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Lorentz Transformation of Fields

  1. Nov 29, 2011 #1
    I am reading Quantum Field Theory by Michio Kaku, and I have come across some equations that I can't quite make sense of. I am hoping somebody can help me out. Please see attached. Thanks.

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  2. jcsd
  3. Nov 29, 2011 #2


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    Wow, a lot of to-the-point questions. I can't use the quote function to try to give an answer to each, because you inserted a picture. Indeed, Kaku's text does not fill in the gaps...

    Do you know some group theory and representation theory, so that you'd be able to understand the answer ?
  4. Nov 29, 2011 #3
    I can try. I know a little. Give me your best shot, thanks.
  5. Nov 29, 2011 #4


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    So [itex]\epsilon_{\mu\nu}[/itex] is a set of 6 independent infinitesimal parameters Kaku uses to describe the Lorentz group locally (i.e. in a neighborbood of the identity element). [itex] L^{\mu\nu} [/itex] are then the generators of the connected component of the identity and can be taken as generators (basis elements) in the Lie algebra of the (restricted or full, doesn't matter) Lorentz group.

    His first equation is axiomatic, it is the definition of the vector field in terms of its behavior under a restricted Lorentz transformation.

    The explicit form of L's (with x and space-time derivatives) is valid in the case of scalar fields, because the spin of the field (angular momentum in the rest frame) is disregarded.

    By physical reasons, U([itex] \Lambda [/itex]) is a unitary operator acting on the representation space of the restricted Lorentz group (actually a double cover of the restricted Poincare group), so by Stone's theorem, the L's should be (essentially) self-adjoint.

    The calculation for the derivative of L([itex] \Lambda [/itex]) is wrong, as L depends on x and captures the implicit dependence of the U operator on x.

    As for point 3), indeed, it's an Euler-MacLaurin series. Turns out that the epsilons are anti-symmetric.

    Point 4) a) can be postulated (see above). b) Yes, all vectors have 4 space-time components. c) All fields, well, you move from the Lorentz group to the spin (1,3) group (SL(2,C)), so that spinorial fields will enter the picture.

    This is off the top of my head. I can give you references such as Weinberg, Vol.1 or axiomatical field theory texts such as: Streater & Wightman (PCT, Spin-Statistics and All That) or better Bogolubov, Logunov, Todorov (1975 better than 1990).
    Last edited: Nov 29, 2011
  6. Dec 2, 2011 #5
    Excellent, thank you very much for the great reply. I am currently brushing up on my group theory. I imagine I will have a follow-up question, but while I continue to digest everything, I just wanted to say thanks!

    I think one of my problems was how I was thinking of the Lorentz Group as a set of 4x4 matrices with real values, where I guess there is a representation in terms of the L operator. I estimate this is where my question will be, though I don't think I can formulate it yet.
  7. Dec 2, 2011 #6


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    The Lorentz group IS (isomorphic to) a group of 4x4 matrices with real entries...
  8. Dec 2, 2011 #7

    Hans de Vries

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    [itex]L^{\mu\nu}[/itex] contains the three generators of rotation [itex]J^i[/itex] and the three generators of boosts [itex]K^i[/itex].
    They form the Poincaré generators together with the four translation operators.

    & P^\mu &=~& -i\Big( ~~~~~~~~~-\frac{\partial}{\partial x^\mu} &\Big)&
    ~~~\mbox{4 translation generators} \\
    & J^i &=~& -i\Big(\, ~~~~x^j\frac{\partial}{\partial x^k}-x^k\frac{\partial}{\partial x^j} &\Big)&
    ~~~\mbox{3 rotation generators}~~~~~~ \\
    & K^i &=~& -i\Big( ~ - x^i\frac{\partial}{\partial x^o}-x^o\frac{\partial}{\partial x^i} &\Big)&
    ~~~\mbox{3 boost generators}~~~~~~

    In the image below you can see how they work. The [itex]\delta[/itex] here is an infinitesimal small parameter.
    You can for instance translate an arbitrary function over an infinitesimal small distance by
    subtracting [itex]\delta\partial f/\partial x[/itex] (The red and blue delta functions)


    If you repeatedly apply the [itex](1-\delta\partial_x f)[/itex] operator then this amounts to an exponential function like
    the one in your book. To translate over a distance [itex]\ell_x[/itex] to the left you do:

    \exp\left(\,i\ell_x P^x\right)\,f(x) ~=~
    \left\{1 +
    \frac{\ell_x }{1!}\,\frac{\partial }{dx } +
    \frac{\ell_x^2}{2!}\,\frac{\partial^2}{dx^2} +
    \frac{\ell_x^3}{3!}\,\frac{\partial^3}{dx^3} +

    The right hand side is just the standard Taylor series. If we write [itex](x-a)[/itex] for the displacement [itex]\ell_x[/itex] and
    let the operators act on [itex]f(a)[/itex] then we get the familiar expression for the Taylor series.

    f(a)+\frac {f'(a)}{1!} (x-a)+ \frac{f''(a)}{2!} (x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+ \cdots

    The rotate and boost operators work in just the same way. The matrix [itex]\epsilon_{\mu\nu}[/itex] contains the three angles
    by which you want to rotate and the three rapidities by which you want to boost just like [itex]\ell_x[/itex] is the
    distance by which you want to translate.

    The above rotates/boosts works on a scalar field, that is they handle the coordinate transformation.
    If you want to transform a (four) vector field then have to operate on the (four) vector parameters
    as well because the vector transforms under a general Lorentz transformation.

    Last edited: Dec 2, 2011
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