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I Why not calculate the "trajectory" of a wave function

  1. Jul 25, 2017 #1
    The classic limit of Schrodinger equation is hamilton-jacobi eqution.

    Wave function's classic limit is ##\exp{\frac{i}{\hbar}S(x,t)}##,##S(x,t)## is the action satisfying hamilton-jaccobi eqution.

    However, a particle travels along single trajectory of ##S(x,t)##,
    Why not make some constrains on wave fucntion to reveal the "single trajectory" from wave function?
     
  2. jcsd
  3. Jul 25, 2017 #2

    mfb

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    That only works in the classical limit, where you can ignore quantum mechanics.

    de-Broglie-Bohm has classical trajectories, but everything is governed by the wave function (or pilot wave), so that doesn't really change anything.
     
  4. Jul 26, 2017 #3
    Thanks.
    How about this way.
    Classic Mechanic :
    1,wave fucntion (##\hbar \rightarrow 0##) is ##\psi \rightarrow \exp{\frac{i}{\hbar}S(x,t)}##;
    2,we get wave function from Hamilton-Jaccobi eqution with boundary condition,
    ##\frac{\partial{S}}{\partial{t}}+\frac{1}{2m}{(\frac{\partial{S}}{\partial{x}})}^2+V(x)=0##;
    3,##\frac{dx}{dt}=\frac{1}{m}\frac{\partial{S}}{\partial{x}}## could define a trajectory ##x(t)## for a particle with initial condition ##x(0)=a##;

    Quantum mechanic:
    1,wave fucntion ##\psi(x,t)##;
    2,we calculate ##\psi(x,t)## from Schrodinger eqution, ##-\frac{{\hbar}^2}{2m}\frac{{\partial}^2{\psi}}{\partial{t}^2}+V(x)=i\hbar\frac{\partial{\psi}}{\partial{t}}##;
    My textbook stop at second step , just calculate wave function and ignore the third step in classic mechanic which make me confused.
    Maybe third step in QM, could be like this -----
    -----3,##(\frac{dx}{dt})_n=\frac{(\frac{\hat{p}}{m})^n\psi}{(\frac{\hat{p}}{m})^{(n-1)}\psi}##,##n \ge 1## is an integer.##\hat{p}=-i\hbar\frac{\partial}{\partial{x}}##

    When ##\hbar \rightarrow 0## , ##(\frac{dx}{dt})_n \rightarrow \frac{1}{m}\frac{\partial{S}}{\partial{x}}##
    This means we define many trajectory for a particle.
     
  5. Jul 27, 2017 #4

    mfb

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    You can define a lot of things, that doesn't mean they have to have useful properties. I didn't study your definition in detail, but if in doubt, it won't lead to a continuous trajectory, or even to ill-defined expressions.
     
  6. Jul 27, 2017 #5

    Demystifier

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    Read about Bohmian mechanics! You can start e.g. from
    https://en.wikipedia.org/wiki/De_Broglie–Bohm_theory
     
  7. Jul 30, 2017 #6
    Thanks.I think Bohmian mechanics is helpful to me.Could help me with another question below?
    ##\psi## is a solution of Schrodinger eqution.
    When ##\hbar \rightarrow 0##,##{\psi}(x,t) \rightarrow {\rho}(x,t)e^{\frac{i}{\hbar}S(x,t)}##
    Define ##\frac{dq}{dt}=\frac{1}{m}\frac{\partial{S}}{\partial{x}}##,
    and ##\delta{(x-q(t))}## means choose trajectories beside ##q(t)##,not the real Dirac function.
    The ##{\psi}_q \rightarrow \delta{(x-q(t))}e^{\frac{i}{\hbar}S(x,t)}## is a solution.
    The linear combination ##{\sum\limits_q}c_q{\psi}_q## is also a solution.
    Too many solutions.
    If we calculates wave fucntion of electron moving around nuclear or the one of electron in the two-slit diffraction experiment,how to get the initial boundary couditon?Is there any example to calculate this kind of thing?
     
  8. Aug 2, 2017 #7

    Demystifier

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    There is no simple recipe hot to find the right solutions (right initial conditions). You must study a whole textbook to learn something about it.
     
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