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Schrodinger equation for close and opens system

  1. Apr 3, 2013 #1
    How do we differentiate the solution of Schrodinger equation for closed and open system.
     
  2. jcsd
  3. Apr 4, 2013 #2

    DrClaude

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    Staff: Mentor

    Welcome to PF!

    The Schrödinger equation is the same. But for an open system, you have to solve it using the density matrix, whereas for a closed system you usually can solve it for the wave function itself.
     
  4. Apr 4, 2013 #3

    kith

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    I wouldn't put it that way. The term "Schrödinger equation" should be reserved for closed systems with their unitarian dynamics where no dissipation and decoherence occur.
     
  5. Apr 4, 2013 #4

    DrClaude

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    I'm not sure that I agree with you, but I did misspeak in my previous post. The time evolution of density matrix is governed by the Liouville-von Neumann equation
    $$
    i \hbar \frac{\partial \rho}{\partial t} = \left[ \hat{H}, \rho \right]
    $$
    That said, you can also approximate an open system using a non-Hermitian Hamiltonian and use that in the Schrödinger equation, and have a non-unitary evolution of the wave function.
     
  6. Apr 4, 2013 #5

    kith

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    True, I didn't think about this. But you will only get dissipation with this, not decoherence. This doesn't change if you use the von Neumann equation (which is derived from the Schrödinger equation). Pure states are still mapped to pure states and the entropy doesn't change.

    In order to take into account all effects in open systems you need more general dynamical equations like the Lindblad equation. Compared with the von Neumann equation it has an additional term D(ρ) which induces dissipation and decoherence. I think it's also more natural than using non-hermitian Hamiltonians because you can derive it from the unitarian dynamics of the combined system "open system + environment" under certain assumptions.
     
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