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Derive Schrodinger equation

  1. Jan 16, 2015 #1
    1. The problem statement, all variables and given/known data
    I am reading Mathematical Concepts of Quantum Mechanics (Stephen J. Gustafson, Israel Michael Sigal. Second edition). The book would like to find an evolution equation which would lead to the Hamilton-Jacobi equation
    $$\frac{\partial S}{\partial t}=-h(x, \nabla S) $$
    in the way the wave equation led to the eikonal one. The book also says that ##\phi (x, t) = a(x, t) \exp( i S(x,t)/\hbar)##. So I express ##S(x,t)## using ##\phi (x,t)## and substitute back to the Hamilton-Jacobi equation, taking ##h (x, \nabla S) = \frac{1}{2m}|\nabla S|^2+V(x)##.

    The book means to take the leading terms when ##\hbar## small compared to a typical classical action ##S## and restore Schrodinger equation. I am kind of lost during the derivation.

    2. Relevant equations
    After the substitution, I have
    $$i\hbar\partial_t \phi=-\frac{\hbar^2}{2m}[(\frac{\nabla \phi}{\phi}-\frac{\nabla a}{a})^2-\frac{2im\partial_t a}{a\hbar}]\phi+V(x)\phi.$$
    Comparing with Schrodinger equation, I figure that the leading term of
    $$[(\frac{\nabla \phi}{\phi}-\frac{\nabla a}{a})^2-\frac{2im\partial_t a}{a\hbar}]\phi$$
    should equal to ##\Delta_x \phi##, but don't know how.

    3. The attempt at a solution
    I am not sure what to search for the problem, but wiki have something on this. A nonlinear variant of the Schrödinger equation is expressed as
    $$i\hbar\partial_t \phi=-\frac{\hbar^2}{2m}\frac{(\nabla \phi)^2}{\phi}+V(x)\phi.$$
    I am not sure what a nonlinear Schrödinger equation is after realizing it's not the same thing as the Schrödinger equation.

    The book's goal seems to be the linear Schrödinger equation. Even though I do see how to obtain the nonlinear Schrödinger equation, I am not sure why ##(\frac{\nabla \phi}{\phi})^2## is a leading term. Could someone help me with this?

    Thanks!
     
  2. jcsd
  3. Jan 16, 2015 #2

    stevendaryl

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    You can derive it in the following way:

    1. Let [itex]\phi = a e^{\frac{iS}{\hbar}}[/itex]
    2. Compute [itex]\frac{-\hbar^2}{2m} \nabla^2 \phi[/itex] in terms of [itex]a[/itex] and [itex]S[/itex] and only keep the lowest-order terms, in powers of [itex]\hbar[/itex]
    3. Similarly, compute [itex]i \hbar \frac{d}{dt} \phi[/itex] in terms of [itex]a[/itex] and [itex]S[/itex] and keep the lowest-order terms in powers of [itex]\hbar[/itex].
    4. Now, use the fact that [itex]\frac{(\nabla S)^2}{2m} + V = -\frac{d}{dt} S[/itex] to show that the results of 2 and 3 are equal (to lowest order in [itex]\hbar[/itex])
     
  4. Jan 16, 2015 #3
    Thanks! It was my first attempt, which never really got carried out. I thought ##\nabla^2\phi## can bring in ##1/\hbar^2## and ##\partial_t \phi## only ##1/\hbar##....... And I just go through it and all is fine.
    And now I kind of want to ask what's the story about the wiki and the equation
    $$
    i\hbar\partial_t \phi=-\frac{\hbar^2}{2m}\frac{(\nabla \phi)^2}{\phi}+V(x)\phi.$$
     
  5. Jan 17, 2015 #4

    stevendaryl

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    Well, if you let [itex]\phi = a e^{\frac{i}{\hbar} S}[/itex], then

    [itex]\phi^* (-\frac{\hbar^2}{2m} \nabla^2 \phi) = -\frac{\hbar^2}{2m} a \nabla^2 a + \frac{\hbar^2}{2m} (\nabla a)^2 - \frac{i \hbar}{2m} (\nabla^2 S) a^2 + \frac{1}{2m} (\nabla S)^2 a^2[/itex]

    [itex]\frac{\hbar^2}{2m} |\nabla \phi|^2 = \frac{\hbar^2}{2m} (\nabla a)^2 + \frac{1}{2m} (\nabla S)^2 a^2[/itex]

    So the difference between them is [itex] -\frac{\hbar^2}{2m} a \nabla^2 a - \frac{i \hbar}{2m} (\nabla^2 S) a^2[/itex].
     
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