Deducing the solution of the von Neumann equation

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Homework Statement


\hat{\rho}(t)=? <br /> |\psi(t)\rangle=U(t,t_{0})|\psi(t_{0})\rangle <br /> \imath\hbar\partial_{t}\hat{p}=[\hat{H},\hat{\rho}] <br />

Homework Equations


<br /> \imath\hbar\partial_{t}\hat{p}=[\hat{H},\hat{\rho}] \Leftrightarrow\imath\hbar\partial_{t}\hat{p}=\hat{H}\hat{\rho}-\hat{\rho}\hat{H}<br />

The Attempt at a Solution



I already know the solution: \hat{\rho}(t)=\hat{U}\hat{\rho}(0)\hat{U}^{+}
But where do I get this from? How do I know that I have to write the time evolution operator multiplied once in front of the density operator and once the Hermitian conjugate after it?

Also, I tried to verify the solution:
\Rightarrow\imath\hbar\partial_{t}\hat{U}\hat{\rho}(0)\hat{U}^{+}=\hat{H}\hat{U}\hat{\rho}(0)\hat{U}^{+}-\hat{U}\hat{\rho}(0)\hat{U}^{+}\hat{H}=[H,\hat{\rho}(t)]
Can't I take any other operator instead of the time evolution operator at this place, since in my attempt to verify the solution the \hat{U} goes away again?

Or is this just guessing as one way to solve a differential equation. Then, still, how do you get the idea?
 
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Why don't you use the definition of the von Neumann density operator ?
 
The definition should be \hat{\rho}=\sum_{i}p_{n}|\psi(t)\rangle\langle\psi(t)|
I can do with that:
\partial_{t}\hat{\rho}=\partial_{t}\sum_{i}p_{n}| \psi(t)\rangle\langle\psi(t)|+ \sum_{i} p_{n}|\psi(t) \rangle\partial_{t}\langle\psi(t)| \Leftrightarrow<br /> <br /> \partial_{t}\hat{\rho}=\frac{1}{\imath\hbar}Hp_{n}|\psi(t)\rangle\langle\psi(t)|+\sum_{i}p_{n}|\psi(t)\rangle\frac{1}{\imath\hbar}H\langle\psi(t)| \Leftrightarrow<br /> \partial_{t}\hat{\rho}=\frac{1}{\imath\hbar}\hat{H}p_{n}|\psi(t)\rangle\langle\psi(t)|+\frac{1}{ \imath\hbar}\sum_{i}p_{n}|\psi(t)\rangle\langle \psi(t)\hat{H}|
 
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