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Statistical operator

  1. Mar 18, 2012 #1
    [tex]
    \hat{\rho} = \begin{bmatrix}
    \frac{1}{3} & 0 & 0 \\[0.3em]
    0 & \frac{1}{3} & 0 \\[0.3em]
    0 & 0 & \frac{1}{3}
    \end{bmatrix}
    [/tex]

    If I have this statistical operator I get
    [tex]i\hbar\frac{d\hat{\rho}}{dt}=0[/tex]
    So this is integral of motion and
    [tex][\hat{H},\hat{\rho}]=0[/tex]
    Is this correct? Tnx for your answer.
     
  2. jcsd
  3. Mar 18, 2012 #2

    fzero

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    Gold Member

    Your operator is a constant multiple of the identity operator, so it commutes with any operator, in particular the Hamiltonian.
     
  4. Mar 18, 2012 #3
    Of course. My example wasn't so good. Suppose I have matrix

    [tex]
    \hat{\rho} = \begin{bmatrix}
    \frac{1}{3} & 5 & 6 \\[0.3em]
    5 & \frac{1}{3} & 6 \\[0.3em]
    5 & 6 & \frac{1}{3}
    \end{bmatrix}
    [/tex]

    Why now [tex]\hat{\rho}[/tex] always commute with Hamiltonian?
     
    Last edited: Mar 18, 2012
  5. Mar 18, 2012 #4

    kith

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    What you have written is the state ρ in a fixed base for a given time t0.

    Compare this with a ket vector, which can be represented as a column vector (1 2 3 ...)T for any fixed time t0.

    In both cases, the states are initial values for the dynamical equations. Not their solutions for all times t.
     
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