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Momentum Operator Construction

  1. Nov 22, 2009 #1
    Quick question about how Schrodinger constructed the momentum operator. I understand that [tex]\widehat{H}[/tex] in the wave equation is the sum of the kinetic and potential operators, using the form H=T+V. The kinetic energy can be expressed by T=p^2/2m, and that's the last part I can follow. Most sources then give the momentum operator as a "substitution" which gives p [tex]\rightarrow[/tex] i[tex]\hbar[/tex][tex]\nabla[/tex]. I have absolutely no idea where this comes from (I've yet to take a formal math class on anything beyond basic calculus, maybe this is the problem), could anyone provide some information on this?
     
  2. jcsd
  3. Nov 22, 2009 #2

    physicsworks

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    [tex]\hat{\mathbf{p}} = -i \hbar \nabla[/tex].
    Actually, it comes from the properties of Fourier transform, so
    [tex]\overline{F(x,y,z)} = \int F(x,y,z) |\psi(x,y,z)|^2 dxdydz = \int \psi^{*}(x,y,z) F(x,y,z) \psi(x,y,z)dxdydz[/tex]

    [tex]\overline{F(p_x,p_y,p_z)} = \int F(p_x,p_y,p_z) |c(p_x,p_y,p_z)|^2 dp_xdp_ydp_z = \int c^{*}(p_x,p_y,p_z) F(p_x,p_y,p_z) c(p_x,p_y,p_z)dp_xdp_ydp_z[/tex]
    where [tex]c(p_x,p_y,p_z)[/tex] is the amplitude of de Broglie wave which has momentum [tex]\mathbf{p}(p_x, p_y, p_z)[/tex]
    can be generalized in case of entire rational function [tex]F(x,y,z)[/tex] (i.e., for example [tex]F(p_x) = \sum_n a_n p_x^n[/tex]) and above equations becomes:

    [tex]\overline{F(x,y,z)} = \int c^{*}(p_x,p_y,p_z) F \left(i \hbar \frac{\partial}{\partial p_x}, i \hbar \frac{\partial}{\partial p_y},i \hbar \frac{\partial}{\partial p_z} \right) c(p_x,p_y,p_z)dp_xdp_ydp_z[/tex].

    [tex]\overline{F(p_x,p_y,p_z)} = \int \psi^{*}(x,y,z) F \left(-i \hbar \frac{\partial}{\partial x},-i \hbar \frac{\partial}{\partial y},-i \hbar \frac{\partial}{\partial z} \right) \psi(x,y,z)dxdydz[/tex]
    Consider [tex]F(p_x, p_y,p_z)=p_x[/tex]. Then,
    [tex]\overline{p_x}= \int \psi^{*}(x,y,z) \left( -i \hbar \frac{\partial}{\partial x} \right) \psi(x,y,z) dxdydz[/tex]
    In other words
    [tex]\hat{p_x} = -ih \frac{\partial}{\partial x}[/tex]
     
  4. Nov 22, 2009 #3
    Alright, thanks a lot, that makes a little more sense now. I'm probably not at the level of mathematical skill I would need to be at to derive that on my own, but I can follow most of your steps. It's so interesting how hbar seems to keep coming up, even in the Fourier transformation. Thanks a lot, that was very helpful
     
  5. Nov 23, 2009 #4

    physicsworks

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    Hbar comes from de Brogile wave :)
     
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