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Deducing the solution of the von Neumann equation

  1. May 27, 2012 #1
    1. The problem statement, all variables and given/known data
    [tex] \hat{\rho}(t)=? [/tex] [tex]
    |\psi(t)\rangle=U(t,t_{0})|\psi(t_{0})\rangle [/tex] [tex]
    \imath\hbar\partial_{t}\hat{p}=[\hat{H},\hat{\rho}]
    [/tex]

    2. Relevant equations
    [tex]
    \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}
    [/tex]

    3. The attempt at a solution

    I already know the solution: [tex]\hat{\rho}(t)=\hat{U}\hat{\rho}(0)\hat{U}^{+}[/tex]
    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:
    [tex]\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)][/tex]
    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 [itex]\hat{U}[/itex] goes away again?

    Or is this just guessing as one way to solve a differential equation. Then, still, how do you get the idea?
     
    Last edited: May 27, 2012
  2. jcsd
  3. May 27, 2012 #2

    dextercioby

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    Science Advisor
    Homework Helper

    Why don't you use the definition of the von Neumann density operator ?
     
  4. May 27, 2012 #3
    The definition should be [itex]\hat{\rho}=\sum_{i}p_{n}|\psi(t)\rangle\langle\psi(t)|[/itex]
    I can do with that:
    [tex]\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[/tex][tex]

    \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[/tex][tex]
    \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}|[/tex]
     
  5. Jun 20, 2012 #4
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