Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Local in space

  1. Oct 11, 2007 #1
    One approach to get a relativistic QM might be to take hamiltonian as sqrt(p2c2 +m2c4). But apparently this does not work because expanding H in inverse powers of c2 will lead to higher order derivative in space. Apparently, one reason this is bad is because higher derivatives mean that the equation is not local in space.

    My question: what does "not local in space" mean? Why does higher derivatives of x make it not local in space?
     
  2. jcsd
  3. Oct 11, 2007 #2
    I've never understood this argument. I think that sqrt(p2c2 +m2c4) is a perfect 1-particle Hamiltonian.

    Eugene.
     
  4. Oct 11, 2007 #3

    Demystifier

    User Avatar
    Science Advisor

  5. Oct 20, 2007 #4
    I can offer some off-hand remarks, but you should check the details of what I say. First, to quickly respond to your question about "not local in space": ultimately, interactions between objects should satisfy causality...objects not within each others light cones should not interact, even in quantum mechanics where "spooky action at a distance" occurs. Look up Bell's Inequalities for a critical look at locality in quantum mechanics. In quantum field theory, the idea of local interactions comes under the guise of the "cluster decomposition principle" (Weinberg mentions it in this talk printed here http://arxiv.org/abs/hep-th/9702027). To get wavefunctions and interaction amplitudes that satisfy locality, we write our Hamiltonians and Lagrangians using "local operators", which we get to now.

    To make it explicit, in quantum mechanics the operator equation we're looking at is
    [itex]\sqrt{\partial^{2}/\partial x^{\mu}\partial x_{\mu}+\cdots}\Psi=E\Psi[/itex]
    where [itex]x_{\mu}[/itex] are spacetime coordinates. In mathematics, a local operator is one that consists of polynomials of x and p(x) acting at the same point x in spacetime (as position space operators acting on wavefunctions these are x and d/dx); the square root in our operator makes it non-local, and there is a whole branch of mathematics that talks about the "spectra" of operators (solutions to differential equations e.g.).
    But what does this mean for us? As you mentioned, we can expand our operator so that it is a polynomial (though an infinite series), but there are two ways of doing this. Expanding in powers of [itex]p^{2}/(mc)^{2} [/itex] leads to an infinite series of higher and higher powers of p (and therefore derivatives in position space). We could truncate this series if [itex]p<<mc [/itex], which amounts to a non-relativistic limit reproducing the NR Schrodinger eqn with a constant potential term. But we are interested in the relativistic arena so that the full expansion must be kept. The problem is, higher derivative equations (above 2nd order) have solutions that generally have instabilities (see mathematics literature on this subject), and stability of solutions is one of the requirements in quantum mechanics (see the boundedness, or square-integrability, issue in its axioms). Related to this is the existence of negative norm states (violating the unitarity axiom), which will necessarily be involved in transititions (interaction amplitudes); the presence of these states can be seen as "negative mass poles" in the propagators used to calculate scattering amplitudes. Furthermore, in quantum field theory, higher derivative terms will appear in the Lagrangian density for the fields, and are necessarily accompanied by larger and larger negative mass-dimension coupling "constants", which is troublesome for renormalizability. None of this is to say that higher derivative theories are un-useful, or don't appear in nature, or that no sense can be made of them. But for the relativistic S.E., these issues are too sophisticated.
    If you instead made an expansion in [itex](mc)^{2}/p^{2}[/itex], we can truncate to the first two terms in the relativistic limit, but the second one involves 1/p, which in position space is [itex](\partial/\partial x)^{-1} [/itex]; this is a non-local operator in position space (so this is the expansion that leads to manifestly non-local terms). Wavefunctions, as solutions of this equation, would have causality problems. Non-locally-interacting theories also show up in physics, but again, these are more sophisticated situations than the relativistic S.E.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Local in space
  1. Local or Nonlocal? (Replies: 12)

  2. Weak localization (Replies: 5)

Loading...