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QM: Density matrix evolution

  1. Aug 30, 2013 #1
    1. The problem statement, all variables and given/known data
    Hi

    The density matrix evolves as
    [tex]
    \dot \rho = -\frac{i}{\hbar}[H,\rho]
    [/tex]
    but is this equation written in the Schrödinger or Heisenberg picture? I'm not entirely sure how to figure this out. In my book it just mentions the equation, not how it is derived (which may have given a hint...)
     
  2. jcsd
  3. Aug 30, 2013 #2
    OK, it is in the Schrödinger picture. The derivation of the equations follows naturally from the Schrödinger equation
     
  4. Aug 30, 2013 #3

    vanhees71

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    [Typos from previous version of this posting corrected!]

    You must be careful with what you mean by the dot. The equation
    [tex]\left (\frac{\partial \hat{\rho}}{\partial t} \right )_{\text{expl}}+\frac{1}{\mathrm{i} \hbar} [\hat{\rho},\hat{H}]=0.[/tex]
    is valid in any picture of time evolution. It's the socalled von-Neumann equation.
     
    Last edited: Aug 30, 2013
  5. Aug 30, 2013 #4
    Thanks for clarifying. The derivation of it is almost trivial in the Schrödinger picture. But when it is used in the interaction picture, people only use the interaction-Hamiltonian and substitute all operators by their slowly-varying versions. How does one justify that the von-Neumann equation is valid in this case? Shouldn't it be derived from scratch in the interaction picture?
     
  6. Aug 30, 2013 #5
    this it not an equation...?
     
  7. Aug 30, 2013 #6

    vanhees71

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    Of course, it must read
    [tex]\left (\frac{\partial \hat{\rho}}{\partial t} \right )_{\text{expl}}+\frac{1}{\mathrm{i} \hbar} [\hat{\rho},\hat{H}]=0.[/tex]
    [I corrected the previous posting for a sign mistake and the missing =0].

    This von Neuamnn equation is valid in any picture of time evolution.

    To make ist more clear one must remember that there is a set of operators generating the algebra of observables that are by definition not explicitly time dependent. In QFT these are the fundamental field operators, in non-relativistic QM (in the first-quantized formulation) it's the position, moomentum, and spin operators, etc.

    The mathematical time dependence of observables and states is only determined up to a time-dependent unitary transformation and this leads to the different choices of the picture of time evolution. The most general choice is the Dirac picture. It turns out that the time evolution is always local in time, and you have two equations of motion for the unitary time-evolution operators for the observable operators and the statistical operator (using [itex]\hbar=1[/itex] for simplification),
    [tex]\begin{split}
    \partial_t \hat{A}(t,t_0)&=\mathrm{i} \hat{H}_1(t) \hat{A}(t,t_0), \\
    \partial_t \hat{C}(t,t_0)&=\mathrm{i} \hat{H}_2(t) \hat{C}(t,t_0). \\
    \end{split}
    [/tex]
    The Hermitean operators [itex]\hat{H}_1[/itex] and [itex]\hat{H}_2[/itex] are arbitrary but constrained by the demand that
    [tex]\hat{H}=\hat{H}_1+\hat{H}_2.[/tex]
    The time evolution of the observable statistical operators are then given by
    [tex]
    \begin{split}
    \hat{O}(t) &=\hat{A}(t,t_0) \hat{O}(t_0) \hat{A}^{\dagger}(t,t_0),\\
    \hat{\rho}(t) &=\hat{C}(t,t_0) \hat{\rho}(t_0) \hat{C}^{\dagger}(t,t_0).
    \end{split}[/tex]
    From this you get the equations of motion in the so chosen picture as
    [tex]\begin{split}
    \frac{\mathrm{d}}{\mathrm{d} t} \hat{O}&= -\mathrm{i}[\hat{O},\hat{H}_1] + \left (\frac{\partial \hat{O}}{\partial t} \right )_{\text{expl}},\\
    \frac{\mathrm{d}}{\mathrm{d} t} \hat{\rho}&= \mathrm{i}[\hat{\rho},\hat{H}_2].
    \end{split}
    [/tex]
    The explicit time dependence refers to the time dependence which does not come by implicit time dependence of the fundamental operators that build up [itex]\hat{O}[/itex].

    Of course, as a function of observables and perhaps explicitly of time the Statistical operator also must fulfill the equation of motion
    [tex]\frac{\mathrm{d}}{\mathrm{d} t} \hat{\rho}= -\mathrm{i}[\hat{O},\hat{H}_1] + \left (\frac{\partial \hat{\rho}}{\partial t} \right )_{\text{expl}}.[/tex]
    Subtracting both expressions for the time derivative of the Statistical Operator one obtains von Neumann's equation:
    [tex]-\mathrm{i}[\hat{\rho},\hat{H}]+\left (\frac{\partial \hat{\rho}}{\partial t} \right )_{\text{expl}}=0,[/tex]
    because [itex]\hat{H}_1+\hat{H}_2=\hat{H}[/itex].
     
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