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Correspondence betw. Hamilton-Jacobi & Schrödinger eqns.

  1. Mar 25, 2010 #1
    Hey everyone,

    I'm trying to understand more about the correspondence between classical and quantum physics, in particular the relation between classical mechanics in Hamilton-Jacobi form and the Schrödinger equation. The former can be obtained from quantum theory by plugging

    [tex]\psi=\psi_0 e^{\frac{i}{\hbar}S}[/tex]

    into the Schrödinger equation, which yields

    [tex]\frac{1}{2m}(\nabla S)^2 + \frac{\partial S}{\partial t} - \frac{i \hbar}{2m}\Delta S=0,[/tex]

    where the last term vanishes in the classical limit. My first question is: What is the interpretation of this quantum term? How is the classical theory affected by it?

    Then, one can also start from the classical Hamilton-Jacobi equation and, with the help of the wave function [tex]\psi[/tex] from above, rewrite it as

    [tex]\frac{\hbar^2}{2m\psi}(\nabla \psi)^2 + i\hbar\frac{\partial \psi}{\partial t} = 0. [/tex]

    In this light, one obtains the quantum equation by "replacing the square of the derivative with the second derivative", citing a phrase I sometimes came across. So my second question is: How can this nonlinear variant of the Schrödinger equation be interpreted, how is its behaviour different from the full quantum theory?

    In summary: What is the interpretation of the operation "replace square of derivative with second derivative" ?
  2. jcsd
  3. Mar 27, 2010 #2
    How about you think about what is being asked and then attempt verbal solutions here; we're not here to help you with questions with no attempt at resolutions.
  4. Mar 27, 2010 #3
    Maybe this is a misunderstanding. This is not related to a homework question. Its simply curiosity, and I hoped someone could comment on this or point me to some reference where this is reviewed in more detail. Obviously, a difference between the above variants is the presence / abscence of the superposition principle. But maybe some more statements can be made I didn't think of.
  5. Mar 27, 2010 #4
    I think you 've missed something in the way. First, the wave function Ψ should read
    \psi=\psi_0(r,t) e^{\frac{i}{\hbar}S(r,t)}
    where r=r (can't find a way to display this properly above).
    Substitute this into Schrodinger's eq. and you 'll get two coupled equations for [tex] S, \psi_0 [/tex]. In the end, set S=p.r-Et and you have the energy of a particle in a potential V:
    E=p^2/2m + V
  6. Mar 27, 2010 #5
    Squeezed, you're right. I forgot to state some assumptions. First, I assumed [tex]\psi_0[/tex] to be simply a constant so this yields only one equation for S (where S=S(r,t); i didn't make the arguments explicit). The ansatz essentially is one for a monochromatic plane wave. Second, my lines above make sense only for a free particle, so no potential for simplicity. In a more general setting, there would similarly arise non-classical terms in the Hamilton-Jacobi equation. It seems to be referred to as the quantum potential. My question is, how does it affect the classical theory?

    The last equation in post #1 becomes the Schrödinger equation by replacing [tex]\psi^{-1}(\nabla \psi)^2[/tex] with [tex]\Delta \psi[/tex]. So it seems that in this picture, this "linearization" realizes the quantum theory. But what about the properties of the nonlinear version?

    Meanwhile, I've come across http://arxiv.org/abs/quant-ph/9703007, it seems related.

  7. Mar 27, 2010 #6
    "In a more general setting, there would similarly arise non-classical terms in the Hamilton-Jacobi equation. It seems to be referred to as the quantum potential. "

    I am not sure what you mean here, so i will show you how to get the so-called Quantum Potential (introduced by D. Bohm). Using
    \psi=\psi_0(r,t) e^{\frac{i}{\hbar}S(r,t)}

    You end up with
    \frac{1}{2m}(\nabla S)^2 + V + \frac{\partial S}{\partial t} - \frac{\hbar^2}{2m}\frac{\nabla^2\psi_0}{\psi_0} =0
    \frac{1}{2m}\psi_0(\nabla S)^2 + \frac{\partial \psi_0}{\partial t} + \frac{1}{m}\nabla\psi_0\cdot\nabla S =0

    Keep V in your equations - as you will see below it doesn't t really matter.

    Note that the definitions
    v = \frac{\nabla S}{m}
    (v=v is a vector)
    and [tex] \rho=\psi_0^2 [/tex] (the density of the wave) give us
    \frac{\partial \psi_0^2}{\partial t} = -\nabla\cdot(\frac{\psi_0^2\nabla S}{m})
    \frac{\partial \rho}{\partial t}=-\nabla\cdot(\rho v)
    (the continuity equation).
    Now, re-order the equation for S
    \frac{\partial S}{\partial t} + V + \frac{1}{2m}(\nabla S)^2 = \frac{\hbar^2}{2m}\frac{\nabla^2\psi_0}{\psi_0}
    The term at the rhs is the quantum potential.

    "My question is, how does it affect the classical theory?"

    For [tex] \hbar ->0 [/tex] you see that this eq. is the Hamilton-Jacobi eq. for S, i.e. particles in classical mechanics represent "ray limits" of the Schrodinger eq.
  8. Mar 27, 2010 #7


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    You might be interested in Section 4.2 of "Introduction to Quantum Mechanics: A Time Dependent Approach" by David J. Tannor on Bohmian mechanics and the classical limit. If you use Squeezed's suggestion and substitute

    [tex]\psi= A(\textbf{r},t) e^{\frac{i}{\hbar}S(\textbf{r},t)}[/tex]

    then the Hamilton-Jacobi equation has an extra term relative to the classical HJ equation which is referred to as the "quantum potential" (in Bohm's form of quantum mechanics). This approach also leads to what is called the hydrodynamic formulation of the Schrodinger's equation.

    There is another reference on this that I want to give you, but unfortunately won't be able to find it until Tuesday or Wednesday. I will post it then.
  9. Mar 27, 2010 #8
    I apologize. I did not clarify what was meant.. I will read through it Gn :S
  10. Mar 30, 2010 #9


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    Here is the reference that I said I would post:

    "Quantum Dynamics with Trajectories" by Robert E. Wyatt, published by Springer.
  11. Apr 1, 2010 #10
    thanks everyone, so far. I will take a look!
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