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Stuck on a complex ODE (conjugate issues )

  1. Jul 27, 2009 #1
    Hi everyone.

    I'm stuck on a problem, banging of head etc. Basically, I've got a time-evolution problem restricted to the positive quadrant of complex space, where the state of the system [tex]\psi \in \mathbb{C}[/tex] is described by the following type of ODE:

    \frac{{d\psi \left( t \right)}}{{dt}} = i\overline {\psi \left( t \right)}

    i.e, the time evolution of the state depends on the conjugate, not the state itself (this is determined by the first principles of my system).

    Numerically, I can easily generate the quiver diagrams of this first difference, and then use it to simulate the time evolution of the system. The norm seems to grow roughly exponentially, while the argument exponentially approaches [tex]\pi /4 [/tex] , which is what I expect.

    My problem is that I am at a loss of ideas for how to actually integrate the expression, in order to obtain the expression for the time evolution of [tex]\psi \left( 0 \right)[/tex] for any given starting point in the space.

    Any assistance on this would be very much appreciated.
    Last edited: Jul 27, 2009
  2. jcsd
  3. Jul 27, 2009 #2


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    Homework Helper

    This is just a pair of coupled DE's. Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal. Your equation can be written

    [tex]\frac{d}{dt}(\psi_R +i\psi_I) = -\psi_I + i\psi_R[/tex]

    which splits up into

    [tex]\frac{d \psi_R}{dt}= -\psi_I[/tex]


    [tex]\frac{d \psi_I}{dt} = \psi_R[/tex]

    which you should hopefully be able to solve, yes?
  4. Jul 27, 2009 #3

    Well, that figures... I was trying to work it while keeping the complex number whole, but completely overlooked treating it as a system of ODEs (and not a complicated one at that).

    Thanks a lot for that wake-up call!
  5. Jul 27, 2009 #4


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    Homework Helper

    In this case, this is likely the easier way to solve it: You could also note that if you take the conjugate of your original DE you get

    [tex]\frac{d \overline{\psi}}{dt} = -i\psi[/tex]

    so taking another derivative of your original equation gives

    [tex]\frac{d^2\psi}{dt^2} = i\frac{d \overline{\psi}}{dt} = \psi[/tex]
    Last edited: Jul 27, 2009
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