Continous and Discrete basis, Multiplication of Density Matrix and Hamiltonian

Suppose I have a wave function $\psi(x)$ in position basis. I can make a density function by simply multiplying $\psi(x)$ and its conjugate $\psi^*(x)$. If I operate the density matrix $\rho(x,y)=\psi(x)\psi^*(y)$ with Hamiltonian $\hat{H}$ I will get, $$\hat{H}\rho(x,y)=\hat{H}\psi(x)\psi^*(y)=\psi^*(y)\hat{H}\psi(x)+\psi(x)\hat{H}\psi^*(y)$$ since the hamiltonian is a diferential operator. At the same time, If I take discrete basis representation, I will get $\langle n|\psi\rangle=V_{1}$ and $\langle\psi| n\rangle=V_{1}^{\dagger}$ which are the column vector and row vector respectively. Now if I operate the Hamitonian matrix $\hat{H}$ on the density matrix I will get, $$\hat{H}\rho=H (V_{1} V_{1}^{\dagger})=(H V_{1}) V_{1}^{\dagger}$$ That means, I need to apply the hamiltonian operator only once on a vector. Why it is not distributed like the differential operator.

$$\hat{H}\rho= V_{1} H V_{1}^{\dagger}+(H V_{1}) V_{1}^{\dagger}$$ Why this equation is not valid in the case of discrete basis ?

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I am not sure if your $\hat{H} \rho(x,y)$ and acting $\hat{H}$ on both $\psi(x)$ and $\psi^*(y)$ is a proper manipulation. Take a pure state density operator for instance, $\rho = | \psi \rangle \langle \psi |$. $\hat{H} \rho = \hat{H} | \psi \rangle \langle \psi |$. The coordinate representation is $\langle x |\hat{H} | \psi \rangle \langle \psi | x' \rangle$, which means the Hamiltonian only acts on $\psi(x)$, not the conjugated one. –  user26143 Nov 22 '13 at 16:30
user26143 is directing you in good direction. In order to apply $\hat{H}$ to functions $\psi$, we need to have explicit definition of $\hat H$. For harmonic oscillator, $\hat H = -\frac{\hbar^2}{2m}\frac{ \partial^2}{\partial x^2} + \frac{1}{2}m\omega^2 x^2$. So when we have function of two variables like $\psi(x)\psi(y)$, the Hamiltonian operator will act only on the part that depends on $x$, treating the rest as a constant: $\hat{H} \left[\psi(x)\psi(y)\right] = \left[\hat{H}\psi(x)\right]\psi(y)$. –  Ján Lalinský Nov 22 '13 at 18:05
Ok I got it ...Thank you user26143 and Jan Lalinsky –  Sijo Joseph Nov 24 '13 at 18:06