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Why replace Hamiltonian with operator?

  1. Dec 30, 2007 #1
    momentum operator in Hamiltonian

    Hello all. I'm in an introductory QM course as a physics major. As I understand it, to quantize a classical system, we just replace momentum in Hamiltonian with momentum operator?
    But why? One answer is that because it works.
    Is there any other reasons why it works or it's just a postulate?
    Last edited: Dec 31, 2007
  2. jcsd
  3. Jan 1, 2008 #2


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    It's just a postulate, but a well-motivated one. There is a formalism of classical mechanics known as the "Hamilton-Jacobi" formalism. You can think of the Schrodinger Eqn as a generalization of that. Unfortunately, they usually don't teach HJ to undergrads, so I don't know if you're familiar with it. Check out Goldstein's "Classical Mechanics" if you want to learn more.

    VERY roughly speaking: the idea that Schrodinger had was to look for wave-like solutions (along the lines of de Broglie). The first step is to look for plane-wave solutions (always the first step, since you build up more general solutions from these using Fourier series). What you discover is that momentum can be represented as a derivative acting on plane-wave solutions (this is true in general wave-mechanics, such as water waves, light,...). It wasn't too much of a stretch to reach the Schrodinger equation.
  4. Jan 2, 2008 #3
    Not only bacause it works...
    As soon as we accepted de Broglie's waves of matter, we may ask a question: what are possible equations that describe those waves?
    If we know DISPERSION RELATION k(w), we may derive the original equation.
    k(w) may be found from experiment or derived from classical mechanics. After that substitution
    1/k -> d/dx
    is a regular mathematical procedure, not a postulate.
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