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Change of Schrodinger equation when we change frame of reference

  1. Nov 21, 2012 #1
    Hi! I think this should be a fairly easy question to resolve, but it's currently giving me fits...

    Suppose we consider a system of three interacting particles where we assume two of the particle (at positions r1 and r2) have infinite mass (such that we can ignore their kinetic energy) and do not interact. Suppose r1 follows the trajectory r1 = v1t and similarly for r2 = v2t.

    Suppose we want to investigate the quantum mechanics of a particle of unit mass located at x that interacts with particle 1 through V1 and particle 2 with V2. I think I'm correct in saying that this corresponds to the time-dependent Schrodinger equation given by

    [tex]
    i \frac{\partial \psi}{\partial t} = -\frac 12 \frac{\partial^2 \psi}{\partial x^2} + V_1(x - v_1t)\psi + V_2(x-v_2t)\psi.
    [/tex]

    But what if we want to consider this equation in a frame of reference where the particle at r1 is at rest? If we let z denote the vector x - r1 = x - v1t, shouldn't the above Schrodinger equation change to...

    [tex]
    i \frac{\partial \Psi}{\partial t} = - \frac 12 \frac{\partial^2 \Psi}{\partial z^2} + V_1(z)\Psi + V_2(z + (v_1 - v_2)t)\Psi
    [/tex]

    I'm currently reading through a paper that claims the final V_2 term should read

    [tex]
    V_2(z - (v_1 - v_2)t)\Psi,
    [/tex]

    and I can't seem to figure out why. Thanks!
     
  2. jcsd
  3. Nov 21, 2012 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Looks like a sign error in the paper.
    Or, if V2(x)=V2(-x), a transformation z -> -z
     
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