Why not calculate the "trajectory" of a wave function

[zhouhao]

The classic limit of Schrödinger 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 function to reveal the "single trajectory" from wave function?

 

[mfb]

Why not make some constrains on wave function to reveal the "single trajectory" from wave function?

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.

 

[zhouhao]

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.

Thanks.
How about this way.
Classic Mechanic :
1,wave function ($\hbar \rightarrow 0$) is $\psi \rightarrow \exp{\frac{i}{\hbar}S(x,t)}$;
2,we get wave function from Hamilton-Jaccobi equation 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 function $\psi(x,t)$;
2,we calculate $\psi(x,t)$ from Schrödinger 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.

 

[mfb]

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.

 

[Demystifier]

Thanks.
How about this way.
Classic Mechanic :
1,wave function ($\hbar \rightarrow 0$) is $\psi \rightarrow \exp{\frac{i}{\hbar}S(x,t)}$;
2,we get wave function from Hamilton-Jaccobi equation 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 function $\psi(x,t)$;
2,we calculate $\psi(x,t)$ from Schrödinger 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.

Read about Bohmian mechanics! You can start e.g. from
https://en.wikipedia.org/wiki/De_Broglie–Bohm_theory

 

[zhouhao]

Read about Bohmian mechanics! You can start e.g. from
https://en.wikipedia.org/wiki/De_Broglie–Bohm_theory

Thanks.I think Bohmian mechanics is helpful to me.Could help me with another question below?
$\psi$ is a solution of Schrödinger 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 function 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?

 

[Demystifier]

Too many solutions.
If we calculates wave function 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?

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.