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[Q]Three Question.

  1. Nov 22, 2008 #1
    Hi. I have three question. First is form of normalized momentum egien function

    Liboff said that of unbounded is [itex] \frac{1}{2\pi}exp[ikx] [/itex] But Why should this

    form be normalized eigen function of momentum in unbounded?

    Second question is whether probability density is equall to particle density.

    Deriving Continuity Equation of QM, Original form of Equation is [itex] \frac{\partial \rho}{\partial t} + \nabla\bullet J = 0 [/itex]

    Replacing [itex] \rho [/itex] to [itex] \Psi [/itex] give probability current density.

    But what does it means? My idea about this quantity is unclear. Please Give me a relavent


    Third question is How to construct Hamiltonian of many body problem.

    Is it right that considering two particle with interaciton of V(x),where x is distance between

    two particle, Hamiltonian of the system(not particle) is [itex] H = \frac{-\hbar^2}{2m}\frac{\partial}{\partial x_{1}} + \frac{-\hbar^2}{2m}\frac{\partial}{\partial x_{2}} + v(x) [/itex]? How can i solve this form?
  2. jcsd
  3. Nov 22, 2008 #2

    Dr Transport

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    Science Advisor
    Gold Member

    The first couple are easy and I'll comment quickly (do not have the time to answer all of them).

    If you normalize a plane wave, you are actually using a Dirac-delta function definition.


    Look part way down the page for Fourier definitions.

    [tex] \Psi \Psi^{*} [/tex] is the probability density the integral over all space is equated to [tex] \rho [/tex].

    This is the basis of QM.

    Hope this helped.
  4. Nov 23, 2008 #3
    Thank you for your help I solved my first question. But How about second and third question? Anyone know about this well?

    I can't imagine real meaning of the Probability current density!
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