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Physical significance behind [H, rho]=0

  1. Jun 4, 2006 #1
    In using the Maximum Entropy Principle to derive the non-equilibrium steady-state statistical density operator in quantum transport, I've seen the following constraint used to let the closed quantum system be in steady-state: [H, rh] = 0

    I still don't understand the physical significance behind this constraint. Can anybody shed some light on it?

  2. jcsd
  3. Jun 4, 2006 #2


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    Well, given that the time derivative of the density operator is given by
    i ihbar drho/dt =[H,rho]

    and the stationarity condition drho/dt = 0, your equation follows...
  4. Jun 4, 2006 #3
    It means it's conserved.
    Last edited: Jun 4, 2006
  5. Jun 4, 2006 #4
    Thanks a million. One more quick question, please. In constraining the average current in a molecular (or mesoscopic) device, again using the MaxEnt Principle, what possibly could have gone wrong to arrive at a zero induced potential drop? I've some educated guesses but I very much like to listen to what you think could have been possibly missed.

    I've banging my head against the wall for sometime now :-)

  6. Jun 6, 2006 #5
    I know this has been answered pretty much, I just wanted to add something.

    If you unitarilly time evolve your density operator [itex]\rho[/itex], you have

    [tex]\rho (t_0+t)=e^{-iHt}\rho (t_0)e^{iHt}=\rho (t_0)e^{-iHt}e^{iHt}=\rho (t_0)[/tex]

    The second equality following from the commutativity of your density operator and H. I.e. the density operator is invariant to time evolution, hence conservation. Just thought I'd add another perspective, though it is equivalent to what Vanesch said.

    Sorry I can't help with your other problem though.
    Last edited: Jun 6, 2006
  7. Jun 8, 2006 #6
    Thanks Perturbation. Just double checking, this is time evolving the density operator in the Heinsenberg picture, right?
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