Physical significance behind [H, rho]=0

  • Context: Graduate 
  • Thread starter Thread starter amrahmadain
  • Start date Start date
  • Tags Tags
    Physical Significance
Click For Summary

Discussion Overview

The discussion revolves around the physical significance of the commutation relation [H, rho] = 0 in the context of the Maximum Entropy Principle applied to non-equilibrium steady-state statistical density operators in quantum transport. Participants explore the implications of this constraint on the density operator's behavior and its relation to conservation and time evolution.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • Amr questions the physical significance of the constraint [H, rho] = 0 in the context of steady-state conditions.
  • One participant explains that the time derivative of the density operator leads to the conclusion that the stationarity condition implies [H, rho] = 0.
  • Another participant asserts that the commutation relation indicates conservation of the density operator.
  • A further contribution reiterates the time evolution of the density operator and its invariance under this evolution, suggesting it supports the idea of conservation.
  • Amr raises a follow-up question regarding potential issues in constraining average current in a molecular device, indicating uncertainty about arriving at a zero induced potential drop.

Areas of Agreement / Disagreement

Participants generally agree on the implications of the commutation relation regarding conservation and time evolution, but Amr's follow-up question introduces uncertainty about the application of the Maximum Entropy Principle in a different context, which remains unresolved.

Contextual Notes

Participants do not fully explore the assumptions or limitations of the Maximum Entropy Principle in the context of the average current question, leaving some aspects of the discussion open to interpretation.

amrahmadain
Messages
3
Reaction score
0
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?

Thanks,
Amr
 
Physics news on Phys.org
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...
 
It means it's conserved.
 
Last edited:
vanesch said:
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...
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 :-)

Thanks,
Amr
 
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:
Perturbation said:
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.
Thanks Perturbation. Just double checking, this is time evolving the density operator in the Heinsenberg picture, right?
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 9 ·
Replies
9
Views
4K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 0 ·
Replies
0
Views
1K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 5 ·
Replies
5
Views
542
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K