system with Hilbert space $\mathcal H$ and density operator $\rho$ and Hamiltonian $H$.  In a system in equilibrium, the density operator commutes with $H$;
\begin{align}
  [\rho, H] = 0,
\end{align}
so we can let $|1\rangle, \dots, |n\rangle$ denote an orthonormal basis for $\mathcal H$ consisting of simultaneous eigenvectors of $\rho$ and $\mathcal H$.  In this basis, the density operator will be diagonal;
\begin{align}
  \rho = \mathrm{diag} (p_1, \dots, p_n)
\end{align}
Moreover, because the density matrix is non-negative and self-adjoint, each diagonal entry $p_i$ is non-negative and real, so the sequence $p_1, \dots, p_n$ can be viewed as a probability distribution, where $p_i$ represents the ensemble-probability that the system is in state $|i\rangle$.  Furthermore, it is straightfoward to show that the von-Neumann entropy expression reproduces precisely the Gibbs entropy;
\begin{align}
  S_\mathrm{vn}(\rho) = S_\mathrm{Gibbs}(p_1, \dots, p_n).
\end{align}
But wait!  There's more!
Agreement with Shannon (information) entropy
In his seminal paper 
"A Mathematical Theory of Communication," 
Claude Shannon introduced a notion of entropy called information entropy, which one often calls Shannon entropy.  For a given probability distribution, $p_1, \dots, p_n$, and for a given positive real number $k$, the Shannon entropy is
\begin{align}
  S_\mathrm{Shannon}(k; p_1, \dots, p_n) = -k\sum_{i=1}^np_i\log p_i.
\end{align}
The Shannon entropy can roughly be thought of as a measure of the information content of the probability distribution.  
It turns out that viewing it in this way, one can formulate statistical mechanics from the perspective of maximizing Shannon entropy subject to constraints as was beautifully discussed by E.T. Jaynes shortly after Shannon's paper;
"Information Theory and Statistical Mechanics"
But notice that the Shannon entropy is the same expression as the Gibbs entropy which we showed above is that same as the von-Neumann entropy provided we appropriately interpret what the density matrix is telling us.
A: Because von Neumann defined it that way. He did probably because trace has nice mathematical properties - the result does not depend on the basis in which the density matrix is expressed, and it is equal to $\sum_k -\rho_{kk} \ln \rho_{kk}$ where $\rho_{kk}$ are eigenvalues of the density matrix, which is similar to the expression $\int -\rho\ln \rho \,dqdp$ used before in statistical physics. There may be other reasons too, but it is important to bear in mind it is just definition.
