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Divergence defined from volume element

  1. Dec 18, 2008 #1

    Fredrik

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    I need some help understanding a definition:
    This is supposed to be an explanation of what the author did on the page before. He had just described how to construct a (complex) Hilbert space from a (real) smooth manifold with a smooth nowhere vanishing volume element, and then moved on to construct operators on that Hilbert space.

    I would appreciate if someone who understands what he's talking about could explain it to me, and maybe translate it to a coordinate independent notation.
     
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  3. Dec 21, 2008 #2

    Fredrik

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    Here's another clue:

     
  4. Dec 25, 2008 #3

    Fredrik

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    OK, let me ask the question in a different way: Suppose we have just constructed a complex Hilbert space from the set of complex-valued functions defined on a manifold M. Is there a natural way to use the manifold to also construct operators on this Hilbert space?

    That seems to be what this author is doing. One operator for each function, and one operator for each vector field. But if we want to construct an operator corresponding to a vector field X, why not just define it as the map [itex]f\mapsto Xf[/itex], where [itex]Xf[/itex] is the map [itex]p\mapsto X_pf[/itex]? Maybe that's what he means by [itex]v^a\nabla_a[/itex], (but probably not, since he mentions Lie derivatives and a volume form), but then what's [itex]\nabla_a v^a[/itex]?

    The text I'm reading is available here. The relevant pages are 14-15.
     
    Last edited: Dec 25, 2008
  5. Jan 5, 2009 #4

    George Jones

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    By the divergence theorem and the compact supports, [itex]\int_M \nabla_a \left( f g^* v^a \right) dV = 0[/itex], so that the product rule gives

    [tex]
    0 = \int_M \nabla_a \left( f g^* v^a \right) dV = \int_M \left( g^* v^a \nabla_a f + f v^a \nabla_a g^* + fg^* \nabla_a v^a \right) dV.
    [/tex]

    Thus,

    [tex]
    \int_M \left( v^a \nabla_a f \right) g^* dV = - \int_M \left( f v^a \nabla_a g^* + fg^* \nabla_a v^a \right) dV.
    [/tex]
     
  6. Jan 5, 2009 #5

    Fredrik

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    Thanks George. I see now that I made a really silly blunder that made me think that that the method you used couldn't work.

    Unfortunately I still don't understand what [itex]\nabla_a[/itex] is. I think the natural way to define a linear operator V for each pair (X,v) where X is a vector field and v a scalar field, is

    [tex]Vf=Xf+vf=X^\mu\partial_\mu f+vf[/tex]

    but this isn't what Geroch is doing. He says that you can define [itex]\nabla_a[/itex] using either a Lie derivative or an exterior derivative. I think I understand Lie derivatives, but I don't understand his definition.
     
    Last edited: Jan 5, 2009
  7. Jan 5, 2009 #6
    Actually your equation is correct and equivalent to (27) in Geroch's notes. [itex]\nabla_a[/itex] represents a covariant derivative, but covariant derivatives of scalar functions are equal to partial derivatives. Covariant derivatives are so frequently used in differential geometry that they become standard currency.

    Best,

    Dave
     
  8. Jan 5, 2009 #7
    If you take v^a to be a unit vector in the time direction on flat space, and choose the scalar function as v=0, then the operator V becomes d/dt, the energy operator. Or if you take v^a as a unit vector in some space direction, and v=0, then V = d/dx, the momentum operator. On the other hand put the vector v^a to 0 and choose the coordinate function v(x) = x. Then V = multiplication by x, the position operator. So the general expression above includes both the mom-energy operator and a position operator, which are the most important examples of operator in quantum mechanics. The idea is to get both types at once, and do the whole thing on a curved manifold.

    Best
    Dave
     
  9. Jan 5, 2009 #8

    Fredrik

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    But Geroch says that it isn't a covariant derivative. His exact words are: "We don't have a metric, or a covariant derivative, defined on M." It's near the top of page 15, just before the stuff I quoted in #1.

    I haven't really tried to understand Geroch's definition of the momentum and position operators since I got stuck on this divergence thing, but I'm looking at it now. I think you got the part about momentum operators right, but I'm not sure about the position operators. His definition looks more complicated, but I haven't tried to understand it yet.
     
  10. Jan 6, 2009 #9
    Oh, I see. The covariant derivative of a scalar function is simply the partial derivative, so without a metric, you still have a covariant derivative of scalar functions. The other required quantity is the divergence of the vector, which can be defined using only a volume form. The volume form is also needed to define the integrals used in the Stokes' theorem (28) and (29). I'm a little sceptical about having a volume form without a metric, since a metric is the obvious dynamical quantity to use to define a volume form. And really, you can't get far in either differential geometry or the exterior algebra without a metric.

    Tell me if you develop another interpretation! However, I still contend that operators id/dx and x on flat space are both of the form (27) and satisfy the Hermiticity condition (30).

    Best,

    Dave
     
  11. Jan 9, 2009 #10

    Fredrik

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    They are and they do, but your definitions still look very different from Geroch's definitions, and it seems that the "eigenfunction" of your x is a Dirac delta, while his definition leads to some other function describing "a particle at the origin". (See the end of section 5).

    I don't understand everything that he's doing, but I'll explain some of the things I do understand, in my own words and my own notation.

    A positive-frequency solution can be written as

    [tex]\phi^+(x)=\int_{\mathbb R^3}d^3p\ a(\vec p)e^{-ipx}[/tex]

    where [itex]p^0=\sqrt{\vec p^2+m^2}[/itex]. We could take the set of such solutions to be our Hilbert space. If we do, it seems natural to define the position and momentum operators the way you did. However, Geroch does things differently. First he rewrites [itex]\phi^+(x)[/itex] as

    [tex]\phi^+(x)=\int_{M_+}dV\ f(p)e^{-ipx}[/tex]

    where [itex]M_+=\{p\in\mathbb R^4|p^0=\sqrt{\vec p^2+m^2}\}[/itex] is the future mass shell, and dV is the volume form on M+. This f is a function from M+ into [itex]\mathbb R[/itex]. Geroch takes the set of such functions to be the Hilbert space.

    Then he makes the observation that if v is an arbitrary four-vector and [itex]\phi^+[/itex] an arbitrary positive-frequency solution, then [itex]i v^\mu\partial_\mu\phi^+[/itex] is also a positive-frequency solution.

    [tex]i v^\mu\partial_\mu\phi^+(x)= \int_{M_+}dV\ f(p) vp e^{-ipx}[/itex]

    So "your" momentum operator acting on [itex]\phi^+[/itex] defines "his" momentum operator acting on f. His definition can be written as

    [tex]P(v)f=vpf[/tex]

    We can also define

    [tex]P_\mu=P(e_\mu)=(e_\mu)^\nu p_\nu =p_\nu[/tex]

    I still don't understand the definition of the position operator, but it's clear that it's a vector field on M+.
     
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