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Relativistic vs. Nonrelativistic KE Operator question

  1. Feb 18, 2016 #1
    Hey, folks. I, on a whim today, started taking a MOOC quantum mechanics course that I have the functional math skills necessary to do but have virtually no background knowledge of quantum to start with and am incredibly rusty on stuff like PDE's; Quite frankly I'm out of my league, but the course just started and I'm going to work my butt off to do as well as I can and at least pass. The first lecture of the course is discussing the time-dependent Schrodinger Equation for a single, nonrelativistic particle:

    [tex]i \hbar \frac{\partial \psi(x,t)}{\partial t} = \bigg( - \frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} + v(x,t) \bigg) \psi (x,t)[/tex]

    And he makes it clear that it's obvious that we can tell that this is the nonrelativistic version because of the kinetic energy operator is p^2/2m. But I don't see p^2/2m, I see what is a kinetic energy operator, [tex]\frac{\hbar^2}{2m}[/tex], but that isn't equivalent to p^2/2m, is it? I'm guessing based on poking around Wikipedia that this has to do with the Hamiltonian/Lagrangian, of which I think I only ever messed with Hamiltonians in Analytical Mechanics, a little bit. How do I understand the jump from one representation to another?
  2. jcsd
  3. Feb 18, 2016 #2

    Vanadium 50

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    The momentum operator is [itex]i \hbar \nabla [/itex].

    Yes, you right when you say you are out of your league. I think you need to start a course or two further back. You will understand QM faster and better by getting the foundations in order than by failing this class, possibly a few times.
  4. Feb 18, 2016 #3
    It's already both a rare opportunity and literally sunk cost at this point and I have a huge wealth of free time for the next months, so I might just try really hard..that being said, I trust your judgment and quite frankly feel like things won't end well. I genuinely had no idea what I was getting myself into, but, eh.
  5. Feb 18, 2016 #4


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