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Question(s) related to Nonlinear Shroedinger Model

  1. Jan 4, 2005 #1
    Relevant to the NLS is the differential equation,

    [tex] \left( -\sum^N_{i=1} \frac{\partial^2}{\partial x^2_i} +c \sum_{i\neq j} \delta(x_i-x_j)\right)f_N = E_Nf_N[/tex]

    How does one show that

    [tex] \left(\prod_{i<j}(\theta(x_i - x_j) + e^{i\Delta(k_j-k_i)}\theta(x_j-x_i))\right)\exp\left(i\sum^N_{j=1}k_jx_j\right) [/tex]

    where [tex]\theta(x)=\frac{|x|+x}{2x}[/tex] and


    is a solution? The textbook just asserts but does not calculate.
  2. jcsd
  3. Jan 4, 2005 #2


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    What have u done???You don't expect us do the 'dirty work' for you... :mad:
    What is the connection between E and 'k'?Why is a nonliear Schroedinger equation??It seems very linear to me...

  4. Jan 4, 2005 #3
    The book's preface claims that calculations are presented "in step-to-step detail". Yet when I encounter problems like this, I suspect that I'm missing something. I've put a lot of hours into reading the book already. So help me out or don't help me, but don't get mad at me.
  5. Jan 4, 2005 #4
    NLS is a second-quantized theory. Its field equation is nonlinear. However, solving for the eigenvalues for finite-particle states leads to (2.87). I don't know why the book has a section on this. It hardly seems relevant to the subsequent development of the book and I've been able to move on without understanding it. But it still arouses my curiosity. Some pdf files available online suggest that NLS is related to Bose condensation.
  6. Jan 4, 2005 #5


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    Okay,maybe i overreacted a little bit,i just did't like the way you presented the problem...
    Can you compute the derivative of the theta Heaviside 'function'??
    Is 'q' a constant??
    And again,what is the connection between E and other variables?Even if u were able to compute that Hessian and the delta Dirac multiplication,u'd still have to equate the two sides,so then u'd need the expression of the energy levels in terms of the 'k','q' and other variables.

  7. Jan 4, 2005 #6
    Apology accepted. I appreciate the importance of presentation, however, typing in all the latex code is a lot of work, so I decided I would present as much information as I believed was necessary at first, and give clarifications as needed.

    Anyway, the answer to your question about the theta function is simple: it's derivative is the delta funtion. But the straightforward differentiation of (2.90) leads to a big mess. f_N is a function of the x_i, E_N is a constant, the eigenvalue of the equation, c is another constant (not necessarily the speed of light), big Delta is a function, q is a dummy variable in its definition. Notice that i occurs both as sqrt(-1) and as index.
  8. Jan 4, 2005 #7
    The field equation is

    [tex] i \frac{\partial\varphi}{\partial t} = -\frac{\partial^2\varphi}{\partial x^2} + 2c|\varphi|^2\varphi [/tex]

    which is the ordinary Shroedinger equation plus self-interaction term.

    The field Hamiltonian is

    [tex] H = \int^L_0 dx \varphi^*(x,t)(-\partial^2_x)\varphi(x,t) + c\varphi^*\varphi^*\varphi\varphi [/tex]

    One looks for eigenstates of the form [tex]\int dx_1...dx_Nf_N(x_1,...,x_N,t)\varphi^*(x_N,t)...\varphi^*(x_1,t)|0\rangle[/tex], and obtains (2.87).
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