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Is there a preferred solution among the solutions to Schrödinger's eq?

  1. Jun 23, 2013 #1
    I am a little confused when it is stated that Schrödinger's equation represents a deterministic evolution of the wave function of a particle. I would be OK with the idea that time evolution went from one state to another state in a deterministic way (even though each state, with respect to its eigenvalues, is not deterministic) if there were only one solution to the equation. However, it has many solutions, so why would there be a preferred solution?
     
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  3. Jun 23, 2013 #2

    Bill_K

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    I don't understand your remark, because leaving aside the issue of what happens during a measurement, the Schrodinger equation itself is deterministic. The time-dependent Schrodinger equation is a first order differential equation in time, and if you're given initial values everywhere at t = 0, the solution at times t > 0 is uniquely determined. There's one and only one solution.
     
  4. Jun 23, 2013 #3
    Thanks for the reply, Bill_K. Your answer then takes me from the frying pan into the fire. I thought --and this is probably wrong -- Schrödinger's equation is a differential equation which has an infinite number of solutions: any linear combination of plane waves.
     
  5. Jun 23, 2013 #4

    Bill_K

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    nomadreid, Just like any differential equation, you have to be given initial conditions before you can write down the solution. You don't know how high a projectile will go until you are told its initial position and velocity. Likewise to solve the Schrodinger equation, you have to be given its initial values. And since it's a partial differential equation, you have to be given the value of ψ(x, 0) everywhere, at the initial time t = 0. Without being told this information, the problem is not uniquely defined. And it's not the fault of the Schrodinger equation especially, that's just the way PDEs work. But given these initial conditions, there is no longer an infinity of different possible solutions - only one.
     
    Last edited: Jun 23, 2013
  6. Jun 23, 2013 #5
    Bill_K. Thanks very much. Initial conditions, right. Makes sense.
     
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